Set Theory

STUDIES I N LOGIC AND THE FOUNDATIONS OF MATHEMATICS

Editors

A. H E Y T I N G, Amsterdam A. M 0 S T 0 W S KI, Warszawa A. R O B I N S O N , NewHuven P. S U P P E S, Stanford

Advisory Editorial Board

Y. B A R - H I L L E L, Jerusalem K. L. D E BOUVl?iRE, SuntaChru H. H E R M E S, Freiburg i. Br. J. H I N T I K K A, Helsinki J. C . S H E P H E R D S O N , Eristol E. P. S P E C K E R, Zurich

NORTH-HOLLAND PUBLISHING AMSTERDAM

COMPANY

SET THEORY

K. KURATOWSKI and

A. MOSTOWSKI Members of the Polish Academy of Sciences

1968 PWN - POLISH SCIENTIFIC PUBLISHERS - WARSZAWA NORTH - HOLLAND PUBLISHING COMPANY - AMSTERDAM

Copyright by PAfiSTWOWE WYDAWNICTWO NAUKOWE (PWN-POLISH SCIENTIFIC PUBLISHERS) Warszawa, Poland 1966 and 1968

This book is a translation of the original Polish “Teoria mnogoici’’ published by PWN-Polish Scientific Publishers, 1966 in the series “Monografie Matematyczne” edited by the Polish Academy of Sciences. Translated from the Polish by M. Mrlnydski The English edition of this book has been publis.hed jointly by PWN and NORTH - HOLLAND PUBLISHING COMPANY

Library of Congress Catalogue Card Number 67-21972.

PRINTED IN POLAND

PREFACE

The creation of set theory can be traced back to the work of XIX century mathematicians who tried to find a firm foundation for calculus. While the early contributors to the subject (Bolzano, Du Bois Reymond, Dedekind) were concerned with sets of numbers or of functions, the proper founder of set theory, Georg Cantor, made a decisive step and started an investigation of sets with arbitrary elements. The series of articles published by him in the years 1871-1883 contains an almost modern exposition of the theory of cardinals and of ordered and wellordered sets. That the step toward generalizations which Cantor made was a difficult one was witnessed by various contradictions (antinomies of set theory) discovered in set theory by various authors around 1900. The crisis created by these antinomies was overcome by Zermelo who formulated in 1904-1908 the first system of axioms of set theory. His axioms were sufficient to obtain all mathematically important results of set theory and at the same time did not allow the reconstruction of any known antinomy. Close ties between set theory and philosophy of mathematics date back to discussions concerning the nature of antinomies and the axiomatization of set theory. The fundamental problems of philosophy of mathematics such as the meaning of existence in mathematics, axiomatics versus description of reality, the need of consistency proofs and means admissible in such proofs were never better illustrated than in these discussions. After an initial period of distrust the newly created set theory made a triumphal inroad in all fields of mathematics. Its influence on mathematics of the present century is clearly visible in the choice of modern problems and in the way these problems are solved. Applications of set theory are thus immense. But set theory developed also problems of its own. These problems and their solutions represent what is known as abstract set theory. Its achievements are rather modest in comparison

vi

PREPACE

to the applications of set theoretical methods in other branches of mathematics, some of which owe their very existence to set theory. Still, abstract set theory is a well-established part of mathematics and the knowledge of its basic notions is required from every mathematician. Last years saw a stormy advance in foundations of set theory. After breaking through discoveries of Godel in 1940 who showed relative consistency of various set-theoretical hypotheses the recent works of Cohen allowed him and his successors to solve most problems of independence of these hypotheses while at the same time the works of Tarsk showed how deeply can we delve in the domain of inaccessible cardinals whose magnitude surpasses all imagination. These recent works will certainly influence the future thinking on the philosophical foundations of mathematics. The present book arose from a mimeographed text of Kuratowski from 1921 and from an enlarged edition prepared jointly by the two authors in 1951. As a glance on the list of contents will show, we intended to present the basic results of abstract set theory in the traditional order which goes back still to Cantor: algebra of sets, theory of cardinals, ordering and well-ordering of sets. We lay more stress on applications than it is usually done in texts of abstract set theory. The main field in which we illustrate set-theoretical methods is general topology. We also included a chapter on Borel, analytical and projective sets. The exposition is based on axioms which are essentially the ones of ZermeloFraenkel. We tried to present the proofs of all theorems even of the very trivial ones in such a way that the reader feels convinced that they are entirely based on the axioms. This accounts for some pedantry in notation and in the actual writing of several formulae which could be dispensed with if we did not wish to put the finger on axioms which we use in proofs. In some examples we use notions which are commonly known but which were not defined in our book by means of the primitive terms of our system. These examples are marked by the sign #. In order to illustrate the role of the axiom of choice we marked by a small circle O all theorems in which this axiom is used. There is in the book a brief account of the continuum hypothesis and a chapter on inaccessible cardinals. These topics deserve a more thorough presentation which however we could not include because of lack of space

PREFACE

vii

Also the last chapter which deals with the descriptive set theory is meant to be just an introduction to the subject. Several colleagues helped us with the preparation of the text. Dr M. Mqczyriski translated the main part of the book and Mr R. Kowalsky collaborated with him in this difficult task. Professor J. EoS wrote a penetrating appraisal of the manuscript of the 1951 edition as well as of the present one. His remarks and criticism allowed us to eliminate many errors and inaccuracies. Mr W. Marek and Mr K. Wibiewski read the manuscript and the galley proofs and helped us in improving our text. To all these persons we express our deep gratitude. KAZIMIERZ KURATOWSKI ANDRZEJMOSTOWSKI

ERRATA Page, lin :

For

Read

18’’

Awl -A into subset

Awl =I onto subsets

232’ 324’

K. Kuratowski and A. Mostowski, Set Theory

CHAPTER I

ALGEBRA OF SETS

0 1.

Propositional calculus

Mathematical reasoning in set theory can be presented in a very clear form by making use of logical symbols and by basing argumentation on the laws of logic formulated in terms of such symbols. In this section we shall present some basic principles of logic in order to refer to them later in this chapter and in the remainder of the book. We shall designate arbitrary sentences by the letters p , q,r, ... We assume that all of the sentences to be considered are either true or false. Since we consider only sentences of mathematics, we shall be dealing with sentences for which the above assumption is applicable. From two arbitrary sentences, p and q, we can form a new sentence by applying to p and to q any one of the connectives: and,

or,

if

... then ... ,

if and only if.

The sentence p and q we write in symbols p A q. The sentence p A q is called the conjuncrion or the logical product of the sentences p and q, which are the components of the conjunction. The conjunction p A y is true when both components are true. On the other hand, if any one of the components is false then the conjunction is false. The sentence p or q, which we write symbolically p v q, is called the dis&nction or the logical sum of the sentences p and q (the components of the disjunction). The disjunction is true if either of the components is true and is false only when both components are false. The sentence i f p then q is called the implication of q by p , where p is called the antecedent and q the consequent of the implication. Instead

2

I. ALGEBRA OF SETS

of writing if p then q we write p + q. An implication is false if the consequent is false and the antecedent true. In all other cases the implication is true. If the implication p -+ q is true we say that q follows from p ; in case we know that the sentence p is true we may conclude that the sentence q is also true. In ordinary language the sense of the expression “if ..., then ...” does not entirely coincide with the meaning given above. However, in mathematics the use of such a definition as we have given is useful. The sentence p if and only if q is called the equivalence of the two component sentences p and q and is written p = q. This sentence is true provided p and q have the same logical value; that is, either both are true or both are false. If p is true and q false, or if p is false and q true, then the equivalence p = q is false. The equivalence p = q can also be defined by the conjunction

(P

+

4) A (4 -+ PI

-

The sentence it is not true that p we call the negation of p and we write l p . The negation l p is true when p is false and false when p is true. Hence l p has the logical value opposite to that of p. We shall denote an arbitrary true sentence by V and an arbitrary false sentence by F; for instance, we may choose for V the sentence 2.2 = 4, and for F the sentence 2.2 = 5. Using the symbols F and V , we can write the definitions of truth and falsity for conjunction, disjunction, implication, equivalence and negatiom in the form of the following true equivalences: (1)

F A F = F,

FAVGF,

VAFGF,

VAVEV,

(2)

F v F r F,

FvVrV,

V V F r V,

VVVZV,

(3) ( F + F ) r V ,

(F+V)=V,

(Y+F)rF,

(V+V)EV,

(4) ( F r F ) = V ,

(F=V)=F,

(V=F)=F,

(V=V)=V,

(5)

1 F r V,

1 V = F.

Logical laws or tautologies are those expressions built up from the letters p, q, r, ... and the connectives A , v ,-t, =, 1which have the

1. PROPOSITIONAL CALCULUS

3

property that no matter how we replace the letters p , q, r, ... by arbitrary sentences (true or false) the entire expression itself is always true. The truth or falsity of a sentence built up by means of COM~Ctives from the sentences p , q, r, ... does not depend upon the meaning of the sentences p , q, r, ... but only upon their logical values. Thus we can test whether an expression is a logical law by applying the following method: in place of the letters p , q, r, ... we substitute the values F and V in every possible manner. Then using equations (1)-(5) we calculate the logical value of the expression for each one of these substitutions. If this value is always true, then the expression is a tautology. Example. The expression ( p A q) + ( p v r) is a tautology. It contains three variables p , q and r. Thus we must make a total of eight substitutions, since for each variable we may substitute either F or V. If, for example, for each letter we substitute F, then we obtain ( F A F) -+ ( F v F ) , and by (1) and (2) we obtain F + F, namely V. Similarly, the value of the expression ( p A q) + ( p v q ) is true in each of the remaining seven cases. Below we give several' of the most important logical laws together with names for them. Checking that they are indeed logical laws is an exercise which may be left to the reader. ( P v 4 ) = (4 V P ) KP v 4) rl = CP v (4 v r)l ( P 4) = (4 A P)

*

"

[ P A (q A r)l= K P A 4 ) A rl

[ p A (q v r)] = [ ( p A q) v ( p A r)] [ p v (q A r)] = [ ( p v q ) A ( p v r)] (PAP)EP (PvP)=P, ( p A F ) = F, ( p v F) "p,

( p A V )= p ( p v V )= V

law law law law

of commutativity of disjunction,

of associativity of disjunction, of commutativity of conjunction, of associativity of conjunction,

first distributive law, second distributive law,

laws of tautology, laws of absorption.

In these laws the far reaching analogy between propositional calculus and ordinary arithmetic is made apparent. The major differences occur in the second distributive law and in the laws of tautol-

4

I. ALGEBRA OF SETS

ogy and absorption. In particular, the laws of tautology show that in the propositional calculus with logical addition and multiplication we need use neither coefficients nor exponents.

[(p -,q ) A (q r)] (p + r) (P v 1PI = v (PA 1 P ) G F P”l-7P 1 ( P v 4 ) = (1 P A 14 ) l ( P A 4 ) = ( 1P v 1 4 ) (P 4 ) = (1 4 1P) (P-,q)‘(lPVq), F+P, P+P, P + V . --f

+

-+

+

law law law law

of the hypothetical syllogism, of excluded middle, of contradiction, of double negation,

de Morgan’s laws, law of contraposition,

Throughout this book whenever we shall write an expression using logical’ symbols, we shall tacitly state that the expression is true. Remarks either preceding or following such an expression will always refer to a proof of its validity. $2. Sets and operations on sets

The basic notion of set theory is the concept of .set. This basic concept is, in turn, a product of historical evolution. Originally the theory of sets made use of an intuitive concept of set, characteristic of the so-called “naive” set theory. At that time the word “set” had the same imprecisely defined meaning as in everyday language. Such, in particular, was the concept of set held by Cantor’), the creator of set theory. Such a view was untenable, as in certain cases the intuitive concept proved to be unreliable. In Chapter 11, $2 we shall deal with the antinomies of set theory, i.e. with the apparent contradictions which appeared at a certain stage in the development of the theory and l) Georg Cantor (1845-1918) was a German mathematician, professor at the University of Halle. He published his studies in set theory in the journal Mathematische Annalen during the years 1879-1897.

2. SETS AND OPERATIONS ON SETS

5

were due to the vagueness of intuition associated with the concept of set in certain more complicated cases. In the course of the polemic which arose over the antinomies it became apparent that different mathematicians had different concept of sets. As a result it became impossible to base set theory on intuition. In the present book we shall present set theory as an axiomatic system. In geometry we do not examine directly the meaning of the terms “point”, “line”, “plane” or other “primitive terms”, but from a well-defined system of axioms we deduce all the theorems of geometry without resorting to the intuitive meaning of the primitive terms. Similarly, we shall base set theory on a system of axioms from which we shall obtain theorems by deduction. Although the axioms have their source in the intuitive concept of sets, the use of the axiomatic method ensures that the intuitive content of the word “set” plays no part in proofs of theorems or in definitions of set theoretical concepts. Sometimes we shall illustrate set theory with examples furnished by other branches of mathematics. This illustrative material involving axioms not belonging to the axiom system of set theory will be distinguished by the sign #= placed at the beginning and at the end of the text. The primitive notions of set theory are “set” and the relation “to be an element of”. Instead of x is a set we shall write Z ( x ) , and instead of x is an element of y we shall write x E y l). The negation of the formula x ~y will be written as x non EY, or x 4 y or 7(x E Y ) .To simplify the notation we shall use capital letters to denote sets; thus if a formula involves a capital letter, say A, then it is tacitly assumed that Ais a set. Later on we shall introduce still one primitive notion: x T R y (x is the relational type o f y ) . We shall discuss it in Chapter 11. For the present we assume four axioms:

1. AXIOMOF EXTENSIONALITY: r f the sets A and B have the same elements then they are identical. l) The sign E, introduced by G. Peano, is an abbreviation of the Greek word dart (to be).

6

I. ALGEBRA OP SETS

Al). AXIOMOF UNION: For any sets A and B there exists a set which contains all the elements of A and all the elements of B and which does not contain any other elements. B I). AXIOMOF DIFFERENCE: For any sets A and B there exists a set which contains only those elements of A which are not elements of B. Cl). AXIOMOF EXISTENCE: There exists at least one set. The axiom of extensionality can be rewritten in the following form: $, for every x, x E A t x E By then A = B, where the equality sign between the two symbols indicates that they denote the same object. It follows from axioms I and A that for any sets A and B there exists exactly one set satisfying the conditions of axiom A. In fact, if there were two such sets C, and C,, then they would contain the same elements (namely those which belong either to A or to B ) and, by axiom I, C1 = C,. The unique set satisfying the conditions of axiom A is called the sum or the union of two sets A and B and is denoted by A v B. Thus for any x and for any sets A and B we have the equivalence (1)

X E Au B ~ ( x E A ) v ( x E B ) .

Similarly, from axioms I and B, it follows that for any sets A and B there exists exactly one set whose elements are all the objects belonging to A and not belonging to B. Such a set is called the difference of the sets A and B and denoted by A-l?. For any x and for arbitrary sets A and B we have (2)

X EA-B

= ( X E A ) A ( x $B).

By means of de Morgan’s law and the law of double negation (0 1, p . 4) it follows that (3)

1( X E A -

B ) f 1( X E A ) v ( X E B),

i.e. x is not an element of A -B if x is not an element of A or x is an element of B. l)

In Chapter I1 these axioms will be replaced by more general ones.

2. SETS AM) OPERATIONS ON SETS

7

Using the operations u and - we can define two other operations on sets. The intersection A n B of A and B we define by A nB

= A-(A-B).

From the definition of difference we have for any x

X E A ~ B ~ ( ~ E A ) A ~ ( ~ E A - B ) , from which, by means of (3) and the first distributive law (see p. 3), it follows that x E A n B 3 ( x E A ) h [ l ( xE A ) v ( x E B)]

= [ @ € A )A l ( X E A ) ]

v [(XE A ) A (XE B)]

EFv[(xEA)A(xEB)]E[(xEA)~(xEB)],

and finally (4)

x E A n B = ( x E A ) h ( x E B).

Hence the intersection of two sets is the common part of the factors; the elements of the intersection are those objects which belong to both factors. The symmetric difference of two sets A and B is defined as (5)

A - B = ( A - B ) u (B-A).

The elements of the set A l B are those objects which belong to A and not to B together with those objects which belong to B and not to A . Exercises 1.Definetheoperations u , n , - b y m e a n s o f : ( a ) - , n , (b)-,u,(c) -,A. 2. Show that it is not possible to define either the sum by means of the intersection and the difference, or the difference by means of the sum and the intersection.

$3. Inclusion. Empty set A set A is said to be a subset of a set B provided every element of the set A is also an element of the set B. In this case we write A c B or B 13 A and we say that A is included in B. The relation c is called the inclusion relation. The following equivalence results from this definition {for every x ( X E A+ X E B ) } = A c B. (1)

8

I. ALGEBRA OF SETS

Clearly from A = B it follows that A c B, but not conversely. If A c B and A # B we say that A is a proper subset of B. If A is a subset of B and B is a subset of A then A = B, i.e.

( A c B) A ( B c A ) + ( A = B). To prove this we notice that from the left-hand side of the implication we have for every x

XEB+XEA,

X E A + X E B and

from which we obtain the equivalence x E A = x E B, and thus A = B by axiom I. It is easy to show that, if A is a subset of B and B is a subset of C , then A is a subset of C: ( A c B) A ( B c C ) + ( A c C),

(2)

i.e . the inclusion relation is transitive. The union of two sets contains both components; the intersection of two sets is contained in each component: (3)

A c A v B,

B c A v B,

(4)

A n B c A,

A n B c B.

In fact, from p

+ ( p v q)

x EA

it follows that for every x -P

[(xE A ) v (xE B)],

from which, by (I), 92, p. 6, X E A+ x E ( A v B), and by (1) we obtain A c A v B. The proof of the second formula of (3) is similar, the proof of (4)follows from the law ( p A q ) + p . From (2), $ 2 it follows that

A - B c A. Thus the difference of two sets is contained in the minuend. The inclusion relation can be defined by means of the identity relation and one of the operations u or n. Namely, the following equivalences hold (5)

(A c B) =(A u B

= B)

= ( A n B = A).

9

3. INCLUSION. EMPTY SET

In fact, if A c B then for every x , x E A + x E B; thus by means of the law (P 4) 4 [(PV 4) 41, [(x E A ) v ( x E B)] 4 (x E B), +

which proves that A u B c B. On the other hand, B c A u B and hence AuB=B. Conversely, if A u B = By then by (3) A c B. The second part of equivalence ( 5 ) can be proved in a similar manner. It follows from axiom B that if there exists at least one set A then there also exists the set A-A, which contains no element. There exists only one such set. In fact, if there were two such set Z1and Z 2 , then (for every x ) we would have the equivalence XEZ,

=XEZ2.

This equivalence holds since both components are false. Thus, from axiom I, Z1= Z 2 . This unique set, which contains no element, is called the empty set and is denoted by 0. Thus for every x X$O,

i.e. ( ~ € 0= ) F. The implication x E 0 -+ x E A holds for every x since the antecedent of the implication is false. Thus 0 t A, i.e. the empty set is a subset of every set. Formula (l), $2, p. 6 implies that X E

( A u 0) = (XE A ) v (xEO)

=( X E A ) v F =

XEA ,

because p v F 3p . From this we infer AuO=A, and from

7F E V A-0 = A .

The identity A n B = 0 indicates that the sets A and B have no common element, or -in other words -they are disjoint.

10

I. ALGEBRA OF SETS

The equation B-A = 0 indicates that B c A . The role played by the empty set in set theory is analogous to that played by the number zero in algebra. Without the set 0 the operations of intersection and subtraction would not always be performable and the calculus of sets would be considerably complicated. 84. Laws of union, intersection, and subtraction

The operations of union, intersection, and subtraction on sets have many properties in common with operations on numbers: namely, union with addition, intersection with multiplication, and subtraction with subtraction. In this section we shall mention the most important of these properties. We shall also prove several theorems indicating the difference between the algebra of sets and arithmetic I). The Commutative laws: A u B = B u A , A n B = B n A. (1) These laws follow directly from the commutative laws for disjunction and conjunction. The associative laws: (2) A u ( B u C ) = (A u B) u C,

A n ( B n C ) = ( A nB) nC.

Again, these laws are direct consequencer of the associative laws for disjunction and conjunction. Formulas (1) allow us to permute the components of any union or intersection of a finite number of sets without changing the results. Similarly, formulas (2) allow us to group the components of such a finite union or intersection in an arbitrary manner. For example: A u (Bu [CU (Du A')]} = [ Au ( D u C)]u (Bu E ) = (Eu C )u [Bu ( Au D)].

In other words, we may eliminate parentheses when performing the operation of union (or intersection) on a finite number of sets. The distributive laws: A n ( B u C ) = ( A n B ) u ( A n C), (3) A u ( B nC ) = ( A u B) n ( A u C ) . l) The theorems given in $ 4 are due to the English mathematician G. Boole (1813-1864), whose works initiated investigations in mathematical logic.

4. LAWS OF UNION, INTERSECTION AND SUBTRACTION

11

The proofs follow from the distributive laws for conjunction over disjunction and disjunction over conjunction, given in 5 1. The first distributive law is completely analogous to the corresponding distributive law in arithmetic. Similarly, as in arithmetic, from this law it follows that in order to intersect two unions we may intersect each component of the first union with each component of the second union and take the union of those intersections: ( A u B u ... v H ) n ( X u Y u ... u T ) =(An

X ) u (A nY )u

... u ( B n T ) u

... u ( A n T ) u ( B n A') u ( B n Y ) u ... u ( H n X ) u ( H n Y ) u ... u ( H n T ) .

The second distributive law has no counterpart in arithmetic. The laws of tautology: A uA =A,

(4)

A nA =A.

The proof is immediate from the laws of tautology ( p v p ) = p and

(PAP) =P. We shall prove several laws of subtraction. A u (B-A) = A u B.

(5)

PROOF.By means of (1) and (2), 92, p. 6 we have X E [ Au (B-A)]

A ) v [(XE B ) A ~ ( X A)], E from which, by the distributive law for disjunction over conjunction

X E [ Au (B-A)]

(XE

[ ( x E A )v (XE B)] A [ ( x E A )v ~ ( x E A ) ]

= ( X E A ) v (XE B), since (x E A ) v l ( x E A )= V , and Y may be omitted as a component of a conjunction. Thus XE [Au

(B-A)]

XE (A

u B),

which proves (5). From ( 5 ) we conclude that the operation of forming difference of sets is not the inverse of the operation of forming their union. For example, if A is the set of even numbers and B the set of numbers divisible by 3 then the set A u (B-A) is different from B, for it contains all even numbers.

12

i. ALGEBRA OF SETS

On the other hand, in case A P. 8,

c By we have

by (5) and (9,0 3,

A u (B-A) -- B,

as in arithmetic. A-B=A-(AnB).

(6)

PROOF. XEA-(A n B ) = ( x e A ) ~T ( ~ E A ~ B ) = ( X E AT)[A( X E A ) A ( X E B ) ] =(XEA)A[l(XEA)V l(XEB)] ~[(xEA)A~(XEA)]V[(XEA)A~(XEB)]

.

= P V [ ( X E A ) A 1( X E B)] = [(X E A ) A 7( X E B)] ~xEA-B.

The distributive law for union over subtraction has in the algebra of sets the following form A n (B-C) = ( A n B)-C. (7) This law follows from the equivalence X EA

n(B-

c)

[(X E A ) A ( X

EB) A

1( X E c)]

E[(xEA~B)A~(~EC)] n B)-C.

From (7) it follows that A n(B-A) =( A nB ) - A

= (B nA)-A

=B

n (A-A)

= B n 0 = 0.

Thus A n (B-A)

= 0.

De Morgan’s laws for the calculus of sets take the following form (8)

A - ( B n C ) = (A-B) u ( A - C ) , A - ( B u C)= (A-B) n (A-C).

In the proofs we make use of de Morgan’s laws for the propositional calculus. The following identities are given without proof.

(9) (10) (1 1)

( A u B ) - c = ( A - C ) u (B-C), A - ( B - C ) = (A-B) u ( A n C ) , A-(B u C ) = ( A - B ) - c .

4. LAWS OF UNION, INTERSECTION, A N D SUBTRACTION

13

The following formulas illustrate the analogy between the inclusion relation and the “less than” relation in arithmetic: (12)

( A c B ) A (c c 0)4 (A u C c B u D),

(13)

( A c B ) A (C c 0)4 ( A n C c B n D),

( A C B ) A (cC 0)-b (A-D C B-c). (14) From (14) it follows as an easy consequence that

(C c 0)+ ( A - D c A-C), (15) which is the counterpart of the arithmetic theorem: x Q y 4 2 - y Q z-x. Exercises 1. Prove the formula: N ( A u B) = N(A)+ N ( B ) - N ( A n B), where N ( X ) denotes the number of elements of the set X(under the assumption that X is finite). Hint: Express N ( A - B ) in terms of N ( A ) and N ( A n B). 2. Generalize the result of Exercise 1 in the following way

N ( A , u A2 u

... u An) = c N(Ai) I

N(Ai n Aj) 1.i

+ i Jc. k N(Ai n A j n Ak)- ...,

where the indices of the summations take as values the numbers from 1 to n, and they are different from each other. 3. Applying the result of Exercise 2 show that the number of integers less than n and prime to n is given by the formula

where p , ,p 2 , ...,p r denote all different prime factors of n.

0 5. Properties of symmetric differenceI) The symmetric difference A - B was defined in Q 2, p. 7 by the formula : A-B = (A-B) u (BAA). (0) I) The properties of symmetric difference were extensively investigated by M.H. Stone. S e e his The theory of representations for Boolean Algebras, Transactions Of the American Mathematical Society N(1936) 37-111. See also F. Hausdorff, Mengenlehre, 3rd edition. Chapter X.

14

I. ALGEBRA OF SETS

The operation (1)

is commutative and associative: A-B

(2)

=B

A-(B-I-C)

IA,

= (A-r-B)'C.

Formula (1) follows directly from (0). To prove (2) we transform the left-hand and right-hand sides of (2) by means of (0): A - ( l e C ) = A - [ ( B - C ) u (C-B)] = {A-[(B-C)

u (C-B)]} u {[(B-C) u ( C - B ) ] - A } .

Using (8), (9), (lo), and (ll), $4, p. 12, we obtain A-(B-C) = {[A-((B-C)]n[A-(C-B)]} = {[(A-B)

u [(B--)--A]

u [(C-@--A]

u ( A n C)] n [ ( A - C ) u ( A n B ) ] }u [B-(C u A)]

u [C-(Bu A)] = [ ( A - B ) n ( A - C ) ] u [ ( A - B ) n B] u [(A- C ) n C ] u ( A nB n C )u[(B-(C u A)] u [C-(B u A)] = [A-(B

u C)]u [B-(C u A)] u [C-(A u B)] u ( A n B n C ) .

Thus the set A - ( B - C ) contains the elements common to all the sets A, B, and C as well as the elements belonging to exactly one of them. To transform the right-hand side of (2) it is not necessary to repeat the computation. It suffices to notice that by means of (1) (A-B)-c

= C-(-A..-l?),

from which (substituting in the formula for A - ( B - L C ) the letters C, A , B for A , B, C respectively) we obtain (ALB)-C = [C-(AuB)]u[A-(BuC)]u[B-(CuA)]u(CnAnB)

= [A-(B

u C)]u [B-(Cu A)]u [C-(A u B)] u ( A n B n C ) .

Thus the associativity of the operation has been proved. It follows from (1) and (2) that we may eliminate parentheses when performing the operation on a finite number of sets.

15

5. PROPERTIES OF SYMMETRIC DIFFERENCE

The operation of intersection is distributive over

A,

that is

A n (BAG') = ( A n B)-(A n C ) . (3) In fact, it follows from (6) and (7), $4,p. 12 that A n ( B I - C ) = A n [(B-C) u (C-B)] = [(A n B ) - C ] u

[ ( An C ) - B ]

= [B n (A-C)]

u [C n (A-B)]

= ( B n [A-(A n C ) ] }u { C n [ A - ( A n B)]} = [(A n B)-(A n C)] u [(A n C ) - ( A n B)] = ( A n R)-I-(An C ) .

The empty set behaves as a zero element for the operation ,:

that is

(4) AAO = A . In fact, (A-0) u (0-A) = A u 0 = A . The theorems which we have proved so far do not indicate any essential difference between the operations and u. However, a difference can be seen in the following theorems. A - A = 0. In fact, A A A = (A-A) u (A-A) = 0. The operation of union has no inverse operation. In particular, we have seen that the operation of subtraction is not an inverse of the union operation. However, there does exist operation inverse to the operation :: for any sets A and C there exists exactly one set B such that A A B = C, namely B = A - C . In other words: (5)

(6)

(7)

A - ( A - C ) = c, A-B= C + B = A - C .

In fact, (2), (4) and (5) imply A - ( A I - C ) = ( A A A ) A C = OLC = c-0 = c,

which proves (6). If A - B = C then AA(A'-B) = A - C and hence B = A'C by means of (6). Thus (6) and (7) indicate that the operation A does have an inverse: the operation A itself. In algebra and number theory we investigate systems of objects usually called numbers with two operations and (called addition and

+

-

16

I. ALGEBRA OF SETS

multiplication). These operations are always performable on those objects and satisfy the following conditions: X+Y = y+x, 0) (ii) x+dv+z) = ( X + Y ) + Z , there exists a number 0 such that x+O = x , (iii) (iv) for arbitrary x and y there exists exactly one number z = x-y (the diflerence) such that y+z = x , (9 x * y =y . x , ( 4 xe(y.2) = (x.y).z,

(vii)

X.(Y+Z)

= (x.y)+(x.z).

Such systems are called rings (more exactly: commutative rings). If there exists a number 1 such that for every x (viii) x . 1=x , then we say that the ring has a unit element. The algebraic computations in rings are performed exactly as in arithmetic. For, in proving arithmetic properties involving addition, subtraction and multiplication, we make use only of the fact that numbers form a commutative ring with unit. Formulas (1)-(7) show that sets form a ring (without unit) if by “addition” we understand the operation A and by “multiplication” the operation n.A peculiarity of this ring is that the operation “subtraction” coincides with the operation “addition” and, moreover, “square” of every element is equal to that element. Using A and n as the basic operations, calculations in the algebra of sets are performed as in ordinary arithmetic. Moreover, we may omit all exponents and reduce all coefficients modulo 2 (i.e., 2kA = 0 and ( 2 k f l ) A =A). This result is significant because the operations u and - can be expressed in terms of A and n. Owing to this fact the entire algebra of sets treated above may be represented as the arithmetic of the ring of sets. In fact, it can easily be verified that: (8)

A u B = A L B - ( A n B),

(9)

A--B = A - ( A n B ) .

17

5. PROPERTIES OF SYMMETRIC DIFFERENCE

Formulas (8) and (4) imply the following theorem: (10)

if A and B are disjoint, then A u B = A-B.

The role which symmetric difference plays in applications is illustrated by the following example. Let X be a set and Z a non-empty family of subsets of X,that is, Z is a set whose elements are subsets of .'A Suppose that (Y

(1 1)

Cz)A(ZEz)

3

(YEI),

(YEZ)A(ZEI) + ( Y U Z E z ) .

A family of sets satisfying these conditions is called an ideal. We say that two subsets A , B of X are congruent moduIo I if A - B E Z and we denote this fact by A&B(mod I ) or by A s B if the ideal Z is fixed. Since 0 EZ, it follows from (5) that A IA, i.e. the relation A is reflexive. (1) implies that ( A & B) + (BG A), i.e. the relation =& is symmetric. Finally, the identity A IB = ( A A C )2 (B- C ) implies that A - B c (A'C) u (B-C), because the symmetric difference of two sets is contained in their union. By means of (1 1) we infer that (A&B)h(B&C)

--f

(A-C),

i.e. the relation & is transitive. Replacing the sign = by the sign in the previous definitions we obtain new notions. For example, two sets A and B are said to be disjoint modulo Z provided A n B i O (see p. 9); we say that A is included in B moduIo Z if A - B & 0, etc. Exercises 1. Show that the set A 1 - A 2 z ... -A,, contains those and only those elements which belong to an odd number of sets Ai (i = 1,2,..., n). 2. For A finite let N ( A ) denote the number of elements of A . Prove that if the sets A l , A z , ..., A,, are finite then

N ( A 1 L A 2 L... L A , , ) N ( A i ) - 2 x N(A8 n Aj)+4

= i

i. j

N ( A i n Aj n Ak) i.j.k

N ( A i n Aj n

-8 I, j . k . 1

n Al)+

...

18

I. ALGEBRA OF SETS

3. Show that (A, v A2 u

v An)-(Bi u B2 v

v Bn) c (Ai-Bi) u

(A, n A2 n ... n An)A(Bi n BZn ... n Bn) c ( A 1 - B I ) v

u( A ~ L B ~ ) ,

... u (An'Bn)

(Hausdorf). 4. Show that for any ideal Z the condition A t B implies

A u C 2 B v C,

A n C = B n C,

A -C 1B - C,

C - A = C - B.

5. For any real number t denote by [t] the largest integer < t. Let A t be the set of rational numbers of the form [nt]/n, n = 1,2, ... Prove that if Z is the ideal composed of all finite subsets of the set of rational numbers, then 1( A , G A, (mod Z)) and A, is disjoint (modulo I ) from A , for all irrational numbers x, y > 0, x # y.

.

$ 6 . The set 1, complement

In many applications of set theory we consider only sets contained in a given fixed set. For instance, in geometry we deal with sets of points in a given space, and in arithmetic with sets of numbers. In this section A , By... will denote sets contained in a certain fixed set which will be referred to either as the space or the universe and will be denoted by 1. Thus for every A A c 1, from which it follows that

(1)

Anl=A,

The set 1-A or -A:

Aul=&

1

is called the complement of A and is denoted by Ac -A=A"l-A.

Clearly, (2)

An--A=O,

Au-A=1.

Since - - A = 1-(l-A), we obtain by (lo), $4yp. 12 the following law of double complementation (3)

--A

=A.

Setting A = 1 in de Morgan's laws ((8), $4, p. 12) and substituting A and B for B and C,we obtain (4)

- ( A n B ) = - A u -By

- ( A u B ) = - A n -B.

19

6. THE SET 1, COMPLEMENT

Thus the complement of the intersection of two sets is equal to the union of their complements and the complement of the union of two sets is equal to the intersection of their complements. It is worth noting that the formulas which we obtained by introducing the notion of complementation are analogous to those of propositional calculus discussed in 0 1. To obtain the laws of propositional calculus (see p. 2-4) it sufficesto substitute in (1)-(4) the equivalence sign for the sign of identity and to interpret the letters A , B, ... as propositional variables and the symbols u , n , -,0.1 as disjunction, conjunction, negation, the false sentence and the true sentence, respectively. Conversely, theorems of the algebra of sets can be obtained from the corresponding laws of the propositional calculus simply by changing the meaning of symbols. From this point of view calculations on sets contained in a fixed set 1 can be simplified by using the operations u , n , Subtraction can be defined by means of the operation - and one of the operations u or n . In fact, we have A - B = A n(1-B) = A n -B and A - B = A n - B = - ( - A u B). The inclusion relation between two sets can be expressed by the identity ( A c B ) = ( A n - B = 0). (5) For assuming A c B and multiplying both sides of the inclusion by - B we obtain A n - B c B n - B and since B n - B = 0, we have A n - B = 0. Conversely, if A n - B = 0, then A = A n 1 = A n ( B u -B)

-.

n B ) u ( A n - B ) = ( A n B ) u 0 = A n B c B. Since ( A = B ) = ( A c B ) A ( B c A), it follows from (5) that ( A = B ) = ( A n - B = 0) A ( B n - A = 0), and, since the condition ( X = 0) A ( Y = 0) is equivalent to X u Y = 0, (6) ( A = B ) = [ ( A n - B ) u ( B n -A) = 01 = ( A L B = 0). It follows directly from ( 5 ) that ( A c B ) zz ( - B c -A). (7) (compare with the law of contraposition p. 4). = (A

20

I. ALGEBRA OF SETS

The system of all sets contained in 1 forms a ring where the operation & is understood as addition and n as multiplication. This ring differs from the ring of sets considered in 0 5 in that it has a unit element. The unit is namely the set 1. In fact, formula (1) states that the set 1 satisfies condition (viii), $ 5 , p. 16 characterizing the unit element of a ring. Hence calculations in the algebra of sets are formally like those in the algebra of numbers. Exercise The quotient of two sets is defined as follows A : B = A u -B. Find formulas for A : ( B u C) and for A : ( B n C) (counterpart of de Morgan's laws). Compute A n (B:c).

5 7.

Constituents

In this section we shall consider sets which can be obtained from arbitrary n sets by applying the operations of union, intersection, and difference. We shall show that the total number of such sets is finite and that they can be represented in a certain definite form (normal form). Let A l , A2, ..., A , be arbitrary subsets of the space 1. Throughout this section these subsets will remain fixed. Let A:=I-Al, for i = 1 , 2 , . . . , n . Each set of the form A?nA$n

... n A >

(i,=Oori,=l

f o r k = 1 , 2 ,... , n )

will be called a constituent. The total number of distinct constituents is at most 2", because each of the superscripts ik may have either one of the values 0 and 1. The number of constituents may be less than 2"; for instance, if n = 2 and Al = l-Az, then there are only three constituents: Az = A: n A:. 0 = A: n A: = A: n A:, Al = A: n A:, Distinct constituents are always disjoint. In fact, if and Sz = A i l n A$ n ... n A'," S, = A:' n A p n ... n A$

21

7. CONSTITUENTS

and if for at least one k < n, ik # jk, for instance ik = 0 and j , = 1, then A$ n Aj$ = 0. Hence S1n s2= 0. The union of all constituents is the space 1. It suffices to notice that 1 = (A! u A ; ) n (A! u A:) n ... n (A: u A:). By applying the distributive law of intersection with respect to union on the right-hand side of the equation we obtain the union of all the constituents. The set Ai is a union of all constituents which contain the component A!. If SI, S,, ... sh are all constituents, then 1 = sl u s, u ... s h . Therefore Ai = (Ain S,) u ( A i n S,) u ... u (Ain &). If S , contains the component A:, then Ai n S, = 0 because A in A : = A in (l-Ai) = 0. On the other hand, if S, contains the component A:, then Ai n S, = S,. Thus A iis the union of those constituents which contain the component A:. Q.E.D. We shall now prove the following THEOREM 1: Each non-empty set obtained from the sets Al ,A 2 , ...,A , by applying the operations of union, intersection and subtraction is the union of a certain number of constituents. PROOF.The theorem is true for the sets A l , A2, ... ,A , . It suffices to show that if X and Y are unions of a certain number of constituents then the sets X u Y, X n Y, X-Y can also be represented as the union of constituents (provided X u Y, X n Y, X-Y are non-empty). Assume that X and Y can be represented as unions of constituents:

x= sl u s2 u

u sk,

-

Y = s1 v

-

._

s 2

u

u sl.

It follows that

xu Y = (slu

..a

u sk) u & ( ?!I

u

u

s).

Thus X u Y is a union of constituents. From the distributive law for intersection with respect to union, it follows that XnY=(SlnS,)u(SlnS,)u

... u ( ~ , n S ) ... u u( S n ~ Sj>u

... u ( s k n S,).

I. ALGEBRA OF SETS

22

si n sj = 0 if Si # gj; otherwise Si n S j = Si.Thus X n Y is a union of constituents

X nY

u Si, u

= Si,

... u S i p ,

or else is empty. If among the constituents Sil,Si2, ...,Sip occur all of the constituents SlyS,,

X-Y

..., s k y then

= X-(X n Y)c

u

(s1

... u Sk)-(sl

u

... u s k )

= 0.

Otherwise, let S j l , S j , , ..., S j , be those constituents among S , ,S2, ... , S k which do not occur among the constituents Si,,Si,, ... ,S i p . We have

X-Y

Y) = [(Si, u ... u sip, u ( S j , u ... u Sj*)]-(Si, u ... u Sip) = ( S j , u ... u Sjq)-(Sil u ... u sip, = ( S j , u ... u S j , ) - [ ( S j , u ... u Sj,) n (Si, u ... u Sip,] = sj, u ... u sj,, = X-(Xn

because

(Sjl u

... u Sj,> n (Si, u ... u SiJ = 0.

Thus X u Y,X n Y and X-Y are representable as unions of constituents. Q. E. D. THEOREM 2: From n sets by applying the operations of union, intersection, and subtraction at most 2'" sets can be constructed. In fact, each such set, with the exception of the empty set, is a union of constituents. Because the number of constituents cannot be greater than 2", the number of distinct unions constructed from some (non-zero) number of constituents cannot be greater than 2'"- 1. Of particular importance is the case where all of the constituents are different from 0. In this case, we say that the sets Al , ..., A , are independent ') . l) The notion of independent sets plays an important role in problems connected with the foundations of probability theory. See E. Marczewski, Independence d'ensembles et prolongement de mesures, Colloquium Mathematicum l(1948) 122-132.

7. CONSTITUENTS

23

THEOREM 3 : I f the sets Al , ...,A , are independent, then the number of distinct constituents equals 2". PROOF.If

S =A:' n ... n A? =A(' n ... n A$ and not all of the equations il =j l , ...,in = j , hold, then S = 0. In fact, if for example, ip = 1 and j p = 0, then intersecting both sides of the last equation in (0) with A; we obtain S = 0. Thus if the sets Al , ...,A , are independent then equation (0) holds if and only if il =j l , ... , in =j,. Q. E. D.

(0)

Example. Let the set D, consist of sequences (zl, ... ,z,) such that each zf equals either 0 or 1 but z, = 0. The sets D 1, ... ,D, are independent. In fact, D k consists of those sequences (zl, ...,z,) for which z, = i,. Thus (il , ...,in) E 05 n ... n 0 : . We shall apply the concept of constituents to a discussion of the following problem of elimination. We introduce the abbreviations To(A) = { A contains at least n elements}, Ti@) 3 { A contains exactly n elements}. Let i, , ...,in, jl, ...,j, be sequences of the numbers 0 and 1. Let P I , ... ,Pn , 41, ...,q, be sequences of non-negative integers. We are interested in finding necessary and sufficientconditions for the existence of a set X satisfying the conjunction of the following conditions: Pk(X n Al), rk(X n A J , ... , P k ( X n An), (9 I'~;(--x nA ~ )T , ~ ; ( - xn A ~ ) ,... , T&-x n A,).

We assume at first that n = 1. Writing i, j , p , q, A instead of il, j 1 , PI

,q1,Al , we obtain the solution:

(ii) [(i =j = 1) A I'jtq(A)] v T:,.&A). In fact, if there exists a set X satisfying (i) and i = j = 1; then A is the union of two sets containing respectivelyp and q elements, and in this case A contains exactly p+q elements. If i = 0 v j = 0 then A is the union of two sets, one of which contains at least p elements and the other at least q elements. Therefore A contains at least p+q elements. Conversely, if condition (ii) is satisfied, then it suffices to choose as X any subset of A containing p elements.

24

I. ALGEBRA OF SETS

Assume that n > 1 and Al , ...,A,, are pairwise disjoint. If there exists a set X satisfying (i), then writing X , = X , s = 1,2, ...,n, we conclude that (iii)

T ~ (n x A,) , A T:;(-x,n A,)

s = I, 2,

for

..., n,

and by virtue of (ii)

[(is =is = 1) A r~,+q,(As)lv ~&+q,(As)y s = 1,2, ...,n. (iv) Conversely, if (iv) holds then for every s (1 s < n) there exists a set X , satisfying (iii). Let

<

X = [(XI n A,) u ( X z n A,) u

... u (Xnn A,,)] u (-Al n -Az n ... n -An).

Therefore

-X = [(-Xl u - A , ) n (-1, u --A2) n ... n (-X,, u -An)] n (A1 u ... u An). Since the sets Ai are disjoint, we have X n A ,

= X, nA,

and --X

n A , = -1, n A,. By applying (iii) we obtain (i).

Next we assume that for all r, s (1 < r, s < n) either A , = A , or A , n A , = 0. We shall designate conditions (i) by W,,Wz, ... ,W,,,V1,V 2, ...,V,,. We shall show that if A , = A , then W, -+ W,, or W, -+ W r y or else W, A W, = F. Indeed, if i, = is = 0, then W, --t W, if p r < p , , and W,-,W, if p , < p , . If i, = 1 and is = 0, then W, + W, in case p r p s , and in case p , < p , , W, A W, 3 F. Finally if i, = is = 1 then W, + W, for p r = p , and otherwise W, A W, = F. Similarly it can be shownthat either V , + V,, or V , --t V,, or V , A V , G F. We conclude that either the conjunction of (i) is false or else we may omit from (i) certain components and obtain an equivalent conjunction in which none of the sets A, occurs more than once. Thus this case is reduced to the preceding case. Now we shall reduce the general case to the case in which the sets A , are either identical or disjoint. For this purpose we note that if M n N = 0, then rj(MV N)

rj(M) V [rj-l(mA r ! ( N ) ]V [r:-2(WA r ; ( N ) ]V ... V [rh(kf) A r;(N)],

25

7. CONSTITUENTS

r b ( M U N ) 3 [ r ; ( M ) A r i ( N ) ] V [ r i - ~ ( M )A r : ( N ) ]V

... V [rA(kf) A rb(N)]. By induction, if the sets S,, ...,s h are pairwise disjoint, then the condition of the form I';(S, u ... u &) can be expressed equivalently as a disjunction of conjunctions, where each conjunction has the form I$:(Sl) A

... A F i ( S h ) .

Represent the sets A, as unions of constituents; then according to the above remark, each of the conditions (i) can be expressed as

a disjunction of conjunctions each of which has the form

r ; : ( x n S,)A

... A . F : ~ ( x ~&),

or respectively,

r,",'(--Xn

s,) A ... Ar,",h(-Xn

&).

Applying the distributive law for conjunction over disjunction, we express the conjunction of conditions (i) as a disjunction, each of which n S,> is a conjunction whose components have either the form or the form I';(-X n S,). Sets occurring in each such conjunction are either identical or disjoint. Thus the general case is reduced to the preceding one. Example. We shall find necessary and sufficient conditions for the existence of a set X satisfying the conditions

X n A n B f 0, XnAQB,

-Xn A n B # 0 , XnB+A.

These conditions can be expressed equivalently as the conjunction of the following six conditions:

r f ( X n A n B), r?(--Xn A n B),

I'f(Xn --A n B),

ry(Xn A n -B), T,"(-Xn A n -B),

r t ( - X n -A n B).

Hence we obtain the desired condition n B ) A Pf(A n - B )

A

I'f(--A n B).

26

I. ALGEBRA OF SETS

In other terms, A n -8 and B n -A have to be non-empty and A n B has to contain at least two elements I). Esercisas 1. Assuming that the set 1 is infinite and that A l , ...,A, are finite, describe a method of obtaining necessary and suflicient conditions for the existence of a finite set X satisfying the conjunction of conditions (i). 2. Let Z be the unit n-dimensional cube, that is, the set of sequences (xl, ...,x,) such that 0 < xi Q 1 ( i = 1,2, ...,n). Let Zm consist of those sequences (xl, ...,x,) E I where 1/2 < x, < 1. Show that the sets Zl,...,Z, are independent. Give a geometrical interpretation for n = 2 and n = 3.

0 8. Applications of the algebra of sets to topology *) In order to illustrate applications which the calculus developed in the preceding sections has outside of the general theory of sets, we shall examine the axioms of general topology and apply the algebra of sets to establish several results. In general topology we study a set 1, called the space, whose elements are called points. We assume, moreover, that to every set A contained in 1 there corresponds a set 2also contained in 1 and called the closure of A. The space 1 is called topological if it satisfies the following axioms (see also p. 119) (1)

AUB=AUZ

(2)

z=A,

(3)

ACA,

(4)

0

-

= 0.

I) The elimination method given above is due to Skolem, Untersuchungen uber die Axwme des Klassenkalkuls ..., Skrifter utgit av Videnskapsselskapet i Kristiania, I Klasse, No 3 (Oslo 1919). z, For more details on topological calculus developed in this section see K.Kuratowski, Topology Z, Academic Press 1966, Chapt. I. For further investigations on this calculus from an algebraical point of view see the paper of J.C.C. Mc Kinsey and A. Tarski, The algebra of topology, Annals of Mathematics 45 (1944) 141-191. In $8 we apply not only axioms I, A, B, C but also axioms (1)-(4). However, we can deduce all theorems givzn in this section from the full axiom system of set theory given in Chapter 11, treating axioms (1)-(4) as assumptions about the operation of closure.

27

8. APPLICATIONS OF THE ALGEBRA OF SETS TO TOPOLOGY

In axioms (1)-(3) the letters A and B denote arbitrary subsets of the space 1. # Axioms (1)-(4) are satisfied if, for example, 1 is the set of points of the plane and if the closure operation 2 consists of adding to the set A all points p such that every circle aroundp contains elements of A. This interpretation will be referred to as the natural interpretation of the axioms (1)-(4). # We shall show how, using only laws of the calculus of sets, it is possible to deduce a variety of properties of the closure operation. -

1 = 1.

(5)

PROOF.For every A we have

Ac 1, and by axiom (3),

-

1

c 1.

PROOF. From B u ( A - B ) = A u B applying axiom (1) we obtain B v A - B = 2 u 3.This implies that A c Bu A Y B and thus - - - A - B c ( B uA - B ) - F = A-B-B~ AT, which proves (6).

-

(7)

ACB+ACB.

PROOF.A c B is equivalent to the equation A u B = B. By axiom (l), XU Z= Z, thus A c F (Cf. 8 3, (9,p. 8).

PROOF.Since A n B c A and A n B c B, theorem (7) implies A n B c 2 and A n B c 2, from which it follows that A n B c A n F. -V A = A and B = B, then A n B = A n B. (9) PROOF.In fact, A n B c A n B by axiom (3), But by (8) and by the .hypothesis of (9), A n B c B= A n B. Therefore A n B = A n B. We call a set closed if it is equal to its closure. Theorem (9) states that the intersection of two closed sets is closed, and axiom (3) that the union of two closed sets is closed. We call a set open if it is the complement of a closed set. By de Morgan’s laws it follows that the union and intersection of two open sets is open.

xn

28

I. ALGEBRA OF SETS

# In the natural interpretation of axioms (1)-(4) closed sets are those sets which contain all their accumulation points (cf. p. 32). Open sets have the property: for every point p contained in the open set A there exists a circle with center p entirely contained in A. # The set ') ht(A) = 1-1-A =A"-"

is called the interior of the set A . The interior of any set is clearly an open set. # In the natural interpretation of axioms (1)-(4), the set Int (A) consists exactly of those points p for which there exists a circle with center p entirely contained in A . # Int(A) c A .

(10)

PROOF.By axiom (3), A" c A", from which, by applying equation (7), $ 6 , p. 19, we obtain 1-A"- c 1-A", hence A"" c A""= A. In particular, the relation Int(1nt ( A ) ) c Int (A) is a special case of (10). This relation may be strengthened as follows: Int (Int ( A ) ) = Int (A).

(1 1)

PROOF. It follows from the definition of Int ( A ) that Int ( A ) = A"--", Int (Int (A)) = [ h t (A)]"-"= [ A C - ~ F - C .

By the law of double complementation we may eliminate two consecutive occurrences of the operation AC;we thus obtain Int (Int ( A ) ) = A"--", and because A'-- = A"- by axiom (2), we obtain Int (Int ( A ) ) = A"-" = Int ( A ) . (12)

Int ( A n B ) = Int ( A ) n Int (B).

PROOF.By de Morgan's laws ( A n B)O-" = (Ac u BC)-", l)

Instead of A w e sometimes write A - .

8. APPLICATIONS OF THE ALGEBRA OF SETS TO TOPOLOGY

29

whence by axiom (1) Int ( A n B ) = ( A n B)c-c = (Ac- u Bc-)c, and a final application of de Morgan's laws gives Int ( A n R) =

n BC-O = Int ( A ) n Int (B).

As a simple consequence of (12) we have: A c B --r Int ( A ) c Int (B). (13) In fact, the assumption A c B gives us A n B = A , from which it follows that Int ( A ) = Int ( A n B ) = Int (A) n Int ( B ) c Int (B).

~-

Int (Int ( A ) ) = Int (A). PROOF.By (10)

Int (Int ( A ) ) c Int (A), whence by (7) and (2) ~-

(14,)

Int(iit(A)) c Int(A).

On the other hand, by (1 l), (3), and (13) Int ( A ) = Int (Int ( A ) ) c h t (Int (A)), and by (7) it follows that

Inclusions (14,) and (143 imply (14). Replacing Int(X) by X"-" in (14), we obtain AC-C-C-C-

-AC-C-

(15) Moreover, substituting Ac for A and applying the law of double complementation we obtain A-c-c-c= A-c-. 1) (16) Equations (15) and (16) show that if we apply in succession the operations of complementation and closure to the set A, then we obtain l) Formula (16) was given by K. Kuratowski in the paper: Sur I'opkrution d'dnulysis Situs, Fundamenta Mathematicae 3 (1922) 182-199.

d

30

I. ALGEBRA OF SETS

only a finite number of sets. Namely, if we start with the operation of complementation, then we obtain the sets A, AC, A C - , A C - C A C - C - Y A C - C - C AC-C-rAC-c-c-c 9

9

9

The next set in this sequence would be Ac-c-2-c-, but by (15) this set equals AC-"-. If, on the other hand, we start by applying the operation -, then we obtain the sets A', A-C, A-C-, A-c-C A - C - C A-C-L-C Y

3

The next set would be

t u t by (16) it is equal to the set

A-C-

Hence by applying the operations of complementation and closure to an arbitrary set A we obtain at most 14 distinct sets. Formulas (17) and (18) will be used in $9. (17)

Zf B = X+-,

then Int[Int(A-B) n B] = 0.

PROOF.Clearly A -B c Bc, whence by (13) and (7) Int[Int(A-B) n B] c Int[Int(Bc) n B].

Thus it suffices to show that Int[Int(Bc) n B] = 0. Since Int(BC) = BCC-C- = B-C- = X-C--Cformulas (12), (16), and (10) that

=x-C-C-

,

it follows by

Int[Int(Bc) n B] = Int[Int(Bc)] n Int(B) = X-c-c-c-c n BCdC BE-C = BC BC-C - 0. -x-C-c (18)

If A = 2 or B =

-~

then Int(A) u Int(B) = Int(A u B).

PROOF.From theorem (13) we conclude that Int(A)

c Int(A u B)

and Int (B) c Tnt (A u B), which implies that Int (A) u Int (B) c Int (Au B). Applying the closure to both sides of the inclusion we obtain by (1) and (7) -

(181)

___

~-

Int (A) u Int (B) c Int (A u B).

For the proof of the opposite inclusion we suppose that, for instance,

i? = B. We apply the identity AuBu[~-(AuB)]=

1,

8. APPLICATIONS OF THE ALGEBRA OF SETS TO TOPOLOGY

31

from which by axiom (3) it follows that A u B u 1 -(A u B) = 1 , whence Bul-(AuB)3

1-A.

Applying closure to both sides we obtain (dnce

B= B):

B u 1 -(A u B) 3 1-A, from which it follows that [1-1-A]uBul-(AuB) = 1, whence Int(A) u B u 1-(A u B) = 1.

It follows from this equation that Int(A) u 1-(A u B)

3

1-By

Int(A) u 1-(A u B)

3

1-B.

and thus __

Adding to both sides of this equation the set 1-1-B obtain

= Int(B)

we

Int(A) u Int(B) u 1- ( A u B ) = 1, thus Int (A) u Int(B)

3

1- 1 -(A u B ) = Int ( A u B).

Applying closure to both sides of the inclusion we obtain by (1) and (3)

Int(A)u Int(B)3 Int(A u B).

( 182)

Inclusions (18,) and (18,) prove theorem (18). Exercises 1. Prove that if the set A is open, then for every set X. 2. Let Fr(A) = 2 n Rove that :

I--A (the boundary of A).

(a) Fr(A u B ) u Fr(A n B) u [Fr(A)n Fr(B)] = Fr(A) u Fr(B), [A.H. Stone]

(b) Fr(A) = (A n

m)u @-A),

32

I. ALGEBRA OF SETS

(c) A u Fr(A) = (d) Fr[Int(A)] c Fr(A), (e) Int[Fr(A)] = A n Int[Fr(A)] = Int[Fr(A)]-A.

3. We call the set A boundary if F A = 1. The set A is called nowhere dense if A i s boundary. Prove that (a) the union of a boundary set and a nowhere dense set is boundary; @) the union of two nowhere dense sets is nowhere dense; (c) in order that the set Fr(A) bs nowhere dense it is necessary and sufficient that A be the union of an open set and a nowhere dense set. 4. Let 1 be a space satisfying besides axioms (1)-(4) the following axiom (where {p} denotes the set consisting of the single element p): {PI = {PI. We say that the point p is an accumulation point of the set A if p E A- {p} (for the plane this condition is equivalent t o the condition that p = lim p n , where pn n= m

E A-{p)).

By A' we denote the set of all accumulation points of the set A, called the derivative of the set A. Prove the formulas: (A U B)'= A' U Be, A'- B' c (A -B)', A" c A', A = A U A', A' = A'. 5. Let 1 denote the space considered in exercise 4. We call the set A dense in itself if A c A'. Prove that (a) if the space 1 is dense in itself, then every open set is also dense in itself; (b) if sets A and 1-A are boundary, then 1 is dense in itself; (c) the sets Int [Fr(A)] and A n Int [Fr(A)] are dense in themselves. 6. Conditions (1)-(3) are equivalent t o the condition

AuAuZ=A~B

.

[Iseki]

0 9. Boolean algebras We shall conclude this chapter with certain considerations of axiomatic character. If we examine the theorems of $52-8, we notice that the symbol E does not occur in the majority of them, though of course it does appear in the definitionsand proofs. This suggests developing aseparate theory to cover that part of the calculus of sets which does not make reference to the E relation. In this theory we shall speak only

9. BOOLEAN ALGEBRAS

33

of the equality or inequality between objects and terms resulting from these objects by performing certain operations on them. We shall base this theory on a system of axioms, from which we shall be able to prove all theorems of the preceding sections in which the E symbol does not occur. This theory, which is called Boolean algebra, has applications in many areas of mathematics'). Let K be an arbitrary set of elements, A and A operations of two arguments always performable on elements of K and having values in K. Finally, let o denote a particular element of K. We say that K is a Boolean ring or Boolean algebra with respect to these operations and to the element o if for arbitrary a, b, C E Kthe following equations hold (axioms of Boolean algebra): (1)

aAb

=b

A a,

(4)

aAa=o,

(5)

aAb=bAa,

(6) (7)

a A (b A c ) = (a A b) A c,

(8)

a A a = a,

(9)

a A (6 A c ) = ( a A 6 ) A ( a A c).

aAo=o,

We define the sum and the difference of elements of K by the equations

A [6 A (a A b)], = a A ( a A 6).

av b =a

a-b

We call a A 6 the symmetric difference of a and b, a A b theproduc, of a and b, and o the zero element'). l) There is a number ?f books with exposition of Boolean algebras. Among them let us mention R. Halmos, Lectures on Boolean Algebras (Princeton 1963) and the monograph of R. Sikorski, Boolean Algebras, 2nd edition (Berlin 1964). 2, The fact that we are using the same symbols for operations in Boolean algebra and for logical operations should not lead to misunderstanding.

34

I. AujEBRA OF SETS

An example of a Boolean algebra is the family of all subsets of a given fixed set 1 where the operations A and A are the set-theoretical operations of symmetric difference and intersection and where o denotes the empty set. We dealt with this interpretation of axioms (1)-(9) in 56l). More generally, instead of considering all the subsets of the space 1, we may limit ourselves to the consideration of any family of subsets K of 1 where the symmetric difference and intersection of two sets belonging to K also belong to K. Such a family is a Boolean algebra with respect to the same operations as in the preceding example. Each Boolean algebra of the type just described is called a jield of sets. We introduce Boolean polynomials. Let x1,x,, ... be arbitrary letters. The symbols (9 0 , (ii) x1,x,, ... are polynomials; iff and g are polynomials then the expressions (iii) ( f ) n (g), (iv) ( f ) A (g) are polynomials. A polynomial is to be understood as a sequence of symbols. Let us suppose that K is a Boolean algebra and that to every letter xi there corresponds a certain element n j E K. We define inductively the value of a polynomial with respect to this correlation. The value of polynomial (i) is the zero element of the algebra K, the values of polynomials (ii) are the corresponding elements in K; if the values of f and g are the elements a and by then the value of polynomial (iii) is a A b and the value of (iv) is a A b. The value of the polynomial .f is denoted by f K ( a , ,a,, ...); clearly f&,, 4 ,...) EK. Let the polynomial f have the form ... (h’) A (h”)..., and the polySimilarly as in $8, our exposition is based not only on the axioms of set theory but also on the axioms of Boolean algebra and, in part, also on topological axioms. As a matter of fact, we can deduce all theorems from the axioms of set theory given in Chapter 11, treating the axioms of Boolean algebra as assumptions about the operations A, A and the element o and the axioms of topology-as assumptions about the closure operation. Similar remarks apply to § 10.

35

9. BOOLEAN ALGEBRAS

nomial g the form ... (h") A (h') ... where the periods denote sequences of symbols which occur both in f and in g , and where h' and h" are polynomials. In this case we say that the polynomial g j s immediately transformable into the polynomial f by means of axiom (1). Similarly we define immediate transformability by means of the remaining axioms (2)-(9). We say that the polynomial g is transformable into f if there exists a finite sequence of polynomials f =fi ,f2, ... ,& = g such that is immediately transforfor each i (1 < i < k) the polynomial mable into the polynomialfi by means of one of the axioms. In this case we write f - g . Clearly, f - f , f - g - r g - f and f - g - h - r f - h . If f - g then f K ( a , ,a,, ...) = g K ( a l a2, , ...) for every Boolean algebra Kand arbitrary elements a ] E K. Polynomials resulting from the expression f l A f 2 A ... A fk (or from the expression fl ~ f A2... A&) by an arbitrary placement of parentheses are mutually transformable into each other by means of axiom (2) (or axiom (6)). For this reason we shall always omit parantheses when writing such polynomials. Moreover, we shall not take notice of the difference in the order of polynomials to which we apply successively either one of the symbols A or A . THEOREM 1: Every polynomial is either transformable into o or into some polynomial of the form s1 s2 ... A sh, where each of the polynomials sj has the form xi,A xi,A ... A xit (il < i2 < ... < i t , t 2 l), and no two components sj, sk (1 \< j < k h) are identical.') PROOF.The theorem is clear for polynomials (i) and (ii). Assume that it holds for polynomials f and g . Iff o (or g o), then (f)A (g) - g and ( f ) A ( g ) o (or (.f) A (g) f and ( f ) A ( g ) 0). At this point we may assume that f - s , A s2 A ... s h and g - t l A t2 A ... A t k . Thus ( f ) (g)"sl A s2 A Ash t l t2 ..* tk. By applying (3) and (4) we eliminate all redundant occurrences of components and thus obtain (f) ( g ) in the desired form. The theorem, tberefore, holds for formula (iii). In the case of polynomial (iv) we apply axioms (9) and (5) and

<

-

-

N

--

l) Theorem 1, as well as Theorem 2, is a scheme: for each polynomial f we obtain a separate theorem.

36

I. ALGEBRA OF SETS

obtain

(n ( g ) A

sh) A

“(Si

ti]

Sh) A tk1

[($I

A (%IA tk).

(S1 A ti) A A ( s p A tq) By means of ( 5 ) and (8) each of the polynomials spA tq is transformable into the product of individual variables. Omitting as in the previous case redundancies we obtain the desired form. The theorem, therefore, holds for formula (iv). Q.E.D. THEOREM 2: Let K be the &Id of all subsets of the non-empty set 1. I f f is a polynomial such that 1( f o ) , then there exist sets Al ,A Z ,... belonging to K such that fK(A1, A 2 , ...) # 0. PROOF.By Theorem 1 we may limit ourselves to consideration of the case where f has the form s1 As2 ... A sj, and where each of the polynomials sj is a product of letters x i . Let n be the number of distinct letters xi occurring in f. We shall prove the theorem by induction on n. For n = 1, f - x i , thus we may choose any non-empty set for the set A i . Assume that the theorem holds for all numbers less than n and that the polynomial f contains exactly n distinct variables. If one of the variables x, occurs in each of the expressions s j , then f x, A g where g contains less than n variables. By the induction hypothesis there exist sets A l , A z , ... such that gK(A1, A Z ,...) # 0. Replacing A , by 1 we leave the value of g unchanged (because g does not contain xp) and we obtain the set 1 n gK(A1, A z , ...) # 0 as the value off. If none of the letters x p occurs in each of the s j , then we substitute in f the symbol o everywhere for some arbitrary x,. Thus we obtain the polynomial g of fewer variables than f and l ( g 0 ) . Hence in this case the theorem follows from the induction hypothesis. 3 : Every equation f = g which is true for arbitrary sets (and THEOREM even for arbitrary subsets of a given non-empty set) is derivable from axioms (1)-(9). PROOF.If the polynomial f n g has the value o for all Al,A z , ... contained in a non-empty set 1, t h e n f a g - o and thus f - g . Therefore polynomial g arises from f by transformation by means of axioms (1)-(9) and by the general rule of logic which states that equal elements may be substituted for each other. e . 0

..a

-

-

-

31

9. BOOLEAN ALOEBRAS

Theorem 3 shows that the equations derivable from axioms (1)-(9) are identical with the equations true for arbitrary sets. Moreover, this theorem provides a mechanical procedure for deciding when an equation of the form f = g is derivable from axioms (1)-(9). Namely, it suffices to reduce the polynomial f A g by the method given in the proof of Theorem 1 and to determine whether or not it is transformed into 0. We introduce an order relation in Boolean algebra by the definition: a

< b = a ~b

= a.

THEOREM 4: a < b = a v b = b. PROOF.If a r\b = a, then a v b = a A b A ( a h b) = an b A a = b. Conversely, if a v b = b , t h e n a A b A ( a h b ) = b . T h u s a A b A b A ( a Ab) = b A b = o and a A o A ( a r \ b ) = 0, whence a A a A ( a h b ) = a A o = a and o A (a A b) = a; that is, a A b = a. We call an element i of the Boolean algebra K a unit of K if (10)

aAi=a

for all elements a E K. It is easy to prove that a unit, i f it exists, is unique. In an algebra with unit we define the complement of the element a by the equation: -a = i A a.

Axioms (1)-(9) are very convenient in most calculations but are seldom used to describe Boolean algebra. In the next theorem we shall present a different system of axioms, which is usually taken as the basis of Boolean algebra. We shall limit ourself to the consideration of Boolean algebras with unit. THEOREM 5: If K is a Boolean algebra with unit, then the following equations hold for all a, b, C E K : =bA

(i) a v b = b v a ,

(i’) a A b

a,

(ii) a v (6 v c) = (a v 6) v c,

(ii’) a A (b A c) = (a A b ) A c,

(iii) a v o = a,

(iii’) a A i = a,

(iv) a v -a = i,

(iv’) a A - a

= 0,

(v) a A ( b v c ) = (aAb)v(CtAc), (v‘) a v ( b A c ) = ( a v b ) A ( a v c ) .

38

I. ALGEBRA OF SETS

PROOF.Equations (i), (i’) (ii), (ii’), (%), (v), (v’) are true for arbitrary sets and thus are consequences of axioms (1)-(9). Equation (iii‘) is identical with (10). We establish (iv) and (iv’) as follows:

=o

by the definition of -a, from (9) and (10) from (8) from (4).

=a

A (i A a) A [a A (i A a)] = a A (i A a) A o

by the definition of sum, from (iv)

=(aAa)Ai

from (11, (21, (3) from (4) from (3).

a h - a = a~ ( i n a )

A (aha)

=a

=aAa a v -a

=oAi

= i

A partial converse to Theorem 5 holds: 0 , i E K ,and if v , A , - are operations defined on elements of K satisfying equations (i)-(v’), then K is a Boolean algebra with respect to the operations a A b = ( a A -b) v (b A -a), a A b and the element 0. The proof of the theorem is not difficult and is left to the reader. Equations (i)-(v’) are most often used as axioms for Boolean algebra. In particular, it is worth noting the symmetry of these equations with respect to the operations v and A . We shall conclude this section by giving an interesting example of a Boolean algebra. Let 1 be an arbitrary topological space with a closure operator (see 5 8, p. 26). We call A c 1 a regular closed set if

THEOREM 6: If K is a set,

A = Int ( A ) . By K we denote the family of all regular closed sets contained in 1. Clearly 0 and 1 belong to K since

-

-

Int(0) = 0 = 0

If A E K then

and

-

-

Int(1) = 1 = 1.

A= A , because

- -

A = Int (A) = Int (A) = A .

39

9. BOOLEAN ALGEBRAS

Thus every set belonging to K is closed (see $8, p. 27). By theorem (18), $8, p. 30, it follows that if A, B E K then

- -

A u B = Int(A) u Int(B) = Int(A u B), which proves that A u B EK. For A E K and B EK let

A @ B = Int(A n B),

A’ = Int(-A),

A o B = (A 0 B’) u (BOA‘).

It follows from this definition and from formula (14), $ 8, p. 29 that if A E K and BEK, then A@BEK,A’EKand AoBEK. THEOREM 7: K is a Boolean algebra with unit with respect to the operations o and 0 . PROOF.It suffices to show that the operations u , @ and satisfy axioms (i)-(v’) of Theorem 5. Axioms (i)-(iii) are clearly satisfied. Axiom (i’) follows from the equation A @ B = I n t ( A n B ) = I n t ( B n A ) = BOA. It is equally easy to show that axiom (iii‘) holds: A @ 1 = Int(A n 1) = Int(A)

= A.

To show that axiom (ii’) holds we apply (12) and (8) from $ 8 and obtain Int[(AgB) n C ] = Int(A@B) n Int(C), A 0 B = Int (A n B)

= Int (A) n Int (B) c

- -

Int (A) n Int(B)

= A n B;

it follows by (lo), $ 8 that Int[(A@B) n C] c ( A @B) n C c A n B n C c B n C. (*) Thus Int{Int[(AOB) n C]} c Int(B n C); that is (see (1 l), $ 8), Int[(A@B)nC]cInt(BnC) c I n t ( B n C ) = BgC. Since (by (*)) Int[(A @ B) n C] c A, we have Int[(A 0 B) n C] c A n (BO C), whence we obtain Int{Int[(AgB) n C]} c Int[A n (BOC)],

40

I. M E B R A OF SETS

and hence Int[(A@B) n C ] c Int[A n ( B O C ) ] . Taking closure on both sides of the inclusion we obtain Int[(A @ B) n C ] c Int[A n (B@C ) ] , that is, ( A @ B ) @ Cc A @ ( B @ C ) .The opposite inclusion is obtained in an entirely similar manner. Thus we may consider the equation (A @ B) @ C = A @ (BO C) as proved. We examine axiom (v). We have A @ ( B u C) = Int[A n ( B u C ) ] = Int[(A n B) w (A n C)]. The sets A, B and C are closed, thus (see (9) 58, p. 27) A n B = A n B and A n C = A n C. By (18) 58, p. 30 we conclude that Int[(A n B ) ' u (A n C ) ] = Int(A n B) U Int (A n C) = ( A @ B )u ( A @ C ) . Thus A @ ( B u C) = ( A @ B ) u ( A @ C ) . We check axiom (v') as follows. From the definitions,

--

A u ( B O C ) = Int(A) u Int(B n C). By (18) 58, p. 30, the right-hand side of the equation equals Int[Au ( B n C)], which equals Int[(A u B ) n (A u C)], that is, ( A uB) @ ( A u C). Axiom (iv') is an easy consequence of theorem (17)§8, p. 30 and of the fact that every regular closed set has the form X-+. Finally we prove that axiom (iv) holds. By (18) $8, p. 30 we have A u A' = Int(A) u Int(-A) = Int(A u -A), since = A. Thus we conclude immediately that A u A' = Int (1) = 1. Theorem 7 is thus proved. # Let us take the plane as the space 1. Every circle together with its boundary is clearly a regular closed set. Since every non-empty set of the form Int(A) contains some circle, we conclude that the Boolean algebra of regular closed sets in the plane has the following property : If A E K a n d A # 0, then there exists B such that B E K , 0 # B c A and B # A. #

41

9. BOOLEAN ALGEBRAS

Exercises 1. From every equation written in terms of variables, the symbols o and i and the operations v, A and - we obtain a new equation by interchanging the symbols o and i and the operations v and A. If the original equation is true in Boolean algebra, then so is the equation obtained from it in this way (Principle of Duality). 2. Show that axioms (ii) and (ii’) are derivable from axioms (i), (V), (iii)-(v), (iii’)-(v’). (Huntington)

4 10. Lattices The concept of lattice is more general than that of Boolean algebra. Let L be an arbitrary set of elements, upon which are defined the operations v and A . We say that L is a lattice with respect to the operations v and A if the following equations hold (axioms of lattice theory) (1) (2) (3)

ava=a, avb=bva, av(bvc)=(avb)vc, a A (a v b) = a,

(4) We call a lattice distributive if

( 5 ) a A (b v c) = (a A b) v (a A c),

a h a = a, aAb=bAa, alz(br\c)= ( a ~ b ) ~ c , av(aA6)=a.

a v (b A c)

= (a v b) A

(a v c).

We introduce an order relation between elements o f a lattice just as we did for Boolean algebras: a
(6) or, equivalently,

a

(7)


3

a h b = a.

Similarly we define the elements o and i (if they exist in the given lattice) as the elements satisfying conditions (8) for all a

a v o = a, E

ahi =a

L.

*)A detailed exposition of lattice theory is given in the book: G. Birkhoff, Lattice theory, 2nd edition (New York 1948). See footnote 1) on p. 34.

42

I. ALGEBRA OF SETS

It is easy to show that o is the smallest element in the lattice and that i is the largest, namely,

(9)

oga
for every a E L. Referring to Theorem 5 , p. 37 we observe that every Boolean algebra with unit is a distributive lattice with zero and unit. The converse does not hold, as is shown by the following counter-example which of itself is important for numerous applications in topology. The family of all closed subsets of an arbitrary topological space is a lattice (with the natural interpretation of the operations: a v b = a u by a A b = a n 6). However, this family is not in general a Boolean algebra, since the difference of two closed sets need not be closed (for example, when the space is the space of real numbers). On the other hand, the following theorem holds. THEOREM: If A is a distributive lattice with o and i and if for every a E A there exists an element -a E A satisfying the equations a v (-a) = i, a A ( - a ) = 0, (10)

then (i) the element -a is unique, and (ii) A is a Boolean algebra with zero and unit with respect to the operations v , A , and -. PROOF.Suppose that the element a’ also satisfies conditions (10). T h e n a ’ = a ’ ~ i = a ’ ~ ( a v - a ) = ( a ‘ r \ a ) v ( a ’ ~- a ) = o v ( a ’ ~ - a ) = a‘ A -a. Similarly, - a = --a A a‘, therefore a’ = -a. For the proof of the second part of the theorem it suffices to show that axioms (i)-(v‘) from p. 37 are satisfied. Axioms (i), (i’), (ii) and (ii’) hold in every lattice, (iii) and (iii’) follow from the assumption that o and i are zero and unit in A, (iv) and (iv’) follow from the assumption that condition (10) holds, and finally (v) and (v’) from the assumption that the lattice is distributive. The concept of Brouwerian lattice l) is intermediate between that of lattice and Boolean algebra. We call a lattice with unit Brouwerian if l) Brouwerian lattices were investigated in detail in the paper: J. C. C. Mc Kinsey and A.-Tarski, On closed elements in closure algebras, Annals of Mathematics 47 (1946) 122-162. The term “Brouwerian algebra” was introduced in this paper because of a close connection between these algebras and logic of Brouwer.

43

10. LA'lTICES

for arbitrary elements a, b 6 L there exists an element of L called the pseudo-diference of a and b and denoted by the symbol a*b such that (u*b c ) = (a b v c).

<

<

The family of closed subsets of a given space considered above is a Brouwerian lattice, where the pseudo-difference of two closed sets A and B is the closed set A - B . We denote by * a the pseudo-complement of a, namely * a = i*u. Notice that, in contrast to the operation of ordinary complementation, the equation (*a) A u = 0 does not hold. This corresponds to the fact that the law of the excluded middle does not hold in intuitionistic logic. In the topological interpretation this means that the nowhere dense set X - A n A , namely the frontier of A , is not necessarily empty. On the other hand, the validity of the equation (*a) v a = i corresponds to the law of contradiction in Brouwerian logic. # Examples

and exercises

1. The set of natural numbers is a lattice with respect to the operation of taking the greatest common divisor as the operation v and the least common multiple for the operation A . The formula a b means that b is a divisor of a. The number 1 is the unit of the lattice, and there is no zero element. 2. We consider euclidean n-space P and the family L, of its linear subsets (points, lines, planes and in general, k-dimensional subspaces where k n) passing through the origin. The family L, is a lattice with respect to the operations v and A defined by: A A B is the intersection of A and B; A v B is the least linear subspace of Lrn containing A and B. For example, if A and B are two planes, then A V B is a 3-dimensional spaceif A n B is a straight line, and is a 4-dimensional space if A n B is a point. The relation is the ordinary inclusion relation. The zero element of the lattice L, is the one-point set consisting of the origin and the unit is the entire space L,. The lattice L, is not distributive but is modular. Nsmely. we call a lattice modular if (a < c) + [a v (b A c) = (a v b) A c].

<

<

<

It is worth noticing that in the lattice L, every increasing sequence a, < a, <... contains at most n + l elements. 3. The set of sentences of an arbitrary mathematical theory is a lattice with respect to the operations of conjunction and disjunction when we identify sentences which are equivalent in terms of the propositional calculus. The zero element of the lattice is the false sentence and the unit element is the true sentence. The formula a < b means that a implies b. #

44

I. ALGEBRA

OF SETS

4 Prove that the formula (a

< C)

-+

[a V (b A

C)

< (U

V

b) A C]

holds in every lattice. 5. Prove that the equations given in (5) are equivalent. 6. Prove that the following fonnulas hold in every Brouwerian lattice (see Mc Kinsey-A. Tarski, op. cit., p. 124): (X S y ) - + ( X - r - Z ~ Y i Z ) A ( Z ' - y ~ Z l , ) ,

(x
A

= ( x l y = o),

y) = (ZLX)

( x v y ) 1.2 = (x 5 2)

V

(Z-f-J2),

" ( y ZZ),

7. The family of compact subsets of a topological space is a lattice with respect to the ordinary set theoretical operations u and n. If the space itself is not compact, then the-lattice does not have a unit; on the other hand, the lattice always has a zero (see p. 139 for the definition of compact space).

CHAPTER I1

AXIOMS OF SET THEORY. RELATIONS. FUNCIlONS

0 1. Propositional functions. Quantifiers We shall begin this chapter, like the previous one, by reviewing certain logical notions. In the propositional calculus, as outlined in 0 1, Chapter I, we dealt with sentences which have constant logical values. Now we shall consider propositional functions. They are expressions which contain variables. If each variable is replaced by the name of an arbitrary element, then the propositional function becomes a sentence. For instance,

> 0,

< 5,

X is a non-empty set are examples of propositional functions. By substitution we obtain, e.g., x

xz

the following sentences: the set of prime numbers is a non-empty set. 25 < 5, 1 > 0, We say that the object a satisfies thepropositional function @(x) if the sentence @(a)arising from @(x) by substituting the name of the object a for the argument x is true. When dealing with propositional functions we shall often assume that the variable x may be replaced only by names of elements belonging to a certain fixed set A. In this case we say that the domain of the propositional function is limited to the set A . For example, the domain of the propositional function x > 0 is limited to the set of numbers (real, natural or rational, etc.). On the other hand, the propositional function x = x has unlimited domain. If every element of the set A satisfies @(x), then we write

A @(XI,

XEA

which is read: for every x belonging to A , @(x).

46

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

For propositional functions with domain limited to the set A we write f', @ ( x ) instead of @(x).

A

x€A

X

Similarly, if @(x) is a propositional function with unlimited domain, then

A @(x) means that for all x,

di(x). For example, the sentence

X

Ax (x = x) is true. The formulas V@<x),

xEA

V@M

are read respectively: f o r some x belonging to A, @(x); and: for some x, @b(x). The symbols /\ and are called quantiJers, is the universal quantifier, the existential quantiJierI). Thus quantifiers are logical operators which allow us to form sentences from propositional functions of one variable and, more generally, propositional functions of n- 1 variables from propositional functions of n variables. In the sequel we assume that A # 0. If A contains a finite number of elements, a,,a,, ..., a,, then the statement that every element of the set A satisfies the condition @(x) is clearly equivalent to the conjunction of the n sentences @(a,),@(a2),..., @((a,) ; and the statement that some element of the set A satisfies the condition @ ( x ) is equivalent to the disjunction of these sentences:

v

v

A

A @ ( X ) S [ @ ( a , ) A @ ( ( a z ) ..*A

A@(Un)],

x€A

V di (x) = [@(al)v di (a2)v

.. v @(an)].

,

x€A

Hence the universal quantifier is a generalized logical conjunction and the existential quantifier a generalized logical disjunction. In the case where A is the empty set we let

A @ ( x )= V xco

and

v @(x) = F.

xro

I) Several authors use other symbols for quantifiers: for example, ( V x ) or ( x ) for the universal quantifier and ( Y x ) or (Ex) for the existential one.

47

1. PROPOSITIONAL FUNCTIONS. QUANTIFIERs

The following theorems (laws of logic) hold : (1)

I f U € A then

A @ (x) xeA

-+

@ ( a ) and @ (a) -+

v

@(x).

xeA

The second formula states that to prove an existential sentence of @ (x) it suffices to find an object a belonging to the set A and the form

v

xs.4

satisfying the condition @(x). Such a proof of an existential sentence is called efective.

A A Y(x)l= [A @ ( x )A A ~ < X > l , 'V P ( x ) v Y(X)l = [V @@) v v 9wl. x

(2)

[@(XI

X

(3)

X

X

x

X

Thus the universal quantifier is distributive over conjunction and the existential quantifier is distributive over disjunction.

(5)

a

v X

[@

(x) A y(x)1

-+

[v

@(XI A

X

v Y(x)lX

By means of simple counter-examples we conclude that the converse implications are not, in general, true. The universal quantifier is not distributive over disjunction and the existential quantifier is not distributive over conjunction. (6)

-IA@(x)l = v Cl@(X)l,

(7)

l[\l@(x>l = A [ l @ W l .

Laws (6)and (7) are called de Morgan's laws. It follows from them that: (6')

1A 1 1@(XI1 =

v @W,

(7')

Hence the existential quantifier can be defined in terms of the universal quantifier (and conversely, the universal quantifier in terms of the existential).

48

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

The first equivalence above shows that the existential sentence

v@(x) X

y@i(x).

can beproved by deriving a contradiction from the assumption X

Such proofs of existential sentences are often used but, in general, are not effective,i.e. they do not give any method of constructing the object satisfying the propositipal function @(x). Sentences can be considered as propositional functions without arguments. Clearly, we have APGP, VP'P.

Moreover, it is easy to check that

A [P v @Wl=[ P v A @(xj1,

(10) (11)

-

v [ P A@(x)l= [P v @Wl. A

X

By means of the law of the propositional calculus [ p + @@)I p v @(x)] it follows easily from (10) that

= [l

A @Wl= A [P

[P

(12) Observe that

[(A@(.I>

(13)

+PI

3

v

+

@Px)l*

[@i(x)

+

PI-

In fact, it follows from (6) that the left-hand side of (13) is equivalent to the disjunction \ l l @ ( x ) vp, and then by (9), to the disjunc-

v p . By (3) we obtain v[l@(x) v p ] or V [ @ ( x ) + p l . X

tion

vl@(x) v X

X

X

X

Propositional functions may involve several variables. For instance: x

>y,

x

E

x,

x2+y2+z2

# 0.

Propositional functions containing two variables are denoted by

@(x,r>,~ ( x , y )*.., .

Laws (1)-( 13) are also valid for propositional functions of several variables. Instead of p in (7)-(13) we can have any function which does not contain the variable x .

49

1. PROPOSITIONAL FUNCTIONS. Q U -

Note the following laws concerning propositional functions of two variables: (14)

A A @(X,Y> = A A @(XtY>,

(15 )

v v @&Y)

X

Y

X

Y

=-

Y

X

Y

X

v v @(XtY).

Thus it makes no difference in which order we write universal or existential quantifiers when they occur together. We usually write instead of A and instead of

A

v

A

XY

X

Y

v v.

XY

X

Y

= A [@(x)

v Yv(Y)l

XY

[V @(XI

(17)

A

v W)l= v v =v X

[@(XI A

YY(Y)l

Y

[@(XI

A

Y(Y)l.

XY

To prove (16) we substitute in (10)

A

Y ( y ) for p and observe that

Y

/\

YQv@(x)

Y

= A [!P(y)v@(x)J by means of the same law (10). Y

The proof of (17) is similar.

PROOF. Using (1) twice we obtain

A @(x,

4

y)

-

@(x, y) and @(x, y )

v @[x, y). These implications hold for any x,y, hence Y

X

By means of (12) and (13) we obtain (18). # As an application of (18) we shall discuss the difference between uniform and ordinary convergence of a sequence of functions. By the definition of limit, the sentence [limfn(x) =f(x)] is equivalent to the

A x

following

n=m

A A v A If,-!-k(x)-f(x)l

s>O

x

k

n

<

50

11. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

Interchanging the quantifiers

A and vk

we obtain the definition of

X

uniform convergence making k independent of x. The fact that k is independent of x and depends only on E is apparent in the above formula with the interchanged quantifiers. # The following diagram gives several other theorems concerning interchanging of quantifiers:

XY

By means of simple counter-examples it is easy to check that none of the above implications can be inverted. In set theory with primitive terms “set” and “membership” (cf. p. 5) a special role is played by a class 3 of propositional functions obtained by applying the operations of propositional calculus and quantifiers to propositional functions of the form

(0

Z ( x ) (ix., x is a set), x ~ and y x =y.

The following clauses constitute an inductive definition of 8: (A) Propositional functions (i) ( a d all the propositional functions analogous to (i), differing from (i) only by the choice of symbols x,y which can be replaced arbitrarily by any other symbols). (B) If @ and !P are propositional functions belonging to R, then @ v Y, @ A Y, @ + !P, @ = Y, 1@belong to R as well. (C) If @ belongs to the class R, then the propositional functions @ and @ (where x may be replaced by any other letter) belong

A

X

X

to the class 8 as well. (D) Every propositional function of the class Z arises by a finite number of applications of rules (A)-(C).

1. PROPOSITIONAL FUNCTIONS. QUANTIFIERS

51

The propositional functions mentioned in (A) are called the atomic formulas of the class 8. Extending the list of atomic formulas but preserving (B), (C) and (D) we obtain classes larger than 8. For instance, if we add new atomic formulas x P y , x Q y , ... (where again, the letters x , y , ... could by replaced by any others), then we obtain the class of propositional functions of general set theory extended by the primitive terms P , Q . ... . This class will be denoted by 8 [ P , Q , ...I. The propositional functions of this class will play a role when we admit primitive notions other than 2 and E.

5 2.

Axioms of set theory

We shall introduce axioms upon which we shall base the rest of our exposition of set theory. The new axioms will allow us to form new sets from given sets, and in this sense they do not differ from the axioms in Chapter I. On the other hand, the essential difference between the new axioms and the old ones is that we shall now deal with sets whose elements are also sets. I) First of all we retain the axiom of extensionality:

I. (AXIOMOF EXTENSIONALITY.) If the sets A and B have the same elements, they are identical. In symbols ,

A [ X € AE X € B ] + ( A = B). X

11. (AxIOhf OF THE EMPTY SET.) There exists a set 0 such that no x is an element of 0 ; symbolically:

v A, ( x 4 PI. P

x

Obviously, there is only one empty set (see p. 9). I) To make the terminology clearer we shall use the term a family of sets instead of a set of sets. Families of sets will be denoted by capital letters printed in boldfaced italics A , B, X, Y,etc.

52

ll. AXIOMS OF SET THEORY. RELATIONS. FLlNCTIONS

11'. (AXIOMOF PAIRS.) ') For arbitrary a, b, there exists a set which contains only a and b. In symbols:

v A { ( x E P )=[ ( x P

= a) v ( x = b)]}.

x

111. (AXIOMOF UNIONS.) Let A be afamily of sets. There exists a set S such that x is an element of S i f and only i f x is an element of some set X belonging to A. Symbolically: X E

(1)

S

=

v [(x€x)(XEA)]. A

X

By axiom I there exists at most one set S satisfying axiom 111 for a given family of sets A. In fact, if X

E s,

3

v v

[(X EX) A

(X€A)]

X

and X E

sz 3

[(X

EX) A (XE A ) ] ,

X

then for any x

x E s, E x E s, and, by axiom I, S , = S,. Axiom I11 states that there exists at least one set S satisfying formula (1). We infer that for any A this set is unique. This set is called the union of the sets belonging to A and is denoted by S ( A ) or U X . XEA

IV. (AXIOM OF POWER SETS.) For every set A there exists a family of sets P which consists exactly of all the subsets of the set A: ( X E P ) = ( X C A). It is easy to prove that the set P is uniquely determined by A. This set P is called the power set of A and is denoted by 2*. V. (AXIOMOF INFINITY.) There exists a family of sets A satisfying the conditions: 0 E A; if X EA, then there exists an element Y EA such that Y consists exactly of all the elements of X and the set X itsev. l)

This axiom can be derived from the remaining ones; therefore we do not give

it a separate number.

2. AXIOMS OF SET THEORY

53

Symbolically,

Thus the set 0 belongs to A, moreover, the set iVlwhose unique element is 0 also belongs to A. Similarly the set N.whose elements are 0 and Nlbelongs to A, etc. VI. (AXIOM OF CHOICE.) For every family A of disjoint non-empty sets there exists a set B which has exactly one element in common with each set A:

For easier reading of the formula, observe that the propositional func[(y E BnX) 3 (y = x ) ] states that there exists an element x tion

vA X

Y

such that the conditions y E B n X and y = x are equivalent. Thus the element x is the unique element of the intersection B n X and the propositional function above asserts that this intersection contains exactly one element. Not all mathematicians accept the axiom of choice without reserve, some of them view this axiom with a certain measure of distrust. They claim that proofs involving the axiom of choice are of a different nature than proofs not involving it, because it predicates the existence of a set B without giving a method of constructing it (i.e., it is not effective). Moreover, among the consequences of the axiom of choice there are many very peculiar ones. On the other I) This is the position of, for example, Borel and Lebesgue (see E. Borel, Elkments de la thkorie des ensembles (Paris 1949, p. 200), contrary to the position of, for example, Hausdorff and Fraenkel, who accept the axiom of choice without reserve attributing to it the same degree of “evidence” as to axioms I - V. A detailed analysis of numerous proofs involving the axiom of choice was made by W. Sierpiriski in his paper L’axiome de M. Zermelo et son rdle duns la thiorie des ensembles et I’analyse, Bulletin de 1’Academie de Sciences de Cracovie (1918) 97-152.

54

II. AXIOMS OF SET THEORY. RELATIONS. FUNCl’IONS

hand, it is unquestionable that many important discoveries in mathematics would never be made without appealing to this axiom. In the sequel we shall mark theorems which are proved using the axiom of choice with the superscript As a rule the axiom of choice will be used only in proofs of theorems which have not yet been proved without the use of this axiom. For each propositional function @(x)of the class W we assume the following axiom’): VIL. (AXIOM OF SUBSETS FOR THE PROPOSITIONAL FUNCTION @.) For any set A there exists a set which contains the elements of Asatisfying the propositional function @ and which contains no other elements. Symbolically this axiom can be written in the following form: There exists a set B such that

A { (x

E B)

= [(xE A ) A @(x)]}

X

(we suppose that the variable B does not occur in @). If @(x) contains free variables different from x, then they act as parameters upon which B depends. Clearly, the set B is uniquely determined by the propositional function @. It is denoted b y { x E A :@ ( x ) } ,which is read “the set of x which belong to A and for which @(x)”. Finally, for every propositional function of the class W in which variables z and B do not occur we have the following axiom: VIb. (AXIOM OF REPLACEMENT FOR THE PROPOSITIONAL FUNCTION @.) If for every x there exists exactly one y such that @(x,y), then for every set A there exists a set B which contains those and only those elements y for which the condition @ (x,y ) holds for some x E A. Symbolically,

{A v A [ @ ( X , Y ) X

I

Y

-0,

=

41) -b

A v r\ [ W B )= v @ ( X , Y ) I . A

B

Y

XEA

The intuitive content of this axiom is as follows. Suppose that the antecedent of the implication holds, namely, for every x there exists exactly one element y satisfying @(x,y). In this case we say that y corresponds to x. The axiom states that for every set A there exists a *) This axiom is dependent on the remaining ones, therefore we do not give it a separate number.

2. AXIOMS OF SET THEORY

55

set B which contains all the elements y corresponding to elements of A and which contains no other elements. For instance, if @(x, y ) is the propositional function Z ( x ) A (y = 2"), then the element corresponding to the set X is the power set 2x. By the axiom of replacement, for every family of sets A there exists a family of sets B which consists of all the power sets ZX where X E A . Observe that we have defined not just one axiom of replacement but actually an infinite number of them. In fqct, for every propositional function belonging to the class R and not involving the variables B and z we have a separate axiom. Similarly, axioms VI; form an infinite collection of axioms. If the uniqueness condition occurring in the antecedent of axiom VIT, is satisfied, then the set B whose existence is asserted in the conclusion of the axiom is unique. The proof follows immediately from axiom I. We call the set B the image of A obtained by the transformation @ and we denote it by {@] "A. Axioms I-VI and all of the axioms VIIo where CJ is an arbitrary , constitute an infinite system of propositional function of the class R axioms, which will be denoted by 2'. Eliminating the axiom of choice from the system Zowe obtain the axiom system denoted by Z. As already mentioned on p. 51 we shall introduce later new primitive notions P, Q, ... into set theory. Then we shall use axioms I-VI and all axioms of the form VII,, where @ is an arbitrary propositional function of the class %[P,Q, ...I. This system, which includes the system Zo, will be denoted by Zo[P,Q, ...I. If we omit the axiom of choice then we shall be dealing with the system which we denote by Z[P,Q, ...I. We can fully appraise the meaning which individual axioms have for set theory only after acquainting ourselves with conclusions which follow from them. At this time we shall be content to make a few remarks of a general character. Axioms 111, IV, VI, and VII are axioms of conditional existence, that is, they allow us to conclude the existence of certain sets assuming the existence of others. Constructions of sets based on axioms III, IV and VII are unique. On the other hand, axiom VI does not assert the existence of a unique

56

II. AXIOh4S OF SET TBEORY. RELATIONS. FUNCTIONS

set: for a given family A of non-empty disjoint sets there exist, in general, many sets B satisfying the axiom of choice. Axioms 11 and V are existential axioms: they postulate the existence of certain sets independently of any assumptions concerning the existence of other sets. We make a few more remarks of a more general nature. Axioms in mathematical theories can play one of two roles. There are cases where the axioms completely characterize the theory, i.e., they constitute in some sense a definition of the primitive notions of the theory. Such is the case, for instance, in group theory: we define a group as a set and operations satisfying the axioms of group theory. In other cases the axioms formalize only certain chosen properties of the primitive notions of the theory. In this case the purpose of the axioms is not to give a complete description of the primitive notions but rather to give a systematization of the intuitive concept. This second point of view is that taken in this book. There arise, of course, problems of a philosophical nature related to establishing the intuitive truth of the axioms. However we shall not treat these problems here. Since the intuitive content of the notion of set is not completely characterized by axioms I-VIT, it is not surprising that the axioms do not suffice for establishing certain results from intuitive set theory. As an example of an intuitively obvious property of sets which cannot be derived from the axioms I-Vn: we give the following: If A is a non-empty family of sets, then there exists a set X such that X E A and X n A = 0 . Some authors accept the so-called axiom of regularity, which states that every family of sets has the property described above. From the axiom of regularity it follows in particular that X E X for no X and, more generally, X , E X * € .. . E X ,E X , for no X , , X,,...,X,. In this book we will not use the axiom of regularity. In Chapter X we shall treat certain other axioms independent of the axioms I-VII. Some historical and bibliographical remarks on the axiomathation of set theory are in order. The first axiomatization of set theory is due to E. Zermelo (Untersuchungen iiber die Grundlagen der Mengen-

2. AXIOMS OF SET THEORY

57

lehre I, Mathematische Annalen 65 (1908) 261-281). The primitive notions of his axiom system were "set" and E, as in the system 2".Axioms I, IV, V and V1 occur in Zermelo's set theory in the same form as in 2". Instead of the infinite collection of axioms VII (of replacement) Zermelo used the axiom of subsets VI'. The ambiguous formulation of this axiom was a source of much discussion. The definite form of the axiom was proposed by Th. Skolem (see the paper of 1922 cited below). Axiom VII was formulated at about the same time by three authors: by D. Mirimanov in Les antinomies de Russell et de Burali-Forti et le problt?me fondamental de la thkoorie des ensembles, Enseignement MathCmatique 19 (1917) 37-52, A. Fraenkel in Z u den Grundlagen der CantorZermeloschen Mengenlehre, Mathematische Annalen (1922) 230-237, and by Th. Skolem in Einige Bemerkungen zur axiomatischen Begrundung der Mengenlehre, Wissenschaftliche Vortrage gehalten auf dem fiinften Kongress der Skandinavischen Mathematiker in Helsingfors (Helsingfors 1922) 217-232. The most influential of these papers was that of Fraenkel, and for this reason axiom VII is often credited to Fraenkel. The axiom of regularity is basically due to Mirimanov (see his paper cited above). Axiom systems of set theory in which the axiom of replacement is written as a scheme of axioms dependent upon an arbitrary propositional function @ (as it is in Z") are called Zermelo-Fraenkel set theories. If instead of the axiom of replacement only the axiom of subsets is assumed, then the system is said to be a Zermelo set theory. A different axiomatization of set theory was proposed by v. Neumann (Die Axiomatisierung der Mengenlehre, Mathematische Zeitschrift 27 (1928), 669-752). Von Neumann introduced into set theory a new primitive notion which entered into his formulations of the axiom of replacement and the axiom of subsets and which allowed him to dispense with referring to the propositional function @. As first suggested by Bernays (A system of axiomatic set theory I, Journal of Symbolic Logic 2 (1937) 65-77), we can take the notion of a class as this primitive notion. Intuitively, X i s a class if it is the collection of all objects satisfying some propositional function. Such a collection is not in general a set (see p. 61).

58

n.

AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

In this intuitive explanation of the notion of class we use such unclear notions as “collection” and “satisfying”. In the formulations of the axiomatic system such notions do not occur. All axioms are stated in terms of the primitive notions “set”, “class”, “membership of an element in a class”, and “membership of an element in a set”. In the literature, systems involving these primitive notions are said to be Godel-Bernays systems. An advantage of these systems over systems of the Zermelo-Fraenkel type is finite axiomatizability; a disadvantage is a less natural choice of primitive notions’). The most famous work based upon a Godel-Bernays type formulation of set theory was done by K. Godel (The consistency of the axiom of choice and of the generalized continuum hypothesis with the axiGms of set theory, Annals of Mathematics Studies, 2nd edition (Princeton 1951)). See also A. A. Fraenkel and P. Bernays, Axiomatic set theory (Amsterdam 1958). In general it seems that the axiomatization of set theory has not yet reached its most perfect form. A critical exposition of different systems proposed as a basis for set theory is given in the book Foundations of Set Theory (Amsterdam 1958) by A. A. Fraenkel and Y.Bar-Hillel. This book contains a complete bibliography concerning the philosophical and mathematical foundations of set theory. The book by Hao Wang and R. Mc Naughton Les systzmes axiomatiques de la thborie des ensembles (ParisLouvain 1953) contains a brief but useful description of the work done in the field. Exercise Show that axioms 111, IV and VII can be replaced by a single axiom stating the existence of the set S({@}“ 2A) for every set A and for every propositional function @ belonging to R. [Hao Wang]

0 3.

Some simple consequences of the axioms Starting with this paragraph, our exposition of set theory will be based upon the axioms of the system 2”. Theorems dependent upon the axiom of choice will be marked by O.

l) Of course whether a notion is natural or not is not its objective property, but depends on psychological and historical motives.

3. SIMPLE CONSBQUBNCES OF THlz AXIOMS

59

THEOREM 1: (THE EXISTENCE OF PAIR.) For arbitrary a, b there exists a unique set whose only elements are a and b.

PROOF.Uniqueness follows from axiom I and existencefrom axiom 11'. The set, whose uniqueness and existence are stated in Theorem 1, is called the urtorderedpair of elements a, b and is denoted by {a, 6). If a = b then we simply write {a}. TH~OREM 2: (THE EXISTENCE OF UNION.) For arbitrary sets A and B

there exists a set C such that (XEC) = [ ( x e A ) v ( X E B ) ] .

In fact, C =

U X.

XdA.BJ

Theorem 2 asserts that axiom A @. 6) is a consequence of the axioms Z. THEOREM 3: (THEEXISTENCE OF UNOXDERED TRIPLES, QUADRUPLES, For arbitrary, a, b, c, ..., m there exist sets: {a,b, c) whose only elements are a, b and c; {a,b,c,d} whose only elements are a, by c and d, ...; {a,b, ...,m) whose only elements are a, by ..., m. ETC.)

In fact, (a,b, c} = {a,b} u {c}, {a,b. c, d } = {a,b, c> u {d), etc. The set l (1) is called an ordered pair. We call a the first term of (a, b) and b the second term of (a,b ) I). THEOREM 4: In order that
PROOF.The sufficiency of the condition is obvious. To prove its necessity, suppose that E (a, b )

and

{c, d> E (a, b),

l) This definition of an ordered pair was given by K. Kuratowski in the paper Sur la notion de l'ordre dans la rMrie des ensembles,Fundamenta Mathematicae 2 (1921) 161-171. See also: N. Wiener, A simplification of the logic of relations, Proceedings of the Cambridge Philosophical Society 17 (1912-1914) 387-390.

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

60

that is (i) { c } = {a> and

(ii) {c> = {a,bl,

or

(iii) {c, d } = {a}

or

(iv) {c, d } = {a,b}.

Formula (ii) holds if a = c = b. Formulas (iii) and (iv) are then equivalent and it follows that c = d = a. Hence we obtain a = c = b = d in which case the theorem holds. Similarly, one can check that the theorem holds for case (iii). It remains to be shown that the theorem holds for cases (i) and (iv). We have then c = a and either c = b or d = b. If c = b then (ii) holds and this case has already been considered. If d = b then a = c and b = d, which proves the theorem. COROLLARY: If (a, b ) = (b, a ) then a = b. By the definition of the set { ~ E A@: ( x ) } axiom VIL implies the following theorem: THEOREM 5: (2)

?€{.YEA:

@(X)}

[@((t)

A

(t € 4 1 .

In particular, if @(x) 3 ( x E A ) (i.e. if the domain of the propositional function @ is limited to A), then t € { x : Q,(x)} = @(t).

(3)

Equivalence (3) leads easily to the following theorems (with the assumption that the domains of Q, and Y are limited to A ) : (4)

{x: Q,(x) v Y(x)} = {x: Q,(x)}u{x: Y(x)},

(5)

{ x : @(x)A Y(x)}= {x: @(x)}n{x: Y(x)},

(6)

{x:

1@(x)} = A + :

@(x)}.

As an example we shall prove (4). For this purpose we apply equivalence (3) to the propositional function @(x)vY(x) and we obtain

(0

t € { x : @ ( x ) v Y ( x ) } 3 [Q,(t) v Y(t)l. According to (3) we have @ ( t ) = t ~ { x : @(x)}

and

Y ( t ) = t ~ { x :Y ( x ) } ,

3. SIMPLE CONSEQUENCES OF THE AXIOMS

61

thus it follows from (i) that t E { x : @(x) v Y ( X ) ) )= t E { x : @ ( x ) } v t E { x : u'(x)} = t E { X : @ ( x ) } u { x : !P(x)},

which proves (4). THEOREM 6: For every non-empty famfly of sets A- there exists a unique set containing just those elements which are common to all the sets of the family A . This set is called the intersection of the sets belonging to the family X. A and is denoted by P ( A ) or

n

XEA

For we have: P(A) = { X E S(A) i ,A, [ ( X EA ) -+ ( X EX)]}. X

If the family A is composed of sets X,,X,, ...,X,,(n finite), then P(A) = X In&n ...nX,,. In case A = 0, the operation P ( A ) is not performable. We conclude this section with a remark on so-called antinomies of set theory. A naive intuition of set would incline us to accept an axiom (stronger than axiom VIL) stating that for any propositional function there exists a set B containing those and only those elements which satisfy this function. The creator of set theory, Cantor, believed (at least at the beginning of his work) that such an axiom was true'). However, it became very early apparent that the axiom formulated in this way leads to a contradiction (to an antinomy). Let us take as an example the propositional function @ ( x ) = (x is a set) A ( x $ x), which leads to Russell's antinomy'). We shall prove THEOREM 7: There exists no set Z such that

A [(X€Z) =@(x)]. X

See G. Cantor, uber unendliche. lineare Punktmannigfaltigkeiten, Mathematische Annalen 20 (1882) 113-121. 2) This antinomy was first published in the appendix to the book of G. Frege, Grundgesetze der Arithrnetik. vol. 2 (Jena 1903). 1)

62

11.

AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

PROOF.If such a set existed, then the equivalence ( x E Z ) = ( X is a set)A(x$x) would hold. From the assumption that 2 is a set we obtain the contradiction 2 E Z = l ( 2 E 2). Interdependence of the axioms. We have shown that axiom A (p. 6 ) follows from Z. Axiom B (p. 6 ) follows from C as well, because A - B = { x E A : l ( x B)}. ~ Axiom C (p. 6 ) follows directly from axiom 11 (the axiom of empty set) or from axiom V (the axiom of infinity). Axiom 11' follows from the other axioms of the system Z. In fact, let A be a family of sets such that O E A and such that there exists at least one non-empty set belonging to A . This family exists by the axiom of infinity. The set {a, b} is the set {@}"A where @ is the propositional function

"

E(x = 0) A (Y = a>l E(x # 0) A (v = bll. Axiom VIL (of subsets) is also a consequence of the other axioms of the system 2. In fact, let A be a set and @(x) a propositional func[(x E A ) -- l @ ( x ) ] , then the empty set satisfies the axiom tion. If

A X

of subsets. Otherwise, let a be an arbitrary element of A such that @ (a). Denote by Y(x, y ) the propositional function [@ ( x ) A (y = x)] v [l@ (x) A (x = a)]. For every x there exists exactly one y such that Y(x, y ) ; namely this element y is x or a, depending on whether @(x) or l @ ( x ) . The set ( Y } " A clearly satisfies the thesis of axiom VI;. Exercises

Show that 1. If X EA , then P(A\ c X c S(A). 2. S(Al u A2) = S(A1) u S(A*). 3. If A l n Az 7t 0. then P ( A l n A*) r> P ( A l ) n P(A&

0 4. Cartesian products. Relations The Cartesian product of two sets X and Y is defined to be the set of all ordered pairs (x, y) such that x E X and y E Y. The existence of this set can be proved as follows. If x EX and y E Y ,

4. CARTESIAN PRODUCTS. RELATIONS

63

then {x, y } c X u Y and { x } c X u Y , whence The set

{t ET

v v (t

xaX YEY

= (x, y))),

where

T = 22x”y~

exists by means of axioms IVY VI‘ and Theorem 2, p. 59. This set contains every ordered pair (x, y ) , where x E X,y E Y , and contains no other elements. Hence this set is the Cartesian product of X and Y. Since there exists at most one set containing exactly the pairs (x, y } , x E X , y e Y , the Cartesian product is uniquely determined by X and Y . This product is denoted by X X Y . If X = 0 or Y = 0 then obviously X X Y = 0. In spite of the arbitrary nature of X and Y,their Cartesian product can be treated in geometrical terms: the elements of the set X x Y are called points, the sets X and Y the coordinate axes. I f z = ( x , y } then x is called the abscissa and y the ord.’nate of z. The fact that the set of points in a plane can be treated as the Cartesian product & x 6 where € is the set of real numbers justifies the use of this terminology. Certain properties of Cartesian products are similar to the properties of multiplication of numbers. Fo; instance, the distributive laws hold:

( x , u x 2 ) x Y =X , X Y U X , X Y , Y X ( X 1 u X,)= Y X X , u Y x X , , (X,-X,) x Y = X , x Y-x, x Y , YX(X,-X,)= rxx,-rxx,. As an example we shall prove the first of these equations: (X,J’) E

(xiU X z ) XY

U X z )A (J’EY) =(XEXiVXEXz)/\(J’EY) = ( x €XIA y E Y ) v ( x EX, A y E Y) (X€X,

= (<X,Y>

x Y ) v ( ( A r>EX2 x Y) 3 (x, y ) E (X, x Y u X,X Y ) . EX1

The Cartesian product is distributive over intersection:

(X,n X , ) x Y

= ( X l x Y )n( X z x Y ) , Y x ( X , n X , ) = ( Y x X , ) n (YxX,).

64

H. AXIOMS OF SET THEORY. RELATIONS. FUNcIlONs

The proof is similar to the previous one. The Cartesian product is monotone with respect to the inclusion relation, that is, (*) If Y # 0, then ( X , c X2) = ( X , X Y c X Z X Y ) = (Y XX,c Y xXz). In fact, let y E Y . Suppose that XI c 1,. Since (for i = 1,2) ( ( x , Y ) E X I X YE) ( x E X ~ ) A ( Y E Y ) , we have the following implication ((X,Y) EX1 x Y ) ((XY u> EX2 x y>; hence X , x Y c XzX Y . Conversely, if X , x Y c X2x Y and y E Y , then (XEX,)+ ( X E X i ) h ( y E Y ) ~ ( ( X , y ) E X ~ X Y ) + ( ( X , Y ) E X z X Y ) ( X EXz) A (J’E y) + ( X EXz), thus Xi cXZ. The proof of the second part of (*) is similar. Using cartesian products we can perform certain logical transformations. For instance, the formulas (p. 49) +

A A @(XY X

Y

X

Y

Y ) =A XY

@iXY Y )

=A @

v v @ ( x ,Y ) = v P ( x , Y ) = v

@(z)

XY

allow us to replace two consecutive universal or existential quantsers by one quantifier binding the variable z = < x , y ) which runs over the Cartesian product X x Y. A subset R of a Cartesian product X X Y is called a (binary) relation. Instead of writing (a, b ) E R, where R denotes a relation, we sometimes write aRb and read: a is in the relation R to by or the relation R holds between a and b. The left domain (DJ(or simply the domain) of a relation R is defined to be the set of all x such that <x,y) E R for some y ; the right domain (0,)-the set of all y such that (x, y ) E R for some x. The right domain of a relation is sometimes called the range, or the counter-domain, or the converse domain. The union F(R)of the left and right domains of R is called the field of R. In geometrical terminology we say that D, is the projection of R on the X axis and D, is the projection of R on the Y axis.

4. CARTFSJAN PRODUCTS. RELATIONS

65

Thus we have

These formulas prove the existence of the sets Dland D,. If the arguments of the propositional function @(x, y ) are limited to the sets X and Y respectively, then the set R = { ( x , y ) : @(x, y ) ] is a relation. Clearly, @(x, y ) = x R y = ( x . y ) E R. Hence it follows from formula (1) that: THEOREM: The projecrion of the set { ( x , y ) : @ ( x ,y ) } on the X-axis is the set ( x :

V@(x,y ) } . Y

The relation ((x, y ) : y R x } is called the inverse of R and is denoted by RQ.Obviously D,(Rc) = D,(R) and D,(Rc) = D,(R). The relation ( ( x , y ) : (xSz A z R y ) } is called the composition of R

v 2

and S and is denoted by R o 5''). Obviously, D,(R o 5') c D, (5') and D r ( R o S) c D,(R). The operation o is associative. In fact, X ( Ro S ) o TY =

V (XTZ

A

Z R o SY)

2

= v/ / ( x T z A z S t A t R y ) z

t

=VV(xTzhzStAtRY) I

E

Z

v [v( x T z ~ z S t ) h t R y ] f

z

=V(xSoTtr\tRy) t

=xRo(SoT)y. Because of the associativity of o we may omit parantheses in expressions of the form R o S o ... oU. We shall prove the formula (ROS)'

=

ScORc,

I) We use this notation instead of the more natural S o R, because composing transformations we normally write on the second place the symbol of the operation which is carried out first (for example, sinoogx)).

66

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

In fact, ~ ( R S)cy o = y R o SX E

V ( Y S ZA Z R X ) 2

=

(xRCz0 z Soy) z

=xScoRcy. Other properties of the operations and

o are given in the exercises.

Examples and exercises # 1. Let X = Y = C (the set of real numbers). The set {<x, y>: x < y) is that part of the plane which lies above the straight line x = y. The set { <x, y>: y = x') is a parabola, its projection on the Y axis is the set { y : (y = x2)}. #

v X

a family of subsets of X x Y. Let F(Z) denote the projection of the set Z (where Zc X x Y) on the X axis and F(A) the family of all projections F(2). Z EA. Prove that 2. Let A be

F[S('4)1 = SlF(PIl# i.e. the projection of a union is equal to the union of the projections. 3. Give an example showing that the projection of an intersection may be different from the intersection of the projections. 4. Prove the formulas (Ru S)c = Rc u Sc, (Rn S)c = Rc n Sc and ( R c ) c =R. 5. Prove the formulas ( R u s)0 T = ( R 0 T ) u (S 0 T), T O ( R U s)= (TOR)u (T 0 S), (Rn S) o T c (RoT ) n (So T ) , T o (RnS) c ( T O R ) n ( T o S ) . 6. Provr that ( X x Y)' = Y x X. Compute ( X x Y) o ( Z x r).

0 5. Equivalence relations Equivalence relations form an important and frequently encountered class of relations. A relation R is called an equivalence relation if for all x, y , z EF(R),the following conditions are satisfied: xRx (reflexivity), xRy 3 yRx (sYmmetrY) x R y A Y R Z+ x R z (transitivity). # Examples 1. Let x , y be straight lines lying in a plane, Let x R y if and only if x is parallel to y. Then R is an equivalence relation. 9

5. EQUIVALENCE RELATIONS

67

2. Let C be a set of Cauchy sequences ( a,, a,, ...,a,,, ...) of rational numbers. The relation R which holds between two sequences if and only if lim (a,-b,,) = 0 is an equivalence relation. 3. Let X be the set of real numbers x such that 0 < x < 1. The relation R which holds between two numbers a, b EX if and only if the difference a--b is a rational number is an equivalence relation. # 4. Let X be any set, K 5 2' and let Z be an ideal (seep. 17). The relation f which holds between two sets X , Y E R if and only if X 2 Y E Z is an equivalence relation. 5. Example 4 can be generalized in the following way. Let K be an arbitrary Boolean algebra and Z any subset of K satisfying the conditions : a < b E Z - + a E Z , (aEZ)A(bEZ)+(av beZ). Then Z is an ideal of K and the relation & (Example 4) is an equivalence relation. We shall now give several theorems which describe the structure of an arbitrary equivalence relation. Let C be any set. A family A of subsets of C ( A c 2 9 is called a partition of C if 0 $ A , S ( A ) = C and the sets belonging to A are pairwise disjoint (i.e. for any X,Y E Aeither X = Y or X n Y = 0). THEOREM 1 : I f A is a partition of C, then the relation RA defined by the formula

XRAY

v

[ ( X E y) A

YEA

( y E y)]

is an equivalence relation whose field is C. The proof of this theorem is left to the reader. 2: If A and B are two diferent partitions of C, then RA THEOREM # RB. PROOF.Suppose that RA = RB; we shall show that A = B. Because of the symmetry of the assumptions it suffices to prove that A c B. So let Y EA and let y E Y . Since S ( B ) = C, there exists 2 E B such that Y E Z . If X E Y then XRAY and hence x R ~ y Because . Z is the unique element of B containing y, we have X E Z .Similarly we can show that x EZ+ x E Y,which proves that Y = Z and hence Y EB.

68

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

THEOREM 3: For any equivalence relation R with jield C # 0 there exists a partition A of the set C such that R = RA.

PROOF.Let Because of the reflexivity of R the elements of the family A are non-empty and S ( A ) 5 C. If Y EA and 2 E A, then for some y , z E C the following formulas hold : ~ ( U E Y F U R Y )A , (uEZ=URZ). U

U

From the symmetry and transitivity of R we infer that if sets Y and 2 have a common element, then they are identical. This proves that the

family A is a partition of C. We show now that R = R i . Suppose that uRv. Denoting the set {zEC: zRu) by Y u we obtain Y , E A and V E Y , ;hence URAOand R c RA. Now suppose that uRAv; then there exist Y in A and an element y such that ( z R y = zeY) and U E Yand V E Y .Therefore uRy and

A 2

WRY;by the symmetry and transitivity of R,it follows that uRv. This proves that RA c R. Hence R = RA, Q.E.D. It follows from Theorems 1-3 that every equivalence relation with field C # 0 defines exactly one partition A of the set C and vice versa. If R = RA then sets of the family A are called equivalence classes of R. The equivalence class containing an element x is denoted by x/R, the family A itself is denoted by C/R. This family is called the quotient class of C with respect to R. # Examples For the relation R of Example 1 each equivalence class consists of all straight lines lying in the same direction (i.e. mutually parallel). For the relation R of Example 2 each equivalence class consists of all sequences of rational numbers convergent to the same real number. Cantor defined real numbers as the equivalence classes with respect to this relation. #

69

5. EQUIVALENCE RELATIONS

A set of representatives of an equivalence relation with field C is a subset of C which has exactly one element in common with each equivalence class. The existence of a set of representatives for any equivalence relation follows from the axiom of choice. Without the axiom of choice we cannot prove the existence of a set of representatives even for very simple relations. Such is the case for the relation of Example 3 I). Exercises 1. Let I = {x: 0 x < 1); for X c Z let X(r) denote the set of numbers belonging to Z and having the form x + r + n where x E X and n is an integer. Show that if Z is a set of representatives for the relation R of Example 3, then a) Z ( r ) n Z(s) = 0 for all rational numbers r, s (r # s); b) Z = Z ( r ) , where the union is over all rational numbers.

<

u r

2. Show that the condition RA c RE is equivalent to the following: every set Y E A

is the union of some family A' c B. 3. Show that if M is a non-empty family of equivalence relations with common field C,then P(M) is also an equivalence relation with field C. 4. Preserving the notation of Exercise 3 prove that there exists an equivalence relation U with field C such that a)REM-,RcU; b) if Vis an equivalencerelation with field C and

A [REM + R

c

VJ,then U c V.

R

5. Assuming in Exercises 3 and 4 that M = (RA,RE), describe P(M) and U.

86. Functions

A relation R c X X Y is called a function ') if (1)

A

XI

[XRYI A XRY'

(Yl = Y2,I.

+

Y,, Y :

Functions are denoted by letters f , g , h , ... . The sets Dl(f)and Dr(f)are called respectively the domain and the range of the func') As was shown by G. Vitali, Sul problem della misura dei gruppi dipunti di una retta (Bologna 1905), every set of representatives for the relation in Example 3 is non-measurable in the sense of Lebesgue. This definition of a function is due to G. Peano, Sulla defnizione di funzione. Atti della Reale Accademia dei Lincei, Classe di scienze fisiche, matematiche e naturali 20 (1911) 3-5.

70

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

tion f. The following terminology will be used: if D l ( f ) = X and D , ( f ) c Y then f is called a mapping (or a transformation) of X into Y ; if, moreover, Dr(f) = Y then f is called a mapping of X onto Y . When D l ( f ) = X , we say that the function f is defined on X . The set of all mappings of X into Y is denoted by Yx.Instead of the formulafEYX we often use the more suggestive formulas f:X - t Y or X T Y . Iff E Y x and x E X , then by the definition of domain there exists at least one element y E Y such that x f y . On the other hand, it follows from the definition of function that there exists at most one such element. Hence the element y is uniquely determined. It is called the value off at x and is denoted by f ( x ) . Therefore, the formula y =f ( x ) has the same meaning as x f y . For f,g belonging to Y x the following obvious equivalence holds: ( f = g ) = A [f(x)=g(x)l. xry E

X

If ordered pairs are identified with points of a plane, the first term with the abscissa and the second term with the ordinate, then it turns out that a function is identical with its graph. DEFINITION: A function f is said to be a one-to-one function if different elements of the domain have different values under the function f :

[ f( X I )

=f

(&)I+

[Xl

= xzl 9

where x1 and xz are arbitrary elements of the domain. 1: I f f E Y x then f” is a function if and only i f f is oneTHEOREM to-one. Moreover, fCeXY1 where Yl is the range o f f and f” is also a one-to-one function.

PROOF.The relationf” is a function if and only if

A

x1. xm Y

bf

A

Yf%

-t

x1 = xzl

-

that is if

A

x1, XI. Y

[u =f(x1> A Y = f ( x z ) - f x ,

= xzl.

71

6. FUNCTIONS

Clearly, this formula is equivalent to (1). The second part of the theorem follows from the formulas for the domain and range of an inverse relation (p. 65).

THEOREM 2: I f f E Y x and g € Z y then the relation g o f is a function Z then X Z Z ) . and g o f € Z X (in other words: if X T Y PROOF.The definition of the composition of two relations implies the equivalences:

v C W Y ) (.Ygz)l = v [ ( f (4 Y ) ( g ( Y )

xg o f z =

A

Y

=

A

=

Y

-g(f(x))

41

= 2;

it follows that

A X, 21.

za

[(xgo f 4 A

(aof2 2 )

-+

z1 = 221

and that every element of X belongs to the domain of g o f . Since the domain of this relation is included in the range of g, g of E Z X . Theorem 2 implies the following formula g of ( x ) =g(f(x))

for

XEX.

THEOREM 3: r f f e y X , g € Z y and the functions f and g are oneto-one, then their composition is also one-to-one.

In fact,

g ( f W ) = s(f(.'I)

-+f(x) =f ( 4

--f

x = x'.

DEFINITION: A one-to-one function whose domain and range are the same set X is called a permutation of the set X . The simplest permutation of X is the identity permutation Ix, that is, the function defined by the formula I,@) = x for all x E X .

THEOREM 4: If f E Y X and f is a one-to-one function, t h e n P o f and f of"= Iy, where Y , is the range o f f .

= Ix

In fact, the equivalence fC(y) = x =f ( x ) = y implies f"(f ( x ) ) == x , thus f" o f = Ix. The proof of the second formula is similar.

72

11. AXIOMS OF SET THEORY. RELATIONS. F"CTI0NS

Let f E Y", g E Z ~q~,E TYand y E Tz.Hence the range of the function 9 o f is contained in T and the same holds for the function y o g . If q~ o f = y o g then we say that the following diagram XT'Y 81

ZTT

1.

is commutative. This diagram shows that starting with an element x E X we can obtain the element q~ of ( x ) = y o g ( x ) in two ways: through an element of the set Y or through an element of the set 2. An example of a commutative diagram will be given on page 79. DEFINITION: A function g is said to be an extension of a function f if f c g . We also say that f is a restriction of g. THEOREM 5 : In order that f c g it is necessary and suficient that D l ( j ) c D l ( g )andf(x) = g ( x ) for all x ~ D , ( f ) . PROOF.Necessity: Suppose that f c g . Then Dlcf)cDl(g), for the projection of a subset is a subset of the projection (see p. 66). If x € D I ( f ) and y =f ( x ) then ( x , y ) EL hence ( x , y ) Eg and y = g(x). Suficiency: Suppose that Qcf) c Dl(g) and f(x) = g(x) for all x e D 1 ( f ) .If (x, y ) Ef then y =f ( x ) = g(x), therefore ( x , y)Eg, which shows that f c g . The restriction f of g for which = A will be denoted by glA. The notion of function should be distinguished from the notion of operation. By an operation we mean a propositional function @(x,y) of two variables satisfying the following conditions:

(w)

A 'V @ (x,v), A X

Y

P(x,Y1) A @(x,uz)

+

@1

= YJl.

X.YI.Yl

These conditions state that for every x there exists exactly one object y such that @(x,y). This object might be denoted by F(x). If, however, the domain of the propositional function @(x,y ) is unlimited, then there may exist no set of all the pairs ( x , y ) such that @(x,y), i.e. there may exist no function f such that @ ( x , y )= [ y = f ( x ) ] . For instance, such a function does not exist if @(x, y ) is the propositional function x = y (see p. 61). On the other hand, the following theorem holds: THEOREM 6: If a propositional function @(x,y ) satisfies conditions (W) and A is an arbitrary set, then there exists a function f A with domain A

6. F U N ~ O N S

13

and such that for arbitrary X E A and arbitrary y LY

=fA(X)l

= @(XX,Y).

Namely, the required function f A is the set { t E A XB:

v Kt

= <X,Y>) A @ ( X , Y ) l )

XY

where B denotes the image of A under the propositional function @ (see p. 55). In particular, if a propositional function @ is of the form ...x ... = y (where on the left-hand side of the equation we have an expression written in terms of the letter x , constants, and operation symbols), then the function fA will be denoted by F [... x . .I. For example, XEA using this notation, Functions of more than one variable

Let X X Y X Z = X x ( Y x Z ) , X x Y x Z x T = X x ( Y x Z x T ) and similarly for any number of sets. If X = Y = 2 then instead of X X X X X we write X3 and similarly for X X X X X X X . Subsets of the Cartesian product of n sets are called n-ary relations. If the domain of a function f is the Cartesian product X x Y , then f is said to be a function of two variables. Similarly, if the domain o ff is the Cartesian product X X Y XZ, then we say that f is a function of three variables. Instead of f ( < x , y ) ) we write f ( x , y ) . ‘THEOREM 7: fl a function f is a one-to-one mapping of the set X x Y onto the set 2, then there exist functions a and By mapping the set 2 onto X and Y respectively, such that f ( a ( z ) , B(z)) = z for every z €2. PROOF.It suffices to take for a the set of pairs ( z , x ) satisfying the [ f ( x ,y ) = z] and for B the set of pairs ( z , y ) satisfying condition

V Y

tile condition

V [ f ( x ,y ) = z]. X

To conclude this section we give a formulation of the axiom of choice using the notion of function. ‘THEOREM 8: If A is a non-empty family of sets and 0 4 A , then there exists a function fE (S(A))” such that f ( X ) EXfor every X EA.

74

11. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

PROOF.Let h = F [{X}Xx]. For X E A we have thus h ( X ) # O XEA

and, moreover, h ( X ) n h ( Y ) = 0 for X # Y . Applying the axiom of choice to the family D,(h) we obtain a set which has exactly one element in common with each set h ( X ) , X EA . As it is easy to show, this set is the required function f. A function with the properties mentioned in Theorem 8 is called a choice function for the family A . Theorem 8 shows that from the axioms 2 it is possible to derive the existence of a choice function for an arbitrary non-empty family of sets not containing the empty set. Conversely, it can be shown that the axiom of choice follows from Theorem 8 and the axioms Z. Exercise 1. Let n

>3

and let

X = A O u... UAn-,,

Bk =Ak+lU

... U A k + n - 1 ,

c k =Ak+lU

where the indices are reduced modulo n. Let tem of functions satisfying the condition fk

( x ) =f k + i ( x )

There exists a function f~ Y k = 0, ...,n-1.

X

for

fk E Y B k ,

... U A k t n - 2 , k = 0 , ...,n - 1,

be a sys-

x Eck+i.

satisfying the equation f k = f l

Bk

for every

$7. Images and inverse images Let A and B be arbitrary sets and R a relation such that R c A x B. For X c A let R'(X) = { y : 'J (xR~)}. X€X

This set is called the image of X under the relation R. Clearly,

R': 2A + 2B. In particular, iff is a function then f ' ( X ) consists of values of the function f on the set X. We shall write f ' ( X ) = {f (x): x E X } . The same symbol will be used for operations, e.g. {(x,y):x~X}, { S ( X ) : X E A } , etc. As we know, there exists neither a function whose value for any x is the pair (x,y> nor a function whose value for any family X is the set S ( X ) . However, every such operation

75

7. IMAGES AND INVERSE IMAGES

determines a function if we limit its domain to an arbitrary given set (see Theorem 6, p. 72). Thus strictly speaking it would be necessary to replace the symbols ( x , y > , S ( X ) , etc., in the formulas {(x,y>: X E X } , { S ( X ) :X E A } by symbols for values of functions with domains Xand A , respectively. It follows from the definition of inverse relation (p. 65) that if Y c B then the image of Y under the relation R" is

R-'(Y) = { x :

v (~R'x)}

=

(x:

Y EY

V(xR~)}. Y €Y

.

This set is called the inverse image of Y under R , If R =f is a function, then f-'(Y) = { x : ( f ( 4= Y ) ) = { x : S ( X ) E Y } ,

v

Y SY

i.e. the following equivalence holds: x Ef-yY) = f ( x ) E Y . If Y reduces to the one-element set { y } , then the set J'-'(Y) is called a coset off determined b p y . Distinct cosets are always disjoint, the union of all cosets is the domain off. We shall now establish several simple properties of images and inverse images. 1: THEOREM (1)

(2)

R c A x B and X,, X , are subsets of A , then R'(X,) u R'(X2) = R'(X1 u Xz), X , c X , + R'(X,) c R'(X2), R1(X, n X,) c R1(Xl)n R1(X2).

(3) PROOF.Formula (1) follows from the equivalence Y E RYXl

u X,)

= v.(C{

EX,)

v ( x E&)l

A

(XRY)}

X

v

[(X

EX')

A

( X R Y ) ]V

X

v [(XEXz)

A

(xR3')l

X

=y E R'(X,) v y E R'(Xz) =y E R'(X,) u R1(X,). To prove (2) it suffices to notice that if ,'A c X , then X , = X , u X,. Thus by means of (1) it follows that R'(X2) = R'(X,) u R1(X2)2 R'(X1).

76

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

Finally, formula (3) follows from the remark that X , nX2c Xi for i = 1,2, whence, by (2), R'(X1 n XJ c R'(X1) and R'(Xl n X2) c R1(X2);and, in turn, we obtain R'(Xl nX2) c R1(Xl)n R1(X2).

THEOREM 2: r f f ~ and p Y , c B, Y2c B, then

f -yy,u Yz) = f-l(Y1)uf-'(Yz), f - v 1 n Y2) =f-'(YJ n f - 1 ( Y 2 ) , f -l(Y1- Y2)=f --1(Y1)--f-1(Yz).

(4) (5)

(6) PROOF.(4) is a special case of (1). Formula ( 5 ) follows from the equivalence x Ef - l ( Y l n Y2)= f ( x ) E Y 1n Yz = (fWE Yl) A (fWE y2) ( X Ef-I(Y1))A (X Ef-'(YZ)) = ( x ~f-l(Y,) nf-'(Y2)). The proof of (6) is similar. Theorems 1 and 2 show that the operation of forming the image under an arbitrary relation is additive, but it is not multiplicative. On the other hand, the operation of obtaining the inverse image is both additive and multiplicative. THEOREM 3: I f f : A + B and i f f is a one-to-one function, then for any X , , X , c A the .following formulas hold

f1W1 nX2) =f '(X1) nf '(-&I,

f '(X1-XJ

=f YXJ

-f '(&I.

For the proof, substitutef" for f in Theorem 2.

THEOREM 4: I f f : A -+ B, Y c f ' ( A ) and X c A , then

f

l(f-l(v) = y,

f-'(f'(n) = X.

The proof of the first formula can be obtained from the equivalence YEf

yf-Yn)= v [(x E f -'m)(Y =f(41 A

=

v [ ( f (4 y) E

A

(Y =f ( x > ) ]3 0,E

n.

The proof of the second formula follows from the implication (x EX)

+

( f ( 4E f '(XI) = x E f - l ( f

-

l(X>)

77

7. IMAOES AND INVERSE IMAOES

In the formula just proved the inclusion sign cannot in general be replaced by the equality sign. For instance, iff is a function of the real variable x and f ( x ) = x2, then for X = { x : x 2 0} we have f-'(fl(X)) # X. But for one-to-one functions we ooviously have f-'(f'(X)) = X. Finally, let us note the following important THEOREM5 : If S c A x B and R c B x C , t h e n ( R o S ) ' ( X ) = R ' ( $ ( X ) ) for every set X c A. PROOF.

v (XROSY) =v v 'v \/

yLZ(RoS)'(X) =

X€X

X€X

[(XSZ) A

(zRJ91

I

[ ( X S Z ) A (ZRy)]

E

2-

xex

= v [(z E S' (x))A ( Z W ] =y E R' (S' (X)). In particular, it follows from Theorem 5 that if f:A+ B and g : B + C , then (gon'(x>=g'(f'(X)) for every set X c A . Exercises 1. Prove that f1(Xl)-f1(X2) c fl(Xl--Xd and fl(Xnf-'(Y)) = f l ( x ) n Y . 2. If g = f l A , then g - l ( Y ) = A n f - l ( Y ) . 3. A value y of a function f is said to be of order n if the set f-l((y}) consists n if all of its values are of n elements. We say that a yunction f is of order n. of order Prove that if a function f defined on a set X is of order n and A c X, then the restriction f l ( f - ' ( f ' ( A ) ) - A ) is of order n-1. 4. We are given a system of r + l disjoint sets A o , A ] , ..., A , included in X and a function of order < n defined on X (n 2 r). Let B = f ' ( A o ) n ... nfl(Ar). Prove that the restrictionfl ( A i n f - ' ( B ) ) is of order n-r. 5. Images and inverse images are used in topology, in particular to define the notion of a continuous function. Let X and Y be two topological spaces and let f: X + Y . We say that f is continuous if the inverse image of any open set in Y is an open set in X. Prove that the following conditions are necessary and sufficient for a function f to be continuous: (a) inverse images of closed sets are closed,

<

<

<

<

<

(b)

6)

f'(4 = f'o,

zcf - ' m I ,

78

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

(d)

f-’o = f -‘@I,

(el f1FWI = 5, where A c X and B c Y. 6. Let f be a one-to-one mapping of X onto Y (hence f”: Y - t X ) . We say that f is a homeomorphism i f f and fc are continuous. Show that each of the conditions arising from (b)-(e) by substituting the equality sign for the inclusion sign is a necessary and sufficient condition for a function f to be a homeomorphism. 7. Show by means of an example that the image of an open set need not be open even though the function is continuous. The dame for closed sets. 8. Prove that the composition of two continuous functions is continuous.

58. Functions consistent with a given equivalence relation. Factor Boolean algebras The construction to be given in this section is one of basic importance in abstract algebra. Let R be an equivalence relation whose field is X , f a function of two variables belonging to Xxxx.

DEFINITON: The function f is consistent with R if

A similar definition can be adopted for a function of arbitrarily many arguments. It results from the equivalence xRy = (x e y / R ) = (x/R = y / R ) that the definition of consistency can be expressed as follows: if x Exl/R and y ~ y ~ /then R , f ( x , y ) / R =f ( x l ,y , ) / R . In other words, the equivalence class f ( x , y ) / R depends on the classes x/R and y / R but not on the elements x , y themselves. This implies that there exists a function Q, with domain (X/R)x (X/R) satisfying for any x , y E X the formula: 9(X/RYY/R)= f (XYY)/R* Namely, 9 is the set of all pairs of the form k , k‘,k” E X/R and

We say that the function

pl

((V,k ” ) , k) where

is induced from f by R .

8. FUNCTIONS CONSISTENT WITH AN EQUIVALENCE RELATION

79

The function k E (X/R)xdefined by the formula k ( x ) = x/R is called the canonical mapping of X onto X/R. The function of two variables k2 defined by the formula W X Y Y ) = <X/R¶Y/R> is also called the canonical mapping of J? onto (X/R)2.A similar definition can be given for functions of three and more variables. THEOREM 1 : If a function f € X X x X is consistent with an equivalence relation R and 9, is the function induced from .f by R, then the diagram

X.X k'l

Lk

(X/R)'--+X/R P

is commutative.

PROOF.For any pair ( x , y ) EX' the following formulas hold: k 0 f ( x ,Y ) = k (f(x,Y ) ) =f( x ,Y)lR, 9, 0 k2(X,Y ) = 9,(k%, Y,) = 9, (XlR>

U P ) =f(x,Y)lR.

Hence k o f = q o k ' . Example. Let X = K be a field of sets with unit U,Zany ideal in K, R the relation A mod I (see p. 17). The set K/R is denoted by K/Z and is called a factor Boolean algebra. The functions f ( X , Y ) = X u Y , g ( X , Y ) = X n Y , and h ( m = U-X are consistent with the relation G (see Exercise 4, p. 18). The functions induced from f, g, h by the relation iwill be denoted by v , A , and -, respectively. Hence (XIR)A (YlR) = WnV l R , -(X/R) = (U-Xl/R.

(WR)v (YIR) = ( X u Y)IR,

THEOREM 2: The set K/Z is a Boolean algebra with respect to the operations v , A , -, with Q/Rand U/R as the zero and the unit element, respectively.

PROOF.It is sufficient to show that the operations v , A , - and the elements O/R and U/R satisfy axioms (i)-(v'), p. 37. For instance, we check axiom (i). Let a = X/R and b = Y / R ;then a v b = ( X u Y ) / R

80

U. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

and b v a = ( Y u X ) / R and hence a v b = b v a . The remaining axioms can be checked similarly.

REMARK.The conditions XAO (mod I) and XEZ are equivalent. This proves that O/R = I. The factor algebras K/Z may have properties quite different from those of K. Thus the construction leading from K to K/Z allows us to build new and interesting examples of rings. Exercises 1. Generalize the example given above by taking any Boolean algebra as K and any subset of K satisfying the conditions of Example 5, p. 67, as I. 2. Let K be the field of all subsets of an infinite set U,and let Z be the ideal of all b i t e subsets of U. Show that every non-zero element of the factor ring K/I can be represented as xvy where x # I and y # I.

0

9. Order relations

DEFINITION 1: A relation R is said to be an order relation if it is reflexive, transitive, and antisymmetric. The last condition means that ( x R ~ ) A ( ~ J R x ) +=( yx) .

A relation which is only reflexive and transitive is said to be a quasiorder relation '). Instead of x R y we usually write x G Ry or x y . We also say that the field of R is ordered (or quasi-ordered) without explicitly mentioning R. It is necessary to remember, however, that an ordering is by no means an intrinsic property of the set. The same set may be ordered by different relations.

<

REMARKS. In former terminology an order relation was understood to be a relation which has the property of connectedness, i.e. # y ) + [ x R y v yRx], in addition to the properties given in Definition 1. The order relations (in our sense) were called partid order (x

l) According to Bourbaki, a proto-order relation (priordre) is a transitive relation satisfying the condition:

.

( x R Y )+( x R X)AdvR Y ) See: Thgorie des ensembles, Chap. 3 , 51, No. 2 (1963).

81

9. ORDER RELATIONS

relations. Order relations which are also connected are called linear order relations and will be considered in detail in Chapter VI. Examples 1. Every family of sets is ordered by the inclusion relation. If it is linearly ordered by this relation, then it is called a monotone family. 2. Every lattice (in particular, every Boolean algebra) is ordered by the relation a b . 3. The set of natural numbers is ordered by the relation of divisibility. 4. A family P is said to be a cover of a set A if A = S(P). A cover PI is said to be a refinement of a cover P2 if for every XEPl there exists Y EP2 such that X 3 Y . The relation R, defined by

<

PzRPl = (P1 is a refinement of P2), is a quasi-order relation in the set of all covers of A . It is not, however, an order relation, that is, there may exist two distinct P1 and P2 such that P I R P z and P 2 R P l . On the other hand, if we limit the field of R to covers which consist of non-empty disjoint sets (such covers are called partitions; cf. p. 67), then R is an order relation. DEFINITION 2: A set A ordered (or quasi-ordered) by the relation < is said to be directed if for every pair x E A and y E A there exists z E A such that x < z and y < z. 5 . Every lattice is a directed set since x < x v y and y < x v y . In particular, the family of all subsets of a given set X, as well as the family of all closed subsets of a given topological space, is directed with respect to the inclusion relation (either c or 3). 6 . The set of all covers of a given set A is directed with respect to the relation R considered in Example 4. For, given two covers P1 and P2, denote by P3 the family of all intersections of the form X n Y where X E P 1 , YE??. It is easy to check that P3 is a cover and PIRP3 as well as P2RP3. DEFINITION 3: An ordered set A is said to be cofinal with its subset B if for every x E A there exists y E B such that x y. Analogously we can define coinitial sets.

<

82

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

# Example. The set of real numbers is cofinal and coinitial with the set of integers. # If an ordered set A contains a greatest element, then A is cofinal with the set composed of this element. The greatest (least) element should be distinguished from the maximal (minimal) element. Namely, an element x of an ordered set A is said to be maximal (minimal) if there is no element y in A such that x < y (y < x). In linearly ordered sets the notions of greatest (least) element and of maximal (minimal) element coincide. This is not always the case for arbitrary ordered sets.

DEFINITION 4: Let A be an ordered set, T any set and let f e A T . An element u E A is said to be the least upper bound of {ft) if ft < u and u is the least element having this property:

(ft < u),

(0 tET

Replacing < by 3 we obtain the definition of thegreatest lower bound. The least upper bound, if it exists, is uniquely determined. For suppose that besides (i) and (ii) we have

Setting v = u in (ii’) and applying (i), we obtain u’ < u. Likewise, it follows from (ii) and (i’) that u u’. Hence u = u’, since the relation < is antisymmetric. The proof of the uniqueness of the greatest lower bound is similar. The least upper bound, if it exists, is denoted in the theory of ordered sets by // J ; , the greatest lower bound by &. If T is a finite

<

tET

A

taT

set T = {1,2, ...,n } and f,= a, fi= b, ... ,f,= h, then the least upper bound of these elements is also denoted by a v b v ... v h, and the greatest lower bound by a A b A ... A h. The greatest lower bound of all elements of A , if it exists, is called the zero element and is denoted by OA or simply 0. Analogously,

83

9. ORDER RELATIONS

the least upper bound of all the elements of A , if it exists, is called the unit element and is denoted by lAor 1. Obviously, a A b < a and a A b b if a A b exists; similarly a < a v b and b < a v b provided a v b exists (in this case A is a directed set). If a < b y then a v b and a A b exist and they equal b and a, respectively. This implies that if a A b and a v b exist for all a , b E A , A is a lattice.

<

DEFINITION 5: An ordering of a set A is said to be complete if for every T and for every f E AT there exist A ft and ft.

v

f ET

feT

Since every lattice is a set ordered by the relation a < b, this definition also explains the meaning of the term “complete lattice”. Examples and exercises

7 . 2x is a complete lattice (for an arbitrary set A‘) with respect to the operations n, u. The existence of the least upper bound is a consequence of Axiom 111, $ 2, the existence of the greatest lower bound follows from Theorem 6, $3. This example will be used in Chapter IV, $ 1. 8. The family of all closed subsets of an arbitrary topological space is a complete lattice under the operations n, u . In this case X,is

v f

the closure of the union of the sets X , ,

r

X,is the intersection of X,.

9. Let A be any set. The family 2A2(i.e. the family of relations with fields included in A ) is ordered by the inclusion relation. Prove that the family of all transitive relations is a lattice and describe the meaning of the operations A and v . 10. The same problem as in 9 for the family of all equivalence relations.

DEFINITION 6: Two sets A and B ordered by the relations R and S respectively are said to be similar if there exists a one-to-one function f mapping the set A onto B and satisfying for arbitrary x , y E A the equivalence XRY

= f(WfO.

84

II. AXIOMS OF SET THEORY. RELATIONS. FUNCTIONS

In this case we say that the function establishes the similarity of the sets A and B (under the relations R and S). # For example, definingf ( x ) = - x for x E C we obtain similarity between the set € ordered by < and the same set € ordered by >. # The notion of similarity is a special case of the more general notion of isomorphism which will be treated in the next section.

5 10. Relational systems, their isomorphisms and types Let A be a set, R,,, R 1 , ... relations respectively of pQyp1,..., p k - l arguments in A; in other words, Rj c APj for j < k. The sequence

( A , & , R I P ,Rk-1) is called a reIationa1 system of characteristic ( p o , p I, ..., p k J and the set A is called the jield of the system. Relational systems are investigated in many branches of mathematics, especially in algebra. We may, for instance, consider a group, as a relational system of characteristic (3) and a ring as a relational system of characteristic (3,3). Boolean algebras (see p. 32) may also be treated as relational systems. In order to simplify our treatment we shall investigate systems of characteristic (2), that is, systems of the form ( A , R), where R c A x A. However, all proofs can easily be generalized to arbitrary systems. DEFINITION: Two relational systems (A, R) and (B, S ) are said to be isomorphic if there exists a one-to-one function f mapping A onto B such that for all x, y E A xRY =f(X>Sf(y). Then we write ( A , R ) z (ByS ) or briefly R z S if no confusion can arise about the sets A and B. The proof of the following theorem is immediate. THEOREM 1; The reIation NN is reflexive, symmetric, and transitive. We shall show that every property of the system ( A , R ) which can be expressed by means of the propositional calculus and quantifiers limited to the field of the relational system, is also a property of every system isomorphic to ( A , R). We say that the property in question is in variant under isomorphism.

10. RJZLATIONAL SYSTEMS

85

Let @ be a propositional function involving free variables x, y. Besides x, y, @ may involve an arbitrary number of other variables u,, u,, ... ,U k - , . Suppose that @ arises from propositional functions of the form (1) ui = u j , (2)

(Ui 3 ~

j

E> Y

by means of operations of the propositional calculus and by means of the quantifiers and A. Thus the variables x and y are not bounded

v

U€X

UEX

by quantifiers. For such propositional functions we have

THEOREM 2 I): If a function f is an isomorphism of the relational systems ( A , R) and
(3)

@I.

v

UrJEX

ak-1 E A . If Y’(A, R, a,, ...,ak-1) then for some a, belonging to A we have @ ( A ,R, a,, a,, ...,ak-1). By the induction hypothesis we thus obtain @(B, S , f (ao),f (a,), ...,f(ak-,)), and it follows that !P (Bys, f ( a , ) , ...,f(ak-1)). Hence we have proved the implication y ( A , R , a l , . - * , a k - l ) y(B,S,f(al), . * * , f ( a k - l ) ) *) Theorem 2 is a scheme: for each propositional function we obtain a separate theorem.

86

U. AXIOMS OF SET THEORY. RELATIONS. FUNCnONS

The proof of the converse implication is similar. Example. The following properties of the system ( A , R ) are by Theorem 2 invariant under isomorphism: 1. Reflexivity: A xRx. xcA

2. Irreflexivity:

// 1( x R x ) .

xeA

3. Symmetry: 4. Asymmetry:

A A

XJEA

[XRY

+

[ x R y -+

YRXI.

-1(yRx)].

%,YEA

5. Antisymmetry:

A

[ ( x R y )A Q R x ) + ( x = y)].

A

[ ( x R y )A ( y R z ) + ( x R z ) ] .

x,yzA

6. Transitivity:

x.y,zeA

7 . Connectedness:

,r\

x.y:A

[ ( xR y ) v ( x = y ) v (y Rx)].

Two isomorphic systems are said to be of the same type. The mere use of the word “type” in this expression does not presuppose that there are objects which we call “types”. All the theorems of set theory could indeed be expressed without using this notion. However, the introduction of the notion of type simplifies the axiomatic treatment of the theory. Moreover, the use of this notion is justified by the fact that Cantor himself developed set theory using this concept. We introduce a new primitive notion TR. The formula a TR ( A , R ) is read: a is the type of the relational system ( A , R). We also introduce a new axiom. AXIOMVTII (OF RELATIONAL SYSTEMS): For every system ( A , R ) where R c A2 there exists exactly one object a such that M TR ( A , R). Moreover, for any systems ( A , R ) and ( B yS ) the following formula holds (a TR ( A , R)) A (j3 TR (B, S ) )

+ [(a = j3)

( A ,R )

Z

(B, S)].

-

The unique object a such that a TR ( A , R ) is denoted by ( A , R) or, if there is no confusion, by El). l) Cantor dealt not with relational types of general systems but with types of linearly ordered sets (cf. $9). His use of the notion of type was imprecise. He defined the type of a linearly ordered set as that property which remains if we

10. RELATIONAL SYSTEMS

81

The object ct is called a relational type if and only if there exists some system ( A , R ) such that a T R ( A , R ) . Exercises 1. Let A have n elements and let r,, be the number of relational types of systems with field A. Show that the number r,, satisfies the inequality

2m2/n!< r,, < 2n2. 2. Prove that r, = 10 and r3 = 104 l).

disregard the nature of its elements but not their order. The single dash over the symbol of set was meant to indicate this single process of abstraction. l) R.L.Davis (Proc. Amer. Math. SOC. 4 (1953)494) states that r4 = 3044 and rs = 291968.

CHAPTER Ill

NATURAL NUMBERS. FINITE A N D INFINITE SETS

In this chapter all theorems will be derived from axioms of Zo(cf. p. 55). As usual, theorems not marked by do not involve the axiom O

of choice in their proofs.

51. Natural numbers Let for any set X,

X’=Xu{X}. The set X’ will be called the successor of X .

THEOREM 1: There exists exactly one family of sets N such that (i) OEN; (ii) X E N + X ’ E N , (iii) if K satisfies (i) and (ii), then N c K. PROOF.It follows from the axiom of infinity that there exists at least one family R satisfying conditions (i) and (3). Let @ ! be the family of all those subsets of R which satisfy (i) and (ii): @ = ( S c R :OEShA(XES-+X’ES)). X

It is easy to show that P(@) is the required family. The elements of N are 0, {0}, (0, {O}}, etc. These sets can be considered as the counterparts of natural numbers 0, 1,2, ..., the operation as the counterpart of +I. We shall prove several theorems of arithmetic which hold for the elements of W ) . l) The reader, who feels it unnatural that in our exposition the role of natural numbers is played by sets, can take a ratural numbers some objects which are in one-to-one correspondence with the sets belonging to N. Such objects may be, for instance, the types of relational systems S(n) = , where n E N.This method

1. NATURAL NUMBERS

89

In order to simplify the notation, the elements of N will be denoted by the letters m, n,p, ... . A set K is said to be inductive if it satisfies conditions (i) and (ii). mEn -,m' c n. (1) PROOF.

Let K = (n:

Anr (mEn

+

m' c n ) } . TO prove (1) it suffices

to show that N c K or, in other words, to show that the set K is inductive. Condition (i) clearly holds. To prove (ii), let n E K and rn En'. Hence m e n or m = n. In the first case, m c n by the definition of K, in the second case, m c n because the sets are equal. Thus m c n', and consequently n' E K, which proves the theorem. The proof of (1) is an example of a proof by induction. (2) n#n. The proof by induction consists in showing that the set {n: n $ n} is inductive. m' = n ' -+ m = n. (3)

PROOF.It follows from m' = n' that m En', thus m ~n or m = n and by (1) m c n. Similarly we prove that n c m. Peano showed that the arithmetic of natural numbers can be based upon the following axioms: (a) zero is a natural number; (b) every natural number has a successor; (c) zero is not a successor of any natural number; (d) natural numbers having the same successor are equal; will be used in Chapter V. where we introduce cardinal numbers. Identifying natural numbers with the elements of the family N , we can base our treatment of arithmetic on the axioms 2 7 only (mostly even on the axioms ,Z without the axiom of replacement); whereas in order to identify them with the types of the system S(n) we would have to appeal to the axioms Zo[TR] and VIII. A detailed analysis of the problem of what axioms of set theory are necessary to justify the laws of arithmetic and analysis was made by P. Bernays in his series of papers under the common title: A system of axiomatic set theory, published in Journal of Symbolic Logic 2 ((1937) 65-77; 6 (1941) 1-17 and 133-145; 8 (1943) 89-106; 13 (1948) 65-79; and 19 (1954) 81-96). See also P. Bernays and A. A. Fraenkel, Axiomatic set theory (Amsterdam 1958).

90

JJI. NATURAL NUMBERS. FINITE AND INFINITE SETS

(e) a set which contains zero and which contains the successor of every number belonging to this set contains all natural numbers. It follows from (i), (ii), (3), (iii) and n’ # 0 that the elements of the set N satisfy Peano’s axioms. (4) For arbitrary m,n exactly one of the following formulas holds: men, m = n , nEm. PROOF.(1) and (2) imply that every two of the above conditions are mutually contradictory. To prove that for every pair m, n one of these formulas holds, we use induction. Let K(n) = {m: m E n v m = n v n e m } . Theorein (4) is equivalent to

A ( N c K(n)), thus it suffices to prove n

that every set K(n) is inductive. The set K ( 0 ) is inductive, for K(0) consists of the set 0 and of those m for which 0 Em and it is obvious that 0 e m + 0 Em’. Suppose that the set K(n) is inductive, i.e. that N c K(n). We shall prove that K(n‘) is also inductive. Condition (i) : n’ E N c K(0) implies that n’ E 0 v n’ = 0 v 0 En’. Since the first two components of this disjunction are false, 0 E n’ and 0 E K(n’). Condition (ii) : Suppose that m E K(n‘), that is, either m E n’, m = n’ or n’ E m. In the second and the third case we obviously have n’ ~ m ’ and hence m‘EK(n’). In the first case either m = n or m En holds. If m = n then m‘ = n’ and hence m’ eK(n’). If m E n then m E K(n) and hence m’ E K(n), for K(n) is inductive by assumption. We thus obtain m‘ E n v m‘ = n v n’em. The third component of this disjunction is false, for it implies n e n , which contradicts (2). Hence we have only the two possibilites, m’En and m’ = n, which since n c n’, prove tbat m‘EK(n’). This completes the proof of Theorem (4). ( 5 ) The set Z = {m: n e m } is identical with the intersection P of all

families K c N such that n’E K and K satisfies (ii). PROOF.Since m c m’, the set 2 satisfies (i). This proves that P c Z , for obviously n’ €2.It remains to be shown that if K satisfies (ii) and n’E K, then Z c K. For this purpose, let L = { n : n E Z 3 nEK}. It

1. NATURAL NUMBERS

91

suffices to show that N c L or, in other words, that L is an inductive family. Condition (i) obviously holds, for O#Z. To prove that L satisfies (ii), suppose that m E L. This means that either m E K or rn $2.In the first case m‘ E K, for Ksatisfies (ii), and we obtain m’ E L . The second case splits into three subcases depending upon whether n E r n , n = m or m E n . The first subcase contradicts the assumption m $2. The second subcase implies m’ = PI’ and hence m’ E K and finally m’ E L, for K c L. In the last subcase we have by (4) either m‘ €12 or m’= n or n em’. If either m’E n or m‘ = n, then by (4) n 4 m’;hence m’ $ 2 and finally m’E L. The condition n Em’ leads to a contradiction, for it implies that n E m v n = m whereas by assumption we have m E n . In ordinary arithmetic the set of all numbers greater than n is defined to be the common part of all sets which contain the successor of n and which contain the successor of every number b which they contain. Theorem ( 5 ) shows that the membership relation E in N is the counterpart of the relation “less than” between numbers. We shall often write m < n or m < n instead of rn e n or rn en’, respectively. The existence of the set N allows us to define in set theory notions analogous to those found in arithmetic and analysis. For example, a function f whose domain is the set N is called an infinite sequence and is sometimes denoted by (Jo, fi, ...,fn, ...). If n E N then a function with domain n is said to be a finite sequence of n terms. The set of all infinite sequences whose terms belong to A is clearly AN, the set of all finite sequences of n terms in A is A”. The set of all finite sequences with terms in A can be defined as

( R c N X A : ( R is a function) A

V (D,(R) = n ) ) .

neN

This definition implies the existence of the set of all finite sequences with terms in A. Exercises l.Showthatifrn,nEN,thenmEnE(mcn) ~ ( m f n ) . 2. Show that if 0 # K c N , then P ( K ) E N n K . 3. Prove that every non-empty family K t N contains an element k such that k n K = 0 (cf. the axiom of regularity, p. 56).

92

lIl. NATURAL NOMBERS. FINITE AND

3 2. Definitions by

SETS

induction

Inductive definitions are the most characteristic feature of the arithmetic of natural numbers. The simplest case is the definition of a sequence cp (with terms belonging to a certain set Z ) satisfying the following conditions : (a) Y ( 0 ) = 2, pl(n‘) = e(pl(n), n) where z E Z and e is a function mapping Z x N into Z . More generally, we consider a mapping f of the Cartesian product Z x N X A into Z and seek a function pl E Z N X * satisfying the conditions :

-

(b) pl(O,a) g @ ) , p l w , a> =S(pl(n, a), n, a), where g E Z A . This is a definition by induction with parameter a ranging over the set A . Schemes (a) and (b) correspond to induction “from n to n+ l”, i.e. pl (n‘) or pl(n’, a) depends upon pl(n) or pl(n, a), respectively. More generally, pl(n’) may depend upon all values y ( m ) where m n (i.e. m en’). In the case of induction with parameter pl (n‘) may depend upon all values q ( m , a), where m < n’; or even upon all values pl(m, b), where m n’ and b E A . In this way we obtain the following schemes of definitions by induction:

<

<

(c)

pl(0)

= ZY

p l w =

0 1n‘, n),

~ ( 0a), = g(a>, pl(n’, a ) = H(yj ( n ’ x A ) , n, a). In the scheme (c), Z E Zand hEZCxN,where Cis the set of finite sequences whose terms belong to Z ; in the scheme (d) g € Z A and H e Z T x N x A , where T is the set of functions whose domain is included in N XA and whose values belong to 2 I). ExampIes of definition by induction (d)

1. The function m f n :

m+O = m y

m+n’

=

(m+n)‘.

l) Scheme (c) could be gzneralized by assuming that the domain of the function h is not the whole set Cx N, but only the set of pairs of the form (c, n) where c E Z n . However, this generalization is not of greater importance.

93

23DEFINITIONS BY INDUCITON

This definition is obtained from (b) if we set Z = A = N , g(a) = a, f@, n, 4 =PI. 2. The function :

(i)

(F) = (3+ n.

(3

= 0,

This follows from (a) if we set 2 = N , z = 0, e ( p , n) = p+n. 3. Let Z = A = Xx,g(u) = Ix and f ( u , n, a) = u o a in (b). Then (b) takes on the form

v @,a) = I x ,

pl(n',

4 = pl (n,a) 0 0.

The function y(n,u) is denoted by a" and is called nth iteration of the function a. We thus have: a'@) = x,

a"'(x) = an(a( x ) )

for x EX, a E.Xx and n E N.

4. Let A = N N .Let g(a) = a. and f ( u , n,a) (b) takes on the form

Y (0, a) = a09

= u+a,,

in (b). Then

Y (n',a) = ~p (n, a) +ant. n

ai. Similarly we

The function defined in this way is denoted by I-0 n

define n u a i ,max at. 1-0

i
It is clear that the scheme (d) is the most general of all the schemes discussed above. By appropriate choice of functions we can obtain from (d) any of the schemes (a)-(d). For example, taking the function defined by H ( c , n,a ) = f ( c ( n , a ) , n, a)

for

a E A , n E N, c e Z N x A

as H in (d), we obtain (b). We shall now show that, conversely, the scheme (d) can be obtained from (a). Let g and H be functions belonging to Z A and Z T x N x Arespectively, and let y~ be a function satisfying (d). We shall show that the sequence Y : Yn= pll(n'xA) can be defined by (a). Obviously Y,,ETfor every n. The first term of the sequence Y is equal to p11(0'xA), i.e. to the set z*

=

((
94

III. NATURAL NUMBERS. FINITE A N D INFINITE SETS

The relation between Y,,, and Y,, is given by the formula Ynt

=

yn u y,l({n‘}x A ) ,

where the second component is

Thus we see that the sequence Y can be defined by (a) if we substitute T for Z, z* for z and let e(c,n) = c u ( ( ( n ’ , a ) , H ( c , n , a ) ) :U E A } for CET. Now we shall prove the existence and uniqueness of the function satisfying (a). This theorem shows that we are entitled to use definitions by induction of the type (a). According to the remark made above, this will imply the existence of functions satisfying the formulas (b), (c), and (d). Since the uniqueness of such functions can be proved in the same manner as for (a), we shall use in the sequel definitions by induction of any of the types (a) - (d).

THEOREM 1: If Z is any set, z E Z and e E Z z X N , then there exists exactly one sequence y satisfying formulas (a). PROOF. Uniqueness: Suppose that y1 and y2 satisfy (a) and let K = {n: y1(n) = y2(n)}.Then (a) implies that K is inductive. HenceN c K

and v1 = y2. Existence: Let @ ( z ,n, t) be the propositional function e ( z , n) let Y (n,z, F ) be the following propositional function

=t

and

( F is a function) A (D,(F) = n’) A (F(0) = z) A

A @ ( F ( m ) ,m, F(m’)). In other words, F is a function defined on the set of numbers < n such mEn

that F(0) = z and F(m’) = e(F(m),m) for all in < n. We prove by induction that there exists exactly one function F,, such that Y ( n ,z, F,,). The proof of the uniqueness of this function is similar to that given in the first part of Theorem 1. The existence of F,, can be proved as follows: for n = 0 it sufficesto take ((0, z ) } as F,,; if n E N and F,, satisfies U(n,z, F,,), then F,,, = F,,u { (n’,e(F, (n),n))} satisfies the condition Y (n’,z , F,,,).

95

2. DEFINITIONS BY INDUCTION

Now, we take as

v [ Y ( n ,z,

A

F

pl

the set of paus (n, s> such that n E N , s E Zand

(s = F(n))I.

Since F is the unique function satisfying Y ( n , z , F ) , it follows that q is a function. For n = 0 we have q(0) = Fo(0)= z; if n e N , then pl(n’) = F,,,(n’) = e(F,,(n),n) by the definition of F,,;hence we obtain q(n’) : e(q(n),n). Theorem 1 is thus proved. We frequently define not one but several functions (with the same range Z)by a simultaneous induction:

d o ) = =,

Y(0) = t,

91(n’) =f(pl(n>,Y(n),n) Y W = g ( q ( n ) , y ( n ) , n ) , where z, t e Z and f , g E Z Z x Z x N . This kind of definitions can be reduced to the previous one. It suffices to notice that the sequence 6, = (q,,,y,,) satisfies the formulas: 60

= ,

On,

= e(@mn),

where we set e(u, n) = ( f ( K ( u ) ,L(u),n), g(K(u), L(u),n)), and K, L denote functions such that K ( ( x , y } ) = x and L ( ( x , y ) ) = y , respectively. Thus the function 6 is defined by induction by means of (a). We now define pl and y by q~(n) = K(6,,) and y (n) = L(6n)m The theorem on inductive definitions can be generalized to the case of operations. We shall discuss only one special case. Let @ be a propositional function such that

A neN A A [W,n, tl) z

A

W

tl,tt

Y

n, t 2 )

+

t,

= t21.

THEOREM 2 ’ ) : For any set S there exists exactly one sequence q such that yo s and G(Yn9 n, y n l ) .

A

nEN

Uniqueness can be proved as in Theorem 1. l) Theorem 2 is a scheme: for each propositional function we have a separate theorem.

96

III. NATURAL NUMBERS. FINITE AND INFINITE SETS

To prove the existence of y , let us consider the following propositional function !P*(n, S , f‘).

( F is a function) A (D~(F) = n’)A ( ~ ( 0= ) S) A

A !D(F(m>, m y~ ( m ’ ) ) .

men

As in the proof of Theorem 1, it can be shown that there exists exactly one function F,, such that !P*(n, S, F,,). To proceed further we must make certain that there exists a set containing all the elements of the form F,,(n) where n E N. (In the case considered in Theorem 1 this set is Z , for the domain of the last variable of the propositional function CP which we used in the proof of Theorem 1 was limited to the set Z . ) In the case under consideration, the existence of the required set Z follows from the axiom of replacement. In fact, the uniqueness of F,, implies that the propositional function

v [F*(%s,II;) P

A

(v = FW)l

satisfies the assumption of axiom VII. Hence by means of axiom VII the image of N obtained by this propositional function exists. This image is the required set 2 containing all the elements Fn(n). The remainder of the proof is analogous to that of Theorem 1. Examyle. Let @ ( S ,t ) be the propositional function t = 2s. Thus for any set S there exists a sequence y such that vo = S and y,,, = 2Pn for every natural number n.

8 3.

The mapping J of the set N x N onto N and related mappings

Using definitions by induction we shall now define several mappings important in the sequel. 1. The mapping J of the set NX N onto N. Let for x, y E N J ( x , Y ) = (X + Y + l

) +x.

THEOREM 1: J is a one-to-one mapping of N x N onto N.

PROOF.Suppose that J(x, y ) = J(a, b). We shall first prove that x = a. In fact, if we suppose x > a, then x = a+r, r > 0. Thus we would obtain (“+r+y+l) 2

+ r = (“+;+I).

3. THE MAPPING J OF N X N ONTO N

91

This implies b > r+y, for )(; is an increasing function. Hence b = r + y + s where s > 0. Substituting this value for b in (1) and letting c =a+r+y+l,

we obtain

(5) +

(3+ r (cz').But this is not true,for =

rz')<

(l) +

r < c, since r< c= (c";) . In the same way it can be shown that x < a does not hold. Hence x = u and we obtain = If y < b then we would have b = y f t , t > 0;

r+;+')("+2bf').

2 ("'g'")

> ("+;+'). Likewise we can- derive a contradiction from the assumption that y > b. Therefore and we would obtain

("+!+')

the function J is one-to-one. Now we shall prove that the range Z of J is identical with N. It follows from J(0,O) = 0 and J ( 0 , l ) = 1 that 0 , l EZ. Suppose that n E Z,i.e. that n = J(x, y ) for some x and y. If y > 0 then n+1 = J ( ~ , y ) + l = If y = 0 then

n=

(i)+ x =

Assuming that x

r'"')

ri'),

> 0 we can write

+x+l = J(xf1, y - l ) ~ Z . thus

n+l =

(xi')+ l .

(xi')+1 in the form

+1 =J(l,x-1); hence n+lcZ. Finally, if x = y = O t h e n n = O and n+ 1 = 1. Hence n + l ~ Z .Theorem 1 is thus proved. THEOREM 2: There exist functions K, L mapping N onto N such that J(K(x), L(x)) = x. Moreover, these functions satisfv the inequalities (2)

K(x) < x ,

L(x) < x.

The existence of the functions K and L follows from Theorem 7, p. 73, the inequalities follow from x J ( x , y ) and y < J ( x , y ) .

<

REMARK:The intuitive meaning of the functions J,K, L can be illustrated by arranging the pairs (Y, y ) of natural numbers into the following infinite array: (090) (031) ((42) *.*

(3)

. . . . . . . . . . . ...

m.

98

NATURAL NUMBERS. FINITE A N D I " I T B

SETS

and then ordering them in the sequence ( O , O ) , (0, I), , <0,2), (1, 1) (2,0), ... (4) The pair ( x , y ) occurs in the J(x,y)th position in this sequence. The nth term of this sequence occurs in the (K(n)+l)st row and the (L(n)+l)st column of (3). 2. The mapping of the set N"+' onto N. We shall define by induction a sequence of one-to-one functions such that the. kth term of this sequence (denoted by zk) is a one-to-one mapping of the set Nk+l onto N. Identifying every one-term sequence with its only term, we let z,,(x) = x for X E N , rk+l(e)= J(-ck(elk'),ekC1) for

eENk+'.

THEOREM 3: The function z k maps Nk+' onto N and is one-to-one. PROOF.For k = 0 the theorem is obvious. Suppose now that it holds for the number k . If e €Nk+'then elk' E Nk+', whence rk(elk')E N and, by definition, rk+l(e)EN. The function z k + l thus maps Nk+l into N . The fact that zk+lis one-to-one follows from the implications: ~

k )l = r k + l (e*> (zk(elk') = rk(e*lk'))A (ek+l= e?+d

+

--f

--f

(elk'

= e*lk') A

(ek+' = e k l ) + (e = e*).

It remains to be shown that for every number n there exists a sequence e E Nk+' such that zk(e)= n. By the inductive assumption there exists a sequence f € N k + l such that ?&(f)= K(n). We let e to be the sequence whose k f 1 initial terms coincide with those off and whose last term is L(n). For this sequence e the following formula holds by definition : rk+l(e)= J ( r k ( f ) , L(n)) = J(K(n),L(n)) = n. 3. The mapping of the set of all finite sequences of natural numbers onto the set N. Let for eENk+l

% ( 4 = J(k,Z k W ) . This function is a one-to-one mapping of the set of all non-empty finite sequences of natural numbers onto N. We have the following: 4: There exists a one-to-one mapping a of the set S of all THEOREM finite sequences of natural numbers onto N which satisfies the condition a(0) = 0.

MAPPING J OF N X N O N T O N

3.-

99

TO prove this it suffices to let b(e) = l+oo(e) for non-empty sequences e and 4 0 ) = 0. 4. The mapping J' of the set N"+l X N"+l onto the set Nn+l and of the set N N x N N onto the set N N . Let n be a natural number. For e, f E N"+l, let J'(e,f) = F [J(ek,fk)] Y

k
+

<

Hence if e and f are sequences with n 1 terms ek and fk (k n) respectively, then J'(e, f ) is the sequence of n 1 terms J(ek,fk) (k < n); and K'(e) and L'(e) are sequences with the terms K(ek) and L(ek) (k
+

Y

keN

K* (9)=

kEN

[K(vk)]

Y

L* (v) =

F [L(qk)]

keN

*

J*(pl, y ) is thus an infinite sequence whose kth term is J(plk,yk), K*(pl) and L*(pl)are infinite sequences whose kth terms are K(Q)k)and L ( Q ) k ) , respectively. THEOREM 5 : The function J' restricted to the set N"+l x N n f l is a one-to-one mapping of this set onto N""; the ,function J* is a one-to-one mapping of NNXN N onto N N . THEOREM 6: For any e E N"+l and for any pl E N N the following formulas hold J'(K'(e), L'(e)) = e, J*(K*(q),L*(Q))) = 9. The proof of these theorems is left to the reader. 5. The mapping of the set N N onto the set ( N N ) N .For k e N and for pl E N N let P)(k) = F [ q J ( k ,n ) ] ; neN

thus the sequence Q ) ( ~ has ) as terms yJ(k,O)yQ)J(k,l),Q)J(k,2), ... . THEOREM 7: The function M. q -+ F [ P ) ( ~ ) ] (which associates with kEN

every sequence pl the sequence Q)(O), pl(l), ping of the set N N onto the set ( N N ) N .

...) is a one-to-one map-

Q)(~),

100

111.

NATURAL NUMBERS. FINIT@A N D INFINITE SETS

PROOF.For every y E N N we clearly have M ( y )E ( N ~ )The ~ . function M is one-to-one. In fact, it follows from M ( v ) = M ( y ) that qdk)= Y ) @ ) for every natural number k. Thus v J ( k , m ) = Y J ( k , m ) for any k , m , and by letting k = K(n) and m = L(n) we obtain v n = ly, for all n. Finally, every element of the set (NN)N,that is, every infinite sequence t whose terms tk are elements of N N for all natural k, can be represented as M ( v ) for some 9.In fact, if q~ is the sequence 9. = tKcn)(L(n)), then pl(k) is the sequence whose nth term is tK(J(k,n)) (L(J(k,n))) = tk(n),i.e. P)(") = tk for arbitrary k. Hence M ( y ) = t. Q.E.D.

5 4. Finite and infinite sets The notions introduced in 0 1and 0 2 of this chapter allow us to derive basic properties of finite and infinite sets from the axioms of set theory. DEFINITION: We say that a set X has n elements and we write =n (where n EN)if there exists a one-to-one sequence with domain n and range X. Such a sequence is called a one-to-one sequence with n terms. A set X is finite if = n for some n EN,otherwise we say that the set X is infinite. A set X has 0 elements if and only if X = 0, for the only one-to-one sequence with 0 terms is the empty sequence. For every p E N the set p hasp elements; in fact, the function I p defined by Ip(x)=x for every X E P is a one-to-one sequence of p terms whose range is p. THEOREM 1: Iff is a one-to-one mapping of t%eset X onto the set Y , then = n if and only i f IY[ = n. PROOF.If e is a one-to-one sequence of n terms with range X , then f o e is also a one-to-one sequence (see p. 71) of n terms whose range is Y . LEMMA:If f is a one-to-one function, D l ( f ) = X u (a}, D , ( f ) = Y u {b} and a ef X,b 4 Y,then there exists a one-to-one function g such that X T Y and such that Y is the range of g. PROOF.Let f ( a ) = a,,f"(b) = b,. I f a, = b then bl = a and the function f maps X onto Y ;thus it suffices to take g =fix. If a, # b then a, E Y and similarly b, E X . In this case, g is defined as follows: g ( x ) =f(4 if x # b,, g(bd = a,.

1x1

1x1

1x1

101

4. FINITE AND INFINITE SETS

Checking that this function satisfies the lemma is left to the reader.

2: Let n E N . The following conditions are equivalent: THEOREM (i) = n'. (ii) There exists a set X,c X and an element a, $Xl such that IX,l = n and X = X,u {al}. (iii) X # 0 and for every X2 and a,, if a, $X2 and X = X2 u {a,} then !X2j= n.

1x1

PROOF.(i) -+ (ii). Let e be a one-to-one sequence of n' terms with range X. Condition (ii) is satisfied when a = e,, X,= X-{a}. (ii) -+ (iii). The condition X # 0 is an immediate consequence of (ii). Letting X, and a, denote respectively the set and element satysfying (ii) we infer that X, u {a2}= Xl u {a,}; and thus there exists a one-to-one function mapping XI u {a,} onto X , u {a2}. By means of the lemma there exists a one-to-one function g mapping X,onto X2 and hence, by Theorem 1 , IX2\ = n. (iii) + (i). Let a be any element of X and let X,= X-{a}. By (iii) we have \X,l = n and thus X, is the range of a one-to-one sequence e of n terms. The sequence e' of n' terms defined b y ek = e, for p ~ n , = n'. ei = a, is one-to-one and its range is X. Hence THEOREM 3: r f 1 x1 = m and IYI = 11, then m n if and only i f there exists a set Y , c Y such that the set X is a one-to-one image of Y,.

1x1 <

<

PROOF.If m n then m c n (see p. 89 and 91). Suppose that Y is the range of a one-to-one sequence e of n terms and Xis the range of a one-to-one sequence f of m terms. The function e o f cis thus oneto-one and maps X onto a subset of Y (for fC1(X) = m c n). Suppose now that there exist a set Y1c Y and a one-to-one function f mapping Yl onto X.To prove that m < n we shall use induction on n. If n = 0 then Y = 0; hence Y1= X = 0 and, moreover, m = 0 and m < n. Suppose that the theorem holds for some n E N and let I YI = n'. Hence Y = Y2u {a}, where IY21 = n and a $ Y,. Since X =f '(Y,), we have f ,(A') =f l(Y2n Y,) ufl({a} n Y,). The set { a } n Yl is either empty or equal to {a}. In the first case we have \Y2n YI = m. The formulas Y2n Yl t Y , and IYll = m show that the inductive hypothesis holds; hence m < n and thus m < n'. In the second case Theorem 2

102

III. NATURAL NUMBERS. FINITE A N D INFINITE SETS

implies m = p’ where p is a number such that IYz n Y,I = p. By the inductive hypothesis, p < n, hence m =p’ < n’. Thus the theorem is proved. THEOREM 4: I f 1 x1= m, IYI = n a n d X n Y = 0, then / X u YI = m+n. PROOF.The proof is by induction on n. For n = 0 we have Y = 0, and the theorem holds. Suppose that the theorem holds for the number n and let IYI = n’. Thus Y = Y , u {a}where a $ Y , and hence X u Y = ( X u Y , ) u {a}, where a $ Y. It follows from the inductive hypothesis that IXu Y,( = m+n and, by Theorem 2, / X u Y , u {a}\ = (m+n)‘. This proves the theorem because (m+n)’ = m+n‘ by the definition of addition (see p. 92). COROLLARY 5: For an arbitrary U the .finite sets contained in U form an ideal. In fact, a subset of a finite set is finite by Theorem 3, and the union of finite sets is finite by Theorems 3 and 4.

THEOREM 6: I f /XI = m and IYI = n, then the set X is a one-to-one image of Y if and only if m = n. PROOF.If m = n, then X is obviously a one-to-one image of Y , because there exist one-to-one sequences mapping the set n onto X and Y. Now suppose that X is a one-to-one image of Y . Then m is a one-to-one image of n. We shall prove by induction that m = n. For n = 0 the theorem is obvious. Suppose that it is true for some n and let m be a one-to-one image of n’. Hence m # 0 and we may assume m = p’. Because p’ = p u ( p } and n’ = n u {n}, p is, according to the lemma, a one-to-one image of n. Hence by the inductive hypothesis p = n and finally m = p ’ = n’, which proves the theorem. COROLLARY 7 : (The so-called drawer principle of Dirichlet.) If 1x1 IYI = n, m > n and f is a functim such that f ’ ( X ) = Y , then the function f is not one-to-one. Dirichlet formulated this theorem as follows: If m objects are put into n drawers and m > n, then at least one of the drawers contains at least two objects. Obviously, our function f is the function assigning to each object the drawer in which it is contained. =m,

4. FINITE AND INFINITE SETS

103

We shall now apply the theorems just proved to draw certain conclusions about infinite sets. THEOREM 8: If a set X is infinite and X c Y , then the set Y is infinite. This is an immediate consequence of Theorem 3. THEOREM 9: If X is infinite and Y finite, then the dzrerence X - Y infinite. This follows from Theorem 4.

is

THEOREM 10: The set N is infinite. PROOF.By way of a contradiction suppose that IN1 = n where n E N . Since n' c N , we infer from Theorem 3 that the set n' has m elements, where m is a certain element of N such that m < n. Because IpI = p for every p E N , we obtain n' n, which contradicts formula (4) (P. 90). 11: The range of a one-to-one infinite sequence is infinite. COROLLARY In fact, such a set is a one-to-one image of N . If it were finite then, by Theorem 1, N would also be finite. From this theorem and from Theorem 8 we obtain

<

THEOREM 12: If a set X contains a .rubset which is the set of terms of an infinite one-to-one sequence, then X is infinite. A set satisfying the hypothesis of Theorem 12 is called a Dedekind infinite set'). Theorem 12 can thus be expressed as follows: I f a set is Dedekind infinite, then it is infinite. The converse theorem is also true, but its proof requires the axiom of choice. 'THEOREM13: If a set X is infinite, then it is Dedekind infinite.

PROOF. Let f be a choice function for the family 2x-{0}. We extend f by letting f ( 0 ) = x , where x is an arbitrary fixed element of X. Thus f assigns an element of X to every subset of X and in particular f(Y)E Y if Y # 0. It was R. Dedekind in his book Was sind rcnd was sollen die Zahlen, Braunschweig 1881, who first called attention to the necessity of defining the notions of finite and infinite sets. The definition proposed there is equivalent to the one above.

III. NATURAL NUMBERS.

104

F"ITE AND INFINITE SETS

Now let us define by induction two sequences A E (2x)Nand a E X N :

a,, = x, a,, = f ( X - A,),

A, = {x}, A,, = A, u {a,,}.

We can prove by induction that the set A , is finite and A, c A, for m n. Hence X - A, # 0 and thus a,, E X - A, and a, E A, for every n E N. The sequence a is one-to-one. In fact, if k <j then ak E A k , but a j $ A k , for, letting j = n', we have Ak c A , and a j = a,,EX-A, c X - A k . Hence a j # ak. The set X is therefore Dedekind infinite, because it contains the subset {a,: n E N } , which is the set of terms of an infinite one-to-one sequence.

<

Exercises 1. I f f E XY, Y is an infinite set and X is finite, then at least one of the cosets off is infinite (Dirichlet 's principle for infinite sets). 2. If 1 x1 = m and IYI = n, then IXx YI = man and lXYl = mn. 3. If lXil = m for i = 1,2.3, IXj n &I = n j k for j , k = 1,2,3, JX,n X, n X31 = nlZ3,then IX, u X , u X31 = n,+nz+nj-nl,-n,j-n31+nlz3. Generalize this formula for the case of an arbitrary (finite) number of sets. 4. Prove that a set X c N is infinite if and only if //V (m < x ) . m xeX

$ 5 . Konig's infinity lemma

Let A and B be sets, B c A , and let f E AB. A finite sequence c E A" (resp. an infinite sequence C E A ~is )called a branch of length n (resp. an infinite branch) if c, =f(cmr) and c , , E B for every m such that m' En (resp. for every m' E N). The element c,, (which need not belong to B ) is called the initial vertex of the branch. If c is a branch of length n (or an infinite branch) and m En (or m EN), then clm is a branch of length m. OKONIGJS INFINITYLEMMA'): Suppose that all the cosets off are finite. If X E A .and for every n E N there exists a branch of length n with initial vertex x, then there also exists at least one infinite branch with initiaI vertex x. I) Denes Konig, Uber eine Schlussweise aus dem Endlichen ins Unendliche, Acta Litt. ac. sci. Hung. Fran. Josephinae, Sectio sci. math. 3 (1927) 121-130.

5. KONIG'S INFINlTY LEMMA

105

PROOF.Denote by K the set of elements u E A such that for every EN there exists a branch of length n with initial vertex x. By assumption x E K. LEMMA:If u E K then f - ' ( { u } ) n K # 0. In fact, suppose thatf'({u}) c A - K . Let e be a sequence of length m whose range is W =f-'({u}). Thus for every je m there exists a number EN such that there is no branch of length n with the initial vertex ei. Let nj denote the least number with this property and let p = (maxnj)'. Hence ni e p for all je m and thus for every element 9 l s m

of the coset f ' ( { u } ) there exists no branch of length p with initial vertex 9. On the other hand, it follows from the assumption u E K that there does exist a branch g of length p' and with the initial vertex u. Letting c j = gjt for jep', we obtain a branch of length p with initial vertex g,. But this leads to a contradiction, for g, ~ f - ' ( { u } ) . Now, let rp be a choice function for the family of non-empty sets of the form K n f ' ( { u } ) where.uEA. Let us extend the function y by defining rp(0) = x. Thus rp assigns an element of A to each set of the form f - ' ( { u ) ) n K . Define a sequence c by induction letting co = x,

cnr = rp(fl({c,,})nK).

We shall show that for every n e N the following formula holds: c,, ~f-' ({c,}) nK . (1) In fact, for n = 0 we have co = x, thus C , E Kand, by the lemma, f-'({cO})nK# 0. Hence by means of the definition of choice function co~~f-l({cO})nK. If (1) holds for some n e N , then by the lemma we obtain f ' ( { c , , } ) n K # 0. Hence cntt = rp (fl ({c,.}) nK) ~ f - ({c,.}) l nK. Formula (1) is thus proved by induction. It follows from (1) that f(c,,) = c,. Therefore c is an infinite branch with initial vertex x. Q.E.D. The proof above uses the theorem asserting the existence of a choice function (see p. 73). Thus we have used the axiom of choice. It is worth noticing, on the other hand, that we have not applied the full axiom of choice but rather a special case guaranteeing the existence of a choice function for an arbitrary family of finite sets. One can

106

III. NATURAL NUMBERS. FINITE AND I"ITE SETS

even show that it is sufficient to assume the existence of a choice function for a countable family of finite sets. It can be shown that it is impossible to eliminate the axiom of choice entirely from the proof of Konig's Lemma. In other words, this theorem is not a consequence of the axioms Z. # Example. Let I be the closed interval [0, 11 = { x : 0 x l } and suppose that Ij,"= { x : j/2" x (j+1)/2"} for j < 2",n EN. Let G be a family of open intervals such that S(G) = I . We say that G covers Ij," if there exists an interval D E G such that I j , . c D.

< <

< <

THEOREM: Let G and Ij,,,be as defined above. Then there exists a number n such that G covers all the intervals Ij,, for j = 1, ...,2"- 1. PROOF.Let A be the family of all the intervals I j , . and let B be the n E N which are not covered by G . family of all the intewals Ij,n+l, Let f(Ij,n+l) be that interval Ik,nwhich contains Ij,n+l. Thus we have defined a function, belonging to AB, whose cosets are finite. To prove the theorem it suffices to show that there exists a number n such that there is no branch with initial vertex I and length n. Suppose the contrary. By means of Konig's lemma there exists an infinite branch, that is, a sequence of intervals Ijn,"such that each interval is contained in the previous one and none is covered by G . Let x be a point common to all the intervals of this sequence. Since I = S(G'), there exists an interval D in G such that X E D .Hence for c D . Thus G covers I j n ,which n , contradicts the some n we have Ijn," assumption that Ijn," E B. # Exercises

1. Give an example of a function f~ NN such that for every n there exists a branch of length n with initial vertex 0 and such that there exists no infinite branch with initial vertex 0 (the cosets o f f cannot be all finite). 2. Prove that if A is a set of finite sequences of the terms 0 or 1 such that [C E A +cln E A] and such that for every n there exists a sequence of at

A n

least n terms in A, then there exists an infinite sequence tp such that for every n the restriction qln belongs to A. Does this theorem hold for an arbitrary set A of sequences of natural terms? 3. Let A be a set of binary relations between natural numbers, A c 2N N, satisfying the following condition:

5. KONIO’S INFINITY LEMMA

107

1. For every n~ N there exists a relation in A whose field contains n. n E N , then R n ( n x n ) E A . Prove that there exists a relation R, with field N such that R , n ( n x n ) A ~ for every n E N. 4. If mankind is to last forever, then at least one man living now will have a male descendant in every generation. [Konig]

2. If R E A and

0 6. Graphs. Ramsey’s theorem Every set G whose elements are unordered pairs of distinct elements is called a graph. The set S(G) is called the field of the graph, its elements are called vertices. If {a,b} E G then the pair {a, b ) is called an edge of the graph. In this case we also say that the vertices a and b are joined to each other. This terminology is motivated by the geometrical interpretation of a graph in case the field consists of a finite number of linearly independent points in This geometrical interpretation is given by a figure consisting of polygonal lines (edges) having no points in common except their vertices. An example of a graph is the set consisting of all the pairs { x , y } where x, y E X and x # y . This graph is denoted by [XJ, and is called the complete graph with field X . If G is a graph with field X, then the difference G‘ = [ q 2 - G is also a graph and is called the complement of G. A graph H is said to be a subgraph of G if H c G. The study of graphs is equivalent to the investigation of symmetric and irreflexive relations, that is, relations satisfying the conditions : xRy +yRx, x non-Rx. In fact, every graph G can be associated with such a relation by letting RG = { ( x ,y ) : { x , y } E G } . Conversely, every symmetric and irreflexive relation R determines a graph G = { { x ,y } : [ x R y ] } where

R

= RG.

RAMSEY’S THEOREM: If G is a graph with infinite field, then either G or G’ contains a complete subgraph with infinite field. PROOF.Let f be a choice function for the family 2S(G)-{0} and let K(x) = {y: { x , y ) } ~ G ,L ( x ) = ( y : ( x , y } ) ~ G ’Thus . the set K ( x ) consists of the vertices joined to x in the graph G and L ( x ) consists of the vertices joined to x in G’.

m. NATURAL

10s

NUMBERS. FINITE

AND INFINITE SETS

Now we define by induction four sequences T E (2S(G))N, AE(~~(G))~, aE(S(G))N and Y E NN.Namely, let To = A.

= S(G),

a. =.f(S(G)),

yo = 0,

T,, = { x : (XE A,) A (the set K ( x ) n A , is infinite)}, a,, -

A,,

J O if TnI# 0, =

Ynr

=

11 if T , , = o .

We consider two cases. Case 1. cp, = 0 for all n E N . In this case we prove by induction that for all n the following formulas hold: (1)

antE Tn,,

12)

The set K(a,t) nA , is infinite,

(3) (4)

A,, c K(a,,) n An,

T,, c A , .

If k > j > 0 then it follows from (3) that Ak c A k - 1 c A,. Since akE Ak-l -Ak (by (1) and (4)), akE A, -Ax. Hence the set of all terms of the sequence a is infinite and the complete graph whose field is this set is contained in G . Thus the theorem is proved for this case. Case 2. There exists j such that y,, = 1. Let jo denote the least among such numbers. Thus T, # 0 for j < jo, which implies aJoE Ti,. The set Ajo is infinite, for it equals either S(G) (if jo = 0) or K(ajo)n Ajo.-l. Both sets are infinite, the former by assumption and the latter because ajoETjo. We define by induction a sequence B of subsets of A j o and a sequence b of elements of A,.

rj)

BO= Ajo, BJ+l =

"

Bj'

bo

=f @ O )

b,+l =

9

if B j # 0, if B j = 0.

The sets Bj form a descending sequence: Bj, c BJ for j e N ; the set Bo is infinite. We shall prove'by induction that each set Bj is infinite. In fact, suppose that this is the case for some j e N . If the set Bj,, were finite, then for every x ~ [ B ~ - L ( ( b , j j - { bwe ~ }would have { b j , x }

6. GRAPHS. RAMSEY’S THEOREM

109

EG, whence the sets K ( b j ) n Bj and K(bj) n A j , would be infinite. This implies bjETj,,, which contradicts the fact that T j ; is empty. It follows from the above that for all j we have bj E B j . If k > j then b k c B k c B j c L ( b j ) ,hence {bk,bj}EG. Letting Y be the set of a11 the terms of b we obtain [ Y ] , c G . The theorem is thus proved, because the set Y is infinite. Example. Let X c N and let G be the set all unordered pairs { x , y } , where x, Y E X and x and y are relatively prime. Applying Ramsey’s theorem we conclude that if X is infinite, then either X contains an infinite subset Y where every two numbers belonging to Y are relatively prime, or else X contains an infinite subset Y where no two numbers belonging to Y are relatively prime.

REMARKS 1. Ramsey’s theorem may be illustrated as follows. Let every edge of the complete graph [XI, where X is an infinite set be coloured either white or black. Then X necessarily contains an infinite subset Y such that all the edges of the graph [yl, are of the same colour. 2. Ramsey’s theorem has a counterpart, also due to Ramsey, in the finite case: For any natural number n there exists a natural number q such that, i f the set X contains at least q elements, then for every graph G c [ X ] , there exists a subset Y c X of n elements such that either [y1, c G or [Yl2 c

G’.

In other words, if G is a graph whose field consists of q elements, then either G or G’ contains a complete subgraph with field consisting of n elements. The theorems given in Exercises 1 and 2 below may also be stated to hold for finite sets. Exercises 1. If G is a graph with infinite field and G = G,uG,U ... UGk, where G i n G j = 0 for i # j , then at least one Gi contains a complete subgraph with infinite field. [Ramsey] 2. Generalize Ramsey’s theorem in the following way. Suppose that X is an infinite set. Let [XI,, be the family of all subsets of X consisting of n elements and let [XI,, = M , u M , u ... u M k where M i n M j = 0 for i # j . Then there exists an infinite subset Z c X such that [Z],cMj for some j .

110

m.

NATURAL NUMLIERS. FINITB AND INFINITE SETS

3. A graph G is said to contain a triangle provided there exist three distinct elements a, b, c such that { a , b } , {b, c } , { a , c } EG. Show that if X has at least 6 elements and G c [ X I z ,then either G or G contains a triangle. 4. Let N =

U Ak

where the sets A k are pairwise disjoint. Show that for any

k
natural number q there exist an increasing infinite sequence 9 of natural num-

bers and a number j o < p such that

n+rl

q~ E A j G for every n l). i=n

l) From the great number of publications devoted to Ramsey’s theorem, let us mention only: F. P. Ramsey, On a problem of formal logic, Proc. London Math. SOC.30 (1929), 2nd series, pp. 264286; Th. Skolem, Ein kombinatorischer Satz mit Anwendung auf ein logisches Entscheidungsproblem, Fund. Math. 20 (1 933) 254-261; R. Rado, Direct composition of partitions, Journal of the London Math. SOC.29 (1954) 71-83; P. Erdos and R. Rado, A partition calculus in set theory, Bull. Amer. Math. Soc. 62 (1965) 427-489. The last paper contains a detailed bibliography of the subject.

CHAPTER IV

GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

In the present chapter our treatment will be based upon the axioms ,Zo as in the preceding chapters. Theorems which are not marked by the symbol are theorems of the system Z. The purpose of this chapter is to generalize the operations of union, intersection and Cartesian product for an arbitrary number of sets. O

0 1.

Generalized union and intersection

Let F be a function from a non-empty set T into the family of all subsets of a given fixed set %. Thus F E ( ~ % )Instead ~. of F(t) we shall write Ft. Let W be the range of F, that is, the family of sets Ft where i ET. The union of the sets belonging to the family W is denoted by S(W) or ( W ) (see p. 52), the intersection is denoted by P ( W ) or (W) (see p. 61). The following notation will be used.

u

n

S(W) =

u F,,

P W )=

t

n Ft. I

It is easy to show that (1)

X E

u F~= v I

( x E ~ t ) ,

XE

t

nF~ 1

(X~F,). t

If the set T consists of the single element a, then

u Ft t€T

= Fa=

f7 F,; teT

on the other hand, if T consists of two elements a and b, then

w. GENERALIZED UNION,

112

INTERSECTION A N D CARTESIAN PRODUCT

Thus these notions are indeed generalizations of the notions of union and intersection of sets to the case of an arbitrary family of sets. It follows from (1) that the following equations hold for arbitrary propositional function @(x, y ) of two variables with a limited domain:

u { x : @(X,Y)}

= {x:

Y

(2)

v @(X,Y)}, Y

n { x : @ ( x , Y ) } = { x : A @wJ. Y

Y

In fact, let Fy = { x : @ ( x , y ) } ; we obtain @ ( z ,y ) = 2 E { x : @(x, y ) }

and thus >€

(x:

v Y

@(x,y)) =

=z E Fy

@(z,y) Y

= V ( Z € F y ) = Z E U F y ~ Z E U { X @: ( x , y ) } . Y

Y

Y

The proof of the second equation in (2) is similar. By (l), the formulas concerning quantifiers given in 0 1, Chapter IT, lead to the following formulas for the generalized operations: (3) (4)

1. GENERALIZED UNION AND INTERSECTION

113

(13)

In formulas (8) and (9) the symbol - denotes complementation with respect to the set X. The proofs of the formulas above follow directly from the respective formulas in $1, Chapter 11. As an example we prove de Morgan’s law (8): XE

-(n Fr) = i nFr) 1[A ( x ~ ~ r ) ] t

E

v

(XE -Ft)

t

3 X E

U(--F,); I

where we apply successively formulas p. 6, (6), p. 47, and (1) above. The diagram on page 50 also leads to formulas for the generalized operations. It suffices to replace the implication sign --r by the inclusion sign c ,and @ by a function F of two arguments having sets as values. In particular, the following important formula holds:

u nF,, n u F,,

(14)

t

s

s

t

This inclusion cannot in general be reversed (see p. 50). THEOREM 1: The union

u Ft is the unique set S satisfying the conditions: I

A (Ft = s)’

(15)

(16) The intersection

A ~[A~~tcaI+(S=X)1. X I nFr is the unique set P satisfying the conditions: A (P= Ft), A { [ A(X=ml ( X 4 . f

(15 ’ ) (16‘)

-+

X

t

114

W. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

u Ft is the smallest set containing all the

In other words, the union

t

sets Ft and the intersection

nFt is the largest set included in each of I

the sets Ft.

PROOF.It follows from (3) and (13) that the union

uFt satisfies 1

conditions (15) and (16). Conversely, assuming that the set S satisfies Ft c S . Setting these conditions we infer from (15) and (13) that

u X = u Ft in (16) and applying (3), we obtain S u Ft. Hence S u Ft. t

c

t

t

=

t

The proof for intersection is similar.

THEOREM 2: (GENERALIZED ASSOCIATIVE

LAWS.)

If

T=

u H,

UEU

where H is a set-valued function with domain U (i.e. HE(^=)'), then

PROOF.Letting S

=

u Ft

and

S,

tET

=

u F,,

tEHU

we reduce equation (17) to the form (19)

s= lJ s,. UEU

By assumption we have S 2 Ft for every t E T , in particular, for every t E H, . Thus S =I S, by Theorem 1. On the other hand, suppose that X 3 S, for arbitrary U E U. If t E T , then there exists u E U such that t e H,, whence it follows that S, 2 F,, and thus X D Ft. Since t is arbitrary, we conclude that X 2 S. Applying Theorem 1 we obtain (19). The proof of (18) is similar.

115

1. GENERALIZED UNION AND INTERSECTION

THEOREM 3: (GENERALIZED COMMUTATIVE LAWS.) tation of the elements of a set T , then

PROOF.Let S =

u F,+,(t). If

t E T,

If q~

is a permu-

then t = cp(qJ"(t)), and because

t

S 2 Fpr(,,, for arbitrary U E T ,in particular for u = ye(t), we have S 3 F,. Conversely, if X is a set such that X I> Ft for t E T , then X I> Fpr(t),because v(t)E T. Thus X 3 S, which shows that S is the smallest set containing all the sets Ft (i.e. S = U F,). The proof of the second formula is similar. 4: (GENERALIZED DISTRIBUTIVE LAWS.)')If THEOREM M= T,, and K = YE^^: A ( Y n T , , # 0 ) } ,

u

UEU

then (20)

(21)

U d J

n u F f = Yu€ K mi, u 17 Ft = n u F,.

UEUteTu

t€Y

u t U t€TU

Y€Kt€Y

PROOF.Suppose that Y E K and U E U . By the definition of the family K we have Y n Tu# 0, thus there exists toE Y nT,,. This implies by (3) that Since this inclusion holds for any u E U (where Y is constant), we infer from Theorem 1 that n F t c uFt.

n

U E UtcTu

t EY

Since Y is arbitrary, we obtain by (3) the following inclusion: (22)

u n F t c n UF,.

YEK t € Y

UEUtETU

To prove the opposite inclusion, suppose that

UF,.

U E UtETU

Y = { t E M : aEF,}. If U E U then by (23) aE F t . Thus there exists tET, such that

u

tETu

') See: A. Tarski, Zur Grundlegung der Boole'schen Algebra I, Fundamenta Mathematicae 24 (1935) 195.

N. GENERALIZED UNION, INTERSECllON AND CARTESIAN PRODUCT

116

a € F,; hence ~ E Ywhich , proves that Y n T, # 0. By the definition of K we have Y E X . It now follows from (24) that A ( u E F ~ that ); t€Y

is, a E (7 F,. This shows that t €Y

a€ U

(25)

f7 F t .

YER t€Y

This together with (22) gives (20). To prove (211, replace Ft in (20) by S - F t , where S = F t . Then we obtain:

u n u (s-m u n teM

=

U€U t€T"

Y€Kt€Y

whence, by de Morgan's laws (8) and (9) and by - ( - F t ) = Ft, we obtain (21). We shall now generalize formulas (1)--(4), $8, Chapter 11, concerning images and inverse images of finite unions and intersections, to the case of arbitrary unions and intersections. THEOREM 5 : Let F E(231>* and let fE Y X . Then

If the function f is one-to-one, then the inclusion sign in (27) can be replaced b y the identity sign.

PROOF.It follows from the definition of image that

V [b

Y € f l ( U Ft)

E

=

v [v x

U Ft) A (Y

=

v (Y

fW)J

( ( x E Fr) A (Y =f

:

= i j [// ((. t

=

E Fr) A

(41

(u =f<.>))]

X

Ef' (FJ) =Y E

u f' (4) >

which proves (26). Similarly, by means of (18) on p. 49, we obtain the following equivalences:

117

1. GENERALIZED UNION AND INTERSECTION

A v [(x E 4) * (Y =f(41 = A (uEflo;t)) = Y E nm), t

+

f

x

whence it follows that (27) holds. If the function f is one-to-one, then using (27) for the inverse funcwe obtain t i o n y and for the sets fl(Ft)

f-l(nmt))=

(m)) nF,,

nf-1

=

t

and by (2), p. 75, it follows

n

f1

f

n (F,)).

(Ff)= f1(

Since (27) also holds, Theorem 5 is proved.

THEOREM 6 : If G E (2Y)Tand f E Y x then

G,) = nf-'(Gt).

(29)

t

t

The proof can be obtained from the following equivalences, which are consequences of the definition of the inverse image (see p. 75).

(uGf) = f ( y ) u Gf = 'v[fO)GI = v Iv ~f-' (G,)] =y u f-'(Gt); €f-l(nG,) =m nG, = ,\ [mGtl

Y ef-'

E

E

E

f

E

E

=A k ~ f - ' ( G , ) ] r yn ~ f-'(Gt). t

t

Formulas (26) and (28) assert the additivity of the operation of forming images and inverse images. Formula (29) asserts that the operation of forming inverse images is multiplicative. The operation of forming images is multiplicative, however, only for one-to-one functions. Examples Let the set 1 be a topological space (see Chapter I, $8).

118

IV. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUm

1. r f F is a function whose values are closed sets (see p. 27), then the intersection P = (7 & is also a closed set. t

PROOF. Since P c F,, we have because 8 = Ft. This implies that

c

for every t ; thus Pc F,, 4 = P, hence P= P, for

Pc n t

P c 3 by Axiom (3), 6 8, Chapter I, p. 26. 2. If G is a f l c t i o n whose values are open sets, then the union

S=

u G, is an open set. f

PROOF.The sets I-Gt are closed, thus the intersection n (l-Gt) t is also closed. By de Morgan’s law (9) the set 1-S is closed; hence the set S is open. 3. I f D is afunction whose values are regular closed sets (cf. p. 38), then the set So= D, is a regular closed set containing all the sets Dt. More-

u t

over, every regular closed set containingall the sets Dt also contains the set So.

PROOF.Clearly So3 Dt, so that Int(So) 3 Int(Dt), thus _______

(9

Int(S0) =3 Int(D,) = Dt.

Since t is arbitrary, we infer by Theorem 1 that Int(S,)

3

u Dt r

and

Int(So) 3

u Dt

= So.

I

On the other hand, Int(So) c So. Thus ~-

Int (So) c So=: So,

which proves that So= Int(SJ. Hence the set So is regular closed. It follows from (i) that So contains each set D,. If 2 is a regular closed set and 2 3 D, for every t, then t

1

4. r f D is a function whose values are regular closed sets, then the set

Po =

is a regular closed set included in each set D*. More-

over, every iegular closed set included in each set D, is also included in Po.

1. GENERALIZED UNION AND INTERSECTION

PROOF.Let X

119

D,. We thus have

=-

t

Po = Int(X) = X"-"- hence

Int (Po)= X"-"-"-"-.

Applying formula (15), $8, Chapter I, p. 29 we have Int(P,,)

=X = Po.

Hence the set Po is regular closed. -~ Since X c D,, we have Int(X) c Int(D,) and Int(X) c Int(D,), that is, Po c Int(D,) = Dt for every t. Finally, if 2 is a regular closed set and 2 c D, for every t, then _ _ _ ~ 2c D,. Hence I n t ( 2 ) c Tnt(X) and Z = Int(2) c Int(X) = Po.

n f

5. As a result of the theorems proved in Examples 1 and 2, it is

possible to define a topological space by taking as primitive notion either that of open set or that of closed set instead of closure. Namely, we may conceive of a topological space as a set with a distinguished family of subsets F. Subsets belonging to the family F a r e called closed sets. We suppose that F satisfies two conditions: (i) I f W c F, then P ( W )E F (that is, the interesection of an arbitrary family of closed sets is closed). (ii) If a family W is finite and W c F, then S(W) E F (that is, the union of a finite number of closed set is closed). We obviously assume that P(0) is the whole space. If we take the notion of open set as primitive, then denoting the family of open sets by G we assume axioms dual to (i) and (Z): (i') If W c G then S ( W ) E G. (ii') If a family W is finite and W c G, then P ( W ) E G. The system of axioms (i)-(ii) is equivalent to the system (1)-(4) given in Chapter I, p. 26. The axioms (1)-(4) are satisfied if we define Aby the formula A= P(W'), where WAis the family of all closed sets containing the set A. Then we have ( A = A> = ( A EF). A similar remark can be made for the system (i')-(Z'). 6. A family R c F is said to be a closed base for the topological space if for every A E F there exists W c R such that A = P(W). A family

120

UNION, INlBRSECI'ION AND CARTESIAN PRODUCT

IV. G E -

R

c F is a closed subbase if the family of all finite unions of the sets belonging to R is a closed base. 7. The notion of open base and subbase can be defined dually replacing P by G (= the family of open sets), intersection by union and union by intersection.

ExerdseS 1. Let F E (2')',

fe

and 35 =

~-I(Y) =

2. Prove that

uFt. Let

=fl

Ft. Prove that

for every Y c 9. UYT'(Y) t

(U Ft)X(UG") t

=

U (FtXG,), t. u

3. Let T be any set and let KC^^. Let the operation DK on Fe(2X)T be defined by the formula XEDK(FE ) {t: x ~ F t } e K . Find K for which the operation DK coincides with the operations of union and intersection discussed above. 4. Show that if Z is an ideal in 2* and K = 2T-Z, then the operation DK is distributive over finite union; that is, 5. Prove that the family of all intervals r < x < s, where r and s are rational numbers, is a base for the space C of real numbers. Prove that the sets { x : r < x ) and ( x : x < r } , where r is rational, form an open subbase for this space. 6. Let X be any set and R be any family of its subsets. Prove that the set 3 can be considered as a topological space with the family R as an open subbase (resp. closed subbase). 7. If X is a topological space and R is an equivalence relation with field X , then X{R becomes a topological space when we assume that a set Uc XIR is open if and only if the union S(U) = UZ is an open set in X. ZEU

8. Prove that

the canonical mapping X -+X/R is continuous if XIR h a the

quotient topology defined in Exercise 7.

8 2. Operations on infinite sequences of sets We shall now consider a special case of the previous operations; namely, where the domain T of the function F coincides with N,that is, where F is an infinite sequence of sets. In analogy with infinite

121

2. OPERATIONS ON INFINlTE SEQUENCES OF SETS

series and products of real numbers, we write m

UFn or UFn n

or Fo u Fl u

...

u F,;

instead of

ntN

n=O

nFn or n

nF,.

00

OFnor Fon Fl n ...

instead of

nEN

n=O

The following formulas follow immediately from formulas (2), Q 1

(j{x: @ ( n , x ) } = (x: v @(n, x)}, w

n=o

n=O

(1)

fi {x: @ ( n , x ) } = ( x : A @ ( n , x ) } , 00

h=O

n=O

where @(n,x) denotes a propositional function of two variables, n is limited to N and x to a given set X. Besides infinite union and intersection we consider the operations Limsup Fn

(limit superior of the sequence Fo,Fly...),

n=m

LiminfF,

(limit inferior of the sequence Fo,Fly...),

n=m

defined as follows m

Limsup Fn n=m

=

w

w

nU

Fn+Ly

LiminfF, = n=m

n=Ok=O

U

w

Fn+k.

n=Ok=O

It is easy to check that LimsupFn is the set of those elements x which belong to Fnfor infinitely many n. Analogously, x belongs to LiminfF, if and only if it belongs to Fn for almost all n; that is, if it belongs to all but a finite number of the F,. It is easily seen that Liminf Fnc Limsup F,. n=w

n=w

(see formula (18), p. 49). If the inclusion sign in (2) can be replaced by the equality sign, that is, if the superior and inferior limits are equal, then their common value is denoted by Lim F,, n=m

122

IV. GENBRALIZBD UNION, O N

ANJJ CARTESIAN PRODUCT

.

and is called the limit of the sequence Fo,Fly ... In this case we also say that the sequence is convergent. This terminology is similar to that used in the theory of real numbers. In order to emphasize this analogy, let us consider the notion of the characteristic function of a given set. Let the set 1 be given and X c 1. The function with domain 1 1 0

(3)

if if

xcX, x~l--X

is said to be the characteristic function of the set Xl). It is easy to show that the sequence Fo,PI, ... of subsets of 1 is convergent if and only if the sequence of the characteristic functions of these sets is convergent to the characteristic function of LimF,,. n=m

It is also easy to show that the following conditions are equivalent2): (4)

Lirn (F,,-A) = 0,

n= m

LimF, = A ,

(4’)

n= w

where the sign A denotes the symmetric difference of two sets. The same equivalence holds for real numbers if we replace F,-A by IF,,-AI. PROOF.Condition (4) is equivalent to the following: every element x belongs to Fn -I A for at most finitely many n. In other words, for every x there exists no such that n > no implies x E F,, = XE A . Suppose that x E Lirn sup Fn, i.e. that x belongs to F,,for infinitely many n. It follows from ( 5 ) that x s A and that x sF,, for all n > no; that is, x E Lirn inf F,,.Thus we have proved that (4) implies

(5)

(6)

Lim sup F,, c A c Lirn inf F,, n=w

n= w

from which (4’) follows by (2). I) See: Ch. de Ia Val& Poussin, Inte‘grales de Lebesgue, fonctions #ensemble, classes de Baire, 2nd ed. (Paris 1936). See: E. Marczewski, Concerning the symmetric direrenee in the theory of sets and in Boolean Akebraq Colloq. Math. 1 (1948) 2W202.

123

6. OPERATIONS ON INFINITE SEQUENCES OF SETS

Conversely, suppose that (6) holds and x E A . Thus x E Lim inf F,, and x E F,, for all n greater than some no. If, on the other hand, x # A , then x $ Lim sup Fn and hence x # F,, from an no on. Hence condition (6) implies that (5) holds for every x if n > n,. We shall still prove several more special laws concerning interchanging the symbols n and v and replacing two operations of the same kind by one such operation. Let F be a function defined on the set N X N , F' a function defined on the set N N x N N and F" a function defined on the set N x N N , where the values of these functions are sets. We use the notation introduced on p. 96-100.

U UFm,n m n

(7)

=

F~(~),~(~,. U F ~ ( ~ ) , ~n (n ~ Fm,n )~ =n P m n P

We shall prove only the first of these formulas. Clearly, Fg(p),L(p) c F,,,,, and thus Fg(p),L(p) c F,, ,,.On the other hand,

uu m

n

if x E U m

u

uu

P

m

n

u Fm,n, then for some m, n we have

x E F,,,, ; hence x E

n

where P = J(m, 4.

FK(P),L(P)

n u Fm,n = U n

(8)

m

n

P,E"

Fm,p(m)*

m

Since q is arbitrary, we have

un

Fm,p(rn)

m

If X E

n u Fm,n,then m

x E F,,

n.

hence x E

c

nU m

Fm,n.

n

for every m E N there exists n such that

n

Letting in) = min ( x E F,, ") we infer that x E

u nFnn, v

m

n

p(m).

n m

Fm,q(m);

N. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

124

In fact, @'"EN", thus Fi,,(m)

u F i , I , whence it follows that

=

I

F;, m

xE n m

,(m)

n u FZ,,. To prove the opposite inclusion, suppose that

c

u F;,

m

v

and let Zm= {v: x E Fi,,}. Hence Z # 0 for every

I

v

rn E N , and so there exists a choice function h for the family consisting of all the sets 2,. Letting h(Zm) = f ( m ) , we infer that f ( m ) N~ N and x E F i , f(m) for every rn E N . It follows from Theorem 7, p. 99 that there exists a sequence pl E N" such that I$") = f ( m ) for every rn E N . Thus x E Fa, ,(m) for every m E N , that is, xE

n F:,

u nFG,

c

,(m)

m

,,,(m).

~m

Using formulas (7)-(10) we shall obtain a formula to be used in Chapter X. In this formula G is a -function of four variables, with domain N Nx N Nx N x N, whose values are sets.

In fact, replacing F:,I in (10) by

nGv,+,m,n(where p is an arbin

trary but fixed function), we obtain

f3 L' f l Gv,p,m,n m v

=

n

U n fGv,+(m):m,n, l v

thus the left-hand side of (1 1) equals

m

n

u u nnGv,,(m),m,n. ~

p

m

Apply-

n

ing (7) and (9) we obtain (11). Exercises 1. Prove that the characteristic function defined by (3) satisfies the following conditions: (a)

(b) fi(X)

= 0, = 1,

(c) f - X W = 1-fx(x), (d) fA n B ( x ) = fA ( x ) .fB ( X I , fA-B ( X I = fA ( x ) -fA

n

B(XI-

2. Prove that if Fo c F, c F2 c

..., then

m

u Fn

n=O

= LimF,. n=m

125

2. OPERATIONS ON INFINITE SEQUENCES OF SETS

m

3. Prove that if Fo3 F,

..., then

3

F, = LimF,,. n = ~

n=m

4. Prove that if Fo= 1, then m

1 = ~ F ~ - F ~ ~ ~ ~ F ~ - ~ ... ~ u ~ U (7( FFn . ~ - F , ) ~ If, moreover, Fo3 F,

3

F22

n=O

..., then m

( F, - F~ ) u(F ~ -F . )u

... u n=o n F n = l-[(Fo-Fl)u(Fz--F~)u

5. Prove that if k, < kz <

..., then

LiminfF, c LiminfFk,, n=m

6. Prove that if

...I.

n=m m

m

n=l

n =1

LimsupFk, c LimsupF,. n=m

n A. n nB.

= 0,

m

n=m

then

m

where Bo = 1. m

7.

u

m

u Bn, where Bl Al and Bn ... uAn-,) for n > 1. Let U n n U c, and let U c.,+, U c,,,,.Prove A.

n=l

n=l

8.

= A,-(A,u

=

=

B ~ , =, A =

n

m

c

m

n

n. m

n,m

that

u k=l (D

A

= LhA,,

where

A, =

n=co

... nBk,,)n(CI,kU ... ucn,k).

9. Prove that (a) Liminf(-A,) = -- LimsupA,, (b) Lim(-A,) = -(LimA,), (c) Liminf(A,nB,) = LiminfA,n LiminfB,, (d) Limsup(A,uB,) = LimsupA,u LimsupB,, m

(e)

n A,

uAn, m

c LiminfA, c LimsupA, c

n=l

n=l

( f ) LiminfA,u LiminfB, c Liminf(A,uB,), (9) Limsup(A,nB,) c LimsupA,n LimsupB,, An), (h) A ILiminfA, c Limsup(A A A LimsupA, c Limsup(A A An); show that the opposite inclusions d o not hold in general. _I_

10. A function f from sets into sets is said t o be continuous if for every convergent sequence F,, F2,... the following identity holds: f (LimF.) = Limf(Fn). n=m

n=m

126

N. OENBRALIZED UNION, INTERSEcIlON AND CARTESIAN PRODUCT

Show that the functions X u Y , X n Y ,

-X and generally

uF,, and nF,, are n

n

continuous with respect to each component.

11. Prove the following condition for a sequence F,, to be convergent: for every sequence of pairs < r n i , Q > such that limmi = limni = co, we have I=m

k m

nI (Fmi

Fnj)= 0.

L

[Marczewski]

12. If K is a family of subsets of N such that the complement of every set in K is tinite, DK(F) = LiminfF,; if K is a family of infinite subsets of N then DK(F) = LimsupF, (see 8 1, Exercise 3). Using this result, generalize the operations Limsup, Liminf for the w e where the argument is a function defined on an arbitrary set T (not necessarily on the set N).

5 3. Families of sets closed under given operations Let X be a fixed set and f a function of an arbitrary number of variables, where each variable ranges over the subsets of X. For simplicity let us suppose that f is a function of two variables; that is, the domain off is the cartesian product 2x x 2x. A family R c 2x is said to be closed under a given finction f if

A [WI ER) A (YzE R)

Y1,Yn

4

(f(Y1,Yz)6R)I-

there exists a f h i l y R1 such THEOREM 1: For each fa@Iy R c that: 1 . R c R1 c 2x; 2: the family R1 is closed under the operation f; 3. the family RI is the least family satisfying conditions 1 and 2, that is, if R’ satisfis the following two conditions

then R1 c R’.

PROOF. Let K be the set of all families R’ satisfying (1). K is a non-empty set, for 2x E K. The required family is the intersection R’.

n

R’EK

The family R1 satisfying conditions 1-3 is uniquely determined. In fact, if R2 also satisfies the same conditions, then R1 c R,, since R1is the least such family. Similarly we obtain R2 c R1.Hence R1 = R2 We denote this family by R*.

127

3. FAMILIES OF SETS CLOSED UNDER GIVEN OPERATIONS

THEOREM 2: For arbitrary families R, R1 and R2 the following conditions hold (i) R c R*, (ii) R1 c R2 + RT c R,*, (iii) R** = R*.

PROOF.Formula (i) follows from Theorem 1 (condition 1). Formula (ii) follows from the fact that R: is a family closed under f and containing R1, thus by miaimality, R t =I RT. Finally, condition (iii) can be proved as follows: (i) implies R* c R**; since R* 3 R* and R* is closed underf, we obtain R** c R* by minimality. Theorems similar to 1 and 2 also hold for the case where there is given not one function f but an arbitrary family of such functions and R* denotes the least family containing R and closed under all these functions. Moreover, the domains of these functions may be sequences of subsets of X. We shall not, however, formulate all of these generalizations. Example 1. Let f denote the union of sets, i.e. f ( Y l ,Y 2 )= Y , u Y , . The least family of sets containing R and closed under f i s denoted by R,. This family consists of finite unions of the form U Y i where k n

# 0 and Y = (Yo,Y 1 ,..., Y n - l ) is a sequence of sets belonging to R ; in other words, Y ER". Similarly, if g is the function defined by g(Yl, Y,) = Y 1 n Y,, then the least family containing R and closed under g is denoted by Rd.

n E N, n

This family consists of all intersections of the form

n Y , where

fen

n E N , nfO, Y E R " .

Example 2. The least family containing R and closed under the operations f and g defined in Example 1 is called the lattice of sets generated by R.

THEOREM 3: The lattice of sets generated by R is identical with the f amih Rsd. More0 ver, RSd= Rd, . PROOF.First we prove the second part of the theorem. Let Z E Rsa, that i s , Z = Y i , where n E N , n # 0 and YiERsfor i < n. We show

n

icn

N. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

128

by induction that Z E Rd,. For n = 1 we have Z = YoE R, and R, c Rd, because R c Rd. Suppose that the theorem holds for n = k and let Z be the intersection of k+ 1 components Yi belonging to R,. In particT j where m e N , m # 0 and T j e R for j < rn. Let ular, let Yk =

u

j<m

Z'

=

n Yi.By the induction hypothesis, 2' E&, k k

where p

E N,

p # 0 and

2 = 2' n

v h E Rd for

h < p . Since Z

u Ti u (2' n T j ) u u =

i<m

=

thus 2' =

u vh h
= Z' n Y,, we have (vh

n Tj).

j < m h
j i m

It follows now from V hn T j E Rd that Z E Rds. We have thus proved that Rd c Rds. In a similar way we prove the opposite inclusion. Now let us show that RSdis the lattice of sets generated by R. It is clear that the family Rsd is included in this lattice, for the operations of union and intersection do not lead out of the lattice. On the other hand, Rsd = (RS& RdS= (Rd,),; thus the family R,d iS closed under both union and intersection. Hence it contains the lattice generated by R. Exercises 1. Let R, be the least family of subsets of X closed under the operation Yl-Y2. Prove that (a) Rd C Rr; (b) if X E R then R, c Rr. Show that the assumption X E R in theorem (b) is essential. 2. Prove that the least field of sets containing R is ( R u cR),d where cR = { X - Y . Y ER } . 3. Let R z , resp. RA, denote the least family containing R and such that for every non-empty family S c R we have Y ERE, resp. Y ERA. Prove that RZA

u

n

YES

YES

= RAE.

Uint. Use Theorem 4, p. 11 5.

0 4. o-additive and a-multiplicative families of sets A family R of sets is said to be u-additive (resp. a-multiplicative) if for every sequence HER* the formula H , E R (resp. H,,eR) n n holds.

u

n

4. 0'-ADDITIVE AND f3-MULTIPLICATIVE

129

FAMILIES

The next theorems follow from Theorems 1 and 2, $3, p. 126 and 127 generalized for the case of functions whose domains are sequences of sets. THEOREM 1: For every family R there exists a unique family B(R) which is a-additive, a-multiplicative, contains R and is the least family with these properties. THEOREM 2: For every family R the following formulas hold: (1) R c B(R);

Performing the operation

u resp. n

r\

on sequences whose terms

n

belong to B(R), we obtain sets belonging to B(R). This remark allows us to obtain a classification of sets belonging to B(R). Namely, for H,, where any family R let Ra denote the family of sets of the form

u

H ERN and Rd the family of sets of the form

n H,,

n

where H e R N .

n

It is clear that R c R' + (R,, c Rb) A (Rd c Ri). We define a a-additive family as a family R such that R = Ru and a a-multiplicative family as a family R such that R = Rd. Since B(R) is both a-additive and a-multiplicative, we obtain (B(R)), = B(R) = (B(R))d.From R c B(R) we have: THEOREM 3 : The family B(R) contains each of the following families Ra, Rad, Ruda, ... Ra, R d u , Rda6, In general, no two of these families are equal; moreover, they do not exhaust the whole family B(R). We shall describe a method which often allows us to decide whether or not an individual set defined by a propositional function belongs to B(R)l). Y

l) See K. Kuratowski and A. Tarski. Les opkrations logiques et les ensembles projectifs, Fundamenta Mathematicae 17 (1931) 240-248; K. Kuratowski, &valuation de la classe borelienne ou projective d'un ensemble de points d I'aide des symboles lo' giques. ibid., 249-272.

130

IV. OE-IZED

UNION, INTERSECTXON AND CARTESIAN P R O D U a

Let @(i,j,...,k,x) be a propositional function where the range of i,j , ...,k is limited to N. Let zi,j,...,k

= {X: @ ( i , j , ...,k,X)},

W = (x: Ln;sa;' ...Qi") @(i,j , ... ,k,x)}, where each of the symbols Q', gl', ...,Q(") is either the universal or the existential quantifier. We have THEOREM 4: If for any i , j , ... ,k E N the set Zi,,, ... ,k belongs to B(R), then W EB(R). PROOF.The proof is by induction on the number of quantifiers. If h = 0, then W = Zj,,,...,k;thus W € B ( R ) by assumption. If the theorem holds for h-1 quantifiers, then each of the sets

Wt = {x: Qy ...L!ih)@(i, j , ...,k, x)} belongs to B(R). Since W =

u Wiwhen Q' is the existential quanti-

iEN

fier, and W =

n Wt when 22' is the universal quantifier, we have in

leN

both cases W EB(R). Q.E.D. We obtain most important examples by taking as R the family F of closed sets in an arbitrary topological space X. In this case, B(R) is called the family of Borel sets of the space X 1 '. # As an example we consider a sequence f, of continuous functions and show that points x for which the sequence fn(x) converges is set. an F,,,, The Cauchy condition for the convergence of a sequence of real can be written in the following form numbers ul ,a2,... ,a,,

...

This implies that the set 2 of points at which the values of the sequence of continuous functions fr ,fi, ..,f n , .. converge is

.

(0

z = {x: Ak Vm Ai [Ifm+i(X)-fm(X)1<

.

I/~I}.

I) After the name of the French mathematician E. Borel who first investigated them.

4. U-ADDITIVE AND U-MULTIPLICATIVE FAMILIES

Letting z k , m. i

131

< 1/h},

= {x: Ifm+i(x)-fm(X)I

we infer from (i) that c

o

m

a

z = kn U flZk,m,i. =l m=II=l Since the set Z k , m , i (for fixed indices) is closed (because the functions considered are continuous), the set 2 is F,,. #. Exercises 1. Prove that (a) the intersection of two F,-sets is an F,-set, (b) the union of an infinite sequence of F,-sets is an F,-set. Prove the analogous properties for F,a-sets and for Borel sets. 2. Prove that every set open in en is an F,-set. 3. Prove that the Borel sets in &n constitute the least a-additiveand u-multiplicative family containing the family G of all open sets. 4. Prove the formulas which are obtained from (1)-(3) replacing B ( R ) by R, or by R,. 5. Give examples of (finite) families R such that R = R, = R,, and examples of families R such that R # R, # R,, = R,,, I).

5 5.

Generalized Cartesian products

As in 0 1, let F be a function whose values are subsets of the set X and whose domain is a set T # 0. DEFINITION: The Cartesian product Ft is the set of all functions f

n:

t ET

whose domain is T and which satisfy the condition f ( t ) E Ft for every t e T . That is, tET

Ft = { f~ d>:

A [ f ( t )E F J } ,

where

@ = XT.

tET

00

If T = N we write

IT Fn instead of n=O

F,. The elements of this ncN

Cartesian product are sequences 9 such that

T,,E

Fn for

n E

N.

l) Exercise 5 is connected with the problem posed by Kolmogorov, see Fundamenta Mathematicae 25 (1935) 578. The following special case of this problem is still open: does there exist a family R such that Ruau # Ra8,d = Roaoi,? See also W . Sierpitiski, Alg2bre de Ensembles, Monografie Matematycme (Warszawa 1951). p. 171.

132

IV. GENERALlZED UNION, INTERSECTION AND CARTESIAN PRODUCT

If all the sets Ft are identical, Ft

=

Y, then we have r[ Ft

=

YT.

t ET

In this case the symbol

n Fr denotes the set of functions with domain

t ET

T and range Y . The set Y T is called a Cartesian power of the set Y . For Y c Ft let Y t denote the projection of Y on Ft. Thus Y' is

n

t ET

the set { f ( t ) : f e y } . Clearly, Y , c Y2 -,Y : c Y : for every t E T . REMARK:Let T = { 1,2}. The Cartesian product

n

Ft and the prod-

t ET

uct Fl x F2 are not identical. In fact, the first product has as elements two-term sequences, the second product, ordered pairs. These two notions are distinct. In practice, however, the distinction between these two kinds of products is inessential, for we can always associate with every pair (x, y ) of Fl x F2 the sequence { (1, x ) , ( 2 , y ) } belonging to F, in a one-to-one manner.

n

tET

If F,,

= 0 for

In fact, i f f E

some t o , then

n F,, then

n Fr

=

0.

tET

f ( t o ) E F t , ; thus F,, # 0.

t6T

THEOREM 1. a set T has aJinite number of elements and Ft # 0 for Ft # 0. every tET, then

n

tsT

The proof is by induction on the number of elements of T . If T is composed of one element, then the theorem clearly holds. Suppose that it holds for the case where T is composed of n elements. Let Tl = T u {u} where a $ T . Suppose further that F, # 0 for t E T I . We Fr # 0. In fact, the induction hypothesis gives shall show that

n n F t # 0; thus let fen F t . Let t O € F a .The set fi = f u { ( a , t , ) } taTi

t ET

t ET

a function belonging to

is

n F t . Consequently, the set fl Ft is non-

teT,

ttT,

empty. Theorem 1 also holds for arbitrary T if all factors Fr are equal. THEOREM 2: Zf Y # 0, then Y T # 0.

133

5 . GENERALIZED CARTESIAN PRODUCTS

In the general case the proof that the Cartesian product is nonempty requires the axiom of choice. 'THEOREM3: If F, # 0 for t E T , then

n

F, # 0 I).

t d

PROOF.A choice function for the family {F,: t E T } is an element of n F , . ttT

In applications of Cartesian products (e.g. in algebra, in topology) we deal mostly with cases where certain operations are defined on the sets F, or where the sets Ft are topological spaces. We discuss first the case where only one operation is defined on each set F,. For convenience, we assume that this operation is binary. In other words, we supFose that besides the function F E (2X)Twe are given a function G such that G, E ( F t f t x F ,for every t E T. The function G induces a binary operation rp on the elements of F,. Namely, we let for f,g E n F t the Cartesian product

n

ttT

tsT

d f ,g) = t I; [Gt (f@I,g (t))] tT

*

Thus q ~ ( f , g is ) that element h of the Cartesian product for which h(t) = G , ( f ( t ) , g ( t ) ) for every t. The operation rp is called the cartesian product of the operations G,. In a similar way we define the Cartesian product of relations. Let R be a function such that (for every t E T ) R, is a relation with field included in F,. The Cartesian product of these relutions is the relation p, whose field is included in n F , , s u c h that ttT

It should be pointed out that e is not the Cartesian product

n R,, because

tsT

e is a binary relation; that is, e is a set of ordered pairs, whereas

R, is a set of functions. However, we can associate in a natural way the t&"

') B. Russell took Theorem 3 as an axiom instead of the axiom of choice. He called this axiom the multiplicati;1e axiom. See A. N. Whitehead and B. Russell, Principia Mathematicu, vol. I, 2nd ed. (Cambridge 1925), p. 536.

134

IV. GENERALIZBD UNION,N-

AND CARTESIAN PRODUCT

Cartesian product n R t with the relation which holds between functions td'

f and g if and only if the function h defined by h(t) = ( f ( t ) , g(t)) (that is, the function F [
rcT

coincides with the relation e defined above. Clearly, the definitions can be applied without modifications to the case where not just one but several operations (or relations) are defined on each F i . Example. Suppose that Ft is a Boolean algebra with respect to the operations v 1 , A i, and the elements 0,, 1,. Let v , A , - denote the Cartesian products of the operations v A ,- , respectively, and let 5, I denote the functions such that E(t) = 0, and ~ ( t=) 1, for all tET.

-,

,

THEOREM 4: The Cartesian product

n F, is a Boolean algebra under

laT

the operations v ,A ,-, 5, I.

PROOF.To prove the theorem it suffices to check that axioms given on p. 37 hold. For instance, we check axiom (iv). Let f E n F t ; it rsT

foIIows from the definition that f v -f is the-function g such that g ( t ) = f ( t ) v-, f ( t ) for every t. Since axiom (iv) holds in F,,we have g(t) = l,, thus g = C . The Boolean algebra Fr is called the direct pro&ct of the Boolean

n

tcT

algebras F1. In a similar way we can define the direct product of groups, rings, and other algebraic systems. &ercim

1. Prove that a Boolean algebra of 2" elements is the direct product of n Boolean algebras of two elements. Hint: Use induction on n. 2. Prove that if the relations R, are (a) refiexive, (b) symmetric, (c) transitive, then their Cartesian product has the same properties. Give an example of a property which does not hold for cartesian products although it holds for the individual factors. In other words, find a function R and a property which holds for all relations Rt but does not hold for their Cartesian product. In the following two exercises we assume that H is a function such that

5. GENERALIZED CARI"

PRODUCTS

135

Ht E FrtXF t , Rt is an equivalence relation in Ft, Q is the cartesian product of the operations Ht and Q is the Cartesian product of the relations R t . 3. Show that if each function Ht is consistent with R t , then p is consistent with Q. 4. Let X be the function such that, for every t, Xt is the quotient function be the quotient function induced induced from Ht by means of Rt and let from Q by e. Show that the following diagram is commutative:

x

0 6. Cartesian products of topological spaces For every t E T , let Ft be a topological space and let CJ denote the closure of the set X c Ft in the space Ft. Hence C is a function such that C , E ( ~ F ~ ) ?for ~ ~every

t.

Clearly, there are many different ways to define the closure operation on the space Ft. In fact, an arbitrary set can be made into &T

a topological space in many different ways. Here we shall discuss one of the special topologies on the space Ft, introduced by Tychonoff I).

n

t ET

Let S be a finite subset of T and let G, be an open set in the space F, for every s E S. We define the neighborhood determined by S and by the sets G, to be the following subset of the Cartesian product n = F,:

n

f ET

=

(f~17:A f ( s ) E G,}. SES

We shall prove that the intersection of two neighborhoods is a neighborhood. In fact, if F is the neighborhood determined by a finite set S and by open sets G, (s E S) and r' is the neighborhood determined by a finite set S' and by open sets Gi (s E S'), then ') A. Tychonoff, ober topologische Erweiterung von Ramen. Mathematische Annalen 102 (1930) 544-561.

136

N. OENERALIZBD UNION, INTERSECTION AND CARTESIAN PRODUCT

Thus r n r' is the neighborhood determined by the finite set S u S' and by the open sets GI', where the sets G:' are defined as follows: G:' = Gt for t ES-S', Gi' = G: for t eS'-S, and G:' = Gt n G: for t e S n S'. We define the closure of a set X c 17 to be the set CX of all f €17 such that each neighborhood T containing f also contains at least one element of X: (*)

Ar [(F is a neighborhood)

A

THEOREM I : The Cartesian producf

(f~r> +

(rnX # O)].

n

Ft is a topological space with

tcT

respect to'the closure defined by

= CX f o r X c

nFt.

tET

PROOF.It is necessary to check that axioms (1) - (4), p. 26 hold. Axioms (3) and (4) clearly hold. Axiom (1). Let f B 2;thus every neighborhood T containing f also contains at least one element of A. Hence r n (A u B) # 0, which implies fE A v B, and thus c A u B. Similarly, 3 c A u B. Hence xuBcA 3 . Now suppose that fe A u B andfg

A. Thus for every neighborhood

r containing f we have r n (A u B) # 0 and for some To containing f we have Ton A = 0. If T is an arbitrary neighborhood containing f,then F n Tois also a neighborhood containing J Therefore r n To n ( A u B) # 0, whence r n Ton B # 0 and hence r n B # 0. This shows that f E 3. Axiom (2). It suffices to show that x'c 2.Let f ~ x and " let 'I be any neighborhood containing f. Thus r n # 0; let g E r nX. Hence r is a neighborhood containing g. Since g EX, we have T nX # 0. This shows that the condition (*) holds; hence

x

fez.

6. CARTESIAN PRODUCTS OF TOPOLOGICAL SPACES

137

Examples of Cartesian products of topological spaces 1. The Cantor set. This is the set C = (0, l}" or, in other words, the Cartesian power of a two-element set. If we define a topology on the set (0,l} by letting 2 = X (the discrete topology), then C becomes a topological space with the Tychonoff topology. co

2f(n)/3"+' to the element f E C

# By assigning the real number n=o

we obtain a one-to-one mapping cp of the set C onto the set of those real numbers of the closed interval [0, 11 whose triadic expansion contains only the digits 0 and 2. # 2. The generalized Cantor set CT is the Cartesian power (0,I}'. The Tychonoff topology may be defined on this set similarly as on the set C. The generalized Cantor set may also be defined as the set of all characteristic functions of subsets of T . In practice, we may identify the set CT with 2.' Therefore at times we shall treat the family 2T as a topological space. Similarly, the elements of the set CTxT = (0, can be identified with the set of all relations on T . In fact, each element of the set CTxT is a characteristic function of a set of ordered pairs of elements of T . THEOREM 2 : The family Kt = { X c T : t E X } is both open and closed in CT;the family ( R c T x T : t R s } is both closed and open in C T ~ T . PROOF.Let I' denote the neighborhood in CT, determined by the set S = { t } and by the open set G, = (1). Then f EI'= f ( t ) = 1; thus'Z consists of the characteristic functions of the sets belonging to the family Kt. Hence this family is an open set. Likewise, the neighborhood determined by the set S and the open set G: = (0) consists of the characteristic functions of the sets belonging to the family 2T- Kt. This shows that the family Kt is closed in CT. The second part of the theorem follows from the first. 3. The Baire space. This is the Cartesian Fower N N or, in other words, the set of infinite sequences of natural numbers. The topology in N N is defined as the Tychonoff topology, where we define the closure operation in N to be = X. If a = (ao,a,, ..., an-l) is a sequence of n terms ( U E N"), then the

x

138

IV. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

set Nu = Nuo,...,an-l = {e: eln = a } is both open and closed in NN. To see this, we notice that Nu is the set of sequences satisfying the conditions y j = a j wherej < n ; it thus coincides with the neighborhood I' in N determined by the set S = (0, 1, ..., n - 1 ) and the open sets Gi = ( a j } for j < n. The complement of I'is open in NN, because it coincides with the union of neighborhoods determined by the sets ( j } and by the open sets G; = N - ( a j } ,j < n. # Assigning the number

to the element y E NN,we obtain a one-to-one mapping of the space N N onto the set of irrational numbers in the open interval (0, 1). In practice, we may identify the Babe space with the set of irrational numbers in the open interval (0,l). # 4. The Cartesian product of a finite number of spaces. The construction described in this section is used both in case where the set T in the formula X = F, is finite and also in the case where T is infinite.

n If T is a finite set, for instance T (0,1, ...,n- l}, then the cartesian products n V t , where for every t the set is open in Ft, form an open base of n Ft. tcT

=

V,

t
t
#In particular, we may take Ft = € where t is the set of real numbers. In this case the space X is the n-dimensional Euclidean space. In the future, we shall make little use of this space because it has not been defined exclusively by means of set theoretical notions (in contrast to the other spaces discussed above). # In later chapters we shall make use of the following theorem. 3: If X = THEOREM Ft is a Cartesian product of topological spaces t

ET

(with Tychonofl topology), 2, c Ft and 2, is a closed set in Ft for every t E T , then the set Z , is also closed. t ET

PROOF.Let P =

n

Z t and let f I# P. For some s E T we thus have

teT

f ( s ) $2,. The element of the subbase of X determined by the one-element set {s} and by the open set Fs-Zs contains f and is disjoint from P.

6. C&l'@SlAN PRODUCTS OF TOPOLOQICAL SPACES

139

Exercises 1. The set of reflexive relations whose fields are included in T is closed in C T ~ T . Prove that the same holds for the sets of symmetric relations, the set of transitive relations and the set of equivalence relations. 2. Prove that each neighborhood in the Baire space contains some neighborhood of the form Na,where a is a finite sequence. 3. Show that every set which is both open and closed in the mire space is the union of a finite number of sets Na. 4. Show that the set (Xc N: X is a finite set) is a Bore1 set in the space CN. 5. Let I be any ideal in 2T. Show that taking as neighborhoods the sets of the form (g: A (g(s) E Gs)]. where S EZ and Gsare open sets in F, for s E S and deseS

fining the closure operation by (1). we obtain the function satisfying the axioms of topology (1)-(4) given on p. 26.

0 7. The Tychonoff theorem A family R of subsets of the set X is said to have the jinite intersection property if every finite subfamily of R has a non-empty intersection. A topological space X (that is, a set with an operation defined on subsets ofXand satisfying axioms (1)-(4), p. 26) is said to be compact if every family of closed subsets of X which has the finite intersection property has a non-empty intersection. This definition implies that a topological space is compact if and only if every family of open sets, whose union is X,contains a finite subfamily whose union is also X. Although the following theorem belongs to topology rather than to general set theory, we give it here because it has numerous applications in mathematical theories in general and in set theory in particular. Moreover, the means used in proving this theorem involve little more than set-theoretical techniques. O

THEOREM1: (Tychonoff.) If for every t E T the space F, is cohzpact,

then the space P =

n Ft is also

compact (relative to the Tychonofl

ffT

topology). In the proof of this theorem we make use of a lemma to be proved in Chapter VII, p. 267.

'

'LEMMA:If Ro is a family of subsets of X with the finite intersection property, then there exists the maximal family R c 2x containing &

140

N. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

which also has the finite intersection property. That is to say, every family of subsets of X containing R and diferent from R contains a finite subfamily with empty intersection. We shall make use of the following two properties of maximal families R with the finite intersection property. (i) If A E R and B E R , then A n B E R . Suppose the contrary. Then the family arising from R by adding to it A n B does not have the finite intersection property. Thus it contains a finite subfamily with empty intersection. Clearly the set A n B belongs to this subfamily. Hence we conclude that there exists a finite Y = 0, which contradicts subfamily R ' c R such that A n B n

n

YER

the assumption that the family R has the finite intersection property. ( i i J I f A c X a n d A n Y f 0foreveryYER,then A E R . In fact, if A c# R, then the family R u { A ) does not have the finite intersection property. Thus there exists a finite subfamily R' c R such Y belongs to R by (i). This that A n 0 Y = 0. The intersection

n

YER

YER

contradicts the assumption that A n Y # 0 for every Y ER. Now to prove the Tychonoff theorem, let Ro be a family of closed subsets of P with the finite intersection property. Let R denote any maximal family of subsets of P with the finite intersection property, containing Ro. For the proof of the theorem it now suffices to show that F 0.

n +

YER

For arbitrary Z c P let 2' denote the projection of Z into Ft and let R' = {y': Y ER}. The family R' consists of closed subsets of 6 . If the j < n , belong to R' and Y,E R, then Y , # 0 because the sets

z,

n

j
family R has the finite intersection property. This implies that

nY;

j
# 0 (see p. 132), whence

nq#0. Thus the family Rt has the finite

jcn

intersection property. From the assumption that every 4 is compact we infer that f-I F#0. It now follows that there is an f E P such that Yc R

for every t E F , f ( t ) E

n y'. We shall prove that f E YnE Rr.

YER

141

7. THE TYCHONOFF THEOREM

r

For this purpose suppose that Y ER and that is a neighborhood containing f. We have to show that T n Y # 0. Let r be the neighborhood determined by a finite set S c T and open sets G, c F, (sES),where clearly f ( s )EG, for SES. Letting I', = { g : g(s)EG,} we have =

r n r,. SES

If Z is an arbitrary set belonging to R,then f ( s ) EF; thus G, n Zs # 0. This means that there exists z,EG, such that for some function g E Z we have g(s) = z, ; hence g E r,. Thus for any Z E R we have r, n z # 0. By (ii) it follows that E R and by (i) E R; that is, r g R,

r,

nr,

seS

which implies n Y # 0. Thus every neighborhood containing f has elements in common with Y , consequently f E Examples 1. The sets C, CT, C T x T are compact. 2. Let @ ( R , x , , ... ,x,) be a propositional function constructed from the propositional functions

(*I

xi

=XI,

(xi,xj)~R

by applying only the operations of the propositional calculus. Such propositional functions are called open. For a,, ,.. ,a,,E T and for an arbitrary open propositional function @ let

Zg= Zg,(al,... ,a,) = { R E T x T: @ ( R ,a , , ... ,a,,)}. The set Zg is both open and closed in C T x T . This fact follows from Theorem 2, $ 6 , when @ is one of the propositional functions (*). For other open propositional functions, this property follows from the relationships between logical and set-theoretical operations as well as from the remark that the finite union, intersection, and complement of sets which are both open and closed are again both open and closed. Now, let Q j be an open propositional function of the variables XI,, xiz, ...,xj,, and let a l l , a j z , ... ,aln, be elements of T EN). Since the set CTxris compact, we have THEOREM 2.

If for every k E N the intersection nZ g j ( a j l ,...,aj,,) i
142

W. GENERALIZED UNION, INTERSECITON AND CARTESIAN PRODUm

is non-empty, then the intersection

n Z o j ( a j l ,... ,a j n j ) is also non-

1sN

empty.

This theorem asserts that if there exist relations Rkwhich satisfy the conditions CDj for j < k (k = 1,2, ...), then there also exists a “universal” relation R which satisfies all of these conditions. Exercises 1. Derive Kiinig’s lemma (p. 104) from the Tychonoff theorem. 2. Show that the Bake space. is not compact.

5 8. Reduced direct products By combining the operation of Cartesian product with forming of equivalence classes we obtain new operations which have found interesting applications in mathematical logic l ) . Let T be any set and F a function defined on T whose values are non-empty sets. Letf, be a function whose domain is Ft X F, and whose range is included in Ft. Finally, let R, be a binary relation with field included in Fr. All arguments below can be generalized for the case where the number of functions or relations is greater than 1. Let I be an ideal in 2=. Define the relation

-I

in P =

n Ft by the

tcT

formula

f

-Ig

= { t : f ( t ) # g ( t ) } €1.

THEOREM 1: The relation wI is an equivalence relation in P. The reflexivity of follows from 0 ~ 1 symmetry , is obvious, and transitivity follows from the remark that for any f,g , h E P, {t:f ( t > # g ( t ) ) = { t :f ( 0 # h ( t ) ) u { t : h ( t ) # g ( t ) > . THEOREM 2: The carte,!ian product rp of the functionsf, is consistent with l) These operations were first defined by J. Lo4 in his paper Quelques remarques, thiorhes et probl2mes sur les classes &finissables d’algdbres. published in the book Mathematical interpretations of formal systems (Amsterdam 1955). See also T. Frayne, A. C. Morel and D. Scott, Reduced direct products, Fundamenta Mathematicae 51 (1962) 195-228.

8. REDUCED DIRECT PRODUCTS

143

PROOF.It is necessary to show that if e', e", d', d" E P, then (e' N l e " ) A (d' ~ [ d "+) [y(e',d') N I y ( e " , d")]. Let h' = y ( e f , d ' ) , h" = y(e",d") and A = { t : h'(t) # h"(t)}. It follows from the definition of pl that h'(t) =ft(e'(t), d'(t)) and similarly h"(t) =fr(e"(t), d"(t)). Hence t E A + [e'(t) # e"(t)] v [d'(t) # d"(t)], whence it follows that

A c { t : e'(t) # e"(t)} u { t : d'(t) # d"(t)} E Z . From Theorems 1, 2 and from 0 5, Chapter 11, it follows that the quotient class P/Z of P with respect to the relation -I and the operation y/Z, induced from y by -[ exist. We still define in P/Z a relation e/Z. It holds between the equivalence classes of two functions e, d E P if and only if { t :( e ( t ) , d ( t ) ) 4 R,} E Z. The set P/Z is called the direct product of the sets F, reduced with respect to Z (shortly: reduced mod I ) . Similarly, the function y/Z (resp. the relation ell> is called the direct product of the functions ft (resp. of the relations R,) reduced mod Z. Let @(x,y, z) be an arbitrary propositional function. The main problem of the theory of reduced products can be formulated as follows: When the sets { t : @(F,, ft,R,)}are known, under what conditions do the set P/Z, the function y/Z and the relation e / Z satisfy the propositional function @? In order to solve this problem, we consider more general propositional functions @(x, y, z, ul,..., uk) of an arbitrary number of variables. Suppose that e l , ...,ek E P and let A0 = { t : @(Ft7 ft,Ri, el(t)7

*'.,

ek(t))}

(clearly, the set A& depends not only on @ but also on the elements e l , ...,ek; we do not write el, ...,ek in the symbol A , in order to simplify the notation). With this notation, the following theorems (i) - (iv) hold:

A a v y $ Z ~A,$ZV

(9

AygZ.

In fact, A m V y= A,uAp; thus (see formula (ll), p. 17) A ~ P ~ Y E Z A (AYE Z), and (i) follows by de Morgan's laws.

= (&€I)

14

"(ii)

IV. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

v @,

If 0 is the propositional function

then

"k

A,$Zr V[Aa$Z]. ek tP

In fact, suppose that A , = { t : @(Ft,ft,Rr,ei(t),* . - , e ~ - i ( t ) ) } # Z .

For t E A, there exists x E F, such that @(Fr,

5,Rt, el(t), - - . , e k - l ( t > 7 x ) .

Let X , denote the set of all these elements x , and let X , = Ft for t # A,. Let ek be a choice function for the family consisting of all the setsX,. We have ekE P and

@(&, 57 R ~el(t)$ ?

(1)

ek(t))

for all t E A,. This implies A , c A , and thus A , #Z. Conversely, if A , these t

# Z then formula (1) holds for all t E A,. Thus for @ ( F t , A, R,, el ( t ) ,

.

*

Y

ek-l(t>).

This implies that t E A@. Hence A , c A , and A , $Z. An ideal Z is said to be prime if for any X c T exactly one of the conditions A E Z, T - X E Z holds. The set {X c T: x $X}is an example of a prime ideal. In Chapter VII we shall prove that every ideal can be extended to a prime ideal. (iii) If 1 is a prime ideal and !P is the formula 7 @, then A&Z= In fact & = T - A , . "(iv) If Z is a prime ideal, then

1(A,$Z).

A,,\~$Z~(A,$Z)A(A~#Z); moreover, if.." is the formula

A uk

Theorem (iv) follows fro= (i)-(iii).

@, then

145

8. REDUCED DIRECT PRODUCTS

A propositional function @(x, y,z, ul, ...,uk) is said to be elementary if it can be constructed from the propositional functions (a) ui = uj, (b) Y(ui,u j ) = uh, (c) € 2 by applying the operations of the propositional calculus and by applying the quantifiers

v, A.

UEX

UEX

The following theorem solves the problem stated above. O THEOREM 3 : If 1 is a prime ideal, @ ( x , y ,z, u l , ..., uk) is an elementary propositional function and el, ...,ek are arbitrary elements of P , then I):

(2)

v/z,@/I,el/z?

@ (p/z,

*

a

Y

ek/z)

PROOF.If @ is one of the propositional functions (a), (b), (c), then (2) holds. In fact, the left-hand side of (2) is equivalent to ej/Z= ej/Z in case (a), to v/Z(ei/Z,eJZ) = eh/Z in case (b), and to (ei/Z, ej/Z)E @ / Z in case (c). The right-hand side of (2) is then equivalent: in case (a) to { t : e j ( t )= e j ( t ) } $ Z , in case (b) to ( t : f t ( e i ( t ) , e j ( t )= ) eh(t)}#Z, in case (c) to { t : (ei ( I ) , e j ( t ) )E R,)$ 1. From the definitions of the set PIZ, the function q / Z , and the relation ,o/Z as well as from the definition of prime idealn it follows that the left-hand and right-hand sides of (2) are equivalent. In turn, it follows from theorems (i)-(iv) that if (2) holds for the propositional functions @ and Y,' then it also holds for the proposi! by applying the operations of tional functions arising from @ and P propositional calculus and by applying the quantifiers and uix

A. UEX

In this way Theorem 3 is proved. 1) Theorem 3 is a scheme, for each propositional function we obtain a separate theorem.

146

IV. QENERALIZED UNJON, INTERSECTION AND CARTESIAN PRODUCT

o

C4: If@D(x, y~, z ) is an elementarypropositional ~ ~ function, ~ then ~

@ ( W ,911, elr) = It : @E, ft ,R,)) 4 1. This corollary follows from Theorem 3 if we assume that the propositional function does not contain the variables 2.4,...,uk. ‘COROLLARY 5: If @(x,y, z) is an elementary propositional function and f o r every t the formula @ ( F c y f t , R tholds, ) then dj(P/I,pl/Z,e/I).

This corollary follows directly from the previous corollary and from the remark that if I is a prime idea1 then T 4 Z. Examples

In the following we suppose that Z is a prime ideal in 2=. 1. If the relations Rt are reflexive, transitive and satisfy the conditions: A (KX, Y> ERtI A Kv,x> ERtI --tx = Y } , x.yaF,

XI

A K<X, Y> ER;) ” ( +,I, YEFt

then the reIation e/I satisfies the same conditions. # 2. If Ft is a field with respect to the operations of addition f, and multiplication g,, then the set PII is a field with respect to the operations q11I and y/Z, where pl and y are the Cartesian products of the operations ft and g, respectively. Similarly, if each of the sets F, is an ordered field with respect to the operations ft and g, and with respect to the order relation R,, then PI1 is an ordered field with respect to the operations y/Zand y / I and the order relation e/Z. These properties follow by Corollary 5 from the remark that propositional functions “ X is a field under the operations D and M” and “X is a field with respect to these operations ordered by a relation R” are equivalent to elementary propositional functions @(x, D, rmp. (X,0,M,R). # These examples show that forming the reduced direct product we can construct from a given family of models of a given system of axioms new models of the same system of axioms. Other applications of reduced direct product will be given in Chapter IX.

~

8. REDUCED DIRECT PRODUCTS

147

Exercises 1. By applying the lemma on p. 139, prove the existence of a prime ideal which contains a given ideal # 2T. 2. Prove that if each of the sets Ft is a Boolean algebra with respect to the operations v t , A t , -* and the elements O t , lf and if for every t the condition

holds, then the set P / I is a Boolean algebra with respect to the operations v / I , AII, --/I and the elements 511, ill, where v, A, - are the Cartesian products of the operations v t , A t and - t , respectively, and l , i are functions such that E(t) = Ot and i(t) = I f for every t. Moreover, P/I satisfies condition (*). Show that the ordinary Cartesian product of Boolean algebras satisfying (*) may fail to satisfy this condition.

$9. Inverse systems and their limits Suppose we are given: Ci) an arbitrary set X, (ii) a set T ordered (or more general, quasi-ordered) by the relation ) ~is , (iii) a function F E ( ~ ~that Ft c X for every t E T;

<;

n

(iv) a function f E

P21, which can be written as

to<*,

F,, + F,,, where to< t l . We assume that the function f satisfies the conditions:

(1)

(2)

f t o t i ~ F z l or

if

to

ftotl:

< t l < t 2 , then -Aotl oftlt2 =Aot2,

(3) At is the identity mapping for every t E T . Then the system U = ( X , T, F,f) is called an inverse system. The Ft consisting of those elements 9 subset of the Cartesian product

n

I ET

that ftot,[9,(t1~1= TOO)

(4)

is called the inverse limit of the system U and is denoted by F,

or

Lim U -

or

Lim (Ffyftotl). tO
148

IV. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

Let f, denote the t-th coordinate of the function F,, defined by function fr :F,

v,

that is, the

fr(v) = pl(Q

(5)

Thus we have (6) Lor,oft, =Lo. Let ( X , T , F,f)and (Y, T , G , g ) be two inverse systems and let h be a function h s nGp, that is, t z7

h , : F, 4 G t . Suppose, moreover, that the following diagram is commutative:

that is, h,,,o Lot,= gtOfl ohtl if Then we can define a mapping h,: F, -+ G, in such a way that the diagram

F,+ hi

I

4

ft

to


F, h*

Gt+G, gt commutes for every t E T . Namely, let y = hm(pl) be defined for every 91 E F, by the condition y 4 ) = ht(v(t)). It is easy to prove that if, for every t , ht is a one-to-one mapping of Ft onto G , , then h, is a one-to-one mapping of F, onto G,. Examples 1. Let the ordering relation on the set T be the identity relation. Then F, = F,. tsT

2. Let the set T be directed, let Ft be the constant function: Ft = X, and let the mapping J o t , : X + X be the identity mapping. Then Fm is the set of all constant functions pl: T --f X .

149

9. INVERSE SYSTEMS AND THEIR LIMITS

To prove this statement, we observe that if to# tl then there exists t2 such that to t2 and tl tz; thus f t O t , [ p l ( t 2 ) ] = v(tz).Hence by (4) y(to) = v(t2),and similarly p(tl) = v(t2).Thus v(to)= q4t1). 3. The set Y x of all mappings a: X + Y can be represented by means of the operation of restriction alA as the inverse limit of the sets Y A where A c X . We take the set 2x ordered by the inclusion relation as the directed set T. We define the function F by the formula F A = Y A for A c X . Finally, we associate with each pair A. c A , the function fAoAl: YA1+ YAodefined by

<

<

fAoAl(a) = @[Ao where

a E YA1.

The role of the set X which occurs in the definition of inverse system YA. is played here by the set S of all partial functions, that is by

u

Let us assign the element @(a)

~nY A ,defined by

ACX

A

to every element cx E Yx. We easily verify that @(a) E Lim(S, 2x, F,f); that is, @: Y x + Lim(S, 2x, F, f). c _

Moreover, @ is a mapping of Y x onto the set Lirn(S, 2,' F, f). For let v ~ L i r n ( S2,' F, f).Then we have vA€Y Af o r x e r y A c X . Define a E Y x b y t h e condition a ( x ) = vlxl( x ) . It is easy to see that @(a) = y ; that is, plA(x) = vlX)(x)for all x E A . Finally, the mapping @ is one-to-one. In fact, if a, # a2, then there exists xo such that al(xo) # a2(xo); that is, a1I{xo}# a z ( { x o ) .Thus ~ @ ~ ~ ~ ~# I l[@@2)]hl X d and @(a # @(a2). It is worth noticing that the argument above also applies when X and Y are metric spaces. In this case 2x denotes the family of compact sets in X and Y Adenotes the family of continuous mappings a: A + Y .

8 10. Infinite operations in lattices

and in Boolean algebras

The theorems in the previous sections of this chapter can be considered as theorems about the lattice 2x (see pp. 41-44). As we know, this lattice is a Boolean algebra and a complete lattice. In a natural

150

W. ff.BNERALIzED UNION, O N

AND CARTESIAN PRODUCT

way the question arises as to whether the theorems in Q 1 can be generalized for the case of arbitrary lattices, or complete lattices, or Boolean algebras. Suppose first that K is any ordered set and f c K T . The following theorems, analogous to Theorems 1-3, p. 113-115, hold: THEOREM 1: The least upper bound g =

bound d =

Aft),if

v ft (resp. the greatest lower

tET

it exists, is the unique element of K satisjying

~ E T

the conditions

A

and

AY;
t ET

t ET

cft

< a)

-P

(8

< a),

(resp. and

If the least upper bound

vfrexists, it is called the supremum of

1 ET

the elements fi, teT. If the greatest lower bound

Aft exists, it is

tsT

called the injimum of the elements fi, t E T.

u H, and for every u U the suprema v A = the supremum v ft = g also exabts, then the supremum

THEOREM 2: If T =

E

t eHu

UEU

g, exist and if

IET

v g u exists

and is equal to g (analogously for infimum>.

UEU

THEOREM 3: if q is a permutation of the set T and the supremum ft = g exists, then the supremum f+,(t)also exists and is equal

v

v

t ET

tET

to g (analogouslyfor injimum). Theorem 1 is a restatement of the definition of the least upper bound. Theorems 2 and 3 can be proved similarly as Theorems 2 and 3, pp. 114-115. The following theorem holds for all ordered sets and is analogous to (3), p. 112. 4: THEOREM

If the

suprema

v ft teT

exist, then for all t E T we have d
= g and the infima

< g.

AET ft = d f

10. WFINITE OPERATIONS

151

IN LATIlCEs

On the other hand, formulas (4)-(11) do not have counterparts for arbitrary ordered sets. THEOREM 5 : I f the set K is a lattice, p , q E K T , and i f the suprema V p t = g , and qt = g2 exist, then the supremum (pt v qJ exists

v

v

t ET

tiT

LET

and is equal to gl v g 2 (similarly for infimum). PROOF.g , v g , 2 pt v qt for all t E T . If

A

( x 3 pt v qt), then also teT

( x 3 p t ), whence x 2 g, ; and similarly x 2 g , ; hence x 2 gl v g,.

tET

The assumption that K is a lattice has been taken in this theorem in order to ensure the existence of pt v qt and g, vg,. 6 : If K is a lattice and the suprema THEOREM

v fr and v ( a~ f t ) t ET

tET

exist, then

(similarly for infimum).

<

PROOF.For every t E T we have a ~ f t a and a AJ

<

thus a ~ f t a A


v ft, whence the required formula follows.

tET

t ET

The inequality sign in Theorem 6 cannot in general be replaced by the equality sign even in the case of complete lattices. However, the following theorem holds. THEOREM 7 : If K is a Boolean algebra and the supremum

then for any a E K the supremum

v (a ~ f t exists ) and

v ft

exists,

t ET

is identical to

teT

V ft (similarly f o r infimum>. PROOF.Since a ~ f < r a A v j t for every t E T , it suffices to show that if A (a A ft < x ) , then a A v ft < x . From the assumptions it follows that -a v (a ft) < -a v x ; hence ft < -a v x for arbitrary

aA

t ET

tET

tET

tET

A

tET. We obtain

\ / A < - a v x , thus a h t ET

vft
152

IV. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

For Boolean algebras the following theorem holds (de Morgan's law). THEOREM8. If the supremum ft = g exists, then so does the

v

infimum

A (-A) t€T

ttT

and is equal to -g

(similarly with supremum and

infimum interchanged). PROOF.Since fr < g, we have - g < -fr for every t E T . If x -fr for every t E T , then ft -x; hence g - x and x - g , whence (-ft) = -g. we obtain

A ttT

<

<

<

<

As the theorems above show, all basic theorems in 0 1 can be generalized for the case of complete Boolean algebras. For non-complete Boolean algebras the theorems hold if we make the additional assumption that all the necessary suprema and infima exist. It is interesting to notice that the distributive law stated in Theorem 7 does not involve the complement sign, yet it can be proved only for Boolean algebras. The general distributive law given in Theorem 4, p. 115, is even more peculiar. We shall prove that Boolean algebras of the form 2x are, in fact, the only Boolean algebras for which this theorem holds. First let us assume two definitions. DEFINITION 1 : A Boolean algebra K is said to be distributive if it is complete and, moreover, if for every set M and for every function f: M --f K and for every partition of M into non-empty sets M = T,,the following identity holds:

u

UEU

where (2)

K = ( Y E ~ A~( Y : nT,,#O)}. u,u

DEFINITION 2: An element a is said to be an atom of a Boolean algebra K provided a E K , a # o and x < a x = 0. A Boolean algebra K is said to be atomic if for every element x # o there is at least one atom a such that a < x. --f

THEOREM 9: Every complete and atomic Boolean algebra K is isomorphic to the field of subsets '=2 where A is the set of atoms of K. Namely,

153

10. INFINITE OPERATIONS IN LATTICES

there exists a one-to-one mapping @ of K onto 2A such that (3)

@itbs T(V.A) = u mf,,, ti-T

(4)

~

f

tsT

t = )n 1 ET

for any set T and for any fimrction f

EK T

PROOF. Let for X E K

@(x)

= {a E A:

a

< x}.

This formula defines a function whose domain is K and whose values are subsets of A. Clearly x < y + @ ( x )c @(y). The function @ is one-to-one. For suppose that x and y are elements of the algebra K such that x n y # 0. We can assume that x-y # 0. B y the assumption that K is atomic, there exists an atom a such that a < x-y. It follows from the formulas a A X < a and a h y a that

<

UAX=O

or

a~x=a,

ar\y=o

or

a~y=a.

and The formulas a A x = o and a A y = o imply

a

= U A ( X - - ~ ) = ( u A x ) - ( u A ~ )= 0 - 0

= 0.

Hence a = 0,which contradicts the fact that a is an atom (Definition 2). Similarly, the formulas a A X = a and a A y = a imply

a

= U A ( X - ~ ) = (UAX)-(UAY>

= a-a

= o,

which again contradicts Definition 2. Thus, either a A x = o and U A Y = U , or U A X = U and U A Y = O . In the former case we have anon < x and a < y , in the latter a < x and a n o n y. Thus in both cases @ ( x ) # @(y). Let f €KT.If a E @(A),then there exists t E T such that a E @(A).

<

u

tET

This implies a d ft


aE

@(v f,). t ET

I) See A. Tarski, Zur Grundlegung der Boole’schen Algebra I , Fundarnenta Mathematicae 24 (1935) 197.

154

W. OBNERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

Hence we have proved that (5)

Now suppose that aE@(Vf,); that is, a < teT

vft.

If a ~ f = t o for

teT

every t, then we have a = a - ( a ~ f t )= a-ft; ,thus

A @-A> = a- teT V ft = .-(a V jJ= a--a = 0, teT

a=

tET

A

because a A

v

ft

= a. But this conclusion contradicts the fact that

tcT

Hence there exists t E T such that o # a A f t < a. This means a ~ f=t a ; thus a and finally a E @(A)c @(A).We have thus

U E A.

u

teT

proved the following inclusion

.

This by (5) implies (3). Formula (4) can be proved even more simply. We have

It remains to be shown that every set X c A can be represented as

@(x) for some X E K .For this purpose, let

T=X, f t = t ,

x=Vfn U;X

(this union exists by the assumption that K is a complete Boolean algebra). According to (3) we have

because a is the unique atom included in a and hence @(a) = {a}.

10.

INFINITE OPERATIONS IN LATIlcEs

155

In this way Theorem 9 is completely proved. THEOREM10: Every complete and atomic Boolean algebra K is distributive. PROOF.According to Theorem 9 there exists a function @ which establishes the isomorphism between K and the field of subsets of some set A . By Theorem 4, 5 1, formula (2) implies the formula

which by (3) and (4) implies

Since the function @ is one-to-one, the formula above implies formula (1). THEOREM 1 1 : If the Boolean algebra K is complete and distributive, then it is atomic '). PROOF.Suppose that K is complete and distributive but not atomic. Let a, be an element # o which does not contain any atom. Let T,, = {u, -u} for ueK and f t = t A a, for t E K . Since K = T,,, we

u

UEK

have equation (l), where M = K and where X is defined by (2) for M = K. It follows from the definition of the set T , that

v f,

=f, Vf-,, = (a0 A U) V

(a, A

-24)

= a,

A (U V -24)

= a,,

teTU

hence

A

v ft

ueu t d U

= 00.

It follows from (1) that there exists a set YoE K such that

Ah#..

tEY,

Let

I)

See A. Tarski, op. cit., p. 195.

156

N. GENEl$4LIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

Since a. contains no atom, b is not an atom. This means that there exists an element c such that (7)

and

o f c d b

c#b.

According to the definition of K (see (2)), YonT, # 0; that is, either C E Y ,or -CE Y o . This implies by ( 6 ) that either b c or b < -c. In the former case we infer that b = c , in the latter case that c -c. Thus either c = b or c = a, which contradicts (7). This contradiction completes the proof.

<

<

# Example. The Boolean algebra K of regular closed sets in the plane is complete but not distributive. In fact, we proved in Chapter I, p. 39 that K is a Boolean algebra 0 and '. Moreover, it was shown with respect to the operations that each element of K, different from 0, contains a non-empty element distinct from itself. The Boolean algebra K is thus not atomic. From Theorems 3 and 4 given on p. 118 it follows that K is complete, from Theorem 11 it follows that K is not distributive. This Boolean algebra K is an interesting example which shows that not all laws which hold for the algebra of sets can be transferred to the theory of Boolean algebras, even in the case of complete Boolean algebras '). #

u,

Exercises 1. Prove that the equation dual t o (1) also holds in atomic Boolean algebras

v Aft

u-U teT,

=

A

vft.

Y E K tEY

Furthermore, show that there exist complete Boolean algebras in which (8) does not hold. I) For more results concerning distributivity in Boolean algebras see R. Sikorski, Representability and distributivity of Boolean algebras, Colloquium Mathematicum 8 (1961) 1-13. In the present section we have dealt with the generalization of the operations of union and intersection. Other set-theoretical operations were also generalized for Boolean algebras. For instance, the operations in cylindrical algebras are generalizations of the operation of Cartesian product. The theory of cylindrisal algebras can be found in the book: L. Henkin, La structure algibrique des thgories mathimatiques (Paris 1955).

157

10. INFINITE OPERATIONS IN LATTICES

2. Give an example of a Brouwerian lattice K such that for some set T,for some function f E K T and for some clement a E K we have

a).CVft)fV(aAft>.

f tT r ET 3. Give an example of a Brouwerian lattice K such that for some set T,for some function f E K T and for some a E K the supremum exists but the supremum

vft

teT

v ( u ~ f t does ) not exist. r

Q 11. Extensions of ordered sets to complete lattices We shall prove that every ordered set can be treated as a subsystem of a complete lattice (that is, of a lattice in which most laws of the algebra of sets hold). We shall also solve a similar problem for Boolean algebras. For this purpose, we fzst introduce a general notion of embedding of one system in another. DEFINITION 1: A relational system ( A , R ) is said to be a subsystem of ( B , S ) if A c B and f', [xRy 5x S y ] (that is, R = ( A X A ) n S). x , Y-A

DEFINITION 2: A system ( B , S ) is said to be an extension of the system ( A , R) if there exists a subsystem (B,, S,) of the system ( B yS ) isomorphic to ( A , R ) . In this case we also say that the system ( A , R ) is embedded isomorphically into the system ( B , S ) and that the function establishing the isomorphism embeds ( A , R ) in ( B , S). THEOREM 1: Every ordered set A can be embedded isomorphically in the family of all subsets of some set (where the famiry is ordered by the inclusion relation). Consequently, A can be embedded in some complete and atomk Boolean algebra. PROOF.Let for a € A O(a) = { x :

Because the relation

x .

< is transitive, we have a < b O(a) c O(b). --t

Since a E 0 (a), we have 0 (a) c 0 (b) --f a E 0 (b) --t a

a


3

O ( a ) c O(b),

< b.

Hence

IV. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

158

and it follows directly that O(a) = O(b) + a = b.

Thus the function a 3 O(a) embeds A in the family of sets O(a) ordered by the inclusion relation. Clearly, this family can be extended to the family 2,' where X = O(a).

u

aEA

In general, the extension described in Theorem 1 does not preserve supremum and infimum; that is, the equation a = b v c does not necessarily imply 0 (a) = 0(b) u 0 (c). We shall consider whether it is possible to extend the set A to a complete lattice preserving suprema and infima. Let the function g~ embed the ordered system (A, &) in the ordered system (ByG).

DEFINITION 3: The embedding y preserves suprema (resp. infima) if for every set T and for every function f E AT such that the supremum (resp. the infimum exists, the supremum ~ ( 5also )

vft

v

Aft)

tET

tcT

teT

exists and

(resp. for infimum:

A ~ ( 5exists ) and tET

We shall now consider a construction which will extend any ordered set to a complete lattice preserving suprema and infima. Let A be a set ordered by the relation For any set X c A, let

<. X+ = { a d : A (x < a ) } , x- = ( a d : A (a < X I } . xsx XEX

The next statements follow from the definitions above:

X

(1)

(2)

C

Y + (x+3 Y')

A

(X-3 Y - ) ;

for any 2 c A we have Z c Z + - and Z c Z-+.

PROOF.By definition,

.

(Z €2)A (Z' €2') -%

(2

< Z'),

159

11. EXTENSIONOF ORDERED SETS

thus (Z€Z)

--t

A (2 Q 2‘) + (zEz+-). Z’€Z+

The proof of the second part of (2) is similar.

z+-+= z+

(3)

z-+-= Z - .

and

The inclusion Z + c Z+-+ follows from (2). If a E Z + - +then and therefore

A (z < a)

zzz+-

// (z < a) since 2 c Z+-. Thus a € Z + .

z €2

The proof of the second part of (3) is similar. We now introduce the notion of a cut for ordered sets. The pair ( X , Y ) is said to be a cut in the ordered set A if X+ = Y , Y - = X. The set X is called the lower section and Y the upper section of the cut. It follows from the definition that

(4)

(.XEX)A(YEY)

+

(x


In fact, every element of X+ is in relation 2 to every element of X . It follows from (3) that ( 5 ) Pairs (2-,Z - + ) and ( Z + - , Z + ) are cuts. Moreover, every cut can be expressed in both of these forms. Finally, it follows from the definition of a cut that If a E A then the pair < { a } - ,{ a } + ) is a cut. (6) We can introduce an order relation between cuts:

<X,Y)Q(U,V)-XcU. In view of showing that the relation < is indeed an order relation, let us prove that { X , Y) < (U,V) = v -=Y. (7) PROOF.Suppose that X c U and V E V . If X E X , then X E U= V - ; thus x v. Therefore V E X += Y,and finally V c Y.In a similar way we prove the opposite implication. Let 9 denote the family of all cuts. 3, is a complete lattice. (8) Let I. c 3, and S= U X, T = U X. <x,x+> EL! < x - . X ) EL!

<

160

IV. GENERAJAZED UNION, INTERSECTION AND CARTESIAN PRODUCT

The cut ( S + - , S + ) is the supremum of ,". In fact, S c S+- and thus for every cut ( X , X + ) E I. If ( X , X + ) < (U,V ) for every ( X , X + ) E I, then X c U and therefore S c U. This implies S+ 3 U+ = V , hence ( S + - , S+> < ( U , V ) . Similarly we can prove that the cut (T-, T - + ) is the infimum of i!. preserving su) ' } x embeds A in (9) The function f(x) = < { x } - , { prema and injima. PROOF. By definition the following equivalences hold:

(x,X + ) < ( S + - , S + )

Suppose that x

x < y = (4-c {y>- = f ( x ) G f W . = T i in the qet A ; then pi 5; x and therefore

v

tET

f(yt)


Let ( X , Y ) be a cut such that f(yJ ,< ( X , Y ) for t e T . Then {pi}c X ,and since v t ~ { y t } -we , obtain y t E X . For any y € Y we have pi < y , thus x < y . This implies x E Y - = X,hence {x}- c X and f ( x ) < (X,Y ) . Thusf(x) = vf(p,) in the set T. I

The proof for infimum is similar. The next theorem follows from (9). THEOREM 2: Every ordered set A can be extended to a complete lattice V preserving suprema and injima'). The lattice 'p constructed above is called the minimal extension of the ordered set A . We now consider the case where A is a Boolean algebra. First, observe that if Z , and Z , are arbitrary subsets of an ordered set A , then (10) ( Z , u Z,)+ = Z: n Z,+, (2,u 2,)- = Zi n Z,. Now let A be a lattice. We prove that if Y , and Y2are upper sections of two cuts in A , then Y , n Y , is the set of all elements y l v y , l) The method employed in the proof of Theorem 2 is due to Dedekind, who introduced the notion of a cut (in linearly ordered sets) in view of defining real numbers. See R. Dedekind, Was sind und was sollen die Zahlen (Braunschweig 1872). McNeille generalized this notion to arbitrary ordered sets (McNeille, Partially ordered sets, Transactions of the American Mathematical Society 42 (1937) 416-460). Recent results in this field are given by G . Bruns, Darstellungen und Erweiterungen geordnefer Mengen, Crelle's Journal 209 (1962) 167-200.

161

11. EXTENSION OF ORDERED SETS

where yi E Y i for i = 1,2. In fact, y , v y , > y i ; thus y1v y , Y~i for i = 1,2. Moreover, y E Y 1n Y2+ y = y v y , where the first component may be understood to be an element of Y , and the second to be an element of Y 2 . Likewise we can prove that if X, and X2 are lower sections of two cuts in A, then X,nX, is the set of all elements of the form x1AX, where xi EX^ for i = 1,2. Combining the above with formula (10) we conclude that if A is a lattice and (Xl,Yl), ( X 2 , Y 2 )are two cuts in A, then

'1

"

(X,u X d + = Yl n y , = {YI JJ2: (Y1 E Y1) A 0 ' 2 E Y,)}, (Y, u Y,)- = X , n X , = {XI A x,: (xl A (x2 E X . ) } .

It follows from the definitions of supremum and infimum in 9 that for any ordered set A and for any two cuts in A we have (12)

(XI y1> v (X29 yz> = <(Xl u &I+-, 9

= ((Y1

(13)

( X l , Yl> A M

2 ,

(11

n Yd-9 YI

u X,)+)

f-l y2>,

Y2) = ( ( Y , u y z r , ( Y , u Yz)-+)

(4 x2,(4

= n n Finally, observe that if A is a Boolean algebra and Z* = { -2: z EZ} for any set Z c A, then (14) if ( X , Y ) is a cut in A , then (Y*, X*) is also a cut in A. &)+>a

The proof of this lemma is left to the reader. Now we shall prove the following

THEOREM 3 : The min;mal extension of a Boolean algebra is a Boolean algebra. PROOF.Let p be the minimal extension of a Boolean algebra A . It suffices to show that 9 is a distributive lattice with zero 0 and with unit I and that for any cut < X , Y ) E Y there exists a cut ( X , , Y , ) E V such that

(15)

<X7y > A (xi,YI>

=1

0,

<x,y > V (Xi, Yi) = I

(see p. 42, Theorem). Clearly, the cut 0 = ( { o } ,A ) is the zero of 9 and I = ( A , { i } ) is the unit. The cut (Y*,X*) defined by (14) satisfies conditions (15).

162

IV. QENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

In fact, the lower section of (X,Y ) A (Y*, X*) is X n Y* (see formula (13)). The only element contained in X n Y* is 0 , because a e X n Y* + ( U E X ) A( - a € Y ) .--) (a -a) --r a = 0. This proves the first part of (15); the second part can be proved similarly. It remains to show that the distributive law holds. Since the inequality (a A c ) v (b A c) < (a v b) A c holds in every lattice, it suffices to show that if (Xl, Yl), {Xz, Yz)and (U,V ) are three cuts in A , then

<

(Vl,

Yl) v W

Z 3

YZ))

A



< ((Xl,

)v ( W Z ,

(U,V>). Applying (12) and (13), this formula can be reduced to the form y1>A

yz> A

( Y , n Y2)-n U c [(Xln V)+n (Xzn V)+]-;

that is, by (11) and by the definitions of Z+ and 2-, (16)

(a&YivYz)

[(aEU)A

Yl€Yl,YZ€YZ

A

A

Xl.SX1, X Z E X Z ,

(b b X1 A U) A (b 3 Xz A U)]

+ (a

< b).

U€U

Suppose now that the elements a and b satisfy the antecedent of the implication (16). For arbitrary x1in Xl the inequality b 2 x1A a holds. This implies -a v b 2 xl. Since x1 is arbitrary, we conclude that the element y = - a v b belongs to Yi. Similarly we prove that y E Y,. Because a satisfies the antecedent of (16), we obtain a y v y = y = - a v b ; hence a a A ( - a V b) = a A b b. This proves (16) and thus the theorem is proved.

<

<

<

8 12. Representation theory for distributive lattices The notion of ideal ') is the basic concept involved in the representation theory for distributive lattices. l) Instead of ideals many authors, especially in recent literature, apply filters, which are subsets of A satisfying the conditions dual to (1) and (2); that is:

(a E I ) A (b E I ) + (a A b E I)

and

(a

>b)

A

(b E I ) -+ (a E I).

All theorems given in the sequel can be formulated in terms of filters by substituting A , v, for v , A, 2 ,respectively.

<

163

12. REPReSENTATION THEORY

DEFINITION: A non-empty set I c A is said to be an ideal of the distributive lattice A if (1)

( a E Z ) h ( b E I )+ ( a v b E I ) ,

(2)

(a < b) A (b E I ) + ( a E I ) .

An ideal I is said to be a prime ideal if I # A and for a, b E A, ( a A b € Z) + [(a€ I ) v (b E I)]. (3) Examples 1. Let A be a lattice of sets (for instance, the lattice of all subsets of an arbitrary set X) and let a be any element of the union S(A). The family Z of all sets M EA which do not contain a is a prime ideal in the lattice A. 2. The family of all finite sets M E A is an ideal in A. # 3. If A is the family of all subsets of the set of real numbers, then the family of all sets of Lebesgue measure zero is an ideal in A. # The set { x : x < a} is an ideal in any lattice. This ideal is called the principal ideal generated by a. We shall make use of the following general theorems concerning ideals. (4) ( a v b E I )+ [aEI)h(bEZ). I n f a c t , a , ( a v b and b < a v b . If a v b E I , then a E I and b E I by (2). a €I +a A b € I . (5) In fact, a A b < a. (6) The set I*(b) of elements x such that x $ i v b for some i E I is an ideal and I c I* (b), b E Z*(b). In fact, if x < i i , v b and y < i , v b , then x v y < ( i l v i z ) v b ; therefore x v y E I * ( b ) , because i 1 v i 2 E I . If x $ i v b and y $ x , then y ,( i v b. Thus the set I*(b) is an ideal. The conditions I c I * ( b ) and b E I*(b) are obvious.

(7) Let the ideals I,, for t E T , constitute a monotone family of ideals such that a € I, and b + I t for every t. Then the union I = I, is

u t

an ideal and a E I, b $I.

164

Iv. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

In fact, if X E Z ~ ,and Y E & , then either both elements x, y belong to It,, or both belong to It,. In any case, ( x v y ) ~ I . If y E It and x y , then x E It; thus x E I. Hence the set I is an ideal and aEZ and b#Z.

<

<

(8) If it is not true that b a, then there exists an ideal I such that aEZ and b$Z. In fact, the set {x: x < a} is such an ideal. Suppose that b < a is not true and let P a $ be the family of those ideals which contain a but do not contain b. (9) If the lattice A is distributive and b < a is false, then every maximal element I of the family P a , b is a prime ideal. In fact, suppose that x A y E I. If x # I then the ideal I is a proper subset of the ideal I * ( x ) . Thus I * ( x ) does not belong to the family pa,b . Since a E I* (x), we have b E I* (x). Thus b < il v x for sohe il E I. Similarly we show that if y $ I, then b < i, v y for some i2 E I. By the distributive law for the lattice A it follows that

< ( i l v x ) r \ ( i 2 v y )= ( i l r \ i 2 ) V [ i 1 ~ y ) V ( x ~ i 2 ) V [ x ~ y ) . .By (5) the elements i 1 ~ i 2iIr\y, , and x ~ belong i ~ to I and by assumption x ~ belongs y to I. Thus the element denoted by the right-hand side of (i) belongs to I. This implies by (2) that b E I, which contradicts I E P ~ ,Hence ~ . the hypothesis that neither x nor y belongs to I leads to a contradiction. Since b $Z, we have I # A. Hence I is a prime ideal. It will t e shown in Chapter VIII, p. 266, that the following theorem is a consequence of Theorems (7)and (8) (and of the axiom of choice). (i) b = br\b

"(10) The family P a , b has u maximal element, that is, there exists I EPa+b such that I is not a proper subset of any ideal belonging to P a . 6 . We have introduced the notions of distributive lattice, Boolean algebra, lattice of sets and field of sets. The relations among these notions are shown in the following scheme: distributive lattice

lattice of sets

J

\

\

J

Boolean algebra with 1

Geld of sets

12. REPRESENTATION THEORY

165

We shall prove that every distributive lattice is isomorphic to a lattice of sets and every Boolean algebra with 1 is isomorphic to a field of sets. 'THEOREM1; Every distributive lattice is isomorphic to a lattice of sets. PROOF.Let A be a distributive lattice. With each a E A we associate the family R(a) of prime ideals satisfying the condition a $ I . This correspondence is one-to-one. In fact, if a # b y then either a < b or b < a is false. By (9) and (10) there exists a prime ideal I such that either I EPo,bor I E Pb,o.In other words, there exists an ideal such that either a E I and b $ I or b E I and a $ I. In the first case we have I E R ( b ) and I $ R ( a ) , in the second case I E R ( u )and 1 $ R ( b ) . It follows from (1) and (4)that I ER(a v b) = a v b $ I

= [ ( a 4 0 v (b 401 ~[IER(U)VIER(b)] = [I E R (a)u R(b)], and from (3) and ( 5 ) that IER(aAb)=aAbq!I

= (a 4 r ) A (b 44 = [ I E R (a)]A [ I E R (b)]

=I E R ( a ) n R ( b ) ; thus R(a v b) = R(a)uR(b) and R ( a A b) = R ( u ) n R ( b ) . These formulas show that the class of all families R(a) is a lattice of sets isomorphic to the lattice A. 'THEOREM2: Every Boolean algebra is isomorphic to a j e l d of sets. PROOF.If the lattice A in Theorem 1 is a Boolean algebra, then it has a zero element o and a unit element i and, moreover, for every a E A there exists an element - a E A such that a A (-a) = o and a v (-a) = i . Under the correspondence a+R(a) the element o corresponds to the empty set and the element i corresponds to the whole set A. Since A = R ( a v -a) = R ( a ) u R ( - a ) and o = R ( a h ( - a ) ) = R ( a ) n R ( - a ) , we have R ( - a ) = A-R(u). The set of all families RCa) is not only a lattice of sets but a field of sets as well. Q.E.D.

166

IV. GENERALIZED UNION, INTERSECTION AND CARTESIAN PRODUCT

We shall give a topological interpretation of Theorem 2. Let A be a Boolean algebra with the zero element o and the unit element i and let P be the set of all prime ideals of A . We assume each of the families R(a) to be a neighborhood of every of its elements. For any set X c P let I E X if every neighborhood of I contains elements of X . 'THEOREM 3. (i) P is a compact topological space. (ii) The sets R ( a ) . are both open and closed in P . (iii) Every set in P which i s both open and closed is identical with one of the sets R(a).

PROOF.The proof that the axioms of topology hold is left to the reader. To prove that P is a compact space, we let K be a family of closed X # 0. sets with the finite intersection property. We shall show that

n

XEK

Let K* be the the family of all finite intersections of the form where n is an arbitrary natural number and X, E K. The family

nX,, I-=n

K* is thus a family of closed non-empty sets.

Let

<

Clearly a, a, E I + a, E I . If a,, a, E I then for some X , ,X , E K* we have R ( a , ) n X , = 0 == R(a,)nX,. Thus [R(a,)uR(a,)]n(X,nX,) = 0. Since XlnX, E K* and R(a,)uR(a,) = R(a, v a,), we conclude that a, v a, E I. Hence the set I is an ideal. We show that i $ I . Suppose the contrary. Then for some X E K * we have R(i)n X = 0. This implies A nX = 0 ; that is, X = 0, which contradicts the hypothesis that K has the finite intersection property. From the formula just proved and from the fact that i $ I it follows by (10) that there exists a prime ideal Zo 2 I. Thus this ideal is an element of P. We shall show that & E (7 X. XLK

Let X be an arbitrary set belonging to K and let R ( a ) be a neigh~ . ~ $ 1which , by the borhood of I,. This implies that ~ $ 1Therefore definition of Z shows that for every Y E K * we have R(a) n Y # 0. In particular, we have R(a) n X # 0. Hence every neighborhood of I.

167

12. REPRESENTATION THEORY

has a non-empty intersection with X , and I, EX= X . This shows that I,€ X.

n

XeK

In this way part (i) of Theorem 3 is proved. The proof of (ii) follows from the fact that the family R(a) being a neighborhood in P is open in P and its complement P -R(a) is also open, because it is equal to R ( - a ) , which is also a neighborhood in P. Finally, to prove (iii) suppose that X is both open and closed in P and let L = {R(a):R(a) c X } . Since every point belonging to the open set X has at least one neighborhood R(a) which is contained in X, we have Y = X and thus X n (P-Y, = 0. Hence the inter-

u

YEL

n

YEL

section of the family composed of the set X and the sets P- Y, where Y E L , is empty. Since this family consists of closed sets, it does not have the finite intersection property. This means that there exists a finite subset {R(a,), R(a,), ...,R(an-,)} of L such that X n ( P - R ( a j ) ) = 0. This implies X = R ( a j ) = R ( v a , ) . Hence

n

j
u

j i n

j
the set X is of the form R(a). The space P constructed in Theorem 3 is called the Stone space of A . The following corollary is a consequence of Theorem 3. COROLLARY 4: Every Boolean algebra with unit is isomorphic to the field of sets which are both open and closed in a compact space I). O

') Representation theory for Boolean algebras presented in this section is due to M. H. Stone. (M. H. Stone, The theory of representations for Boolean algebras, Transactions of the American Mathematical Society 40 (1936) 37-111. This theory was generalized for Boolean algebras with operators by B. J6nsson and A. Tarski (Boolean algebras with operators, American Journal of Mathematics 73 (1951) 891939 and 74 (1952) 127-174). There are a great number of publications devoted to the theory of representation for non-distributive lattices. From recent publications we mention: B. Jbnsson, On the representation of lattices, Mathematica Scandinavical(1953) 193-206;B. Jbnsson, Representations of complemented modular lattices, Transactions of the American Mathematical Society 97 (1960). 64-94; B. Jbnsson, Representations of the relatively complemented modular lattices. ibid. 103 (1962) 272-303;G. Gratzer, A generalization of Stone's representation theorem for Boolean algebras, Duke Mathematical Journal 30 (1963) 469-474.

168

IV. GENERALIZED UNION, INTERSECTION A N D CARTESIAN PRODUCT

Exercises 1. Construct a lattice of sets isomorphic t o the lattice N ordered by the relation of divisibility. 2. Prove that if an ideal Z f. A in a distributive lattice A is maximal (that is, if every ideal containing Z is equal t o either A or I ) , then Z is a prime ideal. The converse theorem is false. 3. Prove that if A is a Boolean algebra with unit, then an ideal Z is prime if and only if for arbitrary a E A either a E Z or --a E Z. 4. Show that in a Boolean algebra with unit the notions of prime ideal and of maximal ideal are equivalent. 5. Show that the family of ideals in a distributive lattice A is, in turn, a distributive lattice with inclusion as the ordering relation.

CHAPJTR V

THEORY OF CARDINAL NUMBERS In this chapter and in the remainder of the book we shall use the axiom system Zo[TR] (see p. 5 5 ) together with Axiom VIII formulated on p. 86. As usual, theorems not marked with the sign O are proved without using the axiom of choice.

8 1. Equipollence. Cardinal numbers We now introduce the notion of equipollence, one of the most characteristic and important notions of set theory l). DEFINITION: The set A is equipollent to the set B if there exists a one-to-one function f with domain A and range B. We write A B, and we say that f establishes the equipollence of A and B. Examples 1. If the set A is finite, i.e., the number of its elements is a certain natural number n, then the set B is equipollent to A if and only if B contains exactly n elements. Thus the notion of equipollence is a generalization to arbitrary sets of the notion, for finite sets, of having an equal number of elements. # 2. Let A be the interval a, < x < a2,B the the interval bl < x < b2. The function

-

is a one-to-one correspondence from A to B. By definition, A

-

B. #

l) The notion of equipollence was first systematically investigated by G. Cantor. See Ein Beitrag zur Mannidaltigkeitslehre, Journal fur reine und angewandte Mathematik 84 (1878) 242-258. It had been known to B. Bolzano; see Paradoxien des Unendlichen, 9 20 (1851).

170

V. THEORY OF CARDINAL NUMBERS

-

THEOREM 1: For arbitrary sets A , B, andC the foIIowing formuIas hold: ( A N B ) + ( B N A), ( A - B ) A ( B - C ) -+ ( A -q. (1) A A , Thus, equipollence is reflexive, symmetric and transitive. PROOF.The function IA (see p. 71) establishes the equipollence of A with itself. If the functionfestablishes the equipollence of A and B then the functionfe establishes the equipollence of B and A (see p. 70). If .f establishes the equipollence of A and B and g establishes the equipollence of B and C, then the composition g of establishes the equipollence of A and C (see p. 71, Theorem 2). The following formulas hold: ( A x B) N ( B X A), (2) ( A x { u } ) N A N AIaI, {a>" {a>, (3) (4)

--

(4 Bl) A (A,

(5)

(6) (7)

-

[ Ax ( B x q

(A,

-

l

-

-

[(Ax B) x CI,

-(4

x &)I,

Bz) + [(A,x 4)

B) + (2A-29, B,) A ( A , n A , = 0 = Bl n B,)

(A

B,) A ( A ,

-

-

(8)

YXx=

(9)

(Y x 2)"

-b

(A1

u Az-B,

u &)>

(YX)T,

(YX x Z X ) ,

( A A B = 0) 3 ( Y A ' B N YAXYB). We omit the proofs of (2)-(7), which are not difficult. On the other hand, we prove the important formulas (8)-( 10). Let fe YXxT; hence f is a function of two variables x and t, where x ranges through the set X and t through the set T and where f takes values in Y. For fixed t the function g, (with one variable x ) defined by g,(x) = f ( x , t ) is a function from X into Y, g, E Y x . The function F defined by F(t) = g , associates with every t E T an element of the set Y x , so F E ( Y ~ ) ~ . If fi and f? are distinct functions belonging to the set Y x x T ,then the corresponding functions Fl and F, are also distinct. In fact, if fi(xo,to) #fi(xo, to) then the elements Fl(to) and &(to) of the set Y x are distinct. (10)

171

1. EQUIPOLLENCE. CARDINAL NUMBERS

Each function F E ( Y X ) = corresponds in the manner described , to the function f defined above to some function fE Y X X Tnamely, by f ( x , t ) = g,W, where gt = F(t). It follows that the correspondence of the function f E YXxT to the function F E ( Y X ) =establishes the equipollence of the set Y X x T with the set (Yx)', which proves formula (8). For the proof of (9) notice that if f E (YxZ)', then f ( x ) is, for every X E X ,an ordered pair
PROOF.Associate with every function f E

If ~1 is a permuta-

Fx the composite funcX

tion g =f o ~ 1 .If fi #fi then for some x EX,fi ( x ) # f z ( x ) . Thus setting Y = vc(x) we obtain fi(vQ) # f 2 ( v O ) , that is, g,Cv> # gzO. Hence the correspondence given by the equation g =f 0 v is one-toone. The function g =fo y belongs to the Cartesian product F+,(x,.

n In fact, if E X then f (~(x)) F,,,,, that is, g(x) FPCx,. Finally, every function belonging to the Cartesian product n x

x

E

can be represented as fo v for some f E take for f the function go vc.

E

nF,.

FP(x)

X

In fact, it suffices to

X

THEOREM 3 : (ASSOCIATIVE LAW.) Let F E(2'9'.

rf X =

u Ty,where

YEY

172

V. THEORY OF CARDINAL NUMBERS

the sets T y are pafrwise disjoint, then

AX

PROOF.Let Gy =

n F,.

-n

(AT').

J-Y

Gy is the set of all functions f with domain

xeTv

Ty such that f ( x ) E Fx for x E T,,.

n

G,, consists of the functions g with

Y EY

domain Y such that g ( y ) E Gy for y E Y. We denote the value g ( y ) of the function g by gy;g, is a function with domain Ty such that gy( x ) E F,. We associate with the function g e Gy the function f defined by

n

YEY

the equation

(0

f(XI

= gy(x),

where y belongs to the set Y,and x c T y . The domain off is X;and for every X E X f , (x) e F,. Thus f E

nF,.

xaT

The correspondence between the functions g and f is one-to-one. In fact, if g(') # g(*)then there exists y E Y such that gp) # gp) and thus there exists X E T ,such that gF)(x) # g F ) ( x ) . Hence by (i), f(')(i)

# f (*) (4. It remains to be shown that to every function f e

nF, there corre-

XET

sponds a function g. For this purpose it suffices to notice that the function g defined at y E Y by the equation gY

=f I T Y

(see p. 72) belongs to the Cartesian product

fl Gyand also satisfies (i). Y

To prove formula (13) it suffices to let F, = A in formula (12) for each x. THEOREM 4: (LAWOF EXPONENTS FOR THE F E ( ~ ' )For ~ . every set H

CARTESIAN PRODUCT.)

Let

PROOF.We define the function of two variables on the set T XH by the equation G
173

1. EQUIPOLLENCE. CARDINAL NUMBERS

in two ways as the union of disjoint sets where This the set of all pairs with second coordinate equal to h and Ht is the set of all pairs with fkst coordinate equal to t. Applying Theorem 3 twice we obtain

For x E H i , x = ( t , h). Thus G, = F, , which shows that

n G,

xeHt

On the other hand, since Hi (Fft

N

FZ:

-

and

= (F,)".

H,

I1(Fi)at ieT

-n

(F:).

teT

Thus by (15) we get (17) For a given h we associate the function g e

n Gx with the function

XET,,

f where f(t)

= g(t, h), obtaining

n

XETh

Gx

-n

r€T

Ft.

Applying (16) we obtain

Finally (14) follows directly from (17) and (18). If the sets A and B are equipollent, we also say that A and B have the same power or that they have the same cardinal number. Of course, in this way we neither define the power of a set nor its cardinal number, but rather we only introduce a new terminology for the old notion of equipollence I). l) Cantor defined the cardinal number of a set A as that property which remains when we abstract from the nature and from the order of the elements of thzset A . To indicate this twofold act of abstraction, Cantor introduced the symbol A to denote the cardinal number of the set A .

174

V. THEORY OF CARDINAL NUMBERS

The new notions are not indispensable. We may formulate all theorems of set theory so that they are statements, not about cardinal numbers or powers of sets, but rather about relationships between cardinal numbers, which can always be stated in terms of the notion of equipollence. On the other hand, many theorems can be stated more intuitively if they are formulated as theorems about cardinal numbers. For this reason it is convenient to introduce this notion. The following theorem can be proved without difficulty. THEOREM 5: For the sets A and B to be equipollent it is neccessary and suficient that the relational systems ( A , A x A ) and ( B , B x B } be isomorphic. We shall denote by 2 the relational type of the system ( A , A x A } . We shall call the cardinal number or the power of the set A . From Theorem 5 and Axiom VIII (p. 86) we obtain the following: THEOREM 6: For arbitrary sets A , B the conditions A are equivalent.

N

- -

B and A = B

This theorem allows us to formulate statements about equipollence as equations involving cardinal numbers. 52. Countable sets

If X is a finite set containing exactly n elements (see p. loo), then Theorem 6, 81 is satisfied if we set x'= n (see p. 102, Theorem 6). In the future we shall identify the cardinal number of a finite set of n elements with the natural number n. The theory of cardinality for finite sets is not essentially richer than the arithmetic of the natural numbers. New notions appear when we turn to infinite sets.

DEFINITION: A set A is said to be countable (or denumerable) if it is either finite or equipollent with the set of all natural numbers. Clearly, two arbitrary infinite countable sets are always equipollent (see $1, Theorem 1). We shall denote the cardinal number of infinite countable sets by a. In Chapter 111, p. 91 we defined a sequence as a function with domain equal to the set of natural numbers. From this definition it

175

2. COUNTABLE SETS

follows that a set is countable if and only if it is the range of a sequence none of whose terms are equal. Speaking somewhat figuratively, a set A is countable if its elements can be “arranged” in an infinite sequence a, ,a2, a3, ... . THEOREM 1: Every countable non-empty set is a set of terms of an infinite sequence. Conversely, the set of terms of any infinite sequence is countable and non-empty. PROOF.The finite set whose elements are a , ,a2,a3, ... ,ak is the set of terms of the infinite sequence: f(0) = a,,

..., f ( k ) = a,,

f(k+ 1) = U k ,

... , f C k f j ) = ak, ...

Every infinite countable set is by definition the set of terms of an infinite sequence. To prove the converse we assume that X is the set of terms of an infinite sequence q~ and that X is an infinite set. Let yo = yo and

or yn+l = yo if there exists no k such that jSn

(yj # Fk).

We prove by induction that for every n there exists a number k such that // (y, # Vk). It follows that A (yn+, # y,) and thus the seiGn

j Qn

quence y has distinct terms. It remains to be shown that every element of the set X is a term of the sequence y. For this purpose we assume that the set {k: f p k is not a term of y} is non-empty and we let ko be the least element belonging to that set. Clearly ko > 0. If i < ko then viis a term of y, say fpi = ym(;).Let m = maxm(i). The smallest number k such that ( v k # y j ) is then f
i<m

ko and thus from the definition of the sequence y we obtain ym+, = ykO,contradicting the definition of the number ko. In a similar manner we establish the following theorems. THEOREM 2: Every subset of a countable set is countable. THEOREM 3: The union of two countable sets is countable.

176

V. THEORY OF CARDINAL NUMBERS

PROOF.Since the case where one of the given sets is empty causes no difficulty, we assume that A is the set of terms of the sequence

ao,al,az, ...,an,... and that B is the set of terms of the sequence

bo, bl, b2, ...,bn, ..The union of A and B is then the set of terms of the sequence ao,bo, ai, bi, az, b2,

.

,an, bn, - -.,

and is therefore countable. From Theorem 3 it follows by induction that the union of an arbitrary finite number of countable sets is countable. A particular case of Theorem 3 is the following 4: The union of afinite set with a countable set is countable. THEOREM

THEOREM 5: The Cartesian product of two countable sets is countable. PROOF.I f A and B are infinite countable sets, then A N N and B N ; and thus A x B N x N. By Theorem 1, p. 170, it follows that N x N N N , and thus A x B N N . If one or both of the sets A and B are finite, then A x B is equipollent to a subset of N x N , that is, to a subset of Nand the theorem follows by Theorem 2. N

N

THEOREM 6: I f the set A is countable, then the set of all finite sequences with terms in A is countable. PROOF.The theorem follows immediately from Theorem 4, p. 98. THEOREM 7: I f y is an infinite sequence whose elements are also infinite sequences, then the set X of elements x which are terms of the sequences y,, is countable. PROOF.By definition, X ={ x :

v mn

(x = ymn)} = (x :

v

(X

=~ J K ( ~ ) , L ( ~ ) ) } .

P

Thus X is the set of terms of the sequence q~ defined by the equation VP

=YK(P),L(P).

THEOREM 8: If A is a sequence whose elements are non-empty countable sets, then the union A , is countable. O

u n

2. COUNTABLE SETS

177

PROOF. Let C,, be the set of sequences p such that A,, i s the set of terms of tp. By assumption C,, f 0 for all n E N . Therefore by the axiom of choice there exists a sequence y such that y , E C, for each n. Thus A , is the set of those x for which there exist m, n E N the union

u n

such that x

= y m nwhich ,

proves the theorem on the basis of Theorem 7.

REMARK: The use of the axiom of choice is neccesary for the proof of Theorem 8. For every countable set A there exists an infinite sequence containing all the elements of A among its terms. Yet there are infinitely many such sequences for a given set A and we have no way of distinguishing between them. In other words, we have no way of associating with every countable set an infinite sequence whose terms contain all the elements of the given set. #Examples of countable sets 1. The set of integers is countable. In fact, it is the union N u N' where N' is the set of integers 0. Because N - N', where the function f ( n ) = -n establishes equipollence, both sets N and N' are countable. It follows that the set N u N' is countable. 2. The set of rational numbers is countable. Indeed, the sequence pl defined by pp = K(p)/(L(p+ 1)) contains all the non-negative rational numbers among its terms and only such numbers. Thus the set of non-negative rational numbers is countable. From this we obtain that the set of negative rationals is countable (see Example 1). Hence the set of all rationals is countable.

<

3. The set of polynomials of one variable with integral coescients is countable.

To every polynomial with integral coefficients there corresponds the unique sequence of its coefficients. By Theorem 6 the set of all finite sequences of integers is countable. 4. The set of algebraic numbers is countable.

In fact, with every polynomial we may associate a finite sequence whose terms are all the roots of the polynomial. We let the first term

178

V. THEORY OF CARDINAL. NUMBERS

of the sequence be that root which has the smaIlest modulus and among those of equal modulus that root which has the smallest argument. Similarly, we let the second term of the sequence be that root different from the first which has the smallest modulus and the smallest argument among the roots having the same modulus. In this way we define by induction the desired sequence. The countability of the set of all algebraic numbers follows now from Theorem 7. We can obtain the same result from Theorem 8. But in this case we have to use the axiom of choice. Exercises 1. Prove that the set of all intervals with rational endpoints (in the space of real numbers) is countable. 2. Prove that in 3-dimensional euclidean space (or more generally, in P ) the set of all spheres with radius of rational length and with center having rational coordinates is countable. 3. Let f be a function with field contained in the set of real numbers. We say that f has a proper extremum at the point a if there exists an interval P containing a such that f ( x ) < f ( a ) for all x E P - { a } or else f ( x ) > f ( a ) for all x E P - { a ) . Prove that the set of proper extrema of such a function f is at most countable. Hint: Use Exercise 1. Generalize the theorem to functions defined on the space Cn (replacing P by an n-dimensional sphere). 4. Prove that every disjoint family of intervals in the space of real numbers is countable. Hint: Use Exercise 1. Generalize the theorem t o families of disjoint open sets in the space en, using Exercise 2. 5. Let Z be a set of points in the plane. We call the point p s Z isolated if there exists an open circle K (i.e. without circumference) such that {p} = Z n K . Prove that the set of isolated points of a given set Z is countable. Hint: Use Exercise 2. Generalize the theorem to the space Ln (replacing circle by ball in the definition of isolated point). 6. Prove that every monotone discontinuous function from the set of real numbers to the set of real numbers has a countable number of points of discontinuity. Hint: Every monotone function has both a limit from the right and a limit from the left at every point; at points of discontinuity those limits are unequal. Apply Exercise 4.

3. HIERARCHY OF CARDINAL NUMBERS

0 3.

179

The hierarchy of cardinal numbers

We shall prove that besides the finite cardinal numbers and the number n there exists infinitely many other cardinal numbers. For this purpose we prove the following very useful theorem. 1. ( O N DIAGONALIZATION ').) If the domain T Of function THEOREM 'F is contained in a set A and i f the values of 1; are subsets of A , then the set Z = {tET: t$F(t)} is not a value of the functiolz F. PROOF:We have to show that for every t E T,F(t) # Z . From the definition of the set 2 it follows that if t E T, then [ t E 213 [ t $ F(t)], Thus if F(t) = 2 we obtain the contradiction: ( t E Z ) = ( t $2). For A = T Theorem 1 has a geometrical interpretation. We may consider the set A x A represented as a square (see. p. 63, Chapter 11). We let R = ( ( x , y ) : y E F(x)}. The set F(x) is the projection onto the vertical axis of those pairs belonging to R which have first coordinate equal to x . The set Z is the projection onto the vertical axis of those points along the diagonal of the square which do not belong to R. It is then geometrically obvious that 2 # F(x), for any x E A ; for if ( x , x ) E R, then x E F(x) but x $ 2 , and if ( x , x) $R, then x $ F(x) but x E Z . This interpretation motivates calling Theorem 1 the Diagonalization Theorem. We apply Theorem 1 to prove that there exist distinct infinite cardinal numbers. 2: The set 2A is not equipollent to A , nor to any subset of A . THEOREM For otherwise there would exist a one-to-one function whose domain is a subset of A and whose range is the family of all subsets of A . But this contradicts Theorem 1. THEOREM 3: No two of the sets A , 2A, 22A,222A (1) are equipollent. Y

l)

This theorem is due to Cantor.

180

V. THEORY OF CARDINAL NUMBERS

PROOF,Let Pk be the kth set in sequence (1) and suppose that there exist k and 1 such that k > I and Pk is equipollent to a subset of P l , The set Pk-1 is clearly equipollent to a subset of Pk, namely to the subset of singletons { x } where xEPk-1. Thus the Set Pk-1 is equipollent to a subset of Pl. Repeating this argument we conclude that each of the sets Pk-l,Pk--2,..., P1+, is equipollent to some subset of Pl, but. this contradicts Theorem 2 because Pl+l = 2'1. THEOREM 4: Let the family A have the property For every X E A there exists a set Y EA which is not equipollent (2) to any subset of X. Then the union ( A ) is not equipollent to any X EA nor to any subset of X E A . PROOF.Assume that (A) -A', c X E A . It follows that there exists a one-to-one function f such that f (A)) = XI. By assumption (2) there is a set Y EA which is not equipollent to any subset of X . As Y c U (A), we have f ' ( Y ) cf'(U (A)); that is, f '(Y) c Xl and consequently Y N f I(Y) c X. The contradiction shows that it is not the case (A) -XI. that Theorems 3 and 4 give us some ideas of how many distinct infinite cardinal numbers exist. Starting with the set N of natural numbers which has power a, we can construct the sets

{

u

u

'(u

u

N , 2N,22N,222N,..., (3) no two of which are equipollent by Theorem 3. In this way we obtain infinitely many distinct cardinal numbers. By the axiom of replacement there exists the family A whose elements are exactly all the sets (3) (see p. 96). By Theorem 2 the family A satisfies condition (2); thus by Theorem 4 the union P = ( A ) has a cardinal number different from each of the sets (3) and from each of their subsets. Again applying Theorem 3 we obtain the sequence of sets P , 2 p , 2ZP,2 z a P , ..., (4) no two of which are equipollent and none of which is equipollent to any of the sets (3). We obtain in this way a new infinite quantity of distinct cardinal numbers.

u

3. HIERARCHY OF CARDINAL NUMBERS

181

We obtain still other cardinals by constructing the family B consisting of all the sets (3) and (4) and by constructing a new sequence

u

Q = (B), 2O, 22Q, ,... We may continue this procedure indefinitely. We see that the hierarchy of distinct infinite cardinals obtained in this way is incomparably richer than the hierarchy of finite cardinals, which coincides with the natural numbers. As a further consequence of Theorem 2 we note the following 5 : There exists no family of sets U which,for every set X, THEOREM contains an element Y equipollent to X . PROOF.By Theorem 2 the set is not equipollent to any subset of the set ( U ) and hence it is not equipollent to any of the (U). sets Y belonging to U because Y E U implies Y c THEOREM 6: There exists no set containing all sets. For otherwise this set would satisfy the conditions of Theorem 5. Theorem 6 shows again (see Chapter 11, p. 61) that we cannot state as an axiom consistent with our axiom system that there exists a set composed of all elements which satisfy an arbitrary propositional function @(x). Theorem 5 is also another indication of the vastness of the hierarchy of cardinal numbers, which is so “large” that it is impossible to construct a set containing at least one set of each power.

u

u

Exercises 1. Prove that the set NN is uncountable. Hint:If rp is a sequence of elements of NN,then the sequence 6 defined by 6, = p,(n)+l is not a term of the sequence 9. 2. Let X be a compact space # 0 having the property: for every finite set S and for every open set G # 0 there exists an open non-empty set G* such that G* c G and n S = 0. Show that y # a. Apply this inequality to show that the Cantor set is uncountablz. Hint:Use the axiom of choice to associate with every open set G # 0 and every finite set S an open subet G* = G*(G, S) c G such that g* n S = 0. Assuming that rp is an in6nite sequence of elements of X, let

z*

Go=X,

prove that

nG, # 0. n

Gn+i =G*(Gn,{Vo. . . . , p n } ) ;

182

V. THFORY OF CARDINAL NUMBERS

31). We say that the sequence b,, b2, ... of natural numbers increases more rapidly an

than the sequence a , , a,, ... , if lim - = 0. Prove the following statements: n-m

bn

(i) for every sequence there exists another sequence which increases more rapidly; (ii) let Z be a set of sequences such that for every sequence a,, a,, ... there exists a sequence 4, b2,... belonging to Z which increases more rapidly than a,, u2, ...; then the set Z is uncountable. Hint: Assume z'= the table

z. We then may represent the elements of Z as the rows of %I,

a1.2,

a2.1. a2,zr

...,al,nn *.., a,,,, ..*

. . . . . . . . . . .. an,zl *..,&,nr ...

Using an appropriate diagonalization argument define a sequence increasing more rapidly than every sequence in the table.

54. The arithmetic of cardinal numbers

We shall define operations of addition, multiplication, and exponentiation for cardinal numbers. The definitions will be chosen so that they will coincide with the ordinary definitions for finite cardinah (that is for the natural numbers) ').

DEPINITION 1: The cardinal number m is the sum of the cardinals

nl and n2, m

= n1Sn2,

if every set of power m is the union of two disjoint sets, one of which has power nl and the other of which has power n2. LEMMA 1 : Given two arbitrary sets Al and A z , there exist sets B1 and Bz such that (0)

A, .-B1,

A2 .-B2,

B1 n B2 = 0.

l) In connection with the problems dealt with in Exercise 3 see the book G. H. Hardy, Orders of infinity, Cambridge Tracts in Mathematics and Mathematical Physics No. 12, 2nd edition (Cambridge 1924). 2, Basic definitions and theorems given in 84 are due to Cantor; see: Beitrage zur Begriindung der tramfiniten Mengenlehre, Math. Annalen 46 (1895) 481-512.

4. ARITHMETIC OF CARDINAL NUMBERS

183

Choose a, and a, such that a, # a, (for instance, a, = 0, a, = (0)). Then, by (4), 0 1, p. 170, the sets B1 = { a , } x A l and B, = { a z } xA, satisFy the desired conditions. 2: The sum nl+n, of any two given cardinals n, and n, THEOREM always exists.

-

PROOF.Assume that A', = n, and A , = n2. Let B, and B, satisfy conditions (0). Then B, u B, is the union of two disjoint sets of power n, and n,. Clearly every set equipollent to B, u B, has the same property. Thus u 3,= nl+n2. Moreover, we have proved that

-

-

X+~=AUB

A ~ B = O .

if

THEOREM 3: Addition of cardinal numbers is commutative and associative: for arbitrary cardinal numbers n,, nz and n3, we have

n r f n , = n,+n1, nl+(n2t n,) = (n1+np)+n. (2) PROOF.If A = nl+n2 then A = Al u A,, where Al n A, = 0, Al = n1 and A2 = n2. Thus A = A, u Al and A = n,+nl, which proves (1); (1)

-

-

-

s

the proof of (2) is similar. Example. By Theorems 3 and 4, 52, p. 175-176 we have

(3)

a+a

=

a,

n+a

=z

a.

DEFINITION 2: The cardinal number m is the product of ni and n,, i s . m =~ t ~ . n ~ ,

-

if every set of power m is equipollent to the Cartesian product A , X A , where A , = nl and A2 = n2. Thus =A,xA,. 3

&A,

-

It is clear that, for arbitrary cardinals nl and n,, the product nl.nz always exists. Definition 2 is a generalization to the case of arbitrary cardinal numbers of the usual notion of muitiplication: for example, we consider the product 3.4 as the number of elements of a set which can be rep-

V. THEORY OF CARDINAL NUMBERS

184

resented as three groups of four elements; that is, a s . the number of elements in the set A X B, where A contains exactly three elements, and B four.

THEOREM 4: Multiplication of cardinal numbers is commutative, associative, and distributive over addition:

n,-n,

(4)

= n,-n1,

tt, * (IIz'It3) = (tt1 *It2)*n3 ,

(5)

+

(nz n3) = tt, * ttz+nl *n3. PROOF. Equations (4) and ( 5 ) are immediate consequences of equations (2) and (4), 0 l, p. 170. Equation (6) follows from the equations (see Chapter 11. 84, p. 63):

(6)

n1.

A , X (A2 u 443) = (A1 X A2) u ( A , X A3J7

[ A , n A3 = 01 + [ [ A lx A,) n ( A , x A,) = 01. THEOREM 5: 1 is the unit for multiplication; namely,

n.1 =n.

(7)

The proof follows from (3), $ 1, p. 170.

Example. By Theorem 5, 62, p. 176 a . a = a,

(8)

a - n = a.

Denote the n-fold product m.m. ... 'nt by m". By Definition 2, m" is the power of the set of all sequences of n-elements ( a , , a2, ...,a,,), where a I , ..., a,, are elements of a set A of power m. In other words (see Chapter 11, $6. p. 170),

(A)" = A?,

Generalizing the example above we obtain the following definition. DEFINITION 3: The cardinal m is the cardinal n raised to p-th power,

rn = no,

-

-

if every set of power m is equipollent to the set AB, where A = tt and B = p. Thus (A)B= AB. - I

-

4. ARITHMETIC OF CARDINAL NUMBERS

185

It is clear that for every two cardinal numbers n and p the cardinal nv always exists.

THEOREM6: For arbitrary cardinals n,t, and q: (9)

nv+q = nv.nq,

(10)

(n*P)q = nq-pq,

(11)

(nB)S = nP-4,

a’ = a , 1” = 1.

These equations follow directly from equations (4), (8)-(10),

0 1.

THEOREM 7: r f A has power m, then the set 2A (which consists of all subsets of A ) has power 2m:

- -

24 = 2‘4.

PROOF.2m is the power of the set (0,1>’4,consisting of all functions f whose values are the numbers 0 and 1 and whose domain is the set A. Each such function is uniquely determined by the set X, of those a for which f ( a ) = 1 (f is the characteristic function of this set, see Chapter IV, p. 122). Under this correspondence to distinct functions fi and fi correspond distinct sets X,, and Xf2. Thus associating with the function f E ( 0 , l}“ the set X, c A, we obtain a one-to-one correspondence between the sets (0, 1jA and 2A.

5 5. Inequalities between cardinal numbers. The Cantor-Bernstein theorem and its generalizations We obtain the “less than” relation beetwen cardinal numbers from the following definition. DEFINITION: The cardinal number m is not greater than the cardinal number n, m d n, if every set of power m is equipollent to a subset of a set of power n. If m < n and m # n we say that m is less than n or that n is greater than m; we write m < n or n > m.

V. THEORY OF CARDINAL NUMBERS

186

For example, (1)

< a, rn < 2"'. n

(2)

For the proof of (2) we notice that m < 2", because every set A of power m is equipollent to the subset of 2A consisting of all singletons of elements of A . Moreover, m # 2m by Theorem 2, 0 3. The following theorem is an interesting consequence of the definition of inequality. O THEOREM 1: r f f is a function defined on the set X and f'(X) = Y , then '.< PROOF.The notion of coset of a functionf (a set of all elements of X which have the same value under f) was defined in Chapter 11, p. 175. Every coset is a set of the form w,= { x E x: f ( x ) = y } .

Since all cosets are distinct and non-empty, there exi ts by the axiom of choice1) a set A containiw exactly one element from every coset. It follows that A is equipollent to the set of all cosets off and thus to the set f'(X). Since A is a subset of X , we conclude that Y Example. The projection of a plane set Q onto an arbitrary straight line has power 2. In this case the cosets of the projection are the intersections Q n L where L is a straight line parallel to the direction of the projection. REMARK: We write m <*n if m = 0 or if every set of power m is the image of every set of power n2). It is an easy consequence of Theorem 1 that {m < n} G {nt <*n}; we saw that the proof of this

-
<

I) The fact that the proof of Theorem 5 requires a new axiom was pointed out by B. Levi as early as in 1902 (R. Instituto Lombard0 di scienze e lettere, Rendiconti (2) 35 p. 863), before Zermelo published his axiomatic set theory with the axiom of choice. A. Tarski has investigated the properties of without referring to the axiom of choice. See A. Tarski and A. Lindenbaum, Communication sur Ies recherches de la thbrie des ensembles, Comptes rendus de la SocWtB des Sciences et des Lettres de Varsovie, Classe 111, 19(1926), 299-330.

<*

187

5. INEQUALITIES BETWEEN CARDINAL NUMBERS

equivalence uses the axiom of choice. Without using this axiom we are not even able to prove the intuitive proposition that the conditions m * a and n < iit are incompatible '). The relation possesses many properties of its arithmetical counterpart. (3) (m A (a + (m < 41, (nt < n)-+(nt+p nfp), (4) (m n)+(m4 awl, (5)

< <

<

( 6)

e

< < (m < n)+(ntQ < np), (nt < n)->(pm < pn).

(7) Law (3) expresses the transitivity of the relation Laws (4)-(7) express the monotonicity of addition, multiplication and exponentiation with respect to < . As an example we prove (3). Let A , B and C be sets of power m, it and p . By hypothesis, A is equipollent to a subset B, of B and B to a subset C, of C. Let f and g establish the equipollences A B, and B N C , . The composition g o f is one-to-one and maps A onto a subset of C,.Thus iit < p . Q.E.D. The laws of monotonicity do not hold for the relation <: for instance, 2 < a but 2+a = a+a = a - a = 2 - a ; similarly, 2 < 3, but 2a = 3a as will be shown in 96. In the arithmetic of natural numbers the laws converse to (4)-(7) are called the cancellation laws for the relation < with respect to the operations of addition, multiplication and exponentiation; these theorems hold in arithmetic provided that p > 1. In the arithmetic of arbitrary cardinal numbers all of the cancellation laws fail to hold: it suffices to let iit = 2, it = 3 and p = a to obtain a counterexample, On the other hand, the cancellation laws with respect to' addition, multiplication and exponentiation hold for the relation <. They follow without dBiculty from the law of trichotomy which we now state but which we shall prove only in Chapter VIII.

<.

-

l) See W. Sierpidski, Sur une proposition qui entraine l'existence des ensembles non mesurables, Fundamenta Mathematicae 34(1947) 157-162. See also A. Uvy, On models of set theory with urelements, Bulletin de l'Acad8mie Polonaise des Sciences, sbrie des sciences math., phys. et astr. 8(1960) 463-465.

188

V. THEORY OF CARDINAL NUMBERS

For arbitrary cardinals

in

and n either m

< n or n < m

I).

In the remainder of this section we shall treat the question of the asymmetry of the relation < . This problem, investigated already by Cantor but not completely solved by him, became the starting point for a number of interesting investigations regarding transformations of sets'). The asymmetry of the relation < is equivalent to the theorem: (0

(nt

< n) A (n < m)

(m = n).

In fact, if (i) holds, then the formulas m < n and n < m never hold simultaneously; otherwise (i) would yield m = a. Conversely, if the relation < is asymmetric and satisfies the antecedent of implication (i), then necessarily m = n, because otherwise the < sings in the antecedent of the implication could be replaced by <,in contradiction to the assumption of the asymmetry of < . To prove (i) in Z we first prove the following more general proposition.

THEOREM 2 : If A and B are sets and f and g are one-to-one, where f E BA and g E AB, then the sets A and B can be represented as unions of disjoint sets A = Al v Az and B = B1 u B,, where fl(A,)

= B1

and

gl(Bz) = A'.

PROOF.We call the element aE A extendable if a E g'(B) and gc(a)~ f l ( A ) For . extendable a we set a* = fc(gc(a)) and call a* the extension of a. Now we construct a maximal sequence of successive extensions starting with the element a. By n(a) we denote the largest natural number l) Tarski showed in his paper Sur quelques th6or2mes qui kquivalent h I'axiome du choix, Fundamenta Mathematicae 5 (1924) 147-154, that cancellation laws for

< are not only consequences of the law of trichotomy but that they are actually equivalent to it in the axiom system B[TR]. Moreover, they are equivalent to the axiom of choice as well. See Chapter VIII, $6. S. Banach, Un thkorame sur les transformations biunivoques, Fundamenta Mathematicae 6 (1924) 236-239.

5. INEQUALITES BETWEEN CARDINAL NUMBERS

189

such that there exists a sequence of n(a) terms constructed by starting with a and taking successive extensions, provided that such a maximal natural number exists. Otherwise, when for every natural number k there exists such a sequence of k terms, we let n(a) = N. The sequence defined by

vo(4

= a,

V j + l W = V j W * for j s n ( 4

is the desired maximal sequence of successive extensions starting from a. For non-extendable a we have n(a) = 1 and vO(a)= a. If n(a) is finite then we let s(a> = v " ~ a , - l ( 4 .

We define A z = {aEA: n(a) = N V ( ~ ( ~ ~ E g ' ( B ) ) h ( g Q ( S ( a ) ) $ f l ( A ) ) ) . ,

A1 = A - A , ,

B1 =f'(A,),

B,

= B-BI.

For the proof of the theorem it suffices to show that g'(B2) =,A,, i.e. that

(8)

b e 4 -+g(b)EA*,

(9) A , = g'(B,). PROOFOF (8). Assume that bEB2; that is, that b $ f ' ( A , ) and let a = g(b). If b $f'(A) then a is not extendabIe, s(a) = a, and by definition we have a E A,. If b E ~ ' ( A )then b ~ f l ( A , and ) b =f(a'), where a' E A2. Clearly a' =f"(gc(a')), that is a' = a* ;thus a* E A,. If n(a*) = N then n(a) = N and U E A , ; otherwise s(a*) = $(a) and again ~ E A , . PROOFOF(^). Assume that ~ E A ,If. a is extendable then a = g(f(a*)). If at the same time n(a) is finite, then s(a) = s(a*) and a* € A 2 . The same is true i f n ( a ) = N, because then n(a*) = N as well. Thus in both cases a* $ A', f ( a * ) $ f ' ( A , ) and thus f ( a * ) E B2. It follows then that a = g ( f ( a * ) ) Eg'(B2). If a is not extendable, then s(a) = a and by the assumption that ~ E A we , obtain aeg'(B). If a were an element of gl(B1) then, by the definition of the set B,, a would have the form g ( f ( a ' ) ) and would be extendable contrary to the assumption. Thus a Eg'(B,). Q.E.D.

Y. THEORY OF CARDINAL NUMBERS

190

As a corollary we have the Cantor-Bernstein theorem :

-

THEOREM 3: I f m

< n and n d m

then m = n.

PROOF.Let A = m and B = n. Since m < n, there exists a one-to-one function f from A onto a subset of B. Since II m, there exists, sirnilariy, a one-to-one function g from B onto a subset of A. By Theorem 2, A = Al u A,, where A l and A , are disjoint; and B = B1 v B,, where B1 and B, are disjoint and where f ‘(Al) = B, and g’(B,) = A,. Thus A l N B, and A, N B,; hence A B. The Cantor-Bernstein theorem can be generalized as follows, Let R be an equivalence relation on the family 2A and let R satisfy the following two conditions: P

<

-

(10) (11)

XRY

-,f EVY X [(fis a one-to-one function) A z€x A (zR~’(z))J,

(XIn X, = 0 = Y l n Y2)A (XIRYl) A (X,RY,) + (Xi U XzRY1 U Y,).

THEOREM 4: If the relation R with field 2A satisjies for arbitrary subsets of A the conditions (10) and (ll), and i f X stands in the reIation R to some subset of Y and Y stands in the relation R to some subset of X, then XR Y ,). PROOF.Assume that X R Y , where Y 1 c Y , and YRX, where Xl cX. From (10) it follows that there exist one-to-one functions f and g such thatf maps X into Yl and satisfies the condition that Z Rf ‘(2)for every 2 t X ; and similarly g maps Y into X , and satisfies TRg’(T) for all T c Y. By Theorem 2, X = X’ v x” and Y = Y’ u Y”, where X’n X” = 0 and Y’ n Y” = 0 and where Y’ =f ’ ( X ’ ) and X” = g’(Y”). Since X’ c X, we have X’Rf’(X’), that is X‘RY‘; simiIarly X‘RY’’. It then follows by (11) that XRY. Q.E.D. We give two examples of relations satisfying conditions (10) and (1 1): I) The Cantor-Bernstein theorem is also called the Schriider-Bernstein theorem. The first correct proof of this theorem, due to F. Bernstein, was published in the book: E. Borel, Legom sur la thgorie des fonctions (Paris 1898). See also J. Konig, Comptes Rendus de l’Acad6mie des Sciences (Paris) 143 (1906). z, Theorem 4 and the examples below are due to Banach; see his paper cited on p. 188.

5. INEQUALITIES BETWEN CARDINAL NUMBERS

191

1. The relation of equipollence on subsets of A. Theorem 4 for this relation is then identical with the Cantor-Bernstein Theorem. # 2. Let A = €". We call two subsets X and Y of A equivalent by finite decomposition if there exist a natural number k and two sequences X,, ...,Xk-l and Yo,... , Yk-lsuch that

Xi nX,

= 0 = Yi nY ,

for

0


Xi and Yi are isometric for i < k. In case X and Y are equivalent by finite decomposition we write X -fin Y.

THEOREM5 : The relation-fin satisfies conditions (10) and (11) and is an equivalence relation on 2A. We omit the proof. # We prove one more theorem about transformations. Like Theorem 2 it is a generalization of the Cantor-Bernstein Theorem. THEOREM 6: (MEAN-VALUE THEOREM.) Let A, B, C, A' and B' be sets such that A 3 C 3 B, A' 3 B', A N A' and B N B'. Then there exists a set C' such that A' 2 C' 3 B' and C C' ').

-

PROOF.It suffices to prove the existence of a function h from A into A' satisfying the following conditions: (12)

the restriction hlC is one-to-one,

(13)

h'(C) 3 B'.

In fact, if h satisfies (12) and (13), then the desired set C' is hl (C). We define h as follows: h(x) =

f(x)

for

XEA-X,

I) This theorem is due to A. Tarski and A. Lindenbaum. See Communication sur les recherches de la thiorie des ensembles, Comptes rendus des skances de la SociCt6 des Sciences et des lettres de Varsovie, Classe 111, 19(1926) 299-330, Theorem 15 on p. 303.

192

V. THEORY OF CARDINAL NUMBERS

where-for the present-X is an arbitrary subset of B and where f and g are one-to-one functions such that f'(A) = A' and gl(B') = B. The function h defined in this way certainly satisfies condition (12) provided that g - ' ( X ) n f ' ( C - X ) = 0. (14) In fact, if h satisfies (14) then h(x') = h(x") holds for no x' and x" such that x' E X and x" E C - X . The function h satisfies (13) if besides satisfying (14) it also satisfies g-'(X) Uf ' ( C - X ) 3 B', (15) because h'(C) = h l ( X ) u h l ( C - X ) = g - ' ( X ) u f ' ( C - X ) . Both condition (14) and condition (15) hold if g'(B'-f'(C-X)) = X. (16) To complete the proof it suffices to show that there exists a set X c B satisfying (16). For this purpose we let F(X) = g'(B'-f'(C-X)) and we notice that X i c X, c B + F(X1) c F(X2) c B.

A function F from 2B into 2B will be called a monotone function on the subsets of B if it satisfies the condition above. Therefore it suffices to prove the following lemma: For every monotone function on the subsets of a given set B, there exists a set X such that F ( X ) = X. We construct X as follows: let K = { X c B: F ( X ) c X } . The family Kis non-empty because B EK. We shall show that the condition F ( X ) = X is satisfied by the set Xo= f X. l XEK

In fact, X E K + Xoc X and thus by monotonicity X E K F(Xo) c F ( X ) c X . It follows that F(Xo)c X for every X E K , and hence F(X0) c

n X =XO,so that F(Xo) c X,. By

XEX

monotonicity it follows

moreover that F(F(Xo)) c F(Xo),and thus P(Xo)E K , so XOc F(XO). We have F ( X o )= X,. Q.E.D. The Cantor-Bernstein Theorem is a consequence of Theorem 6. In fact, if m < n then there exist sets X, Y such that x'= m, Y = n and

-

5. INEQUALITIES BETWEN CARDINAL NUMBERS

193

c Y. If we assume moreover that n < m, then Y contains a subset Z of power m. Letting in Theorem 6 A' = B' = B = Z , A = X and C = Y we conclude that there exists a set C' such that A' 3 C' 3 B' and C' C. Thus C' = Z and C' Y, so Z Y,and consequently

X

m = n.

-

-

-

Exercises 1. We say that a cardinal number n absorbs m if m+n = n. Show: (a) n absorbs m if and only if a .m n; (b) if n absorbs m, then every cardinal larger than n absorbs m; (c) n absorbs m if and only if n absorbs k . m ( k EN); (d) n absorbs m if and only if n absorbs a.m. (Tarski) 2. Without using the axiom of choice show that m < a = m <*a. 3. Without using the axiom of choice show that 7(2'" <*m). (Tarski) 4. Show that 2m> a + 2m> 2'. 5. Show that the closed circle T is equivalent by finite decomposition to the union T u F, where F is an arbitrary line segment disjoint from T l).

<

fi 6. Properties of the cardinals a and c We introduce the following notation: c = 2". The cardinal c is called the power of the continuum. We often meet the cardinal c, as well as the cardinal a, in many parts of set theory and its applications. We shall prove several formulas concerning the numbers n (natural numbers), a and c. (1) c = c+c. In fact (see p. l85), c + c = 2c = 2.2' = 2"' = 2" = c, because = a. (2) n
l+a

In fact, by (2) we have (see (4), $ 5 , p. 187) c < n+c < a+c < c+c, l) For a more detailed discussion of the problems mentioned in $ 5 , see A. Tarski, Cardinul Algebras (New York, 1949), in particular Chapters 1 , 2 and 17.

194

V. THEORY OF CARDINAL. NUMBERS

and by (1) c

< n+c

\< a+c

< c.

Applying Theorem 3, $5, p. 190 we obtain (3). (4)

c = C.C.

Indeed, c = 2" = 2"'" = 2" -2" = c . c, because a+ a = a (see (3), 0 4, p. 183) (5) n.c = a - c = c (for n > 0). By (2) and by (5), 0 4, p. 183 we have the inequalities c \< n - c < a. c < c. c; hence in view of (4)we obtain (5) by applying the Cantor-Bernstein Theorem. By induction from (4) we obtain

(6)

c"

(for n

=c

=c

na = a" = ca (7) In fact (see (lo), p. 185),

c = 2"

< n" \< a" < C"

> 0).

(for n

> 1).

= (2")" = 2"'" = 2" = c,

whence (7) follows by applying the Cantor-Bernstein Theorem. The following equations concerning the number f = 2C are proved similarly:

n+\ = a+f (8)

= c+j

= f+f = f,

n.f=a.f=c.f=f.f=

nC=aC=cC=

f

7 -f C -

(for n > O), (for n > 1).

We shall give examples of sets of powers c and 2'. THEOREM 1: The Cantor set C has power c.

PROOF.C = {O,l>N, so

= 2" by Theorem

7 , p. 185.

THEOREM 2: The set of all infinite sequences of natural numbers has power c.

PROOF.This set is NN.Therefore its power is

aa =

c.

#THEOREM 3: The following sets have the power of the continuum: (a) the set of irrational numbers in the interval (0,l); (b) the set of all points in (0,l) ;

6. CARDINALS

a

(c) the see € of all real numbers; (d) the set of all points of the space

AND

c

195

P,where n is a natural number.

PROOF,(a) follows from the remark on p. 138 and from Theorem 2. (b) follows from the observation that the interval (0,l) is the union of the countable set of rationals in (0, 1) and the set of irrationals in

(0, l), which has power c. (c) holds because function y = 1/2

1 +x-arctgx

is one-to-one and maps the set & onto the interval (0, 1). (d) follows from (c) and equation (6). =i+ e

-

c.

THEOREM4: If A = c, B = a and B c A , then A - B = c. PROOF.By (6) we have that A x A N A; therefore it suffices to show that if M is a countable subset of A x A, then the difference A X A --M has power c. The projection onto A of the points in M constitute at most a countable set, which implies that there exists an element of A which does not belong to the projection of M . The set {(a, y ) : y E A} is disjoint from M and has power c, thus the difference A x A --M has power 2 c. O n the other hand, this set has power c as a subset of A x A. Thus the difference A x A --M has power c by the Cantor-Bernstein Theorem.

<

# COROLLARY 5 : The set of transcendental numbers has power c. PROOF.It suffices to apply Theorem 4 to the case where A is the set € of real numbers and B is the set of algebraic numbers. This corollary, proved by Cantor in 1874, was one of the first applications of set theory to concrete mathematical problems.

THEOREM 6: The set power c. PROOF.

eN of

injinite sequences of real numbers has

= ca = c by (7).

THEOREM 7: The set of continuous functions of one real variable has power c. PROOF.Let r,, r,, ...,r,, ... be an enumeration of all rational numbers. With every continuous functionfof one real variable we associate the sequence of real numbers

(9)

f(r,>f , (r2)

9

-

**

,f (r"),

* *

V. THEORY OF CARDINAL NUMBERS

196

Iff and g are distinct then the corresponding sequence?

f

(h)Y

f (4, ,f @"I, ***

*** Y

g(r,), g ( r A *

-

*

Y

g(rn), * * *

are also distinct. In fact, f # g implies that f ( x ) # g ( x ) for some x ; so if r k , is a sequence of rationals converging to x , then it is not true that f(rk,) = g(rk,) for every n, because in that case, by the continuity o f f and g , we would have

f (x) = lim f (rk,) = limg(rkn)= g ( x ) . n- w

,= w

Thus the set of continuous functions of one real variable is equipollent to the set of sequences (2), which has power c by Theorem 4. On the other hand, the set of continuous functions has power >, c because it contains all constant functions. Thus by Theorem 3, § 5 we obtain Theorem 7.

<

THEOREM 8: The set

&& of all functions of one real variable haspo-

wer 2C. E

PROOF. CCc= cc = 2C by (8). Exercise Prove that the family R of closed sets of the space € has power C. Hint:To prove that 3 C associate with every X E R a family of intervals with rational endpoints disjoint from X and show that the set of all such families has power C. The inequailty ?i 2 C holds because all one element sets belong to R. #

<

Q 7. The generalized sum of cardinal numbers Let T be an arbitrary set, f a function defined on T with cardinal numbers as values. Instead of j(x) we shall also write fx. Assume that the function f satisfies the following condition

there exists a set-valued function F(O) defined on T such that FLo)= fx f o r all x E T. Condition (W) can easily be shown to hold for many functions f. Such is the case, for instance, when has only finitely many distinct values. We shall show in Chapter VII, p. 268 that every f satisfies condition (W), so that condition (W) does not actually affect the generality of our treatment.

197

7. GENERALIZED SUM OF CARDINAL NUMBERS

THEOREM 1: There exists a set-valued function F defined on T such

-Fx

that (1) ( 2)

for

= fx

Fr n Fy = 0

XET,

x # y.

for

Moreover, if F ( I ) and F(') both satisfv (1) and (2), then

u F:')- u Fi'). X

X

PROOF.For X E Tlet Fx = F(j)X { x } ,

where F(O) is any function satisfying (W). If x # y then Fx n Fy = 0, because the set F, consists of ordered pairs with second component x and Fy of ordered pairs with second component y. Moreover,

1.

=

F)

Thus F satisfies conditions (1) and (2). Assume now that functions F") and F(') satisfy conditions (1) and (2). For every x E T the set Qx of one-to-one functions from Fill onto F:') is non-empty. If x f y then Q. n Qy = 0, because every function belonging to QX has domain Fx and therefore is different from every function belonging to Qy. By the axiom of choice it follows that there exists a set Y containing exactly one element in common with each of the sets Ox. Let va be the only element of Y n Qx; then vX is a one-to-one function from the set F$I) onto the set Fi2). It is now easy to show that the function f = (px maps the union =

u

u Fi') onto u F:') X

xeT

in a one-to-one manner. This completes the proof.

X

u

DEFINITION: The sum of the cardinal numbers 1. for x E T is the cardinal F,, where F is any function satisfying (1) and (2). O

X

We denote this sum by xeT

1.

or by

c :.f

The definition is correct since the number

u

Fx does not depend

X

on the choice of the function F satisfying conditions (1) and (2) and

198

V. THEORY OF CARDINAL NLrmBERS

since such a function always exists. However, we cannot prove the existence of such a function without appealing to the axiom of choice, so that the definition of the sum of an arbitrary set of cardinal numbers is based upon the axiom of choice I). If T = {1,2), then It = fl+f2. If T = N, then we shall aIso write tcT

jO+f,+jZ+b+

5

or

* a *

fn

n=O

and speak of the sum of a series of cardinal numbers.

'THEOREM 2: (GENERALIZED COMMUTATIVE ,f = j+). permutation of the set T , then For the proof it suffices to notice that

LAW.)

c

C

Ifp is an arbitrary

_ _ i ( -

,

C fx = U Fx = U Fcp(x)= C fp(xp

where F is any function satisfying conditions (1) and (2). The equations hold on the basis of Theorem 3, Chapter IVY0 1, p. 115 and formula (3).

'THEOREM 3: (GENERALIZED ASSOCIATIVE

LAW.)

If T =

u Tywhere

Y €1

the sets Ty are disjoint, then

C r x = yCe l (xC rxh eTy

XET

j,, that is gY =

PROOF.For y ~ 1 let , g, = x;TY

u F,. Then xeTy

by Theorem 2, Chapter, IV, p. 144. It follows that

which proves Theorem 3. l) The fact that the definition of the sum of an infinite sequence of cardinals requires the axiom of choice was pointed out by W. Sierpidski in the paper: Sur I'miome de M. Zermelo et son rdle dam la Thbrie des Ensembles et Analyse. Bulletin de l'Acad6mie des Sciences de Cracovie, C1. Sci. Math. et Nat., Series A (1918) 112.

7. GENERALIZED SUM OF CARDINAL NUMBERS

'THEOREM 4: (GENERALIZED DISTRIBUTIVE LAW

199

FOR MULTIPLICATION

WITHRESPECT TO ADDITION.) The equation

(c

7x1 *

m

c

=

m)

(fx

holds for every cardinal m. PROOF.Let

5 = nt.

We have

(cix)nt = ruFx)xM X

z(fx.in)m.

and

=

X

X

X

At the same time (see Exercise 2, Chapter IVY0 1, p. 120)

(UFx)xM=

u

(FXXM).

X

X

'THEOREM 5: If gx < fx for x E T, then

c gx < c ix. X

X

PROOF.Let K, be the family of those X c Fx which have power fix. By assumption K, # 0 for every x E T. It follows that there exists a function G (see Theorem 3, p. 133) such that E K,,~

G

that is

GxeKX.

XET

Therefore G , c F, and

which proves that X

'THEOREM 6:

zx= gx. This implies

gx

< c ix. X

If S c T then

PROOF.Let XES,

for

By Theorem 5,

'THEOREM7:

If

fx r=

m for all

c

fx XET

XE

=

T, and n

rn-n.

- then

= T,

200

V. THEORY OF CARDINAL NUMBERS P

PROOF.Let M = m. For every x there exists a one-to-one mapping f of the set F, onto M ; let @, be the set of all such functions. By the axiom of choice there exists a set !P containing exactly one element from each of the sets @., Let fx be the unique element of Y n @., For t~ F, we let

u X

90)= , where x is the (unique) element of T such that t E F,. F, onto M X T and is one-to-one. The function f maps the union In fact, if v (tl) = and v(tZ)=
u

u

N

fx

< m for x E T and n = F,

c

< ma. PROOF. Letting rn, = m for X E T, we have by Theorem 5 c ,f < Em,, 'THEOREM 8: If

and by Theorem 7

m,

then

fx

= men.

Examples 1. We shall calculate the sum k,, where k, is a natural number and n runs through the set of natural numbers. For this purpose we notice that from Theorem 7 it follows that (4)

1+1+1+ ... = 1.a = a,

a+a+a+

... = a - a = a ;

since, by Theorem 5 , m

l+l+l+

... < C k , < a + a + a + n=l

we have by the Cantor-Bernstein Theorem

In particular

+ ... = a,

2 + 2 +2 1 + 2 +3 1!+2!+3!+

+ ... = a,

... = a.

...)

201

g. GENERALIZED SUM OF CARDINAL. NUMBERS

2. It follows from equation (4)that the union of a countable number of countable sets is countable (see Theorem 8, Q 2, p. 176). 3. By Theorem 7, c+c+cf ... = c - a =: c. Similarly, the sum

zfx

XET

where the power of T is c and where each to c - C , that is to c.

rx equals c is itself equal

0 8. The generalized product of cardinal numbers As before, we shall use the axiom of choice and assume that f is a function having cardinal numbers as values and satisfying condition (W) given on p. 196. O

-

THEOREM 1: If the functions F ( * )and T, then

satisfy the condition

= fx = Fi2)for X E

PROOF.As in the proof of Theorem I , § 7, using the axiom of choice, we show that there exists a set which for every X E T contains exactly one function plX which is one-to-one and maps F i l ) onto Fi2). With each function fi E we associate the function fz,where

n

xg T

L ( t ) = vt(fi0)) for t E T. Since fi( 2 ) E Fill for every t E T, p t (5(t)) E y ; ( F i l ) ) = Fr(z)twhich proves Fi2). that f i ( t ) E F j 2 ) ,that is fi E (1)

n

xeT

Iff; # fI'then, for some t , f;( t ) # f ;'(t). Thus, since plt is one-to-one, vt(fl(t)) # vt(f;l(t)), that is f i ( t ) # f y ( t ) . Therefore the correspondence between fi and fi is one-to-one. Finally, if f 2 E Fi2) then the function fi defined by the equation A ( t ) = vF(fi(r))

belongs to nFil) and satisfies condition (1). Thus every function Fi2)corresponds to some function belonging to Fill. belonging to

n

n

V. THEORY OF CARDINAL NUMBERS

202

Thus we have proved Theorem 1. This theorem leads to the following definition. DEFINITION: The prorhsct of the cardinal numbers fx is the power of the Cartesian product Fx , where F is an arbitrary function such

n X

that FX= fx; that is,

n n Fx,

9

where

fx =

F, = f x .

x;T

XET

Just as for generalized sum, the use of the notion of generalized product rests upon the axiom of choice, without which we cannot prove Theorem 1 which is the basis for the definition of product. If T = { 1,2}, then ft = fl fz. For this reason we write To. f l ...,

n

-

t€T

n

-

UD

or

fn

when T = N.

n=O

From Theorems 2,3 and 4 proved in $1, p. 171-172 we obtain directly the commutative, associative and distributive laws:

(where 9 is a permutation of T),

n( n

YEW xsTy

fx)

= n f x xsT

u Ty and TylnTy2 0 for y1 # y z ) ,

(where T =

=

YEU

X

X

If all of the values of the function f are identical, then multiplicatien coincides with exponentiation; that is, (2)

if

fx =

fo for X E T and t =

F, then nix= 1.: X

By equation (13), Theorem 3, tion of sum that (3)

0 1, p. 172, we derive from the defini-

203

8. GENeRALIZBD PRODUCT OF CARDINAL NUMBERS

Finally we note without proof that analogously to Theorem 5, $7. Examples 1. In (2) let fo= a and T = N. Since aa = c (see (7), $ 5, p. 194), we obtain a - a - a . ... = c.

2. In (2) let fo = 2 and T = N. From the equation 2' = c we have 2 . 2 . 2 . ... = c. 3. From (4) we have 2 . 2 . 2 . ... < 2 . 3 . 4 . 5 - ... and 2 . 3 . 4 . 5 . ... .... Thus by Theorem 3, $5, p. 190 we conclude that 1 . 2 - 3 - 4 .... = C.

<

a.a.a-

In the same way it can be shown that k,.k2.k3... = c, if for all n, k, > 1. "THEOREM 2: (J. Konig')).

If 4. <,,f

for all ~ E then T

PROOF.From the assumption that (W) holds (p. 196) it follows that there exists a function F such that Fx= f,. We may assume, moreover, that F, n Fy = 0 for x # y (see Theorem 1, $7, p. 197). Arguing as in the proof of Theorem 5, $7 we conclude that there exists a function G such that G, c F, and 5, = 4,. It follows that G, n Gy = 0 for x # y and that F,-G, # 0. We show first that (5)

cgx 6njx. x

x

Let f be an arbitrary function belonging to

n(Fx-Gx). Such a funcX

tion exists by Theorem 3, Chapter IV, p. 133. For every a E

u G, let X

for for

a$G,, aEG,.

')See J.K6nig, Zum Konfinrcumproblem,Mathematische Annalen 60 (1904) 177-1 80.

204

V. THEORY OF CARDINAL NUMBERS

n

Clearly, fa E F,. If a f b then fa #.fa, because a and b either belong to different sets G,, G, and thenf,(x) == a € G x , f b ( x )= f ( x ) e F X -G,, and thus f.( x ) # fb (x) ; or else a and b belong to the same set G, and then&@) = a # b =fa(x). Thus the functions f. constitute a subset of the Cartesian product Fx equipollent to the union G,, which proves formula ( 5 ) .

n

u

X

It remains to show that

c

gx

X

#

II1., X

For this purpose we observe that every set S equipollent to the union

u Gx can be considered as a union of disjoint sets: S u H,, where H, gx. X

P

=

=

X

Assume that S c

nF,;

let h E H,.

Thus for every t, h(t)E F ~ ,

X

and in particular h ( x ) ~ F , .Hence if h ranges over the set Hx, then the elements h(x) form a set Kx contained in F,, and so (see Theorem 1, 4 5, p. 186) i3], fix= g, < It follows that Fx--Kx # 0 for every x, and hence (Fx-Kx) # 0.

<

n:

E.

Let ~ E ~ ( F , - K Hence ~ ) . pl(x)#K,, and v $ H x , because h ( x ) K, ~ X

for all h belonging to H,. This implies that the function v belongs to none of the sets H , contained in the union S ; that is, v # S and it follows that S # F,, which proves (6).

-

X

Letting in Konig’s theorem g, = 1, fx = 2, and m = T we obtain Cantor’s inequality m < 2m ((2), 4 5, p. 186). Thus Konig’s theorem is a generalization of this inequality.

COROLLARY 3: If mn < m,,, m

for n = 0,1,2 ... and m, > 0, then m

PROOF.By Konig’s theorem

mo+m,+m,+

... < m,.m2*1tt3...,

205

8. GENERALIZED PRODUCT OF CARDINAL NUMBERS

whence

nto+m1+m2+

... < nto.m,-m2 ..

COROLLARY 4: For no cardinal n can nu be represented as the sum of an injinite strictly increasing sequence of cardinal numbers. PROOF. Let n a p. 203)

= mo+nt,+mz+

... . Hence mp < nu

m

and (see (4),

03

By Theorem 3 the sequence mo,ml, ntz, ... cannot be increasing. Thus in particular neither c nor 2c is the sum of an infinite increasing sequence. On the other hand, the number a+2a+2ZU+

...,

and more generally the number

n + 2 ~ + 2 2 ~ +... , cannot be written as a single cardinal raised to the power a.

CHAPTER VI

LINEAIWY ORDERED SETS

$1. Introductioll

The notion of linearly ordered set was introduced in Chapter 11, 81. A linear ordering is also called a total, complete or simple ordering I). The types of relational systems will usually be denoted A(a1though it would be more proper to denote it by E).

8 9, p.

Examples 1. For y, lye N N let q < 1y if @ = y or if the least n such that pl,, # y,, satisfies the conditions rp,, < yn. The relation < linearly orders the set A". If A c N N , then the relation < restricted to A linearly orders the set A. # 2. The set A consisting of all natural numbers of the form 2" is linearly ordered by the divisibility relation, that is, by the relation

{:

(~EA)A(~E:A)/\V(~=~~)} k

3. The set N x N is linearly ordered by the relation R which holds between pairs (m,n} and (p, q } if and only if ( 2 m f 1)/2"

< (2P+ w.

This relation is isomorphic to the relation numbers of the form (2m+1)/2". I)

< in

the set of real

The notion of linearly ordered set is due to Cantor. See Beitrage zur Be-

griindung der transfiniten Mengenlehre. Mathematische Annalen 46(1895) 481-512.

207

1. INTRODUCTION

4. Let P(m) assert that m is an even number. The set N is h e arly ordered by the relation

{(m,n): [p(m)Ap(n) A (m < n)] V [p(m)A

7P(n)] v [l P ( m ) A 1f‘(4 (n < m)]}.

In this ordering every even number precedes every odd number. Of two even numbers, the smaller precedes the larger; of two odd numbers, the larger precedes the smaller. Schematically this ordering can be illustrated b y the sequence: 2,4,6,8, 10,

...

..., 9,7,5,3, 1,

where every number preceding a number x is written to the left of x. 5. The set of complex numbers is linearly ordered by the relation {(x,Y>: “4 < R(Y)l v [R(x) = R 0 1 A [W) < I(Y)l). In this ordering a number x precedes a number y if the real part R(x) of the complex number x is less than the real part RQ) of y. In case the real parts of x and y are equal, x precedes y if the imaginary part I ( x ) of x is less than the imaginary part I Q of y . 6. Let a(n) be the number of distinct prime factors of the natural number n. The set of natural numbers is linearly ordered by the relation: {(x,y>: [a(x) < a Q 1 v [W = .@)I

A

(x
7. The set of concentric circles is linearly ordered by the inclusion relation. #

DEFINITION: An element x is said to be a jirst element of the linearly ordered set A (with respect to the relation R) if x R y for all Y E A . On the other hand, if y R x for all y, then x is said to be a last element of A (with respect to R). Generally speaking, not every set has a first or last element; but if such element exists, then it is uniquely determined. THEOREM 1: In a finite non-empty subset X of a linearly ordered set A there is a Jirst element and a last element.

208

VI. LINEARLY ORDERED SETS

PROOF.The proof is by induction on the number of elements of X . If X has only one element, then the theorem is obvious. Suppose that the theorem holds for subsets with n elements. Let X = ' I u (a} where a $ Y and Y has n elements. Let bl be the first and b2 the last element of Y . Since A is linearly ordered, either a precedes b, or bl precedes a. That element which precedes the other is clearly the first element of Y. Similarly we show that one of the elements a ar-d b2 is the last element of X. In the case of linear order relations we usually speak of similariry of relations instead of their isomorphism. The following theorem shows that the definition of isomorphism can be simplified in the case of similarity. Namely, instead of proving that two formulas x R y a n d f ( x ) S f ( y ) are equivalent it suffices to prove only the implication (1) below. THEOREM 2: In order that two sets A and B linearly ordered respectively by reIations R and S be similar it is necessary and suficient /hat there exist a one-to-one function f which maps the set A onto the set B so tha! x RY f (4S f (Y) (1) for all x , y ~ A . PROOF.Clearly it suffices to show that if x , y E A and f ( x ) S f ( y ) , then x R y . Suppose the contrary: 1( x R y ) . Since the relation R is connected in A , we have either x = y or y R x . In the first case x R y as the relation R is reflexive in A, but this contradicts the hypothesis 1( x R y ) . In the second case, (1) implies f ( y ) S f ( x ) and therefore f ( x ) =f ( y ) because S is antisymmetric. Since f is one-to-one, we infer that x = y , which again contradicts 1( x Ry ) . Hence the theorem is proved. Similar sets are clearly equipollent. The converse theorem holds for finite sets only. THEOREM 3: Two jinite linearly ordered equipollent sets are similar. PROOF.Suppose that sets A and B, linearly ordered by relations R and S, respectively, have n elements. For n = 0 the empty function satisfies the conditions of Theorem 2, consequently it establishes the isomorphism between the relations R and S. +

1. INTRODUCTION

209

Now suppose that Theorem 3 holds for Fets of n elements and let A and B have n + l elements. Let a be the first element of A and b the first element of 3. By assumption, there exists a function fi which establishes similarity between the sets A - { a } and 3-{b}. Let f =f1 v {
In this case we write x < R ~(or x .< y if there is no confusion about the relation R). We also write y > R or ~ y > x. We say that y lies between x and z if x
or

x>y>z.

If x E A and the set { y : x < y } has a first element, then this element is called a direct successor of x (with respect to R). The last element of { y : y < x } (if one exists) is called a d:rect predecessor of x. Each element X E A possesses at most or-e direct successor and at most one direct predecessor. A proper subset X of the set A is said to be an initial segment (a Jinal segment) if x E X implies that every element preceding x belongs to X (every element after x belongs to A’). The set X c A is said to be an interval if the condition x, y E X implies that every element lying between x and y belongs to X . Let o,(X) = { y : (J’RX) A 0) # x)} = {J’: y < X}. The subscript R will sometimes be omitted.

210

M. LINEARLY ORDERED SETS

It is easily seen that O,(x) is an initial segment. However, not every initial segment is of the form OR@). We say that an interval X of a linearly ordered set A precedes an interval Y of A if (XEX)A(yEY)+x
Every family of disjoint intervals is linearly ordered by the relation “X precedes Y or X = Y”. Exercises 1. Let M be a family of subsets of a set Z such that (i) M is a monotone family, (ii) M is not included in any monotone family of subsets of Z different from M. Prove that the relation defined by the equivalence xRy

= {(x

=y) V

‘d [(x $ E ) A ( Y E E ) ] )

EEM

linearly orders the set Z. The family of final segments of this set is identical with M l). 2. Let M be a monotone family of subsets of a set Z. Prove that the family of all sets of the form

uX and n X,where

X d

5 2.

S c M, is monotone.

XES

Dense, scattered, and continuous sets

A set A is said to be densely ordered by an order relation < if for any two elements x , y E A there exists an element z E A between x and y. We say then also that A is dense. In a densely ordered set no element has either a direct successor or a direct predecessor. This property is characteristic for densely ordered sets. In fact, if no element of the set A has a direct predecessor and x, Y E A, x < y, then x cannot be the last element of the set { z : z < y } , for then x would be a direct predecessor of y. Thus there exists z such that x < z < y. Hence the set A is densely ordered. All one-element sets, as well as the empty set, are densely ordered. All other densely ordered sets contain infinitely many elements. A set which is not densely ordered may contain a densely ordered l) See K. Kuratowski, Sur la notion de I’ordre dans la thgorie des ensembles, Fundamenta Mathematicae 2 (1921) 161-171. This theorem indicates the possibility of replacing the theory of linearly ordered sets by the theory of monotone families of sets.

2. DENSE, SCATTERED, AND C O " u o u s

ORDERINGS

21 1

subset. For instance, the set consisting of all positive real numbers and negative integers ordered by the relation <, is not densely ordered, because no element of this set lies between -2 and -1. However, this set does contain a densely ordered subset, namely, the set of positive real numbers. A linearly ordered set which contains no infinite densely ordered subset is said to be sca!tered. For instance, the set of integers and the set composed of all fractions l / n (n = f 1, & 2, ...) are scattered if the order relation is <. Every subset of a scattered set is scattered. THEOREM 1; If A and B are two scattered subsets of a linearly ordered set M, then the union A u B is also scattered.

PROOF.Suppose that there exists an infinite densely ordered subset C o f the set A u B . Since C = ( C n A ) u ( C n B ) , either C n A or C n B is infinite. Let C n A be infinite. Since this set is not densely ordered (as a subset of the scattered set B), there exists a pair al, a, of elements of the set C n A such that al < az and such that no element of C n A lies between a, and az. This implies that for every X€C

(al < x < az) + X E B . Let B1 = C n { x : al < x < az}. This set is infinite, for there are infinitely many elements of C between al and a2. If xl, xze B1 and x1 < xz, then there exists x E C lying between x1 and x, thus x E B,. This implies that the set B1is densely ordered. By (I), Bl c Bywhich means that Blis an infinite densely ordered subset of B. This contradicts the assumption that B is a scattered set. Hence Theorem 1 is proved. A set X contained in a linearly ordered set A is said to be densely ordered in A if, for every two elements x and y of the set A , there exists an element z of X lying between x and y . For example, the set of rational numbers is densely ordered in the set of real numbers, where the order relation is It is clear that if a set X is densely ordered in A , then the sets A and X are both densely ordered. Of course, two sets X and Y densely ordered in A which have neither first nor last elements are always cofwal and coinitial. If a set X is (1)

<.

212

VI. LINEARLY ORDERED SETS

co$nal with Y and Y is coJinal with Z, then the sets X and Z a r e also cojinal. A similar law of transitivity holds for coinitial sets. Let ( X , Y ) be a cut in a linearly ordered set A. The intersection X n Y contains at most one element. In fact, if x, y E X n Y , then x 5 y and y 5 x ; thus x = y. If X n Y = 0, then we say that the cut ( X , Y ) determines a gap in the set A . If X n Y = a, then we say that the element a lies in the cut ( X , Y ) . It can be easily shown that in this case X = {a}-, y = {a}+ and a = x = y . A cut ( X , Y } is said to be proper if

v

X

A

JGY

J

X#O#Y. A set A is said to be continuously ordered if no proper cut in A determines a gap in A. We also say that A is continuous. If (X,, Y,) and (X,, Y,) are cuts in A, then either X , c X z or Xz t X I . In fact, suppose that a e X , - X , and let b E X 2 . By connectedness, it follows that b precedes a, for in the opposite case we would have n E XZ. Thus b E X , and we obtain X z c XI. This implies the following theorem. THEOXEM 2: The minimal extension V :see p. 160) of a linearly ordered set is continuously ordered. In fact, 9 is a complete lattice and, as has been shown before, the ordering in 9 is connected; thus it is linear. The complete linearly or
el

u x

a and b belong to the same component F, and aR,b, l)

This corollary is due to Dedekind. See the reference on p. 160.

213

2. DENSE, SCATI'ERED, AND CONTINUOUS ORDERINGS

or a and b belong to different components Fx1 and F,, and xlQxz. Then the relation S linearly orders the union

uF,. X

PROOF.The reflexivity of S is obvious. If a and b belong to different components of the union F,, then either aSb or bSa, because the relation Q orders the set of indices. On the other hand, if a and b belong to the same component F,, then either aSb or bSa, because the relation R, is connected in F,. Thus the relation S is also connected. If a E F,,, b E F,,, a Sb and b Sa, then xlQxz and xzQxl. This implies that x1 = x2, since the relation Q is antisymmetric. Finally, suppose that a E F,, b E F,,, c E F,, a Sb and bSc. These conditions imply x Q y and y Q z ; thus X Q Z . If x # z, then aSc. On the other hand, if x = z, thm we have x = y = z, because x Q y , y Q Z and the relation Q is antisymmetric. By the definition of S we obtain b R, c. Since the relation R, is transitive, we obtain a R,c and thus aSc. Hence the relation S is also transitive. Thus the relation S linearly orders the set U F , . In the following, by the ordered union of linearly ordered disjoint F, ordered by the resets F, we shall always understand the union

u

u X

lation S defined in Theorem 4. In this case the relation Q ordering the set of indices is assumed to be fixed. We say also that F, is the union of sets F, over the indexing set T. X

Let A be an arbitrary set, linearly ordered by the relation R. For x , y E A let [x,y ] denote the set of those z which equal either x or y , or which satisfy one of the conditions x < z < y or y < z < x . Clearly [x,y ] = p,XI. We shall prove that for arbitrary x , y , z E A

In fact, if t E [x,y ] and t = x or t = y , then clearly t E [x,z] u [z, y]. If x < t < y and t = z or t < z, then t E [ x , z ] .On the other hand, if z < t, then t E [z,y]. Similarly, if y < t < x , then t E [x, z] u [z,y].

214

VI. LINEARLY ORDERED SETS

Let V , be the set of all y such that the set [x,y] is scattered. Clearly V, # 0, because x E V,. We shall prove that the set V, is also scattered. Suppose that on the contrary C c V, and that the set C is infinite and densely ordered. For any cl, c2E C such that c1 < cz we have by (2) [Cl,

c2l

= [Cl, X I u [cz, XI.

Thus the set [cl, cz] is contained in the union of two scattered sets, By Theorem 1, p. 211 this union is scattered. But this is impossible, for between any two distinct elements of the set C there always lies at least one other element of C. Thus the assumption that V , is not scattered leads to a contradiction. The set V, is an interval in the set A. In fact, suppose that y, z E V , and y < t < z. If x =: t or x < t , then [x, t ] c [x, 21. Thus, as a subset of the scattered set [x,z], the set [x,t] is also scattered. On the other hand, if t < x, then [t,x] c p, x]. This means that the set [t, x] is scattered as a subset of the scattered set b,XI. It follows now that t f V,. Thus the set V, is an interval. If x # y , then either V, n V , = 0 or V, = V,. In fact, if z E V , n V,, then the set [x, y ] , as a subset of the union [x, z] u p,z] is scattered. It follows that if u E V,, then the set [u, y ] is scattered because [u,y] c [u, x] u [x, y]. Similarly, from u E V , it follows that u E V,. Thus V, = V,. Let 4 be the family of all the sets V,. This family consists of disjoint non-empty subsets of A and is linearly ordered by the relation e which holds between V, and V , if and only if V , = V, or V, precedes V , (see p. 210). The set A is the ordered union A=UP, PEA

where the family A is linearly ordered by the relation e and each interval P is linearly ordered by the relation R. In fact the union u ( A ) is contained in A and every element x of the set A belongs to V,; thus every x also belongs to one component of this union. We now prove that the relation e is a dense ordering of the family A . Suppose that VxeV, and V, # V,; that is, V , n V , = 0. This implies

215

2. DENSE, SCAlTERED, AND CONTINUOUS ORDERINGS

that the interval [x,y ] is not scattered, that is, for some z lying between x and y one of the sets [x,z] and [z, y], say the first, contains an infinite densely ordered set M. If m,n,p are elements of the set M such that x < m < n < p < z, then the sets [x,n] and [n,y] contain infinite densely ordered subsets. This shows that V,eV,,eV, and V , # V,, # V,,. It this way we obtain the following theorem. THEOREM 5 9 : Every ordered set is the union of scattered sets over a densely ordered indexing set. Exercises 1. Give an example of an infinite set which has a first element, has no last element, and in which every element except the first has a direct predecessor. Moreover, this set should not be similar to the set of natural numbers.

2.Show that if the set X is densely ordered and if sets XIand Xaare continuously ordered and contain subsets dense in themselves and similar to X, then the sets XI and Xa are similar. 3. Prove that the set 0 of those relations R E 2N2 which densely order their fields is a Ga-set (in the space 2”). 4. Show that the union of scattered sets over a scattered indexing set is itself scattered. 5. Prove that if the sets Fx contain neither first nor last elements, and if they are

infinite and densely ordered, then the union

u

Fx is densely ordered (for any or-

xfT

dered set T). 6. Prove that if a set T is infinite, densely ordered and Fx set

+ 0 for each x , then the

u Fx contains a densely ordered subset. Prove that if the union uFx is continuous, then the set T has no gaps.

xeT

7.

XGT

8. Prove that if a set T is continuously ordered and contains first and last ele-

ments and if the sets Fx are continuously ordered, then the union

u Fx

xeT

is con-

tinuously ordered. [Hausdorff] l) Theorem 5 is due to Schiinilies; see Entwickelung der Mengenlehre und ihrer Anwendungen (Leipzig-Berlin 1913) p. 184. A detailed analysis of countable scattered order types is given by Erdos and Hajnal in the paper: On a classification of denumerable order types..., Fundamenta Mathematicae 51 (1962) 117-129.

216

VI. LINEARLY ORDERED SETS

$3. Order types w, 17 and 3, We shall illustrate the notion of order type by means of examples. Order type o.The order type w is the order type of the set N ordered by the relation THEOREM 1: A linearly ordered set A is of type o if and only i f (i) A has a first element ao, (ii) every element x of the set A has a direct successor x*, . (iii) i f a , E X c A and ifthe set X contains the direct successor of every element of X, then X = A. PROOF.Conditions (i), (ii), (iii) are invariable under any transformation which preserves order. Since they are satisfied by the set of natural numbers ordered by the relation 6 , they are necessary conditions for a set A to be of type o. Suppose now that the set A linearly ordered by the relation R satisfies conditions (i), (ii), (iii). Let us define a function f which establishes the similarity between A and the set of natural numbers as follows:

<.

f(0) = a09 f(n+ 1) = [ f W l * . These formulas define by induction the function f. It follows from (1) that the range of f contains a,, and that it contains the direct successor of every of its element. Condition (iii) implies that the range o f f coincides with A. Let us prove that m < n -+f(m)
m # n -+f(m) # f ( n > , m < n -+f(m>Rf(n>. The first of these formulas shows that the function f is one-to-one. Together these formulas show by Theorem 2, p. 208 that the function f establishes a similarity between A and the set of natural numbers ordered by <.

3. ORDER TYPES

0,

r ] .4"D 2

217

Order type 7. Before defining this type, we prove the following important theorem. THEOREM 2: Every two non-empty, denumerable, linearly ordered and dense sets which have neither Jirst nor last elements are similar.

PROOF.Let A and B be sets satisfying the assumptions of the theorem. To simplify notation, we shall use the same symbol to denote the relations ordering both sets. It follows from the assumptions that the sets A and B are infinite. Thus there exist one-to-one sequences a € A Nand b € B N such that a'(N) = A and b'(N) = B. Let us define by induction two permutations q~ and p of the set N such that the mapping f: uq(")4 by(")establishes a similarity between the sets A and B. For this purpose, let pl(0) = y(0) = 0. Now we consider two cases depending upon whether n is even or odd. Case 1: n even'). Let

Case 2: n odd. The definition is similar, but the roles of are interchanged.

We shall prove by induction that if n E N and j (3)

plh) # p l W 9

(4)

Y (4 # Y ( j ) ,

pl

and y

< n, then

( 5 ) 'V(") < 'q(J) < b P ( j ) t 'q(n) 'P(J) bVJ(n) bV(J)' It is clear that these formulas hold for n = 0. Suppose that no > 0 and that (3)-(5) hold for n < no. Let no = n'+ 1. The proof now splits 1)

If there exists no n such that @(n), then the symbol min @(k) shall denote the k

number 0.

218

VI. LINEARLY ORDERED SETS

into two cases according as n’ is even or n’ is odd. We shall consider only the first case. Since the set A is infinite, there exist numbers k such that a, # ap(j) for j < n’. By definition, q~(n’+l) is one of these numbers k. This proves (3), for n = n’+ 1 = no. To prove the remaining formulas, let

P

=(

j

< n’:

Q = ( j G n’: ap(n0) < apcj)}.

< ap(no)},

~p(j)

Thus

(PEP)A (4E Q)

4 alp(q)), and since ( 5 ) holds by assumption for n < n’, we obtain +

* (4 Q>

(P

E

(ap(p)

4

( b ~ ) bpp(g))*

Since the set B is densely ordered, the formula above shows that there exist numbers k such that btp[,,)< bk for every p in P and bk < for every q in Q. It follows from the definition of p that y(no) is one of these numbers k. Thus # by(,)for je P u Q. Moreover, b m o )< bvw 3 ( j E Q) E ap(n0) < avo) and similarly for > ( ~ EuPQ). In this way formulas (4) and (5) are proved. Formulas (3)-(5) show that the function f: a,(,) -+ bp(,,) establishes similarity between the sets (up(,): n EN} and (by(,):n EN}.It remains to be shown that these sets are identical W i t h A and B respectively. In other words, we have to show that every natural number occurs in the sequences q~ and y. We consider only the sequence q. Suppose on the contrary that N--Q]’(N)# 0 and let ko be the least number in this set. Clearly ko > 0. For h < ko let n h denote the unique number such that V(nh) = h and let n be an even number greater than all numbers nh, h < k,. Since ako# a,(j) for all j < n, and for every h < ko there exists j < n such that an = namely j = n h , we obtain k0

A

= min (ak k j
# ap(j)).

This implies ko = v(n+ l), which contradicts ko# pl(N). Theorem 2 is proved.

3. ORDER TYPES

w, q

AND

rz

219

Theorem 2 shows that there exists only one type of sets which are simultaneously # 0, densely ordered, countable and without first and last elements. This type is denoted by 7. # An example of an ordered set of type q is the set of rational numbers ordered by the relation # Another example is given on p. 206 (Example 3). Sets of type 7 have the following property of universality:

<.

THEOREM 3: If B = q and A is any denumerable linearly ordered set, then there exists a set C c B such that A and C are similar. PROOF.We may assume that the set A is infinite. Using the notions introduced in the proof of Theorem 2 we define the sequences and y as in Case 1. However, this time we do not confine n to even numbers only, but we let n be any natural number. Then formulas (3)-(5) are satisfied. We can prove in the same way as in Theorem 2 that yl(N) = N and that the sets {ap("):n e N } and {bV("):n E N } are similar. The first of these sets is equal to A and the second is contained in B. This proves the theorem').

The order type 1. We precede the definition by a theorem. THEOREM 4: Let A and B be sets satisfying the following conditions: (i) A and B are linearly and continuously ordered. (ii) There exists subsets A , c A and B, c B dense in A and B respectively which are both coinitial and cofinal with A and B respectively. (iii) The sets A , and B, are of type q. Then A and B are similar. We shall only outline the proof of this theorem. By Theorem 2 there exists a function fi which maps A , onto B, and preserves order. It can easily be shown that the sets X(a) = A , n {a)- and Y(a) = A , n {a}+ *) Theorems 2 and 3 are due to Cantor. In recent papers those theorems have been generalized for other relations. See: R. FraissB, Sur l'extension aux relations de quelques propri&t& des ordres, Annales scientifiques de 1'Ekole Normale Sup6rieure 71 (1954) 361-388; B. Jbnsson, Universal relational systems, Mathernatica Scandinavica 4(1956) 194-208; B. Jbnsson, Homogeneous universal relational systems, ibid. 8 (1960) 137-142; M. Morley and R. Vaught, Homogeneous universal models, ibid. 11 (1962) 37-57.

220

VI. LINEARLY ORDERED SETS

determine a proper cut in the set A,. Hence the pair

< P , Q> =

(fi(X(a)),f!(Y(a))) is a proper cut in the set B1. Let

Z(U)= ( b E B :

A

(b < y ) } ,

YEP

?(a)

=

(~EB:

A

(x
xsQ

It can easily be shown that the pair (-?(a), p ( a ) ) is a proper cut in the set B. Since B is continuous, there exists an element f ( a ) lying in this cut: it is the last element of the set f ( a ) and simultaneously the first element of the set ?(a). The mapping f satisfies the condition a‘ a“ -+f(u’)
<

<

<.

<

I) See Fundamenta Mathematicae l(1920) 223. S. Tennenbaum proved recently that the Suslin hypothesis is not provable within the framework of axiomatic set theory.

3, ORDER TYPES

0 , ‘1AND

fl

22 1

It has recently been shown that it is not possible to prove in 2 the existence of ordering relations for a set of power 2‘. Exercises 1. Show that every dense and infinite set contains a subset of type 7. 2. Show that every infinite continuous set is of power c. 3. Let r,, r,, ... be an infinite sequence without repetitions consisting of all rationm

a1 numbers. For c =

cj/3j

where

cj = 0

or c j = 2, let Me = { r j : c j

j=o

let C denote the type of the set Me ordered by the relation numerable order type can be represented in the form C l). 4. Lst C, = (c: C =

t}. Show

= 2)

and

<. Prove that every de-

that the Cantor set C is the union

u C,

where

r

the union is over all denumerable order types and where the components of this union are pairwise disjoint. 5. Show that the set C,,is a Ga-set in the space C. 6. Prove that every linearly ordered set containing a dense denumerable subset is similar t o a set of real numbers (ordered by Q). 7. Let M be a monotone family of open subsets of the real line (or generally fo

the space P). Prove that this family is similar to a set of real numbers (ordered by
establishes the required similarity.

0 4. Arithmetic

of order types

We can define operations on order types which are similar to certain operations in ordinary arithmetic just as we did in the case of cardinal numbers. This arithmetic of order types allows us to simplify arguments concerning linearly ordered sets. I)

See K. Kuratowski, Topology I, p. 36.

222

VI. LINEARLY ORDERED SETS

Inverse types. It is easy to show that i f a relation R linearly orders the set A , then so does the inverse relation RC (seep. 65). Of course, the isomorphism of R and S implies the isomorphism of the inverse relations R" and S". The order type of the set A ordered by R" is said to be the inverse of the order type of the set A ordered by R . If the order type of A ordered by R is a, then the order type of A ordered by Rc is denoted by a*. It follows from the equivalence X ( R O )= ~ ~x R y that (1)

a** = a.

Examples 1. If n is a finite order type then n* = n, because every two finite equipollent sets are similar. 2. Likewise q* = q and A* = A. On the other hand, w* # o because a set of type m* (for example, the set of negative integers) possesses a last element whereas a set of type o has no last element. The sum of order types. Let a and ,6 be two order types and let A and B be two sets linearly ordered by R and S such that 2 = a, = B. We assume that A n B = 0. This assumption can always be satisfied, for if A and B are not disjoint, then we can replace them by the sets A x i l ) and B x ( 2 ) which are similar to A and B and disjoint. The sum a+P is defined by a+B=AuB where the set A u B is ordered as follows: all elements of A precede all elements of B and the order in each of the sets A and B is preserved. In particular, if a and B are finite order types, the definition of the sum a+B coincides with the definition of the sum of natural numbers. I t is easy to see that the sum a+@ does not depend on the sets A and B but only upon their order types. Moreover, the following formulas clearly hold:

(a+B)+r

=

a+(B+r),

a+O = a

= o+a.

On the other hand, the commutative law does not hold: for example, w + l # l + w . In fact, l+co = w (thus l + o is equal to the type of

4. ARITHMEnC OF ORDER TYPES

223

the set of natural numbers), whereas w S 1 is the type of a set with a last element. Product of order types. Let a = 2,p = 3. The product a m pof the order types a and is defined by the formula a*/?=AxB, where the set A X B is ordered ,as follows. Let (x,y ) and (x,, y l ) be two elements of A X B. Then (x,y> < <x,,y,) if y 4 y,. If y = y , , then ( x , v> < (x17 Y1> if < For example 1.2 or A2 is the order type of the set of points in the plane ordered as above. It is easy to check that, just as for the sum, the product tl.b depends only upon a and b. For finite order types, the dehition given above coincides with that of multiplication of natural numbers. Moreover, we have for arbitrary order types tl, p, y the formulas: a1 = l a = a,

(aB)y = .%(By),

a0 = Oa = 0.

Similarly as for addition, multiplication is not commutative. For example, co2 # 2w. In fact 2w={1,2}xN=w7

w 2 = N x ( 1 7 2 } =co+w.

The distributive law is satisfied only in the form: a(P+y) = @B+@Y.

In fact, let

u=A,

-

p=B,

-

y=C,

BnC=O.

Then we have (see p. 63)

u AX c

a(/3+y) = A X(BUC) = A x B

-___

=A X

B

U.A

+

x C = tlp tly,

because ( A x B ) n ( A x C ) = 0. Exponentiation of order types in the case of a finite exponent can be defined by induction: aO = 1, @n+1 - a n *a.

224

VI. LINEARLY ORDERED SETS

Exercises I. Prove that ~ + r ] = 7. A+I+rl = A, f l + A # A. 2. Using operations on the order type w , give an example of an infinite linearly ordered set which possesses first and last elements such that every element except the first has a direct predecessor and every element except the last has a direct successor. 3. Prove that (u+p)* = /?*+u*. 4. Prove that q2 = 7. 5. Prove that (wr])*= ( ~ r ] + w )but ~ , wr] # wr]+w. [A. Davis-W. Sierpiriski] 6. Prove that wz is the type of the set of natural numbers ordered by the following relation: m precedes n if either m has fewer prime factors than n, or m has the same number of prime factors as n and M n. 7. Prove that a set A of type A2 does not contain a subset which is denumerable and dens2 in A.

<

0 5. Lexicographical ordering The product of order types is related to lexicographical ordering. In order to define this ordering let us assume that T is a set linearly ordered by the relation Q and let each x E T be associated with a- set F, ordered by the relation R,. We do not assume that the sets F, are disjoint. Let P= F,.

n

XET

Thus P is the set of functions f whose domain is T such that f ( x ) E F, for all x E T. Two arbitrary functions f and g belonging to P determine the set O(f, 8 ) = { x E T: f ( x ) f g ( x ) l .

Clearly O ( f , g ) = 0 if and only iff = g . We now define a relation S in P in the following way: f S g holds if and only if either f = g or the set D ( f , g ) possesses a first element x, and f (xo)~xog(xo). This definition can be written in symbols as follows

fsg

(f=g>

" v { ( f ( x )
If the relation S linearly orders the Cartesian product P, then this product is said to be lexicographically ordered (or ordered according to the principle of first differences).

5. LEXICOORAPHXCAL ORDERING

225

We shall investigate the conditions under which the relation Slinearly orders P . 1: The relation S is reflexive, antisymmetric and transiTHEOREM tive in P . PROOF.The reflexivity of S is obvious. Suppose that both f S g and g SJ If the functions f and g are distinct, then the set D ( f ,g) has a first element x and this element satisfies the conditions f ( x )R,g(x) and g(x)R,f(x). Since, by assumption, the relation R, orders F,, we havef(x) = g(x), which is incompatible with x ~ D ( f g). , Thus f S g and g Sf imply f = g. This shows that the relation S is antisymmetric. Suppose that f S g and g S h. Tff = g or g = h then clearly f S h. Thus we may assume that f # g and g .f h. Hence the sets D ( f , g ) and D(g, h) have first elements x and y respectiyely and these elements satisfy the conditions: f(x) and let R, be the relation ,<. Let f be the function whose value is 0 for even numbers and 1 for odd numbers and let g ( x ) = 1 -f(x). Then the set D ( f , g ) is equal to T and therefore it has no first element. Thus neither f S g and g Sf. THEOXEM 2: ZfT= n or T= o,then the relation S linearly orders the set

P I).

I) In general, Theorem 2 holds when T is an arbitrary well ordered set. See Chapter VII.

226

VI. LINEARLY ORDERED SETS

For the proof, it suffices to show that S is connected in P. For this purpose assume that f , g ~ and P f # g. The set Ddf,g) is non-empty and therefore it has a first element x. Since R, is connected in F,, we infer that f ( x ) R , g ( x ) or g ( x ) R , f ( x ) . This implies that f S g or g S J THEOREM 3: If the sets Al, ..., A , are of types xl, ...,an,then the set A , x A2 x ... x A,, lexicographically ordered is of type c1,,~,-~ ... al.

PROOF.The proof is by induction on n. For n = 1 the theorem is obvious. Suppose that it holds for n and consider the product P = A , x x Az x ... x A,+, of (n+ 1) sets, with lexicographical ordering. Let B = A 2 x A 3 x ... x A,+1. Ordering the set Al X B lexicographically we obtain a set similar to P. Hence it suffices to show that A 1 x B is of type a,,+l.a,. ..:aZ.al. But this follows directly from the definition of the product of types. The definition of anti-lexicographical ordering (ordering by the principle of last differences) is similar to that of lexicographical ordering. The notion of anti-lexicographical ordering rather than that of the lexicographical ordering lies at the basis of the notion of the product of types. # Examples

1. The product 11 is the type of the set of complex numbers ordered lexicographically (where the complex number x + i y is identified with the ordered pair <x, y ) ) . 2. The product q1 is the type of the set of complex numbers of the form r+iy, where r is a rational number and y is a real number, ordered anti-lexicographically. On the other hand, the product 27 is the type of the same set ordered lexicographically. These types are distinct, for a set of type 27 contains continuous intervals whereas a set of type 71 does not. 3. Let T be the set of natural numbers with the usual ordering, let F, = (0,l) for x E T, and let R, be the relation The lexicographical ordering S is isomorphic in this case to the relation in the Cantor set

<.

<

rn

c,,/3" where

C (understood as the set of real numbers of the form n=l

c,, = 0 or c,

=2

for n EN). In fact, associating with each function

227

5. LEXICOGRAPHICAL ORDERING W

YE

Fx the number cf

2f(n)/3"+l we see that cf

=

< cg if and only

n=0

X

if f# g and, moreover, the smallest number no such thatf(no) # g(no) satisfies the inequality f ( n 0 ) < g(n0). 4. Again let T = N (where the order relation is <) and let Fn = N we associate the real number for n e N . With each function

< 1 and each real number x, 0 < x < 1, can be repthis form in exactly one way. In fact, if x = C 2 4 " ) is

Clearly 0 < rf

m

resented in

n=O

the binary representation of x with infinitely many digits different from 0, then the sequence is strictly increasing and ~ ( 0 > ) 0. Assuming J(0) =: ~ ( 0 ) 1 and f ( n ) = y ( n ) - y ( n - 1) - 1 for n > 0, we obtain x=+ In order that rf < rrr it is necessary and sufficient that f # g and that the smallest number no such that f ( n o ) # g(no) satisfies the inequality f ( n 0 ) < g(n0). In this case the relation S of lexicographical ordering is similar to in the set of numbers x , 0 < x Q 1. Hence the type of the relation this relation is A+ 1. #

<

CHAPTER VII

WELL-ORDERED SETS tj 1. Definitions. Principle of transfinite induction

We say that a relations R well orders a set X if R linearly orders X and every non-empty subset of X contains a first element (with respect to the relation R) '). ExampIes 1. Every set of type o is well ordered. # 2 . The set consisting of the number 1 and of all numbers of the form 1- l/n, n = 1,2, ... is well ordered by the relation < . The type of this set is o+ 1. 3. Let a(n) be the number of distinct prime factors of the number n. The relation

{ <xv>: [a (4 < a (Y>l v ( [ a (4 = a (v>lA [x ,< vl) 1 well orders the set of natural numbers. The type of this set is o2(see Exercise 6, p. 224). # 4. The ordered union (see p. 213) S = U F,, where the Fet T x-T

and the component sets F, are well ordered, is also well ordered. For, let Y be a non-empty subset of S. The set of all x such that Y n F, # 0 is a non empty subset of T, therefore it contains a first element x,. Thus the intersection Y n Fxo is a non-empty subeet of the ') The notion of well-ordered set is due to Cantor, who introduced it first in hi, paper Uber unendliche Iineare PunktrnannigfaItigkeiten, No. 5, Mathernat ische Annalen 21 (1883), 8 3. I t is interesting that the notion of linearly ordered set was introduced by Cantor later (in a paper puhlished i n 1895), evidently in the course of systematizing his results in the theory of well-ordered sets.

1. DEFINITIONS. TRANSFINITE INDUCTION

229

well-ordered set Fxo and therefore it has a first element. This element is clearly the first element of Y. 5. The Cartesian product of any finite number of well-ordered sets is itself well ordered by the relation of lexicographical ordering. 6. Every subset of a well-ordered set is also well ordered. The following two theorems are simple consequences of the definition.

THEOREM 1 : In every well-ordered set there exists a Jirst element. Every element except the lust element (if such exists) has a direct successor. THEOREM 2': No subset of a well-ordered set is of type o*. * THEOREM 2": If a linearly ordered set A is not well ordered, then it contains a subset of type w*.

PROOF. Let P be a non-empty subset of A which contains no first element. Let Q(4= { Y : (Y < x)AO,EP))We have Q(x) # 0 €or every x e P . By Theorem 8, p.73 there exists a function f defined for every x E P such that f (x) E Q(x). Let po be any element of P. We define by induction a sequence p l , p2, ...,p,,, ... letting p,, = f ( ~ , , - ~for ) n > 0. Since f ( x ) < x, this sequence is of type w*. Theorems 2' and 2" imply the following.

THEOREM2: In order that a linearly ordered set be well ordered it is necessary and suflcient that it contain no subset of type o*.

THEOREM 3 : Each initial segment of a well-ordered set A is of the form O ( x ) for some x e A . In fact, if x is the first element of the difference A-X, then O(x) = X(see p. 209). THEOREM 4: (PRINCIPLE OF TRANSFINITE INDUCTION ').) If a set A is well ordered, B c A and iffor every X E Athe set B satisfies the condiI) This form of the theorem on transfinite induction was implicitly used already by Cantor; see: Beitriige zur Begriindung der transfniten Mengenlehre, Mathematische Annalen 49 (1897) 336-339. A clear formulation of this theorem is due to Hessenberg; see: Crudbegrife der Mengenlehre (mttingen 1906). p. 53.

VII. WELL-ORDERED

230

SETS

tion

(1) then B.= A. PROOF.Suppose that A - B # 0. Then there exists a first element in A-B. This means that if y < x then y 4 A - By that is, y E B. This shows that O(x) c B. Now it follows from (1) that x E By which contradicts the hypothesis that x q! B. Theorem 4 can be reformulated as follows. Let a subset B of a wellordered set A be called hereditary if it satisfies condition (1). Then Theorem 4 asserts that the only hereditary subset of A is the set A itself. For many propositional functions @ it can be proved that the set { x E A : @(x)} is hereditary; consequently, for such propositional funcholds. This method of proving theorems tions the theorem

n@(x)

XEA

of the form

// @(x) is called the method of transfinite induction. Ofcourse, x d

this method and the method of proof by induction in ordinary arithmetics (which consists in showing that the set { n E N : @(n)} is inductive) are analogous. We shall use transfinite induction to prove several theorems about similarity of well-ordered sets. Let A be a set linearly ordered by the relation R . A functionf which establishes similarity between A and the set f '(A) contained in A , is said to be an increasing function. Such functions satisfy the condition (2)

x i Y -+

f

i f (v).

THEOREM 5 : If a function f defined on a well-ordered set A is increasing, then f o r every x we have x R f ( x ) (that is x < f ( x ) or x =f(x)>.

PROOF.Let B = { x : x R f ( x ) } . Let O ( x ) c B. We show that x E B. In fact, let Y E O(x), that is y < x . By (2) it follows that f ( y ) -< f ( x ) . Since Y E Bywe have y R f ( y ) and thus y < f ( x ) . This shows that the elementf ( x ) occurs after every element y of O(x), that is, f ( x ) E A-O(x).

23 1

1. DEFINITIONS. TRANSFINITE INDUCTION

Since x is the first element of A - O ( x ) , we have x R f ( x ) and finally XEB. Hence the set B is hereditary. Q.E.D. COROLLARY 6 : If the well-ordered sets A and B are similar, then there exists only one function which establishes their similarity.

PROOF.Suppose that the sets A and B are well ordered by R and S and that there exist two functions f and g establishing similarity between A and B. The function gc o f is clearly increasing in A (see p. 208). By Theorem 5 we thus have x R g c (f ( x ) ) for every x € A . Hence g ( x ) S f ( x ) . Considering f" o g instead of g" of we have by the same argument f ( x ) S g ( x ) . This implies f ( x ) = g(x), because S is antisymmetric. 7 : No well-ordered set is similar to any of its initial segCOROLLARY ments. In fact, if the sets A and O ( x ) were similar, then the function f establishing similarity would be increasing and would satisfy f ( x ) E O(x), that is f ( x ) < x . But this contradicts Theorem 5. COROLLARY 8 : No two distinct initial segments of a well-ordered set are similar. For the proof, it suffices to apply Corollary 7 and to observe that, given two distinct initial segments one is always an initial segment of the other. 9 ') : Let A and B be two well-ordered sets. Then either THEOREM (i) A and B are similar, or (ii) the set A is similar to a segment of B, or (iii) the set B is similar to a segment of A. PROOF.Let R and S be relations which well order A and B respectively and let -~ z = ( x E A : V O R ( X )= O s ( y ) } . Y

In other words (see notation on p. 209): X E Z= {the initial segment O,(x) of A is similar to some initial segment O&) l)

of B}.

This theorem is due to Cantor; see Mathematische Annalen 49 (1897) 216.

232

W.WELL-ORDERED SETS

By virtue of Corollary 8, for given X E Zthere exists only one such segment. Thus there exists a function f de!ined on 2 such that this segment is of the form O , ( f ( x ) ) . First we show that either 2 = A or 2 is a segment of A, that is, there exists an a in A such that 2 = O,(a). In fact, let x < ~ ’ € 2 . Since OR(x)is a segment of OR(x’),the function establishing similarity between O,(x’) and O,(f(x’)) also maps O,(x)anto a segment of B. Hence X E Z . Similarly: either f ’ Q = B or else f‘(2)is a segment of B: f ’ ( 2 ) = O,(b). To show this it suffices to observe that .fl(Z) = { w B :

v Iv =f(x)lJ

X€Z

In fact, if O&)

= {YE B :

v ~0,Ol) YEB

=OR(X)}.

is similar to a segment of A then y is of the form

f ( x ) where x E 2. Finally, observe that f establishes the similarity between 2 and f ‘(2). Indeed, we have just shown that x < x’ E Z implies that O,(f(x)) is a segment of 0,(f(x’)), therefore f( x )
By ordinal numbers (or ordinals) w‘e shall understand the order types of well-ordered sets I). Theorem 9, 0 1 allows us to define a “less than” l) The notion of ordinal number was introduced by Cantor. See Mathematische Annalen 21 (1883) 548. Notice that according to our definition the number 0 (the type of the empty set) is an ordinal.

233

2. ORDINAL NUMBERS

relation for ordinals. This relation cannot be defined in a satisfactory way for arbitrary order types. DEFINITION 1: We say that an ordinal a is less than an ordinal j3 if any set of type a is similar to a segment of a set of type p. We denote this relation by u < j3 or /3 > a. We write “ a < p” instead of “ a < /? or a = By’. THEOREM 1: For any ordinals a and /? one and only one of the formulas u < b, a = j3, a > j3 holds. This theorem is a direct consequence of Theorem 9, 5 1. THEOREM 2: If a, p and y are ordinals and ff

if a < p and < y, then

< y.

PROOF.Let A , B and C be sets of types a, j3 and y, respectively. By assumption, the set A is similar to a segment of the set B and B is similar to a segment of C. Thus A is similar to a segment of C. Q.E.D. The following formulas can be proved without difficulty:

(a < B) A (B < a) -, (a = B), (a < j3) A (j3 < r>-, (a < Y>. THEOREM 3: If the well-ordered sets A and B are of types a and and if the set A is sfmilar to a subset B, of fhe set B, then a < /?. PROOF.If this were not so, then we would have j3 < a and then B would be similar to a segment of Bl. This contradicts Theorem 5, p. 230. We now examine sets of ordinal numbers. THEOREM 4: The set W(a) consisting of all ordinals less than a is well ordered by relation Moreover, the type of W(a) is a.

<.

PROOF.Let A be a well-ordered set of type a. Associating the type of the segment O(a) with the element a E A we infer (by the axiom of replacement) that the set W(a) exists and simultaneously we obtain a one-to-one mapping of A onto W(a). It is easily seen that following conditions are equivalent: (i) a, precedes az or a, = az, (ii) O(a,) is a segment of O(az) or O(a,) = O(uZ), (ii) the type of O(u,) is not greater than the type of O(az). This shouts that the relation

< indeed orders

W(a) in type a.

234

W.WELL-ORDERED SETS

THE~REM 5 : Every set of ordinals is well ordered by the relation 6 . In other words, in any non-empty set Z of ordinals there exists a smallest ordinal. PROOF.Let a E 2. If a is not the smallest ordinal of 2, then Z n W(a)# 0. Then in the set Zn W(a)there exists a smallest number /?, as the set W(a)is well ordered (see Theorem 4). At the same time /3 is the smallest ordinal in 2. In fact, if ~ E Z - W ( ~ then ) 5 u; thus 5 > /?. THEOREM 6: For every set Z of ordinals there exists an ordinal greater than all ordinals belonging to 2. PROOF. By the axiom of replacement there exists a set K whose elements are all the sets W(u) corresponding to the ordinals u belonging to 2. W(u)E K E a E Z . Consider the union of all sets belonging to K

s=

ux.

xsK

By Theorem 5 the set S is well ordered by <. Let a be its order type. For a E Z the set W(a)is either a segment of S or identical with S. In any case, u < a. This implies that a < a+ 1 for every a €2.Thus the ordinal o f 1 is greater than every ordinal of Z. COROLLARY 7: There exists no set of all ordinals '). COROLLARY 8: There exists a smallest ordinal not belonging to a given set Z. Let ~ $ (such 2 an ordinal exists by Corollary 7). If a is not the smallest ordinal not belonging to Z, then the set W(a)-Z is nonempty. The smallest number in this set (see Theorem 5 ) is simultaneously the smallest ordinal not belonging to Z. COROLLARY 9: If a set Z of ordinals has the property ( y < 5 E 2) -,( y E Z ) , then there exists an ordinal a such that 2 = W(a). ') Before set theory was axiomatized, Corollary 7 had been considered to be an antinomy. This antinomy was discovered by C. Burali-Forti. See his paper Una questione sui numeri transfniti, Rendiconti del Circolo Matematico di Palermo 11 (1897) 154-164.

235

2. ORDINAL NUMBERS

Namely, this ordinal a is the smallest ordinal among all ordinals not belonging to 2. In fact, if 5 E 2 then 5 < a, because a < 5 would imply CL E 2.Hence z c W(a). On the other hand if B E W(a) then < a and, by the definition of a, 4, e 2. Therefore W(a) c Z.

0 3. Transfinite sequences An ordinal is said to be a limit ordinal if it has no direct predecessor. Thus 0 is a limit ordinal. THEOREM: Each ordinal can be represented in the form I + n where A is a limit ordinal and n is a Jinite ordinal (natural number).

PROOF.Let a be an ordinal, A a set of type a. Every set of the form A-O(a) is said to be a remainder of A. Clearly A--O(al) c A-O(a2)

G

(a,

> a2)v (al = uz).

This implies that there exists no infinite increasing sequence of distinct remainders. Therefore there exists only a finite number of r n ~ N such that there exists a remainder of power m. If n is the greatest remainder of power n, then the segment such number and if X- O(a) is a O(a) has no last element. Thus O(u) is the limit ordinal A. This impIies that a = ilfn.. Q.E.D. By a transfinite sequence of type a or by an #-sequence we understand a function 91 whose domain is W(a).If the values of this function (also called the terms of the a-sequence) are ordinals and if y < < a implies ~ ( y < ) ~ ( p ) then , we say that this a-sequence is increasing. Let 91 be a I-sequence of ordinals where I is a limit ordinal. By The) orem 6 in $ 2 there exist ordinals greater than all the ordinals ~ ( y where y < I. The smallest such ordinal (see $ 2, Corollary 8) is called the limit of the I-sequence y ( y ) for y < I and is denoted by lim pl(y). Y
For example, o = limn = lim n2. new

nxw

We say that an ordinal A is cojinal with a limit ordinal a if A is the

236

VII. WELL-ORDERED SETS

limit of an increasing a-sequence:

An ordinal cofinal with a limit ordinal is clearly ikelf a limit ordinal. The connection between this notion and the notion of cofinality for sets is established by the following theorem.

THEOREM 2: An ordinal 1 is cofinal with the limit ordinal a ifand only if W(1) contains a subset of type a cofinal with W(A). PROOF.Let A be a subzet of W(1) cofinal with W(1) and such that

A= a. For every ordinal E < a there exists an ordinal pl(6) in A such that the set {q: (q E A ) A (7 < pl (t))}is of type 5. The sequence pl (5) is clearly increasing and pl (t)< 1for t < a, because pl (6)E A c W(1). If p < A then there exists an ordinal [ E A such that u , < t. because the sets A and W(A)are cofinal. Thus ,u < t < y ( t ) (see p. 230), which proves that 1is the least ordinal greater than all the ordinals rp([). This proves (1). Suppose in turn that (1) holds. Let A be the set of all terms of pl. We have 7 < 1 for q E A and consequently, since 1 is a limit ordinal, there exists 6 E W(A)such that q < 5. Conversely, if 6E W(A)then < 1 and by the definition of limit there exists 5’ < a such that 5 < pl(t’).This means that some ordinal in A is greater than 6.Hence the sets A and W(A) are cofinal. It follows directly from the definition of limit that

(3)

{ A[P > W 1 J = { P 2 fimP(Y)}.
Y
THEOREM 3: Ifcp and y are two increasing transfinite sequences, iZ is a limit ordinal and 5 = lim y ( y ) , then Yea

237

3. TRANSFINITE SEQUENCES

< I we have 2 p(6), and by

If 6 < E then by (3) we infer that for some ordinal y y(y) 2 6. Since the sequence q~ is increasing, p(y(y))

(3) it follows that limy(y(y))

>~(6).

Y<.l

Applying (3) again we obtain

This together with (4) proves Theorem ?. It follows from Theorem 3 that if a limit ordinal q is cofinal with a limit ordinal and E is cofinal with a limit ordinal A, then q is cofinal with I . We say that an a-sequence q~ is continuous if for every limit ordinal I < CI we have P (4= lim v (7)* Y<1.

THEOREM 4:Let cp be an increasing continuous a-sequence. For a given ordinal y < a, let (5) x,, = lim y n , where yo = y and yn+l = v(yn). n<w

Then ~ ( x , )= x y (when x y < a and yn < c1 for n = 1,2, ...).

PROOF.By (5) we have x,, > yn+l = y(yn), and it follows by (3) that x

> lim p(yn) = p,(lim y.)

'n
= p)(x,,).

ncm

On the other hand, x,, < ~ ( x , , ) ,because the sequence q~ is increasing (see p. 230). Each ordinal [ satisfying the equation ~ ( t=)E is said to be a critical ordinal of the sequence 97. Thus Theorem 4 states that if y, belongs to the domain of an increasing continuous sequence v, then there exists a critical ordinal of this sequence greater than y, provided that this sequence is defined for sufficiently large ordinals. Since no set contains all ordinals, there exists no transfinite sequence consisting of all ordinals. On the other hand, there exist propositional functions @(E,q) (for example, t -- q+l), of at least two free variables, such that the following holds. For every ordinal 7 there exists exactly one ordinal E such that @(q, t).

238

VII. WELL-ORDERED SETS

Such a propositional function defines an operation on the ordinals. I t is possible to define many notions and to prove many theorems about these operations, analogous to those for transfinite sequences. We here state an analogue of Theorem 4 for operations. Let @ be a propositional function with free variables x,y ,p l , ... ,p k . (The case k = 0 where @ has exactly two free variables is not excluded.) For simplicity in the formulas below we shall not write the variables p j explicitly. We assume that small Greek letters range over ordinal numbers. The following abbreviations will be used: (6)

IT{@}

is an abbreviation for

Av X

(7)

R{@}

(8)

Lim{@)

is an abbreviation for

is an abbreviation for

€
min P

(9)

Cont {@}

[@(x, z ) = (z = y)];

Y

A A [W,5) (r < &I; -+

6 4 f

is an abbreviation for

A [(A

is a limit ordinal) 3 @(A, Lim {@})I. e<2.

A

The propositional function in (6) is read: @ defines an operution. The propositional function n{@} is equivalent to the conjunction of two propositional functions

This conjunction will be denoted by

A v ! @ ( x 2y ) , where v ! is X

Y

Y

the quantifier “there exists exactly one y”. The propositional function in (7) is read: the operation defined by @ is increasing, and in (9): the operation defined by @ is continuous. Formula (8) introduces a symbol for the limit of the operation defined by @. In formula (8) the symbol min denotes 0 if there exists no ordinal P

p satisfying the propositional function after the symbol min. U

239

3. TRANSFINITE SEQUENCES

THEOREM 5:

The proof is similar to that of Theorem 4. Namely we define by induction a sequence of ordinals: YO

= min.(@((y,, t

= Y,

t)).

The ordinal E = Lim 7, it the desired ordinal. nca,

The following theorem about operations will be used in 0 4, p. 244-245.

THEOREM 6:

By the axiom of replacement the set ((5, y>: (5 < a) A @(E, y ) } exists. This set is the required sequence. It is clear that this sequence is unique. + n{Y), where Y is the proposiTheorem 6 can be written tional function: (p, is an a-sequence) A @(t, plp3. The unique se-

n{@}

A

Eta

quence pl satisfying the condition Y ( a , y ) is denoted by C,{@}. It should be noted that definitions (6)-(9) as well as Theorems 5 and 6 are schemes; for each propositional function @ we obtain separate definitions and theorems.

5 4.

Definitions by transfinite induction

The theory discussed in this section is similar to the theory of inductive definitions in arithmetic of natural numbers. THEOREM 1: (ON DEFINITIONS BY TRANSFINITE INDUCTION) '). Given a set Z and an ordinal a, let Q, denote the set of all &sequences for 5 < a with values' belonging to Z . For each function h E Z @ there exists one and only one trans$nite sequence f defined on 5 a and such that (1)

f(E)

= hCfl W(01

< for every 5 < a.

I) A particular case of this theorem was already formulated by Cantor; see Mathematische Annalen 49 (1897) 231.

240

W.WELL-ORDERED SETS

PROOF.We show first that there exists at most one sequence f satisfying condition (1). Suppose that g is a sequence defined on the set W(a+ 1) and satisfying the condition (2) g(5) = h[glw(E)I for every 5 a. Let B = ( 5 : f ( 5 ) = g(E)}. If 5 G a and W(E)= By then fIW(5) = gl W([), and by (1) and (2) f(5) = g(E), that is 5 E B. Thus we have shown that the condition W(E)c B implies ~ E B By . the principle of transfinite induction (p. 229) we infer that W(a+l) c B; that is, for every [ a, we have f ( 5 ) = g(E). This means that the functions f and g are identical. We now prove that there exists a function f satisfying (1). Suppose by way of contradiction that for given a such a function does not exist. Clearly we may assume that a is the smallest ordinal with this property; othetwise we can find the smallest ordinal with this property in the set W(a). Thus for each 5 < a there exists a function f E satisfying the condition

<

<

wil

<

(3) &(r)= W O l for every Y 5. It follows from the part of the theorem already proved that for a given E there exists exactly one function & satisfying (3). We now infer that if y G 5 then &I W(y+ 1) =&. Hence &(5) = &(C) provided that C < y. This implies:

[&I W W l = v;I W(Y)l

(4)

for

Y G E.

Let

f ( 5 ) =&(5) for 5 < a

f ( a ) = h(C,), where C, denotes the a-sequence such that C,(E) = f t ( 5 ) for 5 < a. The functionfsatisfies condition (1). In fact, if y E < a then by (3), (4) and ( 5 ) (5)

and

<

(6)

f ( r ) = h[fvl W Y ) l = “I

W(Y)l =h(r) 9

whence f I W(E)=&I W(5) and by (3)

W(5)I = h[fl W(511 for 5 < a. Finally, f ( a ) = h [fl W(a)],because by (5) f I W(a)= C,. In applications of the theorem, the function h is often defined by three formulas: the first gives the value h ( y ) for the void sequence y

f(E)

= h tftl

4. DEFINITIONS BY TRANSFINITE INDUCTION

24 1

(i.e. the value h(O)), the second the value h(v) for sequences F E Z @ whose type is not a limit ordinal (i.e. is of the form 5+ l), the tllird gives the value h(cp) for sequences whose type 1 is a limit ordinal. For instance, the first formula may be of the form

h(0) = A , the second of the form and the third of the form

where I: and G are given functions and A is a given set. Then the sequence f, which exists by Theorem 1, satisfies the conditions f(0) = A ,

I

f(5+

1) = F(f (El),

f(1) =

~ ( u mor)~ ( 2 )

=

rl
G(n ?
m.

Usually when we apply Theorem 1 to prove the existence o f f , we give only these three formulas.

Examples # 1. Derivatives of order a '). Let A be a subset of the real line (or, more generally, A c en).The de;ivative of order a of the set A is defined by transfinite induction as follows ~ ( 0= ) A, ~(a+l= ) A ~ O ' , AW = AW,

n

Y
<

where 3, is a limit ordinal a. In this case we have 2 = 2x, h(0) = 5+1, h(v) =

h ( v ) = [p,(5)]' if cp is of type

nv ( y ) if v is of type 1.#

Y 4.

2. Borel sets of type a'). The family F, of Borel sets of type a is I) We recall that the derivative of order 1, i.e. A', is the set of limit points of the set A . Borel sets were introduced by E. Borel; see Lecoons sur la thiorie des fonctions, (Paris 1898).

242

VII. WELL-ORDERED SETS

defined by transfinite induction: (i) Fo is the family of all closed subsets (of a given space),

depending on whether 5 is an even or an odd ordinal (ordinals of the form I f n , where 1is a limit ordinal, are said to be even if n is even and odd if n is odd). See page 129 for the definitions of 0 and 6. In this case we have

h(O) = &I

and

h(v)

=

(Uv(r)>c7

or

Y<5

h(v) = (Uv(r))a Y
depending on whether the type 5 of ~1 is an even or an odd ordinal. Similarly we define the family G, by the conditions: (iii) Gois the family of open sets, (iv)

G,=(UG,),or G ~ = ( U G for, )o ~
Y<5

depending on the character of 5 (even or odd). # 3. Analytically representable functions of class a '). The set of these functions is denoted by and is defined by transfinite induction as follows: (a) @, is the set of all real continuous functions (of a real variable), @%

(b) @t = (U@,),for 0 < t Y
< a,

where in general d A denotes the set of all functions which are limits of convergent sequences of functions belonging to the class A . In this example:

z =2

m ,

h(0) = Go,

h ( q ) = (Uvfr))u

where 5 is the type of the sequence ~ 1 # .

As another example of an application of Theorem 1 we prove the following theorem. l) This class of functions was first investigated by R. Baire; see Lecons sur les fonctions discontinues (Paris 1905).

4. DEFINITIONS BY TRANSFINITE INDUCTION

243

THEOREM 2: Every limit ordinal of the form A = limy([) is cofinal with some ordinal y

€
< a ').

PROOF.According to Theorem 1, there exists an a-sequence %such that (i) y ( l ) is the smallest ordinal 5 such that y ( 5 ) < 5 < A and

A y(q) < 5, if such an ordinal exists;

w e

(ii) y(5) = A otherwise. We consider two cases: I. There exists no ordinal 5 < a such that y ( f ) = A. In this case the sequence y is increasing by (i) and (by induction) y(E) > p(5) for all t. Since A is the smallest ordinal greater than all the y ( Q , we infer that lim y(5) = A. Thus A is cofinal with a. €
11. There exist ordinals E < a such that y(E) = A. Let y be the smallest such ordinal, that is y ( y ) = A and y(7) < A for q < y. This implies that y is a limit ordinal. In fact, would we have y = 6+1 then y(6) < A and therefore there would exist an ordinal p such that y(6) < p < A and ~ ( 6 < ) p < A, which contradicts y(y) = A. It follows from (i) that for t1< l2< y we have ~ ( 5 , < ) y(E2).This means that yI W ( y )is increasing. Let p = lim y ( l ) . We shall show that p = A. For €
suppose that e < A. The ordinal e is greater than all the y ( q ) for q < y and therefore there exist ordinals < A greater than y ( y ) and greater than y ( q ) for q < y (namely any ordinal greater than both e and y ( y ) and less than A satisfies these conditions). But this contradicts the assumption y ( y ) = A. Hence Q >A. We have e A, because y(E) < A for 5 < y. Thus A = e = lim ~(6).This means that A is cofinal with

<

P
y. Q.E.D.

It is possible to formulate for operations a theorem similar to Theorem 1. As before, let Q, be a propositional function with at least two free variables x,y , p l , ... , p k . Let M ( @ } be the propositional function (with free variables a, y , p l , ... ,pk) l)

We do not suppose that the function

(p

is increasing.

244

W.WELL-ORDERED SETS

(q is a transfinite sequence of type a+ 1) A

A @(?I ~ ( t ) , q ~ ) .

t
This propositional function M{@} will be read:

“ y is the inductive sequence for @ of type a+l”. LEMMA: (#? < a ) A M { @ ) ( a , V ) M{@}(BYPlW(#?+1)). In fact, if ~1 is the inductive sequence for @ of type a+1, then the yE) for every sequence y =: pll W(#?+1) satisfies the condition @(yl W([)>, f < 8; thus it is the inductive sequence for @ of type 1. Further, let Ind {ds} denote the propositional function [M{dS} +

#?+

v c

(q, = a ) ] ; the free variables of this function are a,a,p,, ... y p k . We shall show that by means of Ind{@} it is possible to formulate the tbeorem on definitions by transfinite induction for operations. For simplicity we write Y instead of Ind {@} (see p. 239 for the definition of C,{ Y}j. A

THEOREM 3:

n{@> n{K}; (ii) n{@} A Y(a, a) (0

.+

.+

@(C,{Y},a).

PROOF.(i) Suppose that D{@)and let u be an ordinal. The consequent of implication (i) is equivalent to the conjunction of the formulas: (7)

y(%ai) A y(% az) + (a1 = az);

v Y(a,

(8)

a).

Formula (7) follows directly from the implication

M{@}(a,

A

M { @ )(a,y”) -+

(plf = y f f ) ,

<

which can be proved by showing that if B = { t a : pl‘(E) = ~ ” ( t ) } , then W(E)c B + ~ E for B all 5 u, thus B = W(a+l). For the proof of formula (8), suppose by way of contradiction 7Y(a,a). Hence there exists that there exist ordinals a such that

<

A a

the smallest such ordinal a. Thus for every #? < a there exist an element ab and the inductive sequence y(b) for @ of type b+1. The sequence pl(@) satisfies the conditions @(@@)I W(E),plL@) for E and

<

245

4. DEFINITIONS BY TRANSFINITE INDUCTION

~ $ 8= ) as. By part (i) of Theorem 3 y ( @as well as a, are uniquely determined. It follows from the lemma that y'@1W(E+ 1) = pl") for E @ and thus ac = = yip) for @. The sequence of type a defined by = y(B)(i3)= a8 for @ < a satisfies the condition @IWCB) = y@)IW(B)for every @ < a, and therefore it also satisfies the condition @(@[ WCF),g). Let a be the

<

<

A

@(eY

@
a). The sequence y* such that yf = j7[ unique element such that for 5 < a and such that y z = a is of type a + l and is the inductive sequence for @, that is M { @ }(y*, a). Since y a = a, we now obtain Y ( a ,a) which contradicts the hypothesis. Part (i) of the theorem is thus proved. To prove (ii), suppose that IT{@} and !P(a,al. Thus there exists a sequence y of type u+1 such that M { @ }(a, y ) and ya = a. It suffices to show that y = Ca{Y}or that Y { t ,yc} for every 5 a. But this follows directly from the lemma and from the uniqueness of inductive sequences. Theorem 3 is a theorem scheme: for each propositional function @ we obtain a separate theorem.

<

Examples 1. Let @ = @(C,Y , P ) be the following propositional function: [(C is not a transfinite sequence) A (Y = O)] v [(C is a sequence of type 0) A (Y = P)]v [(cis a sequence of type y+ 1) A (Y = 2'91 v [(C is a sequence of type il where il is a limit ordinal) Clearly

AC v! @(C,Y , P ) . Applying the scheme described in TheY

orem 3 we obtain the propositional function Ind {@Iwith free variables a, P,Y such that Ind {@} (a, Y y P ) The . unique Y satisfying the

A v! U

Y

condition Ind {@}(a, Y, P) is denoted by R,(P).The following formulas are consequences of Theorem 3: (9)

ROV) = P,

Rt+l(P) = 2 R q

RAP) =

u Rc(P).

€
246

W.WELL-OmERED SETS

When P = 0 we write R, instead of R,(P). This set is called the family of sets of rank at most a’). These sets will be examined more closely in $7, Chapter VIII. 2. Let @ = @(C,y , p ) be the following propositional function: ([(Cisnot a transfinite sequence of ordinals) v ( p is not an ordinal)] A (y = 0)) V [(c = 0)A (JJ= I)] V v {[(the type of C is of the form y+ 1) A (y = Cy.p)] [(the type of C is a limit ordinal A) A (y = lim Ct)]}. Clearly

A v! @ ( C , y , p ) ,that is n(@}. Applying C

€
the scheme of

Y

Theorem 3 to this propositional function we obtain the propositional function Ind {@Isuch that Ind {@I(a,y,p). Let nc3 denote the

A v! a

y

unique ordinal y such that Ind {@I(a. y, n). We now obtain the formulas g$= 1, ny+1= ny -n, nA=lim$. E
EKercises 1. Show that it is possible to obtain Theorem 1 from Theorem 3 by an appropriate choice of the propositional function. 2. Construct a propositional function Q which defines an operation CL + ar from ordinals to cardinals in such a way that

as= a,

ay+l= 2’7,

.

a1 =

C at.

€
8 5. Ordinal arithmetic z, The following theorem is a direct consequence of the definitions given in Chapter VI, p. 222-223

TJBOREM 1: The sum and the product of two ordinal numbers are ordinal numbers. I) This notion of rank is virtually due to Russell. He called objects that are not sets the objects of type 0, objects whose all elements are of type n-the sets of type n + l . Russell considered only finite types and “homogeneous” sets, i.e. sets whose elements are all of the same type. See B. Russell and A. N. Whitehead, Principia Mathematiea, 2nd edition (Cambridge 1925). Definitions and theorems given in # 5-7 are due to Cantor; see Mathematische Amden 21 (1873).

5. ORDINAL ARITHMETIC

241

By means of Example 4, p. 228 we obtain the following 2 : The ordered sum of ordinal numbers, where the indexing THEOREM set is well ordered, is itself an ordinal number. The first part of Theorem 1 follows from Theorem 2, by letting the indexing set be a two-element set. Similarly, assuming that the indexing set is of type B and that all components are equal to a, we conclude that is an ordinal. We shall prove some arithmetic laws for the ordinal addition and multiplication. THEFIRST MONOTONIC LAW FOR ADDITION: (a < B)

(1)

+

(y+a

< Y+Bb

P R O O F . L ~ ~ CB==~/ ?, a n d B n C = O . Since a < B , t h e s e t B contains a segment A of type u. The ordered sum C u A , which is of type y+a, is a segment of C u By which is of type y+B. Thus Y+a < Y+B. It follows from (1) (for a = 0) that

!/? > 0) (r+B > Y). (2) Thus the sum of two ordinals different from 0 is greater than the first component. +

THE (3)

SECOND MONOTONIC LAW FOR ADDITION:

(a

< B) -

+

(a+

Y < B+r). -

In fact, assuming that A = a, 3 = p, C = y , A c ByB n C = 0 and applying Theorem 3, p. 233 to the ordered unions A u C and B u C, we obtain (3). In particular, it follows from (3) that B+Y 2 Y. Thus the sum of two ordinals is not less than the second component. On the other hand, from 1 +cu = cu we see that the sum does not need to be greater than the second component (although the first component is not zero). THEOREM 3 : If a 3 b then there exists exactly one ordinal y such that a = /3+ y. PROOF. Let 2 = a , let B be a segment of A of type B and let y = A - B . Clearly, a = B+y. To prove the uniqueness of y suppose

(4)

248

VII. WELL-ORDERED SETS

+

that p+yl = ,f3+yz. By (1) this implies that y1 yz and y2 4 yl. Thus y1 = yz by Theorem 1, p. 233. It follows from Theorem 3 and formula (2) that the inequality a 8 is a necessary and sufficient condition for the equation a = @+x to be solvable. On the other hand, the equation a = x+p is not always solvable, for example we have o # x+2. for every ordinal x . In connection with Theorem 3 we introduce the following definition. The diflerence of the ordinals a and (a >, is defined to be the unique ordinal y such that a = B+y. This ordinal is denoted by a-8. Thus

B)

(5) a = B+(a-B). For instance, 0--n = o,because n+co = w. Similarly 02-o = co2, because o+oz = w2. THE MONOTONIC LAWS FOR ORDINAL SUBTRACTION: (a > all + (a-B > a,-/% (6) (B > 81) -P ( N - B < a - m . (7) In order for the subtraction to be performable we assume in (6) that a1 2 /Iand in (7) that a > 8. PROOF.For the purpose of obtaining a contradiction, suppose that a-/? < al-B. Fromformulas (5)and (1)it follows that a = /l+(a-/?) < 8+(a1-8) = al, which contradicts the assumption a > a,. Similarly, assume that a-B > a-&; then it follows from ( 5 ) and (1) that a = ,B+ (a-p) > B (a-p1). This contradicts B , > ply because @ > p1 implies by (3) that 8+ (a-p1) 2 B1+ (a-B1) = a and consequently we would obtain a > a. The identity 0 - 2 = 0 - 3 = w shows that the symbol < in (7) cannot in general be replaced by <. THE FIRST MONOTONIC LAW FOR ORDINAL MULTIPLICATION : (8) (a < 8) -,(ya < y8) for y > 0. In fact, y/? is the type of the Cartesian product C x B ordered antilexicographically, where 3 = j3 and y. Let A= a and let A be a segment of B. The Cartesian product C X A ordered antilexicographicdly is a segment of C x B. Q.E.D.

+

c=

249

5 . ORDINAL. ARITHMETIC

THE

SECOND MONOTONIC LAW FOR ORDINAL MULTIPLICATION :

(a G B) (aY G BY). (9) To prove (9), we consider sets A , B, C of types a, Byy respectively, and we assume that A c B. Thus A X C c B X C which proves (9). It follows from the identity 1- w = 2.w that the symbol < in (9) cannot be replaced by <. +

LEFTDISTRIBUTIVITY

OF ORDINAL MULTIPLICATION WITH RESPECT TO

ORDINAL SUBTRACTION :

(10)

,8

for

a @ - y ) = ap-ay

y.

PROOF.Since multiplication is left-distributive over addition (see p. 223), we have by (5) aB = a[y+(B-y)] = ay+a@-y). This implies (10) by the definition of subtraction. We now prove a theorem about division of ordinal numbers.

THEOREM4: I f B iy an ordinal > 0, then for each ordinal exist ordinals y and

e such that

(1 1)

a=By+e

and

a

there

e
The ordinals y and e are then uniquely determined. PROOF.Since 1 < 8, we have by (9) a ,< ptc.If a = pa, then it suffices to let y = a and e = 0. Therefore suppose that a < Pa. The product @a is the type of the set B X A where = ,8, 2= a. It follows from the hypothesis c1 < /?a that the set B x A contains a segment O ( @ , a)) of type tc. Since

0, x> E O() = { ( x < a) v [(x = 4 A 0)< b)l), we have ((b, a>) = ( B x oA(a)) u (OB(b)x { a ) ) , where every element belonging to the first component precedes every element belonging to the second component. Since the first component is of type ,8.OA(a) = By and the second of type OB(b)= e, we have a=By+e where y < a, e < B. In this way we have proved that there exist ordinals y and e satisfying conditions (1 1). To prove uniqueness, suppose that

(12)

BY+&?

= BYl+el,

e<

B Y

el

< 8.

250

W.WELL-ORDERED SETS

Let y

> yl. Then y = yl+ ( y - y l ) and By+@ = B h + (Y-Yi)I = BYi+B(Y-Yi)+e.

+@

(13) Since y-y1 2 1: we have by (8) (14)

2 B.

B(Y--Y1)

It follows from (12) and (13) that BY1

+el 3 BY1 +B

(7 -711,

whence by (14) (1 5 )

BYl+@l

z BYl+B.

As B > el, we obtain by (1) /?yl+B > Byl+el. But this is impossible, because (15) implies ,8y1+el > Byl+el. Thus the hypothesis y > y1 leads to a contradiction. Similarly it can be shown that y1 > y does not hold. Hence y = yl. This formula and (12) imply that By+@ = by (1) this formula implies e = el. Thus Theorem 4 is completely proved. The ordinal y in (11) is called the quotient and e the remainder. The following theorem is a consequence of Theorem 4. THEOREM 5 : (THEEUCLIDEAN ALGORITHM FOR ORDINAL NUMBERS.) For any two ordinals uo and u1 different from 0, there exist a natural number n and sequences u2, ...)un, ..., Bn such that ul > a2 > ... > un > 0 and a0

= .,B,+a2,

a1 = a2Bz+%, un-2

* * a ,

= an-1 B n - 1

+an,

an-1

= UnBn.

PROOF.According to Theorem 4 and to the theorem on inductive definitions, there exist infinite sequences y and y such that and, for j > 1,

Vo = u07

YO =z

91 = u l ,

Vj-1

v1 =

= VJYI+I+V~+I

and


~ j - 1

Q?i+l = yj+l = 0

if

Vj

if

# 0, y j = 0.

251

5. ORDINAL ARlTHhmTrc

Since there exists no infinite decreasing sequence of ordinals, all terms of Q, from a certain term on are equal to 0. Let n‘ = min (vf= 0). J

Clearly, n’ > 0, because by assumption yo and y1 are # 0. To prove the theorem it suffices to let n = n’- 1, aj = qf for 2 j < n’ and /II = yj+l for 1 < j < n’. Now we shall prove several formuIas concerning the operation of taking limits (p. 235). We assume that A is a limit ordinal and that tp is an increasing A-sequence. Then we have lirn[a+plG)] = a+lim ~ ( 5 ) . (16)

<

€4

€4

p = limtp(5). If 5 < A , then q(F) < p and therefore C
dinal a+p is the smallest ordinal greater than all ordinals a+v([) for 5
€4

/3 = lim tp(E). For €4 < a - p . Let C < a.p. By Theorem that 5 = cry+@ < aj3 and e < a.

PROOF.We may clearly suppose that a # 0. Let

5 < A we have tp(5) < /I, thus a.&) 4 there exist ordinals y and e such If y > /3 then we have 5 2 a/?+@ 2 up, which contradicts the hypothesis. Thus y < p, which implies that for some 5 < A we have y q ( 8 . Hence 5 d a.tp(O)+e a.v(5) f a = a*[v(5)+11, and C d &.Q,(E+1), because the function q~ is increasing by assumption. Since A is a limit ordinal, we have 5+1 < A and the formula 5 < a.v(E+l) shows that C is not greater than some ordinal of the form a.p(r) where 7
<

<

252

W.WELL-ORDERED SETS

every p). It follows from (8) and (17) that the function p ( 6 ) = a.8 possesses the same property. Theorem 4, 0 3, p. 237 implies that there exist critical ordinals for the function s and for the function p . A critical ordinal for the function s is 5 = a v o , and, more generally, every ordinal of the form cl.w+@ where e is an arbitary ordinal. In fact, = a(l+o)+e

s(a.w+e) = a+a.o+e

= cl.o+e.

One of the critical ordinals for the functionp is lim a", where a" = u-cl

... a . All critical ordinals for the function p

n<m

will be determined

n

in the following section. Exercises l)

1. Determine whether lirn [p(n + y(@]

= limq(8

+ lim y(4) holds.

2. Determine whether lim [p(@].a= [limp(Q .a] holds.

3. Prove that if al+/31 = a+/3 and B1 > /3, then a1 < a. 4. Show that for every ordinal CL there exists a finite number of ordinals B such that the equation a = E+/3 is solvable for 5 (each such ordinal is called a remainder of a).

5. Show that if a sequence q is increasing, a = Iimq(5) and if €4

GC

then there exists an ordinal p < A such that @([) is constant for p ihs constant is equil to ths smill:st rerniinder of a ').

= p(E)+e(E),

<5
and

5 6. Ordinal exponentiation The operation of ordinal exponentiation is defined by transfinite induction as follows: (1)

(2)

yo

=

1,

Yc 'Y,

(3)

where il is a limit ordinal. l) More material related to the exercises can be foundin Sierpihki's books; in particular see Cardinal and Ordinal Numbers, 2nd edition (Warszawa 1964). z, See A. Hoborski, Fundamenta Mathematicae 2 (1921) 193.

253

6. ORDINAL EXPONENTIATION

We say that ya is the power of y, y is the base and a the exponent. It follows from the definition that if y 1, then

>

tl

(4)

ya < yfl.

< p -+

We shall prove that y'+q

(5)

= 5 . 1.

Y Y

PROOF.Given an ordinal E, let B denote the set of those c ~ W ( y +1) for which yC+c = ye. y'. We shall show that if [ y, then

<

W(c)cB+I;rB.

(*>

In fact, the following three cases are possible: (i) 5 = 0, (ii) c is not a limit ordinal; (iii) I; is a limit ordinal > 0. In case (i), I; E B, because yS+O = ye = ye- 1 = ye. yo. In case (ii), I; = Cl+ 1 where E W(C); thus, by assumption, C1 E B. Hence we have ,ye+cl = y C -ycl and therefore - €+(51+1) = ( C t 5 1 ) + 1 = = e -Y Y Yetr1 . = (y€.ycl) Y (YC1* Yi

el

=

which shows that thus

yrl+l

=

Y'.Y',

I; E B. Finally, in case (iii), 5 + C is a limit ordinal, yet5 = lim ya. a<E+t

Applying Theorem 3, p. 236 to the functions p(a)

= ya

and y(a)

= .$+a we infer that

lim y"

= lim ye+a,

a<€+[

Since for a :c

yb+c

a
we have

YE+".

I

a<

CIE: W(c),

C

it follows that a~ By i.e. ye++"=

y ' . ~ ' . Thus y'+c

=

lim (yc. ya) = y' lim ya a
= y'. ye,

a<:

which implies that I; E B. Hence implication (*) is proved. By induction it follows that B = W(q+ 1). Thus q E By which proves (5). (y9q = y".

The proof is analogous to that of (5). Let B denote the set of those (ye)' = yEc. It suffices to show that implication (*) holds. As previously we consider cases (i), (ii), and (iii). In case

5 E W ( ~ +1) for which

254

W.WELLORDERED SETS

(i) we have 5 E B, because (y?O = 1 = yo = y". In case (ii) we have 5; = cl+ 1, where tl satisfies the condition ( ~ ~ = ) ~ ytC1. 1 This formula in turn implies (f)S = (y€)C1+l = ("/)51y' = y"'y' = yssl+c = ye(cl+l) = y5 which shows that 5; E B. Finally, if 5; is a limit ordinal > 0, then

Since Theorem 3, $2, p. 236 implies that lim 7'" = lim yq = y", acC

qctc

we have 5 E B. Hence formula (6) is proved. y

(7)

> 1 + y' 2 E .

This formula follows from (4) by Theorem 5, p. 230. The operation of ordinal exponentiation allows us to find all critical ordinals of the function p ( t ) = a.5 (compare p. 252). Namely, these critical ordinals are all ordinals of the form am+O where Q is any ordinal. In fact, = Cll+(m+a) = a(l+m)+s = am+^ (am+O) = a. The function f(5) = y' is-according to (3) and (4)-increasing and continuous (on every set W(ct)). Thus this function possesses critical ordinals by Theorem 5, p. 239. According to this theorem, those critical ordinals can be obtained as the limits of the sequences c ~ , where go is an arbitrary ordinal and a"+' = 7"". For example, assuming that y = w , a. = 1 we have a1 = w ,

a2 = om,

a3 = wmm,

...

The limit of this sequence, E

= lim an, nca,

is the smallest critical ordinal of the function o', i.e. the smallest ordinal satisfying the equality (8)

we = E .

Such ordinals

E

are called epsilon-ordinals.

255

6. ORDINAL EXPOrJENTIATION

Exercises 1. Let ilbe a limit ordinal. Show that if the function p is continuous on the set satisfies the conditions ~ ( 0= ) 1, p(1) = y, p ( a + 8 ) = p ( a ) . y @ ) for a, p, a+p < A, then p(6) = y' for 6 < 1. 2. Show that if W F = a+/? and p # 0, then /3 = 0 6 . 3. Show that for every ordinal a there exists an epsilon-ordinal greater than a.

w(A) and

$7. Expansions of ordinal numbers for an arbitrary base

The operation of ordinal exponentiation can be used in order to represent ordinal numbers in the form similar to decimal expansion of natural numbers. For this purpose, we shall first prove the following theorem. THEOREM 1: If y and Q such that

> 1 and

a = yq.B+e,

1

< CI < y',

then there exist ordinals q,B and

B
O
e
PROOF.Let C be the smallest ordinal such that u < y:. Clearly 0 < C

< 5. If 5 were a limit ordinal, then we would have yc = A
0


yq

< CI < yqf?

By virtue of Theorem 4, $ 5 (p. 249) there exist ordinals /? and such that CI

=

e

rq.B+e, e < y".

If /3 2 y then we would have CI 2 y'J-y+e 3 yq++'. Therefore and the ordinals P,q and e satisfy the theorem.

B
THEOREM 2: If y > 1 and 1 < CI < yq, then there exist a natural number n and sequences ,P2, ... , Bn and ql ,v2, ... , rln such that (1)

u = ~ ' ~ B ~ + y ' l * / % f+ Y q n B n ,

(2) q > q l > q 2 >

... >q.,

o
for

i=l,2,

..., n.

The proof is almost a repetition of that of Theorem 5, p. 250. Namely,

2 56

W.WELL-ORDERED SETS

we define by induction three sequences vyy,6 such that

vo = min (ct < yc), c

To= a , qlj =

and

Y*'+16,+l+qI+l

if

qlI

if

qlj = 0.

qj+1

6o= 0,

< Y~'l+lY

Yj+1 < w j ,

6I+l


# O , and vj+1 = vj+l = 6 j + l =

0

The existence of these sequences follows from the theorem on definition by induction (see p. 239) and from Theorem 1. Clearly qlI = 0 from a j on. Now we let n* = min(qI = 0), n = n*-1 and qj = yI, Isj = aj I

for 1 < j c n * . Formula (1) for ordinals pj and yI satisfying condition (2) is called the expansion of an ordinal number a for the base y. The ordinals pj are called digits and the ordinals qi exponents of this expansion. If y = o,then the digits are natural numbers. Examples a = 02+co.5+9 is the expansion of the ordinal a for the base o. To expand the same ordinal for the base 2 it suffices to notice that o = lim 2" = 2m,thus n<m

o2= (2m)2= 2m*2

and

w. 5 = 2m+2+2*.

Therefore 02+o.5+9

= 2m4+24+2+2m+23+20.

In a similar way we obtain corn = 2m2.

For epsilon-ordinals, the expansion for the base o is 8 = o*.Thus an epsilon-ordinal E cannot be represented in the form (1) for y = o With exponents smaller than E . Two ordinals represented in the form (1) can be compared with respect to their magnitude by means of the following theorem. THEOREM^: I f q > & > yq

... > 5 p a n d y > 6 n f o r n < p , then

> y"16,+y5"2+

... +yEp6p.

251

7. EXPANSIONS OF ORDINAL NUMBERS

PROOF.From the assumption it follows that 7 2 & + l and ~ - 8 ~

> 0. This implies yq

Since ~

€

2 y" y = y"6,+ y'qy -81)

> y"y, 1

we have

a y h 8 , +y".

y"82+ybS. 2 Thus yq 2 yE18, y'"2 ~

€

1

+

+

Repeating this operation, after p steps we obtain the inequality yq 2 y'16,+yfa82+

... + y f p 8 p + y f p ,

from which the required inequality follows directly (since

If

THEOREM 4:

a = y"p1+

= y"'1+

where

> 0).

... +f"'-lpi-l+y"ipi+ ... +y"-'pi-1+y%9i+

> 7 2 > .*. > q n , O< ...ypn< 7,

~1

qi-1

... +y4npn, ... + y h Y p y

>t i >

0<6i,

-.*,8p

>tpy

< y,

and y"pi # y W i , then [(qi > ti)v (Ti = t i ) A (pi > 831PROOF. Suppose that qi > &. It follows from Theorem 3 that y""+ ... +yqnp > pi > yE'8i+yC'+1&+l+ ... + y w p p , (a > C)

therefore a > [, because by (l), p. 247 we have

... + y ~ i - l p i - l + y q ~ @ i +... + y q n p n > y"'1+ ... + y " - q ? i - l + y w i + +y'P8p = r. In turn, suppose that qi = Ei and pi > Oi. It follows from Theoa = y"pl+

1..

rem 3 that

y"(pi-si) > yqi > y4+181+1+ ... +y"Pp. This implies yqi8i+yvi((Bi-Oi) > yCiOi+YC'+lai+l+ *.* + y e p 8 p g , or y"pi > y w i + y€'+1&+1+ ... y W p .

+

Therefore we have y'ipi+ ... +Yqnpn >y'i6i+ which shows that a > [.

+yEp6p

258

VII. WELL-ORDERED SETS

Finally, if either Ti < Ei or qi = ti and pi < 61, then applying an analogous reasoning we obtain the inequality a < C. This concludes the proof of Theorem 4. Theorem 4 shows that expansions of ordinals possess properties analogous to those of expansions of natural numbers. In both cases, comparing the magnitudes of two numbers, we consider the first non-identical components of their expansions and we compare the exponents. In case the exponents are equal, we compare their coefficients. THEOREM 5: For a given base, every ordinal number can be represented in the form (1) in exactly one way. PROOF.In fact, by virtue of Theorem 4, if yqlpl+

... +y"n/$,

...

= ~ ' 1 6 ~ + +yE~P2op

(where ql > qz > ... > q n , 51 > 5 2 > ... > tpy0 < PI, ..., p n < Y, 0 < 61, . . . , 8 p< y), then n = p and q k = &, p k = 8k for k < n . Theorem 4 allows us to establish a connection between the notion of power of ordinal numbers defined in $6, and the notion of lexicographical ordering introduced in $ 5, Chapter VI. THEOREM 6: The power yq is the order type of the set of those functions belonging to the lexicographically ordered set W(y)"('J)which have values f 0 only for a finite number of arguments. PROOF.According to Theorem 4, $2, p. 233 yq = W(y7). To each ordinal CI E W(yq)there corresponds a unique expansion YfllBl+yq~B2+

***

+Y".il,

where q > ql > qz > ... > qnand 0 < pl, ...,pn'< y. In turn, this expansion can be associated with the function f a E W ( ~ ) " defined (~) as follows h(qi)= pi for i = 1,2, ..., n,

L(t) = 0

for 5 f vl,qz, *.., q n Conversely, each function g E W ( Y ) ~which ( ~ ; assumes values different from 0 only for a finite number of arguments can be associated with ) that g =fa. the ordinal a ~ W ( y q such Finally, it follows from Theorem 4 that fa precedes fs in the lexicographical ordering of the set W(y)"(q)if and only if a < C.

7. EXPANSIONS OF ORDINAL NUMBERS

259

As another application of expansion (l), we shall establish a characteristic property of powers of the ordinal w. if

DEFINITION: An ordinal e is said to be a remainder of an ordinal a e # 0 and there exists an ordinal a such that a = a + e.

THEOREM 7: In order that every remainder of an ordinal a be equal to a it is neccessary and su8cient that the ordinal a be a power of the ordinal cr). PROOF.If every remainder of the ordinal a equals a, then in the expansion a = co'11pl+w~2/9~+... + c r ) ' l n p n

we have n = 1 and p1= 1, i.e. ct = ~ " 1 . Suppose that a = cop and let e be a remainder of a. Thus for a certain o we have

(3)

a =a+@

and

o < a,

i.e.

u<

cop.

Let us expand a for the base w: a = w'h+wvvz,+

We have 7 < /3, i.e. q+1

< /3.

... +wvknk.

Let z = wB-wv+l. We obtain

+

I)+ wfl = wq(n+ 1) w q + l + z = wq(n+l+m)+z = wv.w+z = wv+l+ z = wB

cop
and we infer that cop = o f wp, that is, a = a + a. According to (3), o+e = a+a, which shows that = a. Q.E.D.

e

Using expansions for the base w we can define two operations on ordinals, called natural addition and natural multiplication'). These operations have more properties in common with operations of addition and multiplication of natural numbers than the operations of ordinal addition and multiplication considered before. In order to define these operation let us consider two ordinals

... + d k n k , ,8 = coclml+w~2mz+ ... +mclml.

a = coq1n,+wq2nz+

Upon completing these expansions by powers of w with the coeffi') G . Hessenberg, Grundbegrife der Mengenlehre (Gottingen 1906), pp. 591-594.

260

VII. WELL-ORDERED SETS

cients 0, we obtain expansions with the same powers of o: (4)

= w"pl+ WE2p2+

+wEhPl, 1

... + W B h q h .

@ = ws1ql+o"qq2+

(5)

.

The natural sum of u and @ is defined to be a(+)@ = o"(P1+41)+

+

o"((PZ+q2)

+w"(Ph+qh).

*.'

The natural product a( . ) p is defined to be the ordinal arising by formal multiplication of the expansions (4) and ( 5 ) as though they were polynomials in o:multiplying two powers of o we take the natural sum of the exponents and the terms obtained in this way are ordered according to their magnitude. Natural addition and natural multiplication are commutative. Examples 1. A natural sum may be different from the ordinary sum, for instance: 03+w~+~.2+1,

[o2+W+11(+)[w3+01=

= o3+w.

[o2+W+l]+[W3+o]

2.

[o'+ O+

3.

[OD+'+

+

1]( .)[o'

.

W]

= 05+w4+03 2+ w2+

O.

11( )[ o m + l + w"+ w ]

om+

- go.2+2

fOD.2+1.

2+

(9.2+

(-p+ 2 +

O"+l. 2+wD+o.

4. Expansions (4) and (5) can be rewritten as a = w"pl(+)w'2P2(+) =

.*-

(+)ufEhphi

41(+I4 2 (+I **

qh

-

Exercises 1. Show that the sum a ( + ) p is an increasing function with respect to a as well as with respect to p. 2. Show that for every ordinal y there exist at most finitely many pairs a,B such that a(+)B = y. 3. Prove that if €

< mua and q < mua,

then E.7

< coma.

7. EXPANSIONS OF ORDINAL NUMBERS

261

Conversely if an ordinal 5 satisfies the condition

(6 < O . h < 0 -+ (6.q < 0. (6) then there exists an LY such that = UP'. REMARK: The ordinals 5 satisfying (6)for all e, r ] are called the principal ordinals of multiplication '). tj 8.

The well-ordering theorem

Well-ordered sets owe their importance mainly to the fact that for each set there exists a relation which well orders it. This theorem, called the well-ordering theorem or Zermelo's Theorem ') is equivalent to the axiom of choice on the basis of the axioms Z[TR]. In this section we shall prove this equivalence and formulate several theorems equivalent to Zermelo's theorem. Some applications of these theorems will also be given.

THEOREM 1: If A is any set such that there exists a choice function for the family 2A-{0}, then there exists a relation well ordering A . PROOF.Let f be a choice function for the family 2A-{0}; we can extend this function to the whole family 2A letting f(0) = p where p is any fixed element which does not belong to A. Now let C denote the family of relations R c A x A well ordering their field. In virtue of the axiom of replacement, there exists a set consisting of all ordinals where R E C . Let a be the smallest ordinal greater than every ordinal R of this set. According to the theorem on definition by induction, there exists a transfinite sequence q . ~of type a such that Pls =f(A-{%I :7 < El). If yc#pthenpcEA-{~,,:r]
262

W.WELL-ORDERED SETS

the definition of a. Therefore there exists the smallest ordinal?!, such that vS = p . This implies that A = { v , : q < p}, thus A is the set of all terms of a transfinite sequence of type p whose all terms are distinct. Consequently there exists a relation well ordering the set A into type ,& Q.E.D. REMARK: In the proof above we used only ordinals of the form where R c A x A and the ordinal a. It is therefore easy to reformulate the proof in such a way as to eliminate the notion of an ordinal. We simply replace by the family of all relations S c A x A which are similar to R and a by the family of all well-ordering relations R c A x A . The proof thus modified can be based on axioms Z and is independent of the axiom of replacement l). Another method of eliminating ordinal numbers consists in replacing them by the so-called von Neumann’s ordinals to be discussed in $9. The proof of Theorem 1 obtained by this modification is also based only on the axioms Z but this time with the axiom of replacement. The converse to Theorem 1 is also true: THEOREM 2: r f there exists a relation well ordering a set A, then there exists a choice function for the family 2*-{0}. In fact, for X E - (0) ~ we ~ define f ( X ) to be the first element of X in the given well-ordering. The following corollary is a direct consequence of Theorem 1. COROLLARY 3: For every set A there exists a relation well ordering A .

Now we shall formulate the so-called maximum principle, which is often used in place of the well-ordering theorem. Let A be an ordered set. We say that A is a closed set if for every linearly ordered subset B of A the supremum x exists in A . An

v

XEE

element x is said to be maximal if there is no y E A such that y

> x.

l) Such a proof can be found in many books; see for example A. A. Fraenkel, Abstract Set Theory (Amsterdam 1953), pp. 309-315.

263

8. THE WELL-ORDERING THEOREM

THEOREM 4: I f A is an ordered closed set and there exists a choice function for the family 2A-{0}. then there exists a maximal element in A1). PROOF.As in the proof of Theorem 1, we extend f by letting f ( 0 ) = p where p is any element not belonging to A. Let a be the ordinal defined in the proof of Theorem 1. By theorem on transfinite induction there exists a sequence y of type a such that

v

vc =f ( { xE A : x > ll4 V,,})Y where the symbol after the sign f denotes the empty set if not all y,,, 7 < E, belong to A or if the supremum y,, does not exist.

v

W E

Since A is closed, we see that if y~ # p , then v,, # p for every ory,, exists and is < vc. If dinal 7 < 6 and the supremum k =

v

ll
we had ~l~ f p for every 5 < a, then there would exist a transfinite sequence of type a with distinct terms belonging to A. But this contradicts the definition of a. Consequently, there exist ordinals [ such that cpE = p. If?! , is the smallest such ordinal, then there is no X E A ve. Hence the supremum pC is a maximal elesuch that x >

v

€
E d

ment of A. Q.E.D. O COROLLARY 5 : (The so-called ZORN MAXIMUM PRINCIPLE.) In every closed, ordered set there exists a maximal element. The following theorem is a special case of Theorem 4.

THEOREM 6: If A is a family of sets with the following property

(*I

(uX) A for every monotonic family B E

c

A

XEB

and if there exists a choice function for the family 2 A - { O } , exists a maximal element in A ? ) .

then there

I) M. Zorn, A remark on method in transfinite algebra, Bulletin of the American Mathematical Society 41 (1935) 667-670. See also the next reference. . *) Theorem 6 was proved by K. Kuratowski in his paper Une mdthode d'dimination des nombres transfinis des raisonnements mathdmatiques, Fundaments Mathematicae 3 (1922) 89.

264

W.WELL-ORDERED SETS

For the proof it suffices to notice that the family A is ordered by the inclusion relation and the union X is the supremum .of B.

u

XFB

Theorems 4 and 6 show that the exlstence of maximal elements follows from the axiom of choice. We now show that, conversely, the axiom of choice follows from the existence of maximal elements, 7: If for every family of sets A satisfying condition (*) THEOREM there exists a maximal element, then for every fami& Z of non-empty sets there exists a choice function. PROOF. Let A be a family of functions f such that (i)The domain off is a family C’ c 2. (ii) f (X)E X for all XEC’. Let us recall that a function f with domain C’ is the set of pairs (x,f ( X ) ) where X EC,. Thus the formula fi c fi means that C’, c C’, and f i ( X ) =f,(X) for all XECf,:

that is, the functionf, is an extension of the functionf,. We shall show that the family A satisfies condition (*). For this purpose, suppose that B is a monotonic family included in A and let F denote the union B. The elements of the set F, are pairs of the form (X,y ) where X E Z . In fact, each component f of the union F is a set of such pairs. If (X, y 1 ) € F and ( X , y z ) ~ Fthen , there exist fi andf, such that (X,y l ) ELEB and (X, y z ) ~ fE B. i This implies that y 1 =fi ( X ) and y z =f, Since the famiIy B is monotone, we have either fr cfi or fi cfi. By (1) we infer that in both cases, y I =fi(X) = f i ( X ) = y z . Thus the set I: satisfies the condition

u

(a.

[(X, Y l ) E FI A [(X,Y2) E PI -b (Yl = Yz), i.e., F is a function. The domain of this function is a family included in 2.

u C’, i.e.

f EB

If (X, y> E F then there exists a function f~ B such that (X,y) ~ f . This implies that y =f ( X ) ; hence y = F ( X ) =f ( X ) E X . Therefore the function F belongs to the family A. This shows that this family satisfies condition (*).

265

8. THE WELL-ORDERING THEOREM

fo

By assumption there exists a maximal element in the family A . We shall show that C,, = 2. Suppose the contrary. Then there exists an element X of the difference 2- Cf0which is a non-empty set. Thus there also exists an element x E X. Letting f =fo u {(X,x)>, we obtain focf,fo # f and f E A . But this contradicts the hypothesis that fo is a maximal element in A . Hence the function fosatisfies the condition fo E X for all X E Z . Q.E.D. Applications 1. Extension of an order to a linear order. 'THEOREM8: For every relation Ro ordering the set A there exists a relation linearly ordering A which contains R,. PROOF.Let R be the family of relations ordering A and containing R,. It can easily be shown that this family satisfies condition (*) of Theorem 6, therefore it contains a maximal element R. We shall show that R is the required extension of R, to a linear order. Since R is by definition an order, it suffices to show that the relation R is connected. Suppose on the contrary that there exist elements a, b E A such that l ( a R b) A l ( b R a). We shall show that the relation R' = R u { ( x ,y>: ( x R a) A (b R y ) } = R u S orders A. This will provide a contradiction since R' 2 R and R' # R. Clearly x R'x for every x E A. To show that the relation R' is transitive, suppose that x R'y and y R'z. One of the following cases holds :

(a

(0 ( x RY) 0,R 4; (ii) ( x R y ) h ( y Ra) A (b R z ) ; (iii) ( x R a) h (b R y ) h ( y R z); (iv) ( x R a) h (b R y ) A ( y R a ) v (b R 2). Case (iv) is impossible: it implies b R a, which contradics the hypothesis. In case (i) we obtain x R z since R is transitive. Therefore we also have x R'z. In cases (ii) and (iii) we obtain ( X R a) A (b R z ) by the transitivity of R, thus X S Z and, consequently, x R ' z . This shows that R' is transitive.

266

VII. WELL-ORDERED SETS

Finally, in order to prove that R' is antisymmetric, suppose that x R'y and y R'x. We now have the cases analogous to (ixiv) where z is replaced by x. Cases (G)-(iv) are impossible, case (i) implies x = y. Thus Theorem 8 is proved.

2. Let A be a distributive lattice and I, its ideal not containing an element b (see p. 163). The family of all ideals Z 3 I, of the lattice A not containing the element b satisfies condition (*) of Theorem 6. Therefore there exist maximal elements of this family. In particular, there exist maximal elements in the family Pa.6 of those ideals which contain a but which do not contain b (provided that a n o n a b , (see p. 164). If, in particular, A is a lattice with unit i and I is an ideal different from A (hence I does not contain i), then there exists at least one maximal ideal different from A and containing I. Such an ideal is prime (see p. 168, Exercise 2). 3. Let m 2 a, and let M be a family of a power < itt such that each element of M is a subset of A and has the power m. Then there is a set Z such that O

O

(2)

z==

ZcA,

Z=m,

-

A-Z=m

and Z n X # 0 # X - Z

for every X E M .

PROOF. In virtue of the well-ordering theorem there is a smallest ordinal a such that there exists a sequence of type a without repetitions composed of all the elements of A. Let xE denote the 6th term of this sequence and let ME be the 5th term of a sequence of type a (not necessarily without repetitions) which contains all the elements of M. We define by transfinite induction two sequences p and q of type a. Namely p e is the first term x, belonging to Mc- S,, and qc is the first term x,, belonging to (ME- { p J )- S, where

s, = {pq: q < 5 ) u {qq: q < 5). The elements p Eand q, exist, because the set SE is of power and M Eis of power nt.

< ni

267

8. THE WELL-ORDERING THEOREM

The set Z consisting of all p e where E < a satisfies condition (2). # The theorem above has an interesting topological application l). Let A = € (thus m = c) and let M be a family of non-empty perfect sets (a perfect set is a set identical with its derivative). Such sets are of power c2). Hence there is a set 2 which has a point in common with every perfect subset of the set & and whose complement possesses the same property. It can be proved that such a set is non-measurable in the sense of Lebesgue. # 4. Let X be an arbitrary set, R* a family contained in 2x. We shall prove that the set A of all families R with the finite intersection property satisfving the condition R* c R c 2x possesses property (i) (see p. 139). Suppose that B c A and that the set B is linearly ordered by R EA . Clearly, it suffices to show that inclusion. We shall show that

u

REB

this union possesses the finite intersection property. Let EN and Xi E B for i < n. For each i there is a family Ri E B such that XiE Ri. Since B is linearly ordered, one of these families, say R,, contains all the remaining families. This implies Xi€Ro for i < n and since R, has the finite intersection property, we obtain # 0.

n i

Q.E.D. It follows from Theorem 6 that for every family R, c 2x with the finite intersection property there is a maximal family R with the finite intersection property such that R c 2x and R contains R,. # 5 . Hamel's basis 3). A set X c & is said to be independent if for any finite sequence x,, x l , ..., x , - ~ of distinct elements of X the equation r, xo+ rl xl+ ... +rn-l x , - ~ = 0 is satisfied by rational numbers r,, ..., r,+ if and only if all these numbers are equal to 0. An example of an independent set is {fi,

d?}.

I) See F. Bernstein, Zur Theorie der trigonometrischen Reihen, Leipziger Berichte 60 (1908) 329. See K. Kuratowski, TopoZogy I , p. 514 (Warszawa 1966). ') G. Hamel, Eine Basis aller ZahZen und die unstetigen L6sungen der Funktionalgleichung f ( x + y ) = f ( x ) + f ( y ) , Mathematische Annalen 60 (1908), 459-462.

268

VII. WELL-ORDERED SETS

It is easy to show that if B is a monotone family of independent X is also an independent set. By Theorem 6 this implies sets, then

u

XEB

'THEOREM9 : There exists a maximal independent set. Such a set is called a Hamel basis for &. If H is a Hamel basis, then every number x # 0 can be uniquely represented in the form

(3)

ribi ,

x= i
where n E N,bi are distinct elements of the basis and ri rational coefficients different from 0. For if there existed a number x not having such a representation, then the set H u { x } would be independent, contrary to the assumption that H is maximal. On the other hand, if there were two representations rib; = ryby, then the elements i
j<m

of the set {b;, ...,b;+, b:, ..., b;-J would not be independent. O COROLLARY 10: There exist non-continuous functions of 'the real variable x satisfying for all x , y the equation

In fact, let H be a Hamel basis and let x0eH. Denoting by f ( x ) the number ro such that in the expansion (3) x, occurs with the coefficient ro we obtain the required function. This function is not continuous, because it takes only rational values and is not constant. # The theory given in this section enables us to prove the theorem mentioned on p. 196. O THEOREM 11: If 1 is a function defined on the set T whose values are cardinal numbers, then there exists a function I; defined on T such that = it for every t E T.

PROOF. Let t E T. Since it is a cardinal, there exists a set X such that it = y. According to Corollary 3 there exist an ordinal a and a relation R such that R orders X into type a. Let a, be the smallest ordinal with this property. Now we define the function F by Ft = W(a,). Exercises 1. A family A of sets is said to be inductive if i f has the following properties:

8. THE WELL-ORDERING THEOREM

269

(i) if XEA, then every finite set Y c X belongs to A, (ii) if every finite set Y c X belongs to A, then X EA. Show (without the axiom of choice) that the maximum principle is equivalent to the theorem: every inductive family possesses a maximal element I). 2. Show (without the axiom of choice) that the maximum principle is equivalent to the following theorem: every linearly ordered subset Z of an ordered set A (that is, a set with the property x , y E Z 4 [(x < y ) v (y Q x)] is contained in the maximal linearly ordered set included in A ”. 3. Show (without the axiom of choice) that the maximum principle is equivalent to the following theorem: for every family F of non-empty sets there exists a maximal family of disjoint sets contained in F *).

0 9. Von Neamann’s method of

elimination of ordinal numbers

In this section our .exposition is based exclusively on the axioms of

z.

We shall show that it is possible to define sets possessing exactly the same properties as ordinal numbers. We shall establish a one-toone correspondence between those sets and types of well-ordering relations. DEFINITION 4, : A set A is said to be an ordinal number in the seme of yon Neumann (briefly: a VN ordinal) if it has the following properties : 1. Every element of A is a set. 2. If XEA then Xc A . 3. I f X , Y E A then X = Y or X E Y or Y E X . 4. If 0 # B c A then there exists an X such that X EB and Xn B = 0. 1) This formulation of the maximum principle was given by Teichmiiller in the paper Braucht der Algebraiker das Auswahlaxiom?, Deutsche Mathematik 4 (1939) 567-577. This paper also gives other forms of the maximum principle. Garrett Birkhoff, Lattice Theory, Second Edition (New York 1948), p. 42. *) R. L. Vaught, On the equivalence of the axiom of choice and a maximal ptinciple, Bulletin of the American Mathematical Society 58 (1952) 66. A great number of other theorems equivalent t o the maximum principle, as well as many applications of the axiom of choice and Zermelo’s theorem, can be found in the book: W . Sierpidski, Cardinal and ordinal numbers (Warszawa 1965). See also H.Rubin and J. Rubin, Equivalents ofthe axiom ofchoice (Amsterdam 1963). 3 J. v. Neumann, Zur Finfiihrung der transjiniten Zahlen, Acts Litt. Ac. Scientiarum r. Univ. Hung. Franc. Joseph., Sectio sc. math. 1 (1923) 199-208.

270

VlI. WELL-ORDERED SETS

Examples of’ VN ordinals: (i) the empty set No = 0, (ii) the set Nl = { 0 } , (iii) the set N2 = (0, { 0 } } = {No,N l } , (iv) the set N3 = { N o , N l , N 2 } , (v) the set N , = {No,N l , N z , ...}, (vi) the set = N,+{N,}.

We prove several properties of VN ordinals. 5 . If A is a VN ordinal then there exists no finite sequence of sets xi, ...>XkSuch that x k E X 1 E X 2 E ... EXk-lEXkEA. PROOF.Suppose that there exist sets X,, ...,X , with this property. Let B = {X,,..., &}. Since A, we have X k c A by 2 and thus Xk-1 E A. By the same argument, X k - 2A~ and so forth. Therefore all the sets X,, ...,Xk belong to A and consequently B c A . None of the sets Xi satisfies condition 3. In fact, for i > 1 we have X i - , € X i n B and for i = 1 we have XkEXlnB. Q.E.D. 6. If A is a VN ordinal and M EA, then M is also a VN ordinal. PROOF.We prove that M satisfies 1-4. (i) If X E M , then we also have X EA because 2 implies M c A, and thus X is a set. (ii) Suppose that X E M and Y E X . We have

Y E X EM EA, which in view of the inclusion M c A implies that Y E X EA ; consequently Y EA by 2. According to 3 we have either Y = M or M E Y or Y EM . In the first case we obtain

M E X EM EA , and in the second

MEYEXEMEA, which contradicts Theorem 5 . Thus Y E M .As Y is arbitrary we infer that X c M . Therefore the set M satisfies condition 2. (iii) If X , Y EM , then X , Y EA because M c A. Thus 3 implies that either X = Y or X E Y or Y E X . (iv) Suppose that 0 # B c M . The set B is thus a non-empty sub-

9. VON NEUMANN’S METHOD

271

set of A and in view of 4 it contains an element X such that X n B = 0. This shows that M itself satisfies condition 4. 7. r f A and B are VN ordinals, then ( A E B) = ( A c B) A ( A # B).

PROOF.If A EB, then A c B by 2 and A # B since otherwise we would have B EB, contrary to 5. Suppose that A # B and A c B. The set B-A is therefore a non-empty subset of B and according to 4 there exists a set X E B-A such that X n ( B - A ) = 0. Now it suffices to show that X = A , because X = A together with X E B imply A EB, which proves the theorem. The condition X E B implies X c B. Since X n (B- A) = 0, we get X--A = 0, that is, X c A. Suppose that A-X# 0. Hence there exists a set Y such that Y EA-X and Y n A - X = 0. Since A-X c B, we infer that Y E B , and according to 3 we have either Y E X or X E Y or X = Y. But Y E X is impossible, for Y EA-X; similarly X E Y is not the case, because it would imply Xq! A-X, that is (in view of X EA ) X E X EA , contrary to 5 ; finally, we cannot have X = Y since Y E A and XEB-A. Hence we have proved that A-X = 0, i.e. A c X , which shows that A = X. 8. Each VN ordinal is well ordered by the inclusion relation. PROOF. It suffices to show that if A is a P N ordinal, then (a) the inclusion relation is connected in A ; (b) every non-empty subset of A has a first element. Now (a) follows from 3,6 and 7; and (b) follows from 4 and 7. 9. If A and B are VN ordinals, then A n B is also a VN ordinal. PROOF. We show that A n B satisfies conditions 1-4. (i) From X EA n B it follows that X EA ; thus X is a set. (ii) From X E A ~ itB follows that X E A and X E B; hence X c A ; X c B and consequently X c A nB. (iii) From X , Y EA n B it follows that X , Y EA ; thus X = Y or X E Y or Y E X . (iv) From 0 # M c A n B it follows that 0 # M c A ; hence there exists an X such that X E M and X n M = 0.

272

w.WELL-ORDERED sm

10. If A and B are VN ordinah, then either A c B or B c A.

PROOF.In fact, suppose that A # A n B # B. From 9 and 7 it follows that A n B E A and A n B E B . This implies A n B E A n B , which contradicts 5 because A n B is a VN ordinal. Hence A = A n B or B = AnB. 11. If A and B are distinct VN ordinals, then either A is a segment of B or B is a segment of A. Consequently, these sets are not similar (with respect to the inclusion rehtion). PROOF,Suppose that A # B; by 10 we have either A c B or B c A. Suppose that the former holds. It follows from 7 that A E B, which shows that the elements of A precede (in the set B) the element A . Hence the set A is a segment of B, consequently it Cannot be similar to B (see Corollary 7 , p. 231). 12. For every relation R well ordering its $eld there exists exactly one VN ordinal ordered by the inclusion relation simihly to R. PROOF.Let Z be the field of the relation R and H the set of those z E 2 for which there exists exactly one V N ordinal N, satisfying the condition: N, is ordered by the inclusion relation similarly to the segment O(z) of the set 2;in this case we write N, N O(z). Suppose that O(x) c H. We shall show that X E H. By 11 there exists at most one V N ordinal similar to O(x). Hence it suffices to show that there exists at least one such V N ordinal. Let N, = (2 < X ) A (x= N,)}.

(x''V z

We are going to show that N, is a VN ordinal. In fact, condition 1 clearly holds. If N, B N, and Y E N ,,then Y is a V N ordinal. Moreover, since Y E N , O(z), we see by 11 that P is similar to a segment of N,; hence Y is similar to a segment O(t) of the set O(z). Since t < z < x, we obtain t e O ( x ) and, consequently, ~ E : Hand Y - N t . In view of 11 this implies Y = Nt and finally Y EN, , because Nt EN,. Thus N, c N,;that is, N, satisfies condition 2. Condition 3 follows directly from 10 and 7. Now let B be a non-empty set contained in N , and let zo be the

-

273

9. VON NEUMANN'S METHOD

smallest element of 2 such that N,,E B. If there were Y E N , , such that Y E B , then we would have Y = N, where z i z,, , which contradics the definition of zo. Hence the set N . is a VN ordinal. If z1 i z2 i x , then N,, O(z,) and N,, N O(z2).Thus N,, is similar to a segment of N,,.This implies by 11 that N,, c N,, and N,,# NZ2.Therefore N. O(x), which proves that x E H. By induction we now infer that H = Z . The set

-

N* = {

N

x:vx ( x E Z ) A ( X = N,)}

is a V N ordinal ordered similarly to Z . The proof is analogous to that . carried out for the set N It follows from properties 12, 11, 8 that V N ordinals indeed satisfy all the requirements for ordinal numbers. In connection with the reasoning just given it is worth mentioning that in the proof of property 12 we made essential use of the axiom of replacement. Without this axiom the existence of the sets N, and N* could not be proved.

CHAPTER VIII

ALEPHS AND RELATED TOPICS In this chapter we shall discuss application of the theory of wellorderings to the arithmetic of cardinal numbers. fj1. Ordinal numbers of power a I)

The cardinal number of an ordinal 5 is the power of any set ordered in type 5. We denote this cardinal number by Thus i= W(5,.

-

Ordinals of power a can be treated as the types of well-ordered sets of natural numbers. This fact implies the following theorem. 1: All ordinals of power a form a set. THEOREM

DEFINITION 1: The smallest ordinal greater than every ordinal of power a will be denoted by Q. The existence of the ordinal 9 follows from Corollary 8, p. 234. THEOREM 2: (5 < Q) = (g< a). PROOF.If 2 < a, then 5 < 9 by Definition 1. Conversely, if 5 < 9, then there exists an ordinal 5 such that E < C and = a, thus 2 <‘f:= a.

a

-

DEFINITION 22): K1= i.e. K1= W(J2). In Theorem 2, letting 6 = 52 we obtain K1 non-< a. On the other hand, a < K1 and thus we have the following corollary. I) Called by Cantor wnabers of the second clam According to Cantor, the first class consists of h i t e numbers. 2, t( is the Hebrew letter aleph.

275

1. ORDINAL NUMBERS OF POWER a

COROLLARY 3: K1> a. It follows that the set of ordinals 5 such that W(Q), is uncountable.

F<

a, that is, the set

THEOREM 4: If nt < K1 then m < a. In other words, there is no cardinal which lies between a and K1. PROOF.Let m < K1. Then there exists a set M c W(Q) such that = m. Let 5. Thus 5 Q. Moreover, t 252 as otherwise g= and m = K1. Therefore, 5 < Q and thus by Theorem 2, 5 < a and thus m d a. Q.E.D. The following form of the induction principle holds for ordinals

a=

a

<

5 < sz.

5: Let the set A of ordinals satisfy the conditions: THEOREM

(1)

OEA,

(2)

(3)

5eA

-+

(5+1)~A,

if pl is an increasing sequence and

if

pl(n) E A

for n E N, then

[limpl(n)]E A. n

Then W ( Q ) c A .

PROOF.Suppose that the theorem does not hold. Let a be the least ordinal such that a < s? and a ?$A. From (1) it follows that a # 0. If a is not a limit ordinal, then a = 1 for a f in A, whence by (2) t f l ~ A , and U E A , contrary to the definition of a. I t remains to examine the case where a is a limit ordinal. Since 5 < a, there exists a relation R which well orders the set N of natural numbers into type a. We define a sequence k o , k l ,... of natural numbers by induction: let ko be the first element of the set N with respect to~the relation R and let k,+l be the least number k > k , such that OR(~.) < O,(k) (where OR@) denotes, as usual, the segment of N determined by k ) . Such a number k exists because a is a limit ordinal. __ Let pl(n) = OR(k,). Thus p ( n ) < pl(n+ 1). Moreover, pl(n) < a for n = 0, 1,2, ... , because O,(k.) is a segment of a set of type a. If 5 < a then there exists a number rn such that OR(rn)= 5. Since the sequence k o , kl,... is increasing, it follows that for some n, OR(m) c OR(kn),

[+

27 6

whence 5

W I . ALEPHS A N D RELATED TOPICS

< p ( n ) . Hence a = limp(n). On the

other hand, p(n)


n<w

implies p ( n ) E A for n = 0 , l ,2, ... and thus by (3), U E A . But this contradicts the definition of a. Clearly the set WCQ) satisfies conditions (1) and (2). In fact, a more general theorem holds for W(52). THEOREM 6 : If 5 < 52 and 7 < Q, then E f q < Q and t - 7 < Q. For 5 f r l = z+ij and 5.);1= Cij; since a, a+n = a - g< a and = a.a, it follows that E+q < a and 5 . 7 a. Using the axiom of choice we shall show that the set W ( Q ) satisfies condition (3).

<

<

‘THEOREM 7:

If y (1) < p (2) < ... and p (n) < 52,

then limp (n) < 52. n i o

PROOF.Let a = limp(n). Then W(a)= n
u W(p(n)). The set W(a), n

being a countable union of countable sets, is countable (see p. 176). Thus a < 9. Application. Let F, be the set defined on p. 242. ‘THEOREM 8: The family Fa is identical with the family B of Bore1 sets (i.e. the least a-additive and cs-multiplicative family containing all closed sets). PROOF.By transfinite induction with respect to a it is easy to show that for every a (in particular for a = Q), Fa c B. It remains to show that the family Fa is a-additive and a-multiplicative. For this purpose let X be a sequence of sets such that X , €Fa for all n. Thus for all n there is an ordinal a, such that XnE Fan; we may assume that an is the least ordinal greater than an-l such that X,eFan. By Theorem 7 it follows that there exists p such that an < p < Q for every n, moreover, we may assume that is, for instance, odd. Then

Q.E.D. Ekercises 1. Prove (without using the axiom of choice) that if aB

< Q.

CL

< Q and B < Q, then

277

1. ORDINAL NUMBERS OF POWER 0

# 2. Let @, be the family of analytically representable functions of class u (see

p. 242). Prove that the union

u @,

is the least family of real functions such that

a
(i) every continuous function belongs to the family, (ii) iff. belongs t o the family for n E N and if f ( x ) = lim f n ( x ) for every x, n +ca

then f belongs t o the family. Discuss the role played by the axiom of choice in the proof of this theorem. 3. If { X a ) a < n is a sequence of type P of closed subsets of the space t or N N and if X , 2 Xu+,for all u < 9, then there exists an ordinal 52 such that x,= X, for all u > B. Hint: Denote by No, N l , N,, ... a sequsnce whose terms constitute an open subbase of the space. Associate with the ordinal u where Xu # Xa+l the least number m such that N m n X a # 0 = Nm-X,+, and show that with distinct u are associated distinct natural numbers. 4. Prove that if A is an arbitrary subset of & (or of N N ) , then for every transfinite sequence of derivatives of A (see p. 241) there exists a term A(,) such that A(@)= A(,,) for all u > B, where B < 52. Hint: Notice that all derivatives A(,,) are closed sets and apply Exercise 3. 5. Prove the following theorem of Cantor-Bendixson: Every set A c & (or A c N N ) is the union of a perfect set and of a countable set. Hint: Show that the difference A -A' is a countable set (see Exercise 5, p. 178). #

-=

6. Show that 2 , ' > K, without using the axiom of choice. Hint: The set 2NxN is the union Z U UZg, where 2~is the set of relations well-ordering their fields in type are not well-orderings.

8 2.

E
E and where Z is the set of relations which

The cardinal K(m). Hartogs' aleph

We now generalize the construction carried out in cardinal a to the case of an arbitrary cardinal m.

01

for the

THEOREM 1: For every cardinaZ rn there exists a set

< nz}.

c.

Zjm) = ( 5 : E

PROOF.Let A' = in. Every relation R whose field is contained in A is a subset of A x A , that is, R c 2AxA.Therefore there exists a set R of all relations R c 2AxAwhich well order their fields. Associate with every relation R E R its type. By the axiom of replacement we obtain the set Z(m) of ordinals such that

f€z(m)-+Z < m.

278

VIIX. ALEPHS

AND

RELATED TOPICS

<

m,then there exists a relation R ordering a subset Conversely, if of A in type 5. Thus [ ~ Z ( n t ) . Q.E.D. THEOREM 2:

If E E Z(m),

then W(E)c Z(m).

For 7 < E implies that

< j.

___3

DEFINITION: K(m) = Z(m). By this definition we have correlated with every cardinal nt an aleph Z(m). This operation is referred to as Hartogs’ aleph function’).

-

3: K(m) non THEOREM

< m.

PROOF.The set Z(m), as a set of ordinals, is well ordered by the relation (see p. 234). Let [ = 2 0 . Suppose that K(m) my that is 5 m. Then 5 E Z(m) and thus by Theorem 2 the set W(5) is a segment But this is impossible because by Theorem 4, p. 233 -of Z(m).“(5) = 5 = Z(m), and no set is similar to its segment (p. 231).

<

<

<

COROLLARY 4: m < m+K(m). The inequality is obvious and equation is impossible as it implies m. that K(m)

<

<

THEOREM 5 : If there exists an ordinal E of power m ythen K(m) > m. PROOF.By Theorem 2, W(5)c Z(m) and thus K(m) 2 = m; hence by Theorem 3, K(m) > m.

r

THEOREM 6: For every set X of ordinal numbers there exists an ordinal a such that T < for every 5 E X .

a

PROOF.Let S

=

U W(E), m

=

5 and

-

a = Z(m). Since

-

5 = W(5)

€EX

and W(5)c S for every 5 EX, we have by Theorem 5,

F=

-~ ( 6-<) s

THEOREM 7: K(m)

< 2“m)

=m

< K(m)

- ( m= ) a.

=~

< 22m2.

PROOF.Let A’ = m and let X be the family of those relations R c A x A which well order their fields. Clearly, X c 2AxA.The set X is the disl) F. Hartogs, vber das Problem der Wohlordnung, Math Ann. 76 (1915) 442. Theorems 3 and 5 are due to Hartogs.

2. THE CARDINAL

u(m). HARTOOS’ ALEPH

279

joint union (2)

where X , is the subfamily of X consisting of relations of type a. To every subset Y c Z(m) there corresponds in a one-to-one manner the X , = F(Y) c X , therefore the family of subsets Y is of power union

u

aEY

<2X = 22m2.Thus 2N(m)< 22m2.Q.E.D. P

Exercises 1. Using the axiom of choice show that

K(m) < 2m2.

(3)

Hint:From the axiom of choice it follows that there exists a set T,containing exactly one element from each of the sets &(see formula (2)). From the identity mf = nt which we shall prove in $6, p. 292, we conclude a stronger inequality, namely

2. In the definition of the set Z(m) and of the cardinal K(m) replace the relation by (see p. 186) and prove for so defined K*(m) the theorems analogous to Theorems 3-5. [Lindenbaum]

<

<*

83. Initial ordinals The ordinal p is said to be an initial ordinal if p is the least ordinal .$ such that 2 = @; that is: (1)

y

< p + y < ij.

For example, the ordinals (o and SZ are initial ordinals. We shall also denote these ordinals respectively by w oand col in agreement with the notation for initial ordinals which we shall introduce in this section. THEOREM 1: For every infinite cardha1 m,the type p of the set Z(m) (that is, of the set of all ordinals of power < m) is an initial ordinal. Since 5 < y E Z(m) implies 6 E Z(m), it follows that (see p. 234) Z(m) = W(p). On the other hand, since p $ W(p), we have p $Z(m), whence -pl non m. From m 2 7 it follows 9, > 7; thus (1) holds. The proof of the following more general theorem is similar.

<

280

Vm. ALEPHS AND RELATED TOPICS

THEOREM 2: If m is a function deflned in a set X and m, >, a for

every x E X , then the ordinal

U Z(m,)

is an initial ordinal.

XfX

For a given initial ordinal 9) we shall denote by P ( v ) the set of all initial ordinals y < cp and we let

-

(2)

4 v ) = P(cp>

*

DEFINITION 1: The ordinal L ( Q ) ) is said to be the index of the initial ordinal v. Clearly L ( W ) = 0, and r(S2) = 1. THEOREM 3: I f y and 9 are initial ordinals and if y < 9, then ~(y)

< 4P). PROOF.By assumption, y E P(v), and it follows that P(y) is a seg- nient of the set P(p); thus P ( y ) < P ( v ) . THEOREM 4: To distinct initial ordinals there correspond distinct indices. Theorem 4 follows from Theorem 3. THEOREM 5: Every ordinal a is the index of some initial ordinal. PROOF.Suppose that a is the index of no initial ordinal. Assume, moreover, that a is the least ordinal having this property. We shall show that these assumptions lead to a contradiction with Theorems 1 and 2. In fact, if a = /l+1 then let y be such that t ( y ) = /?,and let y ___ = Z @ ) . By Theorem 1, 9 is an initial ordinal; moreover:

W'(v) = zm = W Y ) u {Y: (?T = 5%

whellce = P(Y) u {Y). Thus ~(9) = ~ ( y ) + l= a. I t remains to consider the case where a is a limit ordinal > 0. By assumption, to every 5 < a there corresponds exactly one ordinal Y E (by Theorem 4) such that i(yt) = 5. Let 91 = Z(&).By Theorem 2,

u

€
y is an initial ordinal. Moreover, yc E P(v) for E < a. Finally, if y E P(v), then y E Z(&) for some 5 < CI and thus y < which by Theorem 3 implies ~ ( y < ) 5f 1 < a: therefore y coincides with one of the ordinals yt, 5 < a. Thus P ( y ) = a, that is, ~ ( q= ) a. On the basis of Theorem 5 we have the following definition.

28 1

3. INmAL. ORDINALS

DEFINITION 2: Let w, denote the initial ordinal whose index is equal to a; that is, &(ma)= a. In this way we have associated with every ordinal an initial ordinal. Every initial ordinal is associated with its index and distinct initial ordinals are associated with distinct indices. It is easy to check that the following theorems hold. 6: (a < p) = (o,< as)E For < Op). THEOREM THEOREM 7: Z(G,) = W(O,+~),in other words, K@,) = G,+,. and U Z(G,) = W ( o A ) when , 2 is a limlt ordinal. €4

THEOREM 8: If 1 is a limit ordinal, then ol= lim 0,. €4

Thus the function oEis continuous, and it follows that the ordinal w1 is cofinal with 2. 9: Every initial ordinal is of the form m y for some y . THEOREM PROOF.Assume that a > 0 and that

w, = ~ ~ ~ . n , -+ ~ r ~. ... n+ o , +Y fnk, .

where

y1 > y,

> ... > Y k .

By Theorem 4, p. 257 it follows that

(3)

my1

< o, < wy+zl+

1).

Since y1 # 0, y1 = 1+S for some 6; thus o r t . ( n l + l ) = wl+d.(nl+l) = o.cod(nl+l).

The power of the ordinal 5q equals the power of qE, because one is the power of the set W(E)x W(q) and the other is the power of the set W(q)x W(5).Therefore

- -

;UYi(nl+l)=(nl+l)w.od=w.wd=coyi,

and by (3) we have

01"1 <

z.

Since or1

0,

< w,, it follows by (1)

= WYI,

since o, is initial.

THJZOREM 10: Every limit ordinal A # 0 is cofinal with an initial ordinal a, where a is the least ordinal among ordinals cofinal with A.

282

WI. ALEPW

A N D RELATED TOPICS

PROOF.We shall prove that the least ordinal a cofinal with 1 is an initial ordinal. Let y < a. It suffices to show that 7 # Z (see formula (1)). Suppose on the contrary that 7 =5. Then there exists a sequence y of type y whose values are exactly all ordinals 5' < a; thus a = lim 9 (0. tcv

It follows then by Theorem 2, p. 243 that the ordinal a is cofinal with some p < y. Thus il is cofinal with p, which contradicts the definition of a. For every limit ordinal a we shall denote by cf(a) the least 5 such that a is cofinal with oE.Thus, for example, cf(w) = 0,

c f ( Q = 1,

Cf(O@U,= 0,

c f ( o a ) = 1.

Clearly, for every limit ordinal a we have

cf(03 < a and c f W < a , since a < o, and thus a is cofinal with oefor some 5 shall prove that, for every ordinal a,

(4)

"(5) (see p. 285).

cf(aa+l)

< a. In Q 4 we

= a+ 1

$4. Alephs and their arithmetic

Every cardinal number of an infinite well-ordered set, that is, any power of an ordinal 2 o (see 0 1, p. 274) is called an aleph. The axiom of choice implies that every infinite cardinal is an aleph (see Corollary 3, p. 262). However, many theorems about alephs do not require the use of the axiom of choice; in particular, the law of trichotomy (see p. 232) holds for alephs. For every ordinal a, let

-

K, = Ga, i.e. K, = W ( o a ) . In particular KO= a, K1= (see Definition 2, p. 274). By Theorem

a

7, p. 281 we have (0)

K(KJ

= K,+l-

THEOREM 1: If a < p then K, c N,. Theorem 1 is an immediate consequence of Theorem 6, p. 281.

4. ALEPHS A N D "HEIR ARlTHMETIC

283

<

THEOREM 2: If m K,, then m is an aleph. For m is then the power of a subset of a well-ordered set. 3: The cardinal K,+' is the direct successor of K,; that is, THEOREM there exists no cardinal m such that K, < m < Nu+'. PROOF.If m < then nt is the power of a subset of the set W ( O , + ~ and ) , thus is the power of well-ordered set. Thus m is an aleph: m = K p and by Theorem 1, K, < m c K,+' implies that a < B < a+ 1, which is impossible. The proof of Theorem 4 is similar. 4: If a is a limit ordinal and if K, < m for every 5 < a, THEOREM then m 4 K,. Theorems 3 and 4 imply that the hierarchy of alephs is in a certain sense complete: it cannot be enriched by the introduction of new cardinals comparable with the alephs. THEOREM 5 '): Kq = K, for every a. PROOF.Let 5 < w, and 7 < w, where 5 # 0 or # 0. Expanding the numbers 6 and q at the base w (see Theorem 5, p. 258) and adding whenever necessary terms with coefficient 0, we can represent E and 7 uniquely in the form: 5 = wR.m1+~~~.rnz+...+wYk.mk, '

(1)

7 = wY~.nl+w~'~.nz+...+wYk-nk,

<

where y1 > '/z > ... > Y k and for all i k either mi > 0 or ni > 0. By Theorem 9, p. 281, w, is of the form wL. Since E < w, and 7 < w,, it follows by (1) that (see p. 257) A > yl. Let @(O, 0) = 0 and let

@(E, 7)= wyl'J(ml,nl) +...$wYk'J(mk, nk). (2) Tn this way we have associated with every pair of ordinals 5,q smaller than w, an ordinal @(E,q)which is smaller than w, (since y1 < A). We prove that (3)

[@(5, 7)= @(5,7)1+ (5 = 0A (7 = 7).

l) This theorem was first proved by G. Hessenberg; see Grundbegriffe der Mengenlehre, p. 593 (Gottingen 1906).

284

WI. ALEPHS AND RELATED TOPICS

This formula is obvious in the case where Qi(6, q) = 0 = @([, z), for then 5 = q = 0 = 5 = z, so suppose that @(t, q) = Qi(5, z) > 0. Then 6 > 0 or q > 0 and [ > 0 or t > 0. Let l and t be represented as

...+o d h ' P h ,

[ = w4*p~+wss'p2+

(4) t =08"4~fw4'q2f...+08h'qh,

where for every i

- (5)

@([,

< h either p i > 0 or qi > 0. Then = w81'J(PI~ q1)+

-**+wdh*J(ph, qh).

Since all coefficients in the expansions (2) and (5) are positive and since expansion at the base w is unique (see p. 258), it follows that

k = h,

y i = di

and

J(rni, ni) = J ( p i ,qi)

for

i = 1,2, ..., k.

<

Thus m i= p i and ni = qi for i k and hence by (1) and (4), 6 = [ and 7 = t. Thus (3) holds. Finally, every 6 < w, is a value of the function @. For assume 8 = coy1 r, wya. r2+ ...+ w Y k . rk, where rl, r,, ...,rk are positive; it suffices to take as 5 and q the ordinals (1) where mi = K(rl) and ni = L(ri) for i = 1 , 2 ,..., k . I f 8 = O i t s u f f i c e s t o l e t t = q = O . Therefore the function Qi establishes a one-to-one correspondence between elements of the set W[w,)and ordered pairs of elements of W(w,). Thus the sets W(w,) and W(w,)x W(w,) are equipollent. Q.E.D. COROLLARY 6: K,+K, = Nrnax(=,p) = K,.N,, where max(a, ,6) = a if a 2 ,6 and max(a,B) = ,6 if@ > a; PROOF.Assume that max(a, ,6) = a. Then K, KB and

- +

<

K, < N,+N, < N,+N, = 2N, K, < K,.K, < K,.N,

< N,-K, = K, , = K,,

whence the desired identities follow by the Cantor-Bernstein Theorem (p. 190). THEOREM 7: W(CO,+~)W(w,) = PROOF. Let W ( O , + ~= ) A and W(w,) = B. The difference A - B is a well-ordered set and thus its cardinal number is an aleph, which we shall denote by N,,. The identity A = ( A - B ) u B implies that

285

4. ALEPHS A N D THFJR ARITHMETIC

= K,,+K, = Kmax(y,cx), thus a+ 1 = max(y, a). As a < a+ 1, it follows that y = a+ 1.

'LEMMA 8:

u Ft < cz,where F is any function whose values are

tcT

tfT

sets. For

C z = [(t,x): m

(6)

teT

on the other hand, the set

X E F t ) J ;

u Ft

can be obtained from the set

td'

{ ( t , ~ ) : ( ~ E T ) A ( ~ E F by , ) ) the mapping f defined by the formula f ( ( t , x ) ) = x , and thus it has power less than or equal to that of the latter (see Theorem 1, p. 186). 'THEOREM 9: cf(co,+l) = a+1. Let 1= C~((W,+~). Then the ordinal w B is cofinal with and there exists a transfinite sequence ~1 of type cop whose limit is w,+~.It follows that if 6 < then there exists an ordinal 5 < w, such that 5 < p(t) or, in other words,

(7) _c

W ( m )<

Because W(cos)= K,, and because for all 5

< cu, we have

<

K,+1, that is W m ) Nu, it follows by (7), the lemma and Theorem 8, p. 200 that

< <

,8. At the same time (see (4). Thus u f l < max(a, p) and a + l p. 282) cf(co,+l) a + l , that is /3 a+l, and thus p = a+1. 'THEOREM 10: For arbitrary a,

<

PROOF.Since (5

< a)

--f

(K, < K,) and

-<

NU,

it follows by Theorem 8, p. 282 that

c

€
Hb

< K,.K,

= Nu,

286

WI.ALEPHS A N D RELATED TOPICS

which implies (8), because

CK,

Ha<

&
'THEOREM 11: Ifa = P + l , then

CK, = K,.

(9)

&
On the other hand,

if.

is a limit ordinal

PROOF.Let a = /?+1. Then (5

> 0, then

< a) 3 (5 < /?)and by (8)

Suppose that a is a limit ordinal; then

mw,) =

u W(o,>; &
-

hence (by Lemma 8)

K a = W(w,) = U

W

&
O

B

Crn)= CK,, &
{
which implies (lo), because by (8)

The following more general theorem can be proved in a similar manner. 'THEOREM 12. If a is a limit ordinal, rp is a transfinite increasing sequence of type M and A = limq(E), then &
Remark. It is interesting to note that the aleph K1 and more generally, the aleph*, can be defined without appealing to the notions of ordinal or cardinal numbers. In order to formulate the appropriate theorem, we introduce the following definition: set A contained in

5. ALEPHS AND THEIR ARITHMETIC

287

the Cartesian product X" = XXX X ... x X is finite in the direction of the k-th axis provided that for every element ( x l ,... ,X k - 1 , x k + l , ... ,x,) belonging to X"-' the set {xk: (XI 3

..

*

xk-l

2

xk 9 xk+l,

---

3

X n ) E A}

is finite (more figuratively: every straight line parallel to the kth axis intersects A in a finite number of points). The following theorem holds I): A neccesary and suflcient condition for the set X to be of power < K, (n finite 2 0) is that the set Xn+2be representable as a tlnion Al v ... v An+2,where Ak is finite in the direction of the k-th axis. $ 5 . The exponentiation of alephs

First of all we make note of the following elementary theorem. THEOREM 1: If.

< By then K,"B = 2"s.

PROOF.From the inequality 2 < K, < 2Nuand from the laws of exponentiation for cardinal numbers (p. 185 and p. 187) we obtain 2x8

< #!B < 2Na"B

=ps,

which implies the desired equality. 'THEOREM 2: (THE HAUSDORFF RECURSION FORMULA 2).)

Nus u+l = K , " f l . K u + l * PROOF.We shall examine two cases. Case I: a+l Q B. Then by Theorem 1 we have

1) This theorem is due to W. Sierpiriski (Sur quelques propositions concernant la puissance du continu, Fund. Math. 38 (1951), 1-13). See also K. Kuratowski, Sur une caractt!risation des alephs, ibid. 14-17. A further generalization is given by R. Sikorski, A characterization of alephs, ibid. 18-22. F. Hausdorff, Der Potenzbegriff in der Mengenlehre, Jahresberichte der Deutschen Mathematiker Vereinigung 13 (1904) 570.

288

MII. ALEPHS AND RELATED TOPICS

< KO< 2's, we also have K,Hfl.Ka+1 = 2HB.Ka+1< 2's.2xs = 2%9+v= 2 4 .

On the other hand, since Ha+] (3)

By (2) and (3) it follows that

Nus a+l

= 2Ns= K:a

< K : f l . K a + l < 2's

=

ql,

which implies (1). . Case 11: p < a f l . In this case, Kp4,+1

(4)


= H:;y

=

"XI.

It remains to show that the opposite inequality also holds. For this purpose we consider the set W ( O , , , ) ~ ( , ~that B ) is, the set of all transfinite sequences of type cog whose terms are less than mu+1. We shall show that

w ( m a + , ) ~ ( ~cs '

(5)

u

W(E)W(ws).

t<wa+t

For if y is a transfinite sequence of type m8 < mu+,, then (see Theorem 9, p. 285) the set of its term is not cofinal with W ( W , + ~ ) . Thus there exists an ordinal 6 < for which W(5) contains all the terms of the sequence y; hence y E W(,$)W(o,d. Thus inclusion ( 5 ) holds. As W(5)= $ < K, for 5 < ma+,, it follows that < H;@.

-

m)

Therefore from (5), by Lemma 8, p. 285 and by Theorem 8, p. 200 we obtain

<

=mW(ma+l)w(oij

m w o ~ )"

E<wa+i

which together with (4) proves (1). 'THEOREM 3. (THETARSKIRECURSION FORMULA').) If p is an increasing sequence of a limit type a and I = lim p(t) and fi < cf(a), then Cia

K;s

=

C Kz$),in particular, #f@= C #$:

€
for every #
for every limit ordinal a and

€
A. Tarski, Quelques th6orPmes sur les alephs, Fundamenta Mathematicae 7

(1925) 7.

289

5. EXPONENTIATION OF ALEPHS

PROOF.By assumption, cf(1) = cf(a). Thus no transfinite sequence of type wB with values in W(1) is convergent to 1, hence for every such sequence there exists 6 < a such that this sequence belongs to W(p, (5))w(~a). Therefore which implies that

The opposite inequality follows from the remark that HA2 Z, which implies that Hi’ = KA*K:’ CKfiK’ 2 K ’ >,cK$,. : C

a

‘THEOREM 4: (THEGENERALIZED HAUSWRFF FORMULA.) For finite n, ‘a+n uB

(8)

= K:’.

HE+,.

PROOF.For n = 1, (8) coincides with (1) Assume that (8) holds for a particular n. Replacing in (1) a by a+n we obtain

Kz$,+l = K:B+;Ka+n+l, thus by the induction hypothesis and by Corollary 6, p. 284 it follows that

K Z $ ~ + ~= Kt’.Ku+n.Ku+n+l=

K:’.Ka-tn+l-

Thus (8) holds for n + l , and hence it holds for arbitrary finite n. Setting c1 = 0 in (8) and using Theorem 1 we obtain the following theorem. 5 : (THEBERNSTEIN FORMULA’).) For Jiniie n ‘THEOREM K;fi = 2ufi* K,. (9)

Examples 1. (9) implies that KrO=2WO.K,. F. Bernstein, Untersuchungen nus der Mengenlehre, Mathematische Annalen 61 (1905) 150. More results concerning exponentiation of alephs are given by Heinz Bachman. Transfinite Zahlen. Ergebnisse der Mathematik und ihrer Grenzgebiete, (Berlin 1955), PP. 143-155.

VIII. ALEPHS AND RELATED TOPICS

290

2. Theorem 1 implies that Kfl = 2x1.

3. In formula (6) let a = mi = Q, /? = 0. Since cf(ol) = 1, the hypotheses of Theorem 3 are satisfied and we conclude that

We now give a theorem in the proof of which we shall use Konig's theorem (p. 203): 'THEOREM6: If a is a limit ordinal and j3 2 cf(@), then for every cardinal m

K, # mu,.

(10)

PROOF. Let 5 = c=(ct).Then the ordinal cod is cofinal with a; that is, there exists an increasing sequence y of type ma such that limp(C) = a. tcwg

By applying Theorem 12, p. 286 it follows that (1 1)

Ka =

c K,co.

C<WC

By Kokig's theorem we have

On the other hand, since KdC)4 K, and Ot= Kc

< K,,

we obtain

Formulas (ll), (12) and (13) imply that

(14)

K,

< K",B.

If there existed a cardinal m such that K, = mV, then we would have NU@= mxsv = mu@= K,, contrary to (14). We conclude that (10) holds. 'COROLLARY 7: If ?!, 2 cf(a), then K, < Kz@. From Theorem 6 it follows, in particular, that for no m and j3 do

5. EXPONENTIATION OF ALEPHS

29 1

the equations K, = mu@,K ,, = mUs, N, = mu@hoId (see p. 254 for the definition of 8); for (see p. 282) cf(co) = cf(wm) = cf(&) = 0. It also follows from Theorem 6 that if KO, = mu@, then p can assume only one of the values 0,1, ...,n- 1. Zn fact, cf(w,) = n, and m*s does not hold. thus for p 2 n the equation K,,

COROLLARY 8: K, # 2H.. For K, = 2x0 implies that KfSp = 2No’No = 2H0 = K, , whereas by Corollary 7 we have Q > K,. We shall conclude this section by evaluating the power of the set

-

P n ( M )= { X c M: X

= n},

where

For m and n finite the power of P n ( M )is

-M

= m.

( 3.

< z,

‘THEOREM 9: If M is an infinite set of power nr and if n then P,,(M) has power mn. PROOF. Let 2 be a fixed set of power n. To every X E P , ( M ) there corresponds in a one-to-one manner a non-empty family C(x> of func= X; clearly CCX’)n C ( X ” ) = 0 for tions f E M Z such that f’(2) X’ # X“. By applying the axiom of choice we conclude that the power of Pn(M) is not greater that = mn. Conversely, to every function f e M Z there corresponds the set Af = { ( z , f ( z ) ) :(zEZ)}.This set has power n, because the set Z has power a. Moreover, Af c 2 x M and the set Z XM has power n. m = rn. If f’#.f” and, for example, f’(zo) #f”(zo), then Aft # A p t , as (zo, f ’(zo)) E Aft- Apt. It follows that the set P , ( Z x M ) has power 2 rnn and thus Pn(M) has power >, mn for the sets P n ( 2 X M ) and P,(M) are equipollent.

rz

‘COROLLARY 10: mg. power

If IF=111 >, a,

then the set { X c M.

E< m}

has

2

rim

5 6.

Equivalence of certain statements about cardinal numbers and the axiom of choice

From the well-ordering theorem and thus indirectly from the axiom of choice (see p. 262) it follows that every cardinal is an aleph. Thus

292

VIII. ALEPHS AND RELATED TOPICS

the laws of arithmetic for cardinal numbers coincide with those for alephs and we have the following theorem. 'THEOREM 1: (1)

A [(in < n) v (n < m)] (law of trichotomy), A [m2 = in], in #N A [(m-n = m+n = m) v (nten = m+n = n)], in. n $ N A [(n? = n2) + (m = n)],

m, n $ N

(2) (3) (4)

ni, n # N

A

(5)

[(m < n) A ( P < q> -+ (m +P < n+q)I,

in, n , v . q$N

/A> [(m < n) A (P < 9)

(6)

in, n , v , q#N

(7)

A W + P< n+v) A [(in-p < it-9) -+ n , v#N

+

-+

(map

< n*q)l,

(nt < n)l,

m.n,p#N

(8)

m < n].

nip

Seemingly each of the laws (1)-(8) is a special consequence of the axiom of choice. We shall show, however, that each of these laws in conjunction with the axioms Z[TR] and VIII implies the axiom of choice.

THEOREM 2 l): If for every pair of infinite cardinals m and n either < n or n < m, then there exists a choice function for every family o non-empty sets.

m

x.

PROOF.Let X be an arbitrary infinite set and let m = By assumption either m < K(m) or K(m) in. The second case is impossible by Theorem 3, p. 278; in the first case m is an aleph and thus there exists a relation well ordering X. Therefore the hypothesis of the theorem implies the well-ordering theorem, and thus the existence of the desired choice function (see p. 262).

<

REMARK: Theorem 2 can be expressed more concisely: formula (1) l)

This theorem was first proved by F. Hartogs; see his paper cited on p. 278.

6. EQUIVALENCE WITH THE AXIOM OF CHOICE

293

for all infinite m and n implies the axiom of choice. We shall employ this sort of abbreviated language; in the statements of other theorems the word “implies” will be taken to denote the existence of a proof which does not use the axiom of choice.

LEMMA 3: (3) (2) -+ (4). In fact, if m$ N , then by (3), mz = m. If m yn # N , then by (2) m2 = rn and nz = n, whence m2 = nz + m = n. --f

LEMMA 4:

If

p $ N and p . K ( p ) = p + K ( p ) ,

then p is an aleph’).

PROOF.Let A‘ = p , B’= K ( p ) . By hypothesis there exists a partition A x B = Pu Q into two disjoint sets P and Q such that P”= p , = K(p); thus e’ is an aleph and there exists a relation well ordering Q. There are, a priori, two possibilities. I. There exists U E Asuch that {a} x B c P. However, since {a) X B = K(p), it would then follow that Ktp) < p contrary to Theorem 3, p. 278. Thus case I is impossible. 11. For every a E A, ({a} x B ) n Q # 0. As there exists a relation well ordering Q, there exists a function which associates with every a E A the earliest element q(a) of ({a} x B ) n Q. Moreover, as q(a) is of the form (a, b ) where b E B, it follows that q(a’) # q(a”) for a‘ # a“. Thus the function q establishes a one-to-one mapping of A onto a subset of Q; hence A” < and z i s an aleph. Q.E.D.

-

THEOREM 5 : (4) implies the axiom of choice. PROOF.Let t be an arbitrary infinite cardinal. Let p = EX’, m = p + K ( p ) and n = p . K ( p ) . Clearly pz = ( p y = .EN0 = p ,

which implies that p

< p + l < p - p = pz = p

and thus

P f l =P. Since 2K(p)

= K[p) = K(p)+l

= [K(p)I2, we obtain from these for-

l) This lemma as well as Theorems 3-7 are due to A. Tarski. See his paper: Sur quelques thkordmes qui equivalent d I’axiome du choix, Fundamenta Math. 5 (1924) 147-154.

294

mulas that

WI. ALEPHS AND RELATED TOPICS

+

m2 = [P K ( P ) ] ~ = PZ +2P

*

+

K (PI "(PN2

= P+P[KNWl+"P) = P+P.K(P)+K(P) = P[l+K(P)l+K(O) =P.W)+K(P)

= (P+ 1 ) W ) = P-K(P)

= d"WIZ = nz.

Thus by (4) we have m = n, that is, p + K ( p ) = p.K(p), and thus p is an aleph by Lemma 3. Since E < p , E is also an aleph. Q.E.D. COROLLARY 6 : Each of the formulas (2) and ( 3 ) implies the axiom of choice. THEOREM 7 : Each of the formulas (5) and (6) implies the axiom of choice. PROOF.Let P be an infinite cardinal. Clearly

E < E+K(F) and K(t) < f+K(f). If the strict inequality were to hold in both of these formulas, then by ( 5 ) we would obtain the false inequality f+K(E) < E+K[E). Thus either I= E+K(f) or K(f) = E+K(f). But the first equation implies f >,K(E) which contradicts Theorem 3, p. 278. Thus the second equation holds, which implies E < K(E) and thus E is an aleph. It follows that ( 5 ) implies the axiom of choice. The proof of the second part of the theorem is similar, except that instead of E+K(E) we examine the product E.K(E). THEOREM 8: Each of the formulay (7) and (8) implies the axiom of choice. PROOF.Let E be an arbitrary infinite cardinal. Let nt = K,.E; then m + m = m. Now let n = K(m) and p = m. Hence m + p = in and

6. EQUIVALENCE WITH THE AXIOM OF CHOICE

295

= m+K(m), which implies that m+p < n+p. If the equation m+p = n+p were true, then the inequality m K(m) would also be true, in contradiction to Theorem 3, p. 278. Hence m + p < ~ t + q and by (7) m < n, that is m < K(m). Thus t < K(m), which implies that t is an aleph. The proof of the second part of the theorem is similar; we replace m by txoand discuss the products m - p and n a p instead of the sums m+p, n - v 9. It is not known whether the formula

n+p

A [m+nt

= m]

m4N

is equivalent to the axiom of choice in the system of axioms Z[TR] and VIII. From Theorem 1 (proved with the axiom of choice) it follows that for every infinite cardinal m and for n E N , m" < 2'". Specker has shown that the weaker theorem m"non-2 2" can be proved without the axiom of choice. We shall use this fact in Chapter IX, p. 333. THEOREM 9 2):

Ij m is infinite and n E N, then m"non-> 2m.

PROOF.Assume that Am==m and that there exists a one-to-one function F mapping 2A into A". We shall show that there exists a function f which associates with every transfinite sequence pl of distinct elements of A an element of A which is not a term of y. The theorem then follows from the existence of such a functionf as follows: Let a be the least ordinal of power K(m). By theorem on inductive definitions there exists an a-sequence y such that yc =f(y 15) for 5 < a. By construction, ye is not a term of y I 5 and thus yc # y,, for 5 # q. We conclude that Z = K(nt) in contradiction to Theorem 3, p. 278. It now remains to construct the function f. For this purpose we

x>

l) A detailed discussion of the theorems about cardinal numbers equivalent to the axiom of choice is given in the book H. Rubin and J. Rubin, QuivaZents of the axiom of choice (Amsterdam 1963). Most of these theorems are due to Tarski. 2, See E. Specker, Verallgemeinerte Kontinuumhypothese und das Auswahlaxiom. Archiv der Mathematik 5 (1954) 332-337.

296

Vm. ALEPHS AND RELATED TOPICS

choose an integer ko > 1 such that 2k"> k: and ko distinct elements of A : a,,, ... , ak,-l. If v = (v,,, ... , P ) k - ] ) with k < k,,,then we put f(tp)=aj where j = mini (ai # vs for s < k ) . Assume now that y = ( y o , ..., y k - 1 ) with k Z k , distinct terms. Denote by S ( y ) the set {a),, ..., y k - 1 ) . There are 2k subsets of S ( y ) ; we may represent them as {yi: i e Z } = S ( y , Z ) where Z is contained in (0, 1, ... , k- l}. By the definition of F each F(S(y, Z ) ) is an n-termed sequence. Since there are only k" sequences in A" whose terms are elements of S ( y ) and since 2k > k", we infer that there is at least one Z such that not all terms of F ( S ( y , Z ) ) belong to S ( y ) . We order sets 2 similarly to the lexicographical ordering of their characteristic functions and choose the first Zo with the property stated above. Now we define f(y) as the first term of F ( S ( y , Z o ) )which does not belong to S(v). Finally assume that the type of g~ is an infinite ordinal a. Since Z is an aleph, there exists a one-to-one mapping of the set W(a) onto the set W(a)". Let the ordinal 5 < a correspond to the sequence (5(O), ..., .$"-l)) under this mapping. Let HE= 0 if (yeco), ..., ye@-l))does not belong to the set of values of the function F, let Ht = F e ( ~ ' B..., ( ~ pp-1)) ), otherwise, and let XO= { y ~ :y~ $ He}, (ao, ..., = F(Xo).If all of the elements a j were terms of g ~ , then for some to< a we would have (a,, ...,a,& = (ve$o),...,~ ~ r - 1thus ) ) ;this sequence would belong to the set of values of F, which implies that HE,,= Xo and ye,,e X o= yEoE HE,,.On the other hand, from the definition of the set Xo it follows that ye€Xo= ye$ He, for every E. Thus some aj is not a term of g~and it suffices to set f(v) = a j where j is the least index such that aj has this property. COROLLARY 10: If m is an infinite cardinal then m + 1 < 2 m . PROOF.From the fact that 2 m > m it follows that 2 m 2 m f l . If m+ 1 = 2 m then, since m2 >, m 2 2 m+ 1 , we would conclude that m2 >, 2m, contrary to Theorem 9. Exercises 1. Derive the Cantor inequality m < 2m and its strengthened version f . m < 2m (for finite f and infinite m) from Theorem 9. [Specker] 2. Prove that the following formulas are equivalent to the axiom of choice:

297

6. EQUIVALENCE WITH THE AXIOM OF CHOICE

A

[(m-n = m)v(m.n

= n)].

[Leiniewski]

m.n#N

3. Prove that the following formula is equivalent to the axiom of choice:

A

[(m <*n)v(n <* m ) ~

m.n#N

(the relation

<*

is defined on p. 186).

[Lindenbaum]

Hint: Replace X(m) by X*(m) (defined in Exercise 2, p. 279) in the proof of Hartogs' theorem.

4. Prove that the following formula is equivalent to the axiom of choice:

[Tarski] Hint: Let m = 2(@'), q = X(m) and show that formula mP = mq would imply m > X (m).

mp = m

Q

mq and that the

$7. The exponential hierarchy of cardinal numbers

In Q 4 it was shown that addition and multiplication of alephs is an entirely straightforward matter which reduces to the comparison of alephs. The theory of addition and multiplication of arbitrary cardinals has the same banal character provided that we accept the axiom of choice. Moreover, as was shown in 06, even very specialized laws of cardinal arithmetic imply the axiom of choice. On the other hand, as was indicated in 9 5 , the exponentiation of alephs is not simple even assuming the axiom of choice. In this section we shall consider certain cardinals for which the laws of exponentiation are relatively simple. In the next chapter we shall examine the hypothesis according to which all cardinal numbers belong to this class. The sets R, satisfying the following inductive definition were defined already on p. 246: R,, = 0,

-

R,+l = 2Ra,

RA =

u RE

( A a limit ordinal).

E
Let aE = Ra+t. The cardinals aE constitute the exponential hierarchy of cardinals.

298

MII. ALEPHS AND RELATED TOPICS

In order to derive arithmetic laws for these cardinals we must first establish several properties of the sets R,. 1: (a) R , c R , f o r u < P ; (b) X E R , + X C R , . THEOREM PROOF.The proof is by induction with respect to P. It sufficesto show that if Po 2 0: and if the theorem holds for all P 4 Po, then it also holds for Po. The case Po = 0 is obvious. If Po = j3+1 and ci < Po, then ci P and thus by assumption R, c R, and from (b) we obtain R, c RB+l. If Po is a limit ordinal and CL < Po, then R, c RB, because Rpois the union of all Re for i! < Po. Part (a) is proved. Assume now that X e R B o ;then Po # 0 and Po either has the form b+ 1 or else it is a limit ordinal. In'the first case XE2RB, that is, X c RB and thus X c Rb, by (a) proved above. If Po is a limit ordinal, then there exists P < Po such that XER, and thus, by assumption, X c R, and by (a), XER,,. Thus part (b) holds. THEOREM 2: R, x R, c Ra+z. PROOF.The elements of R. x R, are pairs ( X , Y ) where X , Y ER,. As {X, Y } c R, and {X}c R,, thus {X, Y } E Ra+l and { X } E R,+l. It now follows that { {X}, {X,Y } ) c &+I, and thus { { X } , {X, Y } } E Ra+2. Q.E.D. THEOREM 3: If ci is a limit ordinal then R, x R, c R,. The proof is analogous to that of Theorem 2. THEOREM 4: R, 3 Nu, where Nu is the a-th V N ordinal. PROOF.For ci = 0, R, = N, = 0. If & 3 Nu then NuE 2Ra = Ra+l and thus { N , } c R , + l . AS Na c R,, it follows that N,u { Nu} c RE+, c Ra+l. If u is a limit ordinal, and if the theorem holds whence for i! < a, then since Nu c N,, N , c U R e = R,.

<

u

z,,

€
€
THEOREM 5: = 2"for n < w ; a, = No. PROOF. The first part of the theorem follows from the definitions by induction. To prove the second part of the theorem it suffices to prove that there exists a one-to-one mapping v, of the set R,, into N where Vn+1 is an extension of v,. Define yo by v(0) = 0; for XER,+*let V,,+~(X) = v,(X)if X ER,;

299

7. EXPONENTIAL HIERARCHY OF CARDINALS

if X E R , , + ~ - R ,then let v,+l(X) = j
2""y", where Yo,Y l , ..., Yk-l

are the elements of X ordered so that v,(Y,-l) < v,(Y,) for 1 < j < k. It is easy to show that the functions v,, are one-to-one and that Y , , + ~ is an extension of v,. THEOREM 6: = 2aa;a, < ,a, for a < p. The theorem follows from the definitions and from Theorem 1. THEOREM 7: a2 = a,. PROOF.The theorem is obvious for a = 0. If it holds for the ordinal p, and if a = p+1, then it also holds for the ordinal a, because a, = = 2's; and thus a: = 2as8+as,which implies that a: = a, as a# < as+ap < a; = as. Finally, if a is a limit ordinal, then by Theorem 3 it follows that af a,; and because af >, a,, we conclude that l: = a., COROLLARY 8: aa+as = = amax(a,p). The corollary follows from Theorem 7 with a proof similar to that of the analogous theorem for alephs (see p. 284). 'THEOREM9: Ifa is a limit ordinal then a, =

-

<

cat.

I
PROOF.On the one hand,

-

-

a,.a by Theorem 4, since Na = Z. But On the other, a, = a,.a, since a,.a C at, the desired equation follows by the Cantor-Bern€
stein theorem. REMARK.The axiom of choice is used in this theorem because the very definition of the infinite sum of cardinal numbers demands the use of this axiom. We shall now establish several laws concerning exponentiation of the cardinals at. THEOREM 10: Ifa PROOF. =

< /3

then at0 = a,,+l.

aY < (2aa)afl

= 2aaafi= 2"s

- as+1.

300

vIII. ALEPHS AND RELATED TOPICS

THEOREM 11: I f c t + l

> ,I!? then

a:$1

= au+l.

= 2'" = au+l. at!$ = We lack a full description for powers of aE in the case of the power aF :, where CI is a limit ordinal and ,I!? < a. It turns out that it is not possible to prove any simple formula for such powers. Certain fragmentary results in this area are collected in Theorem 18 below. First we shall establish certain relations between the hierarchy of alephs and the exponential hierarchy. OTHEOREM 12: If I is a limit ordinal 0, then (1) HA+, CAI+, for n = 0,1,2, ... PROOF.Suppose that there are numbers 3, and n for which the inequality (1) does not hold. Among them there exists a least one, say A,,; there also exists a least ordinal no such that for I = A,, and n = no the inequality (1) does not hold. The case where A,, = 0 and no = 0 is impossible since KO = a = u,. We shall show that no = 0. For if no = n , + l , then by the definition of no it follows that PROOF.

-

>

<

Kle+nl< which by (3), p. 279, implies that

aAo+nl,

N(Klo+nl) Q 2aAo+nly and thus

K (NAo+nJ


= aio+no.

But since K(KA0+.,) = KAo+nl+l= KAo+no (see p. 282), we infer that KAo+no

<

aAo+no,

which contradicts the choice of n., Thus the assumption that no # O leads to a contradiction. Every ordinal 5 < 5 can be represented in the form y + n where y is a limit ordinal less than A,, (see Theorem 1 , p. 235). Thus by the choice of 5 we have

Kt = K y + n which implies that

<

ay+n

<

~ A O ,

7. EXPONENTIAL HIERARCHY OF CARDINALS

301

and thus by Theorems 4 and 8,

Finally since KA,= &
Kc,we conclude that KAoQ an,, in contradiction

to the definition of &.-Thus the assumption that there exist numbers 1 and n for which (1) does not hold leads to a contradiction. Theorem 12 provides an estimation of alephs “from above” by means of the exponential hierarchy. No estimation of the numbers at by means of alephs is possible, even for 5 = 1. The well-ordering theorem implies that for every ordinal a there exists an ordinal n(a) such that aor= K,,,. The axioms of set theory yield only fragmentary information about the ordinals n(a). The following theorem is obvious. ‘THEOREM 13: a < p -+ n ( a ) < n(p). ‘THEOREM 14: The function n is continuous (on every set of the form W(a>>. PROOF.Let a be a limit ordinal and let 1 = lim ~ ( 5 ) Applying . €
Theorem 12, p. 286 we obtain

and thus A = n(a). The following is an easy corollary of Theorem 14. OCOROLLARY 15: If a is a non-zero limit ordinal, then cf(a) Q n(a). PROOF.Let cf(a) = 6 and let q~ be an increasing sequence of type u)d such that a = lim rp(5). Then €<ma

n(a) = lim n(q~(5)) = lim y ( 0 , €+Jq

€<ma

where the composite sequence y = n o q~ is increasing. Thus n(@) 2 >, 6 . ‘THEOREM 16: n ( y f 1 ) is not a limit ordinal, then n(y+l) > y ; if n(yf1) is a limit ordinal, then cf(n(y+ 1)) > y. PROOF.The first part of the theorem is obvious since the function n is increasing.

302

W I . ALEPHS AND RELATED TOPICS

Assume that n ( y + l ) is a limit ordinal and that 6 = cf(z(y+l)). If 6 < y were true then by Corollary 7, p. 290 K,(y+l) < N$y+l) would also be true, that is X

ay+l < ayil

= 2w:* 12"v z a Yfl,

which is impossible. Thus 6 > y. Q.E.D. Using the ordinals n ( a ) we are now able to evaluate the power where a is a limit ordinal. THEOREM 17: Ifa > 0 is a limit ordinal, then

B a:

PROOF. Assume that B < cf(a). Clearly it suffices to show that a@ : a,. By Theorem 3, p. 288,

<

c' c"

The last sum can be separated into two sums and where in the first sum 5 is such that n(5+1) and in the second such that B < n(E+l). From Theorem 1, p. 287 it follows that the [th component of the sum is equal to 2'@; and as K, a@< a c f ( a ) a,, we have

<

c'

<

-

C' < /!?.a, < a,.a.

==

<

a,.

<

The 5th component of the second sum = a X 1 = 2'@@ 2aa't+1=at Thus because K, < K,(t+l, =

+2,

It follows that X

<

a+, a, = a., Assume now that is an increasing sequence of type convergent to a and that cf(a) @ ~ ( a )Then . aa8

a@ !

<

On the other hand,

a,+l

< < = a,ab < 2 4 = aa+l.

=

= 20a = 2Wa) = 2t'y

xn(q€l

acf(a)

=

7. EXPOMNTIAL HIERARCHY OF CARDINALS

303

which proves Theorem 17. 'COROLLARY 18: If CI is a limit ordinal then

1

a,+1

if if

a8+1

if

aa

uy=

n(P) < Cf(a)* cf(@ n(B)

B<

<

< n(a>,

Q.

Corollary 18 shows that the difficulty in giving geners formuAdsfor the powers aD; is caused by the lack of knowledge about the values of the function 3t. 58. Miscellaneous problems of power associated with Boolean algebras

We shall illustrate how one calculates the cardinality of a set by solving the problem of the cardinality of the set of prime ideals of a complete atomic Boolean algebra. As we know such an algebra is isomorphic with the field 2y of all subsets of some set Y (see p. 152). We shall assume that r'= M and that m is infinite.

'THEOREM 1: The field of sets 2y contains 22" prime ideals'). First we reduce the proof to the following lemma: 'LEMMA1: There exists a family S C of~ power ~ 2" such that every finite subfamily S, c S is independent (see p. 22). With every function f E (0, l}" we associate the family

scf) = { It is clear lies. If k is a Zj # Y .

u

i
~ [ [ ( Z E S ) ~ c f ( ~ ) = O ) ] V [ ( Y - ~ ~ S ) ~1>1])* (f(Z)= that distinct functions are associated with distinct faminatural number > 0, and if Z j E : S ( f ) for i < k, then Indeed, assume for example that Z t e S for i < p and

l) This theorem was first proved (for m = xo) by G.Fichtenholz and L. Kantorovitch, Studia Mathernatica 5 (1935) 69-99. A simpler proof was given by F. Hausdorff, uber zwei Siitze von G. Fichtenholz und L. Kantorovitch, Studia Mathematica 6 (1936). Tarski's papers: Sur les classes d'ensembles closes par rapport d certaines opdrations Pldmentaires, Fundamenta Mathematicae 16 (1930) 181-304 and Zdeale in vollstandigen Mengenkorpern, ibid. 32 (1939)45-63 contain an extensive discussion of similar problems.

304

VIII. ALEPHS A N D RBLATED TOPICS

that Y - Z i e S for p

< i < k.

then letting Wi = Y - Z i for p

If the equation


n W,= 0, contradicting the fact that

uZi = Y were true,

k k

we would have

n (Y-ZJ

n

icp

the family composed of the

p
sets Zo,...,Zp-l and W,, ...,Wk-l is independent. Let Z( f) be the family composed of sets 2 c Y having the following property: there exists a finite number of sets Z1,..., 2 , ~ S ( f ) such that 2 c 2,u u 2,.It is clear that

...

Thus the family Z(f) is an ideal. Since no finite sum Z,u ,.. uZ, of elements of S(f) is equal to Y , it follows that Y $Z(f). Finally by definition we have S(j') c Z(f). By 2, p. 266 there exists a maximal (and thus a prime) ideal J ( f ) containing I ( f ) (see Exercises 2, p. 168). Ifft #A, then JV;) # JY;,. If for exarnpleficz) = 0 andfi(2) = 1, then 2 E SUl)c J ( f i ) and Y - 2 E SV;)c J ( f 2 ) and thus the ideals Jcfi) and J(&) are distinct, because otherwise Y = 2 u ( Y - Z ) ~ l ( f i ) in contradiction to the definition of prime ideal. Thus the set of all prime ideals is at least of the same power as the set of all functions f E (0, l}", that is, at least of power 22m. On the other hand, every prime ideal is contained in the family 2y and thus the set of all prime ideals has at most power 22m. It remains to prove the lemma. For this purpose we shall accomplish a series of reductions. Let X be an arbitrary set and let R be a family of subsets of X. We shall say that R is (a) independent, (b) weakly independent, (c) very weakly inhpendent, if respectively (a) every finite subfamily R1c R is independent, (b) MMi # 0 for arbitrary distinct M,Mo, ,Mk-l E R, ick (k = 0,1,2, ...),

u

...

8. MISCELLANEOUS PROBLEMS OF POWER

305

(c) M l - M 2 # 0 for arbitrary distinct M I , M2 E R. We say that the cardinal m satisfies condition (A), (B), or (C) provided there exists a set X of power m and a family R c 2x of power 2m satisfying the condition (a), (b), or (c), respectively. It is clear that if m satisfies condition (A), then for every set X of power m there exists a strongly independent family R c 2x of power 2m. Similar theorems hold for conditions (B) and (C). The lemma above is equivalent to the statement that every infinite cardinal m satisfies condition (A). We now show that conditions (A), (B), (C) are mutually equivalent. The implications (A) + (B) + (C) are obvious. Thus it suffices to show that (1)

(B)

+

(A),

(2) (C) (B). The scheme of proof of (1) and (2) is as follows: we assume that there is a set X of power m and a family R c 2x of power 2" satisfying condition (b) (respectively (c)) and we then construct a set K of the same power m as well as a family S c 2K of power 2" satisfying condition (a) (respectively (b)). PROOFOF (1). Let X be a set of power m and R a weakly independent family of power 2m consisting of subsets of X . Let K denote the family of all finite subsets of X . Clearly K = K,,, where K,, is the -+

u

-

-

n

family of all subsets of X which contain exactly n elements. Moreover, K,, = m"= m (see p. 291), and thus K = m .n = m. For T E R ,let X ( T) = { Z E K :ZnT # O } , and denote by S the family of all subsets of K having the form WT). We shall prove that every finite family { X ( T , ) ,. . . , X ( T k ) ) ,where T , , ... ,Tk are distinct elements of R, is independent. We shall show, for example, that the set X ( T , ) n ...n X ( T J n [ K - X ( T , + , ) l n ... n[K--X(Tk)]

is non-empty.

306

VXII. ALEPHS AND RELATED TOPICS

Indeed, from the assumption that R is weakly independent it folk

lows that TI-

u Ti # 0 f o r j

i=p+l

=

1 ,2, ... , p . If xi is an element of this

difference, then the set {xl ,... ,x,} clearly belongs to X ( T j ) and to K - X ( T i ) for j = 1 , ...,p, i = p + l , ..., k. From the independence proved above it follows in particular that X ( T ) # X(T') for T # T'. Therefore S is a family of power 2m consisting of subsets of K and every finite subfamily s,c S is independent; that is, the family S satisfies condition (a). PROOFOF (2). Let X be a set of power m and let R be a very weakly independent family of power 2 m consisting of subsets of X . For every set 2 let 2' = 2";thus 2' is the family of all finite sequences

u

ncN

whose terms belong to 2. Let K = X'; clearly K =

K

u X" and as x';

= mn = m

it follows that

nsN

= m.a = m. Let S = {M': M ER);clearly S is a family of subsets of K and s'= R' = 2m since the mapping M + M' is one-to-one. We shall now show that the family S is weakly independent. Let MyMo, ... ,Mk-1 be distinct elements of R and assume that M'c U M ; . ick

From the assumption it follows that M - M i # O for i < k ; let xi E M - M i for i
PROOFOF THE LEMMA. Let X and X' be two disjoint sets of power m and let a one-to-one function f map X onto X'. For M c X let M* = M u [ X ' - f ' ( M ) ] . Let S be the family of all sets M*. The family S consists of subsets of the union K = X u X', where I?= m +m = m and S = 2m because the mapping M + M* is one-to-one. If M , # M 2 and M l , M 2 c X , then M: #M,*. For X E M ~ - M ~ implies that x E M:-MM*, whereas x E M,--M, implies that f ( x ) E MT- M:. Thus S is very weakly independent, which proves that the cardinal m has property (C); hence, by (1) and (2), m also has property (A). This completes the proof of the lemma.

-

8. MISCELLANEOUS PROBLEMS OF POWER

307

Exercises 1. Prove that if a Boolean algebra is finite and has 2" elements, then it contains exactly n prime ideals. Hint: Every prime ideal has the form (x: X A U = o), where u is an atom. 2. The Boolean algebra K with unit of infinite power m contains at least 1 and at most 2"' prime non-principal ideals. Hint: Let at denote any element of K such that thc ideal { x : x < ut) is prime; then there exists a proper ideal I containing all elements -ut. 3. Construct countable Boolean algebras with resp. n , a and 2a prime, nonprincipal ideals (where 11 E N). 4. We say that a subset X of the Boolean algebra K is a base for the ideal I if (1) X c Z, (2) for every a E Z there exist finitely many xl, ...,x, elements of X such that a < x,vx,v ... vx,,and (3) no proper subset of X satisfies conditions (1) and (2). Prove that an ideal is principal if and only if it has a one element base. 5. If K = 2 N and I is a non-principal prime ideal, then no base for Z is countable.

CHAPTER IX

INACCESSIBLE CARDINALS. THE CONTINUUM HYPOTHESIS

In this chapter we shall deal with two rather unrelated topics. First we shall discuss certain properties of the so-called inuccessible cardinal numbers. As we shall see, these cardinals are situated immensely high in the hierarchy of cardinal numbers. Although we can examine the properties of inaccesible ordinals within the framework of the system Z"[TR], the existence of such numbers cannot be deduced from these axioms. In fact, each proposition regarding the existence of an innaccessible number demands its own axiom. The second topic to be discussed is the continuum hypothesis and several of its consequences. This hypothesis, formulated by Cantor, concerns the relationship between the exponential hierarchy and the hierarchy of alephs. It has been shown that this hypothesis is independent of the axioms .Zo[TR], VIII. Opinions vary among mathematicians as to whether this hypothesis should be rejected or accepted as a new axiom of set theory. The :motive for treating these two topics in one chapter is that they give some idea of the nature of propositions indecidable in the axiomatic system of set theory.

8 1.

Inaccessible cardinals

On p. 279 we defined initial ordinals as numbers of the form w,. We shall now divide initial ordinals into regular and singular ones. The ordinal w, is singular if it is cofinal with an ordinal smaller than itself, that is, if there exists a sequence q~ of type ,!I, where ,!I is a limit ordinal < corn, such that lim ple = 0,. An ordinal w , is regular if it is not singular.

E-=P

309

1. INACCESSIBLE CARDINALS

is regular for every a. 'THEOREM1: The ordinal = limp16 where j3 is a limit ordinal < o,+~,then by PROOF.If

<

letting y = cf(B) we would have y < a + l and C~(CO,+~) y < a contrary to Theorem 9 on p. 285. The initial ordinal o,is called weakly inaccessible if it is regular and if the index a is a limit ordinal. For example, the ordinal w is weakly inaccessible. On the other hand, the numbers o,, wQ, w, are not regular and thus are not weakly inaccessible. 2: If a > 0 , then the ordinal w, is weakly inaccessible if THEOREM and only if a = w, = &a). PROOF.Assume that a > 0 and that w, is weakly inaccessible. The ordinal a is then a limit ordinal # 0; let y = cf(a). If a < w, then, since w, = limw,, it follows that w, is not regular. Thus a = w, since

p < wB for

t-za

all j3. Let y = cf(a). By the definition of y there is an increasing sequence pl of type my such that a = lim y ( & , thus w, = lirn €<my

c<wy

if y < a then w, would not be regular. Conversely, assume that a = w, = cf(a) > 0. Thus a is a limit ordinal and to show that w, is weakly inaccessible it suffices to show that it is regular. Assume thus that is a limit ordinal < w, and that o, = lim pl(t). If y = cf(j3) then j3 is the limit of an increasing seE
quence y of type m y . Thus o, = lim pl(y(5)) and cf(a)

<


On the

€<my

other hand, y < w y and w y /?,because y ( 5 ) 2 5 for 5 < m y . Thus we have the inequality cf(a) y w y < j3 0,; hence from our initial assumptions it follows that a = j3. Thus w, is cofinal with no ordinal < w,. Q.E.D. REMARK: The equation a = w, does not characterize inaccessible ordinals. In fact, if j3 is a sequence of type o defined inductively by the formulas Po = 0, ,9n+l= oBnthen the ordinal a = limp,, satisfies the

< <

<

n

equation a = w, (see p. 237), but co, is not regular since cf(a) = 0. See also exercise 3, p. 314. We now introduce the notion of weakly and strongly inaccessible cardinals.

310

M. INACCESSIBLE CARDINALS

The cardinal m is weakly inaccessible if it is an aleph, m = K, and W , is a weakly inaccessible ordinal'). We shall characterize weakly inaccessible cardinals in terms of the theory of cardinal numbers. 'THEOREM3: The cardinal m is weakly inaccessible if and only i f it is inJinite and satisjies the conditions: (i) if n < m y &en there exists a p such that n < p < m; and (ii) if I is a set of power < m and f a function whose values jx < m . are cardinals < m, then if

c

XEI

PROOF.Let m = K, be a weakly inaccessible cardinal. Then ct is a limit ordinal, which proves condition (i). Let I be a set and f a function satisfying the antecedent of the implication (ii). We can assume that , where y ( x ) < a for I = W(OA),where 3, < a and that fx = x

< COL.

If there exists a largest ordinal among the ordinals ( x ) ,say p, then

Otherwise letting p = lim y ( x ) we have p


by the assumption of

X<WA

inaccessibility, thus again

c

.

K,(x,

X<WL

Thus in either case

fx

< KAK, < h.

< m, which proves that condition (ii) holds.

X€Z

Conversely, assume that m is an infinite cardinal satisfying conditions (i) and (ii). Let m = K,. By (i) it follows that a is a limit ordinal. If a= were m o t regular then there would exist a limit ordinal /3 < o, and an increasing sequence y of type /? such that o,= limy ( l ) Letting .

I = W(#O

and

fe =

-

w(oqI(c+l))-W(OGD0)

C-=B

for

6 €1

we obtain (see pp. 284-286)

in contradiction to (ii). Thus Theorem 3 is proved. l) This notion was first introduced by F. Hausdorff in his paper Grundzuge einer Theorie der geordneten Mengen, Mathematische Annalen 65 (1908) 443.

311

1. INACCESSIBLE CARDINALS

The cardinal m is strongly inaccessible') if it is weakly inaccessible and satisfies the formula (iii) n < m + 2 n < m . Clearly condition (iii) implies (i). The existence of weakly inaccessible cardinals which are no tstrongly inaccessible presents a problem which is not decidable on the basis of the accepted axioms. The generalized continuum hypothesis, which we shall treat in 0 6, implies that every weakly inaccessible cardinal is also strongly inaccessible. 'THEOREM 4. The cardinal m = Ka is strongly inaccessible if and only i f it is weakly inaccessible and n ( a ) = a (see p. 301 for the definition of the operation n). PROOF.If m is weakly inaccessible and n(a) = a , then n < m implies that n < KO or n = KBfor some p < a. In the first case clearly 2" < in, in the second case 2n 2afl= = Kx(B+l); since z is strictly increasing (see p. 301), it follows that 2n < H,,,, = K, = m. Thus m satisfies (iii) and hence is strongly inaccessible. Conversely, if m is strongly inaccessible and if a = 0, then it is clear that n ( a ) = a. On the other hand, if a # 0 then a is a limit ordinal and thus a = limp. Since the operation z is monotonic and con-

<

B
tinuous (p. 301), we have n ( a ) = limn@). It follows from (iii) that B
< a +-Kn(B)< K, and thus n(p) < a for p < a, hence limn@) = a. B
Therefore n ( a ) = a. Q.E.D. 'THEOREM 5 2): In order that m = K, be strongly inaccessible it is neccessary and suficient that m satisfy the condition (iii) and

l) This notion was first introduced by A. Tarski. See his paper Uber unerreichbare Kardinalzahlen, Fundamenta Mathematicae 30 (1938) 68-89, reference 1) on page 87. The first publication using this notion was the paper of W. Sierpitiski and A. Tarski, Sur une propriitd caractiristique des nombres inaccessibles, Fundarnenta Mathematicae 15 (1930) 292-300. An equivalent definition of strongly inaccessible cardinals was also given by E. Zermelo, uber Grenzzahlen und Mengenbereiche, Fundamenta Mathematicae 16 (1930) 29-47. Theorems 5 and 6 are given in Tarski's paper cited above.

312

M. INACCESSIBLE CARDINALS

PROOF.Assume that m = K, is strongly inaccessible. It suffices to show that m satisfies (iv). Moreover, we may assume that m > KOand thus, by Theorem 2, that a = o, = cf(a). Let g < m. Either g is finite or g = Ke where 5 < a. In the former case it is clear that mF = m. In the latter we apply Corollary 18, p. 303 and Theorem 4, p. 311. From ct = cf(a) we obtain 5 < cf(a) and thus mg = ~ t=t ate = a, = K, = ni.

Hence mr

=

m for all g < m and thus mg = m - m = m,

2

F a t

which proves (iv). Conversely, assume that m is a cardinal number and that conditions (iii) and (iv) are satisfied. We have to show that m is strongly inaccessible, that is, that conditions (i), (ii) and (iii) are satisfied. Because condition (iii) is satisfied by assumption and because (iii) implies (i), it remains to show that (ii) is satisfied. Thus let f be a transfinite sequence of cardinals less than m and let 1 be of type w Bwhere < a. Applying Konig's theorem (p. 203) we obtain

2

fy

< m"B

< 2 mF = m,

rim which proves (ii). OTHEOREM 6: Let m and n be cardinal numbers and m 2 No. The cardinal m is strongly inaccessible and larger than n if and only if there exists a family R of power m such that: (1) There exists a set A of power 2 n such that A E R , YCOB

(2)

(3)

X E Y E R+ ( x is a s e t ) A ( x c R), XER+~XER,

PROOF.Assume that m = K, is strongly inaccessible and that m > n. If a = 0 then the desired family is clearly Rm (see p. 297). Assume that u # 0; we shall show that the family R, satisfies all of the desired conditions. The family R,has power m because o + a = a, and thus a, = Rm+u = E, and a, = K, by Theorem 4, p. 311.

-

313

1. INACCESSIBLE CARDINALS

-

Clearly condition (1) is satisfied if n < KO.If n = K,, where /l< u then &+,= afl implies that Rm+scontains a subset of power Ks(see p. 300). However, since co+p < o+a = a, we have Rmotfl€ Rlotgtl c R,, and thus some element of & has power 2 n. Thus (1) is satis5ed. If XEY E R,, then X ER, as was shown on p. 298; thus (2) holds. Condition (3) follows from Theorem 1, p. 298 because X E R , + (XERB),and XERs + 2 X c R, + 2 X E RB+l c R,.

v

,
Finally to prove (4), assume that X c R, and that there doss not exist a function mapping X onto R,. For every Y E X there is the least ordinal q(Y) such that Y E Rqm.If the set of all such q(Y) were cofinal with a, then there would exist an increasing sequence of type oCfla) convergent to 01 whose terms are ordinals of the form q(Y). Since . cf(u) = 01 = o,,the set of terms of the sequence would have power m = K,. Thus we would obtain X 2 m = K, = a, = Rm+, = R,. And then there would exist a function mapping X onto R, in contradiction with the assumption. Thus we have shown that there exists an ordinal y c a such that q(Y) < y for every Y E X ,that is X c R,, X ~ R , , + ~ a n d X E R,. Assume now that R is a family satisfying (1)-(4) and let m = R". From (1) and (2) it follows that m > a. From (3) it follows that m satisfies (iii) and thus (i). It remains to show that m satisfies (ii). For this purpose notice that (4) implies that every subset of R of power less than m belongs to R. Conditions (2) and (3) imply that every element of R is a set of power c m contained in R. Thus R is identical with the family of its subsets of power < m. Applying Corollary 10, p. 291 we conclude that rng = m and thus by Theo-

-

I

_

-

x=

t<m

rem 5, p. 31 1 m is strongly inaccessible. Exercises 1. Let &(A') and P d X ) be respectively the families of unions and intersectionsof less than 8, sets bslonging to X. Prove that if w, is strongly inaccessible then sa(pa(X)) = pa(Sa(x)).

[Sierpihski-Tarski]

2. Show that the infinite cardinal m is strongly inaccessible if and only if for every function f whose values are cardinal numbers < m and whose domain Z has f, < m holds. [Tarski] power < m the formula

n

XSI

314

M. INACCESSIBLE CARDINALS

3. Show that a cardinal m = Ha is weakly inaccessible and greater than N, if and only if ct is a limit ordinal and cf(ct) = a. Hint: If j3 is an increasing sequence of type m c f ( a ) such that a = lim Be, then a > 85 > 5 for 5 < m c f ( a ) and thus

0 2.

ct

e<wcf(a)

> w c f ( a ) = ma.

Classification of inaccessible cardinals')

The strongly inaccessible cardinal K, is said to be inaccessible of class 1 if the set of strongly inaccessible cardinals KB < K, is ordered by < in type a. The cardinal K, is called inaccessible of class 2 if the set of cardinals KB < K, which are inaccessible of class 1 is ordered by in type a. Clearly in a similar manner we can define inaccessible cardinals of class 3, 4 and so forth. We shall define the 6th cardinal of the class 7. We shall say that a is the 6-th cardinal of the class 7 (in symbols M(5, 7,a)) if there exists an infinite sequence pl of type q+1 such that K,; (i) yo is the set of all strongly inaccessible cardinals (ii) if ( 7 then vC+1 is the set of all cardinals NEB < K, such that Kpe y C and such that the type of the set {K, E ~ Q :y < p } is equal to /?; (iii) if 3, < q + l is a limit ordinal then y1 = fys; l

<

<

<

(iv) {KpE y q : /? < a} has type 5;

CV)

KaEVq.

E
<

The cardinal K, is inaccessible of class q if there exists ,$ a such that M(5, q, 4. The proof that the definition above coincides with that given earlier for the cases 7 = l, 2,3, ... is left to the reader. Hyper-inaccessible cardinals, also introduced by Mahlo, are defined in a different manner. In order to explain the principle behind the classification of hyper-inaccessible cardinals assume that K, is the first strongly inaccessible aleph. Then a = lim 5 where none of the alephs Keis inaccessible. €
2. CLASSIFICATION OF INhCCESSIBLE CARDINALS

315

Thus a can be represented as the limit of an increasing continuous infinite sequence y of type z < a such that each of the alephs Ky(€l, 6 < z is accessible. Clearly such a representation can also be given for the second strongly inaccessible cardinal, the third, etc. We shall call the strongly inaccessible cardinal K, hyper-inaccessible of type 0 if for every increasing continuous sequence of ordinals 9 of a such that lim vC= a at least one of the cardinals K,,, 5 <3! , type

<

€
is strongly inaccessible. Replacing the words “strongly inaccessible” at the end of the definition above by the words “hyper-inaccessible of type 0” we obtain the definition of the hyper-inaccessible cardinals of type 1. Similarly as for inaccessible cardinals of class 17 we can define hyper-inaccessible cardinals of type E where E is an arbitrary ordinal. Exercise 1. Show that the first inaccessible cardinal of class 1 is not hyper-inaccessible. Hint: Consider the sequence of critical numbers for the operation defined by the formula W E = 7.

Q 3. Measurability of cardinal numbers Abstract measure theory arose in connection with the theory of measure for subsets of the Euclidean plane. The discussion which evolved around this problem’) led to the formulation of a new hypothesis regarding the existence of certain cardinal numbers. It is interesting to note that this hypothesis is incomparably stronger than the existential hypotheses associated with the classification of Mahlo.

DEFINITION. The cardinal m is said to be measurable if for every set A of power m the Boolean ring 2A contains a a-additive prime Y =A. ideal I such that

u

Y€I

The following theorem gives an equivalent formulation of this definition. l) This problem was formulated by S. Banach. See S. Banach and K. Kuratowski, Sur une giniralisation du probkme de la mesure, Fundamenta Mathematicae 15 (1929) 127-131.

316

M. INACCESSIBLE CARDINALS

THEOREM1: For a cardinal m to be measurable it is necessary and sujicient that there exist a set A of power m and a function defined on all elements of the family 2A having values 0 or 1 and having the following properties: (i) f ( 4= 1, (ii) f ( { a > )= 0 for ~ E A , (iii) i f X , is a sequence of dajoint subsets of A then

PROOF.Let Z be a o-additive prime ideal in 2Asuch that

u Y = A and

YE I

let f ( X ) = 0 or 1 depending on whether XEZ or X$Z. Equation (i) follows from the fact that ,441 (see p. 163). equation (ii) from the fact that each singleton {a} is contained in some YEZ. If X, is a seX,EZ by the quence of disjoint subsets of A and X,EZ for all n, then o-additivity of Z and we get (iii) since f ( X n ) = 0 for every n. If, for some no,Xno$Z,then A-X,,~ZbecauseX,,n(A-X,,) =OeZ. FromX,,nX, = 0

u

m

it follows that X,EZ for n #no. Since

u Xn#Z, both

the left and

n=o

the right-hand sides of equation (iii) are equal to 1. Conversely, assume that f is a function satisfying (i)-(iii) and let Z = {XcA: f ( X ) = 0). From (i) it follows that A gZ and thus Z # 2A. If Y c XEZ, then f ( Y ) = 0 since 0 =f ( X ) =f ( Y u (X-Y)) =f ( Y ) + f ( X - Y ) and thus f ( y ) = 0. I f X1€Zand &EZ, then

f(XIUX2) =f((Xl-Xz)u(Xz--X~)u(XlnXz)) =f (XI -&I

+f (X2 ---&I +f (XI nX2) = 0 >

because each of the sets X l - X 2 , X2-Xl, X1n& belongs to Z. Thus Z is an ideal. Let Z$Z. In order to show that Z is a prime ideal we have to show that A-ZEZ. But A = Zu(A-2) and thus by (iii) f(A) = f ( Z ) f ( A-Z) and hence f(A - Z ) = 0, i.e. A -ZE: 1. From (ii) it follows that

+

U Y DaUz A( a } = A .

Yti

3. MEASURABILITY OF CARDINAL NUMBERS

317

It remains to be shown that the ideal Z i s a-additive. For this purpose assume that X,EZ for nE N. Since m

W

W

U Xn = nU (xn- kU xk>= nU X, n=o =o
m

and thus m

m

U X~EZ, that is

n=o

U X.EZ. n=0

A function satisfying conditions (ii) and (iii) is said to be a a-additive, zero-one measure defined on A. If in addition the function satisfies condition (i), then it is said to be non-trivial. In the sequel we shall speak simply of a measure defined on A instead of “a-additive zero-one measure” since we shall not consider any other kind of measure. If Z is an ideal in 2A and all elements of a set X$Z have some property W, then we say that almost all elements of A have the property W. Similarly, if a function is defined on a set X#Z, then we say that it is defined almost everywhere.

THEOREM 2: If n < m and m is non-measurable, then n is non-measurable. PROOF.Let A c By n, B = m and let f be a measure on A. Extending f to 2B by the formula g ( X ) = f ( A n X ) we obtain a measure on B. By assumption g is a trivial measure and thus g ( B ) = 0 and g ( X ) = 0 for all Xc B. Hence g ( A ) =f(A) = 0. The cardinal KO is clearly non-measurable. Thus by Theorex 2 the non-measurable cardinals form a “segment” in the class of all infinite cardinals. The following theorems are intended to provide some information about the extent of this segment. The ideal Z is said to be m-additive if for every family R c I of power m y U Y E I .

x= -

YER

318

IX. INACCESSIBLE CARDINALS

uY A and if m is a non-measurable cardinal then Z is m-additive. PROOF. Let g =in and let F€ZM;we have to show that u F(m) EI. 'THEOREM3,). If Z is a a-additiveprime ideal in 2A such that

=

Y d

mEM

We assume, temporarily, that F(m,) nF(m,) = 0 for m, f m2. We deF ( ~ ) E ZThe . family J is an ideal fine the family J c 2Mby Z E J Z

u

nrez

in 2M.Indeed, if Z 1EJ and Z 2EJ then

u F(m)u u F(m)E Z and thus

mdl

U

mEZ,

F(m) E I, whence 2,u 2, EJ. If Z , c Z2EJ then

mcZ,WZ*

O=

UF

mEZl

and thus

u F ( 4 €1,

meZI

u F(m)eI

and

Z,EJ.

mEZl

We now show that ( Z, nZ2 = 0) --f [ ( Z,E J ) v ( Z 2 € J ) ] . Let Z1nZ2 = 0. By the assumption that F(m,)nF(m2)= 0 for m1 # m2,it follows that

[uF ( m ) ] n [ u F(m)] mEZ,

= OE&

thus one of the sets

meZ.

u F(m),

mEZ,

i = 1,2, belongs to Z and hence one of the sets Z i , i = 1,2, belongs to J. The ideal J is a-additive. For if Z , E J for EN, then F(~)EZ mEZn for n E N and thus F(m)EZ, whence Z , EJ.

uu

u

neN meZ,

Finally, we show that

u

n

uY

YE.?

= M . For

F(m)E I by assumption,

u F(n)EI, which implies that { m } e J . Since M = u { m } c u Y , it follows that u Y M .

therefore

meM

n4ml

=

YEJ

Y d

If the ideal J were different from 2* then it would be a prime a-adY = M and thus 5 would be ditive ideal satisfying the condition

u

YEJ

measurable, contrary to our assumption. Thus M E J and

u F(m)E Z .

m&l

l) S . Ulam, Zur Masstheorie in der allgemeinen Mengenlehre, Fundamenta Mathematicae 16 (1930) 140-150.

3. MEASURABILITY OF CARDINAL NUMBERS

319

The general case, where we do not assume that F(ml)nF(m2)= 0 for m, #m,, is reduced to the case just considered as follows. Let < well order M and let F'(m) = F(m)F(n). Then F'(ml)

u

nem

nF'(m2) = 0 for m,

# m2. Thus

u F'(m) E I ,

which proves the

mcM

theorem, because

u F'(m) u F(m). =

mEM

mcM

'COROLLARY 4: I f M" is non-measurable, f is a function defined on M such that f, , is a non-measurable cardinal for all m c M , then ,f is meM a non-measurable cardinal.

PROOF. If A' =

f,, then A = rntM

u A,,, where z,,, ,f =

and A,, nA,,

meM

= 0 for rn, f m , . Restricting to A,,, the measure f defined on A we obtain a measure on A, and thus this measure must be trivial. Thus f ( A m )= 0 for m E M and by Theorem 3 it follows that f(A) = 0.

'THEOREM 5 9 : If m is a non-measurable cardinal, then 21" is also a non-measurable cardinal. PROOF.Assume that 2 m is measurable and let m = K,, A = (0, l}w(mu). Thus A is the set of sequences of type co, whose terms are 0 or 1. Let . Z be a u-additive prime ideal in 2* such that Y =A .

u

YE1

For every 5 < o, we have A = A P ) n A r ) where AS) = {P)E A: P)C = i } . Clearly ALo)nAL1)= 0 and thus exactly one of the sets ALo', A f ) belongs to the ideal Z. Let it be that of the numbers 0 or 1 for which A p ) EI. By Theorem 3 it follows that A ~ ) E and Z thus Af-'E) $ I , On

u

eem,

n

€<-,

the other hand, it is easy to check that this intersection contains exactly one element, namely the sequence P)such that P)€ = 1-iE for 5 < co,. Thus we obtain a contradiction since every one-element set belongs to I. 'COROLLARY 6: Every cardinal less than the first strongly inaccessible '. cardinal is non-measurable1 The corollary follows directly from Corollary 4 and Theorem 5. I)

See S. Ulam, op. cit., Theorem 1, p. 146.

') This is a result of S. Ulam and A. Tarski. See S. Ulam, op. cit., p. 150.

320

0 4.

M. INACCESSIBLE CARDINALS

The non-measurability of the first inaccessible aleph

The problem of the measurability of the first inaccessible cardinal proved to be more difficult than the problems discussed in the preceding section. The first results concerning this problem were obtained by Hanf and Tarski in 19601). The method discovered by Tarski allowed him to establish the non-measurability of many inaccessible cardinals ') but it has not provided a solution to the question as to whether there exists at least one measurable cardinal. In this section we shall sketch the simplest of Tarski's results '1. 'THEOREM 1: The first strongly inacessible cardinal is non-measurable. PROOF.Let m = K, be the first strongly inaccessible cardinal. Assume Y that there exists a prime #-additive ideal Z in W(v) such that

u

Y€I

= W(v). Consider the set K of all ordinals w,, c1 < v and let 2 = KW@) be the set of sequences of type Y , whose terms are these ordinals. Finally, let Z/Z be the Cartesian product KW(")reduced mod Z (see p. 142). Thus the elements Z/Z are equivalence classes of sequences T E Z ,where ~1 is equivalent to y if vC = pc for almost all E . We denote by q ~ * the class of all sequences equivalent to T. The set Z/Z is ordered by the relation 4 defined by the formula

q* -_ 3 y* = {t:Qlc > yt) E Z (see p. 143). We shall show that the relation 4 is a well-ordering. In fact, assume that there is a sequence y(")(n E N) of elements of 2 such that #"+')* 4 #")* for n E N. Thus for every n E N there is a set X,, EZ such that g$'+l) < q ~ pfor ) E$X,,. From the #-additivity of the ideal it follows that X, = X#Z and thus @+l) < py) for all n E N and for all E E X . For fixed &, 4 X we then obtain a strictly decreasing sequence of ordinals, which is not possible.

n

'1 A. Tarski, Some problems and resuIts relevant to the foundations of set theory. Proceedings of the 1960 International Congress for Lopic, Methodology and Philosophy of Science (Stanford 1962) 125-135. *) A. Tarski and J. H. Keisler, From accessible to inaccessible cardinals, Fudamenta Mathematicae 53 (1964) 117-199. 3, This proof is due to J.H. Keisler; see bis paper Some applications of the theory of models t o set theory. Proceedings of the 1960 International Congress for Logic, Methodology and Philosophy of Science (Stanford 1962) 80-86.

4. NON-MEASURABILITY OF THE ALEPH

321

We shall denote the class containing the constant sequence (the sequence defined for 5 < v by the equation cpt = 0,) by the symbol a,*. These classes form a proper segment of the set Z/Z. In fact, if y* 4 w.* then there exists a set X E Z such that yc < o, for all 5 $ X . %X,g = (5: (EE W(Y)- X) v (ye = up)}.The sets X,,fl < c1, form a family of power E < 'i; = m ; if all these sets X, belonged to the ideal Z then by X, E Z and thus W(Y)= X u X' E Z which contradicts Theorem 3

u

u

Bca

Pca

the assumption about 1. Thus, for some t9 < a, X, $ Z. It then follows that for almost all 5, cpe = wp and consequently cp* = w;. Thus the classes OX form a segment of the set Z/Z. In order to show that this segment is proper, we consider the function cpo defined by yo@ = w bfor 5 < Y. Clearly, {t:yo(& < ma} = W(0r).Thus if Q) 4: o:, then the set W(v)- W(a)belongs to Z. By a method similar to that used above we show that this assumption leads to a contradiction. Namely, we consider the family of power Z < rn composed of sets {t}where 5 belongs to Z since cp* F w*. Using the well-ordering theorem we infer that for every e there exists exactly one ordinal q(e) such that 2"Q= K4(e). Let P be the set of ordinals 5 < Y having the property (96 = w) v (there exists

e such that wp < Q):< w ~ ( ~ ) ) .

If EEP and q+ # w then by e(5) we denote the least ordinal such that we(€) < 9:

G

oq(Q(6)) *

If 96 = o,we put e(E) = 0. Case I. P$Z. We define a sequence y by letting y6 = ~ ~ ( for 6 ) 5 E P and ye = 0 for t $ P. For almost all [, y E< cpe and consequently y* 3 y*. This implies that there exists /?< Y such that yt = 0, for almost all E. Thus e(6) = /Ifor almost all E and hence ~ ~ ( ~ (= 6 )us@) )

322

X. 1NACCESSrSI.E CARDINALS

and

ve < ogV),which implies that

q(p) < v since

q* 3 (o,(~))*. On the other hand, K, is inaccessible and wehave a contradiction with the

definition of v*. Case 11. P EL For almost all

AP

(1)

E, we have ye # o and

< Ye)

[(oP

--f

(wg(P)

< cod]'

For E$P let 9, = cog(,). From (1) we obtain and thus

[(E) is a limit ordinal and [ ( E )

C(5) > 0 and

= lim q(e). P 4 ( 0

We define a sequence y by

cf(C(E1)

E V ,

for for

%=(o

~ E P . If y, = c ( E ) for some E$P, then the ordinal [ ( E ) would be strongly yt < Y. inaccessible (see Exercise 3, p. 314) in contradiction to Consequently yt < [ ( E ) for almost all 5 and < qc, and thus the sequence @€ = o$€has the property that 6* 4 pl*. Thus there exists p < v such that = o8 for almost all 5, that is, y, = p. Let X be the set of 5 for which yc = ,6. In the sequel, 5 will always denote an element of X . As has been shown above, the ordinal ((5) is the limit of an infinite sequence of type o8< v . Using the axiom of choice we associate with every ordinal 5 E X an increasing sequence u(0 of type cog such that its terms belong to Kand rpa = lim af).

[(n <

9-8

For fixed q

< w B the

z, = af)

for

sequence z defined by

EEX

and

z, = 0 for E$X

satisfies for all E E X the inequality z, < y , , thus z* 3 y*. It follows that there exists an ordinal a, < v such that for almost all 5 a(€)=

(3) 1 %' Let X, be the set of those t < v which satisfy (3). Then W(v)-X, E I for all q < op; since w8 < v and KO is non-measurable, we obtain

u ( W V ) -xl) I . E

tl< m,g

323

4. NON-MEASURABILITY OF THE ALEPH

Thus

Equation (3) then holds for every 7 and for every E €Xo. A sequence of type wB < v of ordinals a, less than v cannot be convergent to v because cf(v) = v .Thus there exists an ordinal orsuch that a, < coy < Y for all 7 < o+; hence by (3)

- lim a$€)= lim a,, < coy "- w m B wmb for all E E X ~We . conclude that q* -?, (coy)*, but this contradicts the definition of y . This completes the proof of Theorem 7. By a method similar to that used in the proof of Theorem 7 it can be shown that many other inaccessible cardinals are also non-measurable. For example, the first hyper-inaccessible cardinal and the first hyper-inaccessible cardinal of type 1 are non-measurable. It is not known whether the hypothesis: There exist measurable cardinals is consistent with the axioms of set theory. If this hypothesis will prove to be consistent, then it will be an extremely strong existential proposition asserting the existence of large cardinal numbers. Properties of non-measurable cardinals often play a role in abstract algebra and topology I). We prove below a theorem indicating their significance for the notion of compactness. Let m be an infinite cardinal number. DEFINITION: The topological space X is said to be m-compact if every family R of open sets which covers X (that is, X = Y)

u

Y ER

contains a subfamily R1c R of power

< m which also covers X .

I) A detailed discussion of these applications can be found in a paper of H. J. Keisler and A. Tarski, From accessible to inaccessible cardinals, Fundamenta Mathematicae 53 (1964) 225-308.

324

M. INACCESSIBLE CARDINALS

An example of an m-compact space is any space of power < m with the discrete topology (i.e. such that every singleton {x} is open). The notion of m-compactness coincides with the ordinary notion of compactness when m = KO.The Tychonoff theorem (see p. 139) states that the Cartesian product of an arbitrary number of KO-compact. spaces is itself KO-compact.We shall show in the following theorem that this property of the cardinal KO is exceptional. 2: If m is a non-measurable cardinal and lit > KO,then there THEOREM exists a Cartesian product of subsebof N (with the discrete topology) which is not m-compact. LEMMA3: If the space X is m-compact, then there exists an accumulation point for every set Y c X of power m, that is such a po'nt x that for every non-empty open set V containing x the intersection Y n V is infinite. Otherwise it would be possible to cover the space X with a family R of open sets such that for all VER the set Y n V is finite. Then there would also exist a family of power < m with this property, but this implies Y m-KO= m. PROOFOF THEOREM 2. Let m be a non-measurable cardinal and let M be a set of power nt. Let I = N M and let T = np?'(Mwith ) y ranging over I. Then the elements of T are functions g mapping I into N and the neighborhood of g determined by the elements tpl, ..., tpk E I consists of functions g' such that g'(tp,> = g(tp,) for j k. All neighborhoods, of this form constitute a base for the space T . For arbitrary m E M , let g, be the function defined on the set I by the equation g,(y) = tp(m). It is clear that g , E T and that gm8 # gmtI for m' # m". Thus the set Y consisting of all functions g, has power lit. We shall show that if there were an accumulation point a of Y then there would exist a tr-additive prime ideal J of subsets of M such that 2 = M . Because we assume that m is non-measurable, it follows P

<

<

u

ZEJ

that the set Y does not have accumulation points, which by Lemma 3 completes the proof of the theorem. Therefore, assume that a is an accumulation point of Y and let 2, = { m E W q(m) # a ( y ) ) for ?€I. Let J be the family consisting of all the sets Z,.

325

4. NON-MEASURABILITY OF THE ALEPH

Every subset Z of the set M can be represented in the form (mE M: yo(m) = 0 } , where qoE (0, l}" c I. It follows that Z = Z,, if a(qo)# 0 and Z = M-Z,, if u(yo) = 0. Thus for arbitrary Z c M either Z E J or M - Z E J . Examine now the neighborhood of u determined by yl, ..., v k . By assumption this neighborhood contains infinitely many functions g,,, (mE M ) , that is, there exist infinitely many elements m such that g m ( y i )= u(yi); equivalently, yi(m) = d ( q i ) and thus m$Z,,. In other words, the set (M-z,~) is infinite for arbitrary tpl, ...,yk.

n

igk

It follows that every subset of the set Z , belongs to J. For otherwise for some y we would have the inclusion M - Z , c Z , and thus the intersection ( M - Z , ) n ( M - Z # ) would be empty. Similarly, for arbitrary q a n d y , Z , u Z , E J , since otherwise Z , u Z , = M - Z , for some 6, and then ( M - Z , ) n ( M - Z , ) n(M-&,) = 0. Finally M $ J since M = 2, would imply that M - Z , = 0. Thus J is a prime ideal in the ring of subsets of M . m It remains to show that J is o-additive. Assume that Zqi 4 J. The

u thus u Z,, I= 1

complement of this union is of the form Z,,;

.= M .

Let

iEN

C(m)=rnin(meZpi). Then E e N M = I and u(E) is a natural numIEN

ber n. From the definition of E it follows that (E(m)= n) = (E(m) = a(Q) --f ( ~ E Z , , and ) thus (m$Z,) --f ( ~ E Z , ~ hence ); (M-Z,,)n ( M - Z c ) = 0 in contradiction to the property of the sets Z , proved m

above. Thus

u Z,,

E J.

Q.E.D.

i-1

The fact that the Cartesian product of m-compact spaces need not be nt-compact has several algebraic consequences'). The problem whether it is consistent to assume that there exists a ') See J. to$, Linear equations and pure subgroups, Bulletin de l'Acad6mie Polonaise des Sciences. SCrie des Sciences Math. et Phys. 7 (1959) 13-18; A. Ehrenfeucht and J. Lo& Sur Ies produits carthiens des groupes cyeliques, ibid. 2 (1954) 261-263. Cf. also J. Mycielski, a incompactness of NU and Two remarks on Tychonoff product theorem, ibid. 12(1964) 437-441, and the paper of KeislerTarski quoted on p, 323.

326

M. INACCESSIBLE CARDINALS

cardinal m > KOsuch that the Cartesian product of m-compact spaces is always m-compact has not been solved. Moreover, it is not known whether the least such number would coincide with the least measurable cardinal').

0 5. The axiomatic introduction of inaccessible

cardinals

It can be shown that the existence of inaccessible cardinals cannot be deduced from the axioms Zo[TR] of set theory. On the other hand, it is possible to strengthen this system by new axioms guaranteeing the existence of inaccessible cardinals. AXIOMOF TARSKI').For every set M there exists a family R of sets such that (i) M E R , (ii) X E Y A Y E R-+ X E R , (iii) X E R --f 2 X ~ R , (iv) ( x c R) A ( f l ( x ) R ) +XER.

A

+

f eRx

'THEOREM 1: Tarski's Axiom implies that for every cardinal rn there exists a strongly inaccessible cardinal > m. Indeed, such a cardinal is R" where R is any family satisfying (i)(iv), where it4 = rn. We shall write In(a) instead of there exists a family R of power K, satisfying (ii)-(iv). Tarski's Axiom does not allow us to conclude the existence of cardinals of the first Mahlo class and thus of cardinals of higher classes. L6vy3) has formulated an axiom scheme from which we can deduce the existence of such cardinals.

-

1) In connection with this problem see J. LOB, Some properties of inaccessible numbers in Infinitistic Methods (Warszawa 1960), pp. 21-23; D. Monk and D. Scott, Additions to some results of Erdos and Tarski, Fundamenta Mathematicae 53 (1964) 335-343 and the paper of Keisler and Tarski cited on p. 323, in particular Theorems 2.30 and 5.11. *) See the paper cited on p. 31 I . 7 A. Levy, Axiom schemata of strong infinity in axiomatic set theory, Pacific Journal of Mathematics 10 (1960) 223-238.

3 27

5. AXIOMATIC INTRODUCTION OF INACCESSIBLE CARDINALS

LBVY’S

SCHEME:

n{@> A R{@} A Cont

--f

‘dB [@(a,P>

A

a.

(I7,R, Cont are defined on p. 238).

In (811;

Gvy’s scheme states that every increasing and continuous operation defined for all ordinals has for at least one argument a a value /? such that K, is strongly inaccessible. Tarski’s axiom follows from Uvy’s scheme. Namely, letting 0 be the propositional function p+E = C we conclude from LCvy’s scheme that for arbitrary p there exists an inaccessible cardinal > I. Using Uvy’s scheme we can prove the following theorem. OTHEOREM 2: There exist cardinals of the .first Mum0 class. PROOF.Let @ be the following propositional function with free variables x and y : [(x is not an infinite increasing sequence of ordinal numbers) A A

0 = o)] V

v ( ( x is an infinite increasing sequence of ordinals) A th the type a of the sequence x is a limit ordinal)^ A 0 = lim XJ] v [(the type a of the sequence is of the form p+ 1) A €
A

{Y = min [(5 > xp1 A hi5)1>]). 6

From Tarski’s axiom, which is a consequence of LCvy’s scheme, it follows that @(x,y), that is, @ defines an operation. Using the

p, v! X

Y

construction described on p. 244 we obtain a propositional function Ind {@}, which we shall abbreviate as Y , such that

n{y>,

p ( a , ~+ ) @(Ca{y},a)*

Denote by pa the unique ordinal B such that Y ( a , B). Then we have @(plW(E),pE) for every ordinal 5 ; thus by induction with respect to 5 and by the definition of @ we infer that pE

< FlS

pA = lim p E €
for 5 < % if 1 is a limit ordinal,

pa+1 = min [(5 > pa1 A In (01 6

Thus we may define the operation 5+pt

-

as follows: p6+lis the

3 28

M. INACCESSIBLE CARDINALS

least ordinal a > pe such that K, is strongly inaccessible; if il is a limit ordinal, then p~ = lim pd. €4

Now apply L6vy's axiom scheme to the propositional function O ( t , 7) defined by q = pot. Clearly A R ( @ }A Cont ( 0 )and thus there exists an ordinal CI such thai In(p,,J that is, such that K,,,, = lit is strongly inaccessible. This cardinal belongs to the first Mahlo class. Indeed, cf(pm,) = cf(a), because pm, is cofinal with a. Since in is inaccessible, cf(a) = ,urn,.The set of ordinals t < p,, such that In(5) is clearly of type < pa; on the other hand, this set contains all ordinals of the form p,c+l where 5 < a and thus its type is 3 a 3 cf(a) = p,,. Thus this set has the type p,, and it follows that Kcm, belongs to the first Mahlo class. Similarly we can deduce from Uvy's axiom scheme the existence of cardinals of an arbitrary Mahlo class. On the other hand, this axiom scheme is not sufficient to conclude the existence of hyper-inaccessible cardinals. L6vy has formulated another axiom scheme from which we can deduce the existence of arbitrary large hyper-inaccessiblecardinals'). Undoubtedly the process of adding new axioms which guarantee the existence of larger and larger cardinals can be extended without limitation.

n{@>

$6. The continuurn hypothesis Cantor conjectured that the cardinals K1 and a, are identical. This conjecture, referred to as the continuum hypothesis, has been proved to be independent of the axioms of set theory. The continuum hypothesis is a particular case of the generalized continuum hypothesis, according to which the hierarchy of alephs is identical with the exponential hierarchy: (1) a, = K, for all ordinals a. This hypothesis is also independent of the axioms of set theory. We shall denote the generalized continuum hypothesis by (H). 'THEOREM1 : In the system consisting of axioms Z"[TR] and VIII, the hypothesis (H) is equivalent to: l)

See the paper cited on p. 326.

6. THE CONTINUUM HYPOTHESIS

(c)

329

r f m is any cardinal number then there exists no cardinal F such that m < F < 2"'.

PROOF.If m = K, then (H) implies that 2m = 2Na= 2"" = a,1 = K a + l , and thus no cardinal lies between nt and 2'". Suppose now that there is no cardinal between m and 2m. We shall prove by induction that (1) holds. For a = 0, (1) is obvious. If (1) holds for an u , then it also holds for u + l since

a,+l = 2"" 3 K a f l > K, = a,; would lie between 2'" and a,. Finally, if and if a,+l # Ka+l, then (1) holds for all 6 < I where I is a limit ordinal, then

The generalized continuum hypothesis implies a simplification of the laws of exponentiation of cardinal numbers.

'THEOREM 2: (H) implies: (a) n(u) = u for all a, (b) i f u is a limit ordinal and /I< u, then a

a/ =

Qa {aa+1

B < cf(a>, cf(u, < B < a.

for for

PROOF.Clearly n(0) = 0. If z(u) = u, then K,~,+ll = aacl by definition and thus by (H) K,(a+l) = which implies that n(u+ 1) = u 1. Finally, if I is a limit ordinal and if n(u) = u for all c1 < I , then

+

Hqa)= C

tca

=

~5 =

and thus n ( I )= I . Formula (b) is a consequence of (a) and of Theorem 17, p. 302.

'THEOREM 3 : (H) implies that if

then

330

IX. INACCESSIBLE CARDINALS

(i) s,,&+~ = a?, (ii) ~ , + l ,= ~ (iii) s,+~,,= ap (iv) sa,$ = a, (v) s,,, = (vi) sa,s = as

PROOF.(i)

i f /3 < y f l , fi is a if ,5 > y f 1, j3 is a i f a and ,5 are limit if a and /3 are limit

if

a and

/3 are

limit ordinal, limit ordinal, ordinals and 8 , < cf(a), ordinals and cf(a) < p < ct,

limit ordinals and

> ct.

< aa implies that

(ii) By definition,

(iii) Similarly we have Sr+l,,

= ay+1+

-v

, L hE+1 Y+1<5
=

a.,

(iv) From Theorem 2 we obtain (v) Similarly, s a , p = a,

+ c

at:

= a,+p-cf(4.a,,l

= Ua+l.

cf(a)<&
(vi) From the elementary formula .a.;

= ac+l for CI

< E it follows that

on the other hand, it is clear that

The hypothesis (H) was used only to prove (iv) and (v). Using Theorem 2 we can further obtain explicit values for s , , ~ + ~ . In a similar way we can also calculate the sum t,,, = C a;p. We €
(0

t,,l,, = a?.

6. THE c o N T I m HYPOTHESIS

33 1

t,,D= as+l, if u is a limit ordinal and /l 3 ur (ii) ta,s = a, [f c1 is a limit ordinal and /l < a . (iii) Formula (i) reduces the evaluation of t,+l,g to Theorem 2. The following theorem shows that by accepting (H) we do not have to distinguish between weakly inaccessible and strongly inaccessible cardinals. 'THEOREM 5 : The hypothesis (H) implies that every weakly inaccessible cardinal is strongly inaccessible. PROOF.If conditions (i) and (ii) from p. 310 are satisfied, then m = K, where a is a limit ordinal, thus n < ilt implies that n = K, where E < a. Consequently, by (H) 2 n = < K, = in and it follows that m satisfies condition (iii) from p. 311. There are many simple properties of cardinal numbers which are equivalent to the hypothesis (H). We shall give several examples'): 'THEOREM 6: The hypothesis (H) is equivalent in the axiom system ,Z'[TR] to each of the following formulas: (1)

A

(2)

/I 64% < Nu%),

( e l = Kutl), N

(3)

Remark: Formula (3) states that for every set X of power the family of all subsets Y c X which are not equipollent with X has the same power as X. For sets X whose power is an aleph whose index is a limit ordinal such a formula is not true (provided we accept (H); see Theorem 3 h)).

PROOF.(H) t (l), since (H) -+ (2), because

l) See H. Bachman, Transfinite Zahlen, Ergebnisse der Mathematik und ihrer Grenzgebiete, Springer 1955, p. 157. This book contains many other similar equivalences.

332

M. INACCESSIBLE CARDINALS

and 'a K.+P = (2Xa+l)Xa = 2Na+l = N ~ + ~ .

i ( H ) - i 7 2 ) . Assume that Nu+,< 2'". Then Ka+l and thus 2.' K:$ K,Kq, (2'3'a = 2'2,

<

<

<

which implies 7 2 ) . (H)-+(3). Indeed, by (HI,

K:$I K

< Ha+,< 2'"

= s"+~, a+l and by Theorem 3 (i)

U

we easily obtain that ~ , + l , = ~ +a,+,~ = K,+,. (3)+(H). Indeed, (3) implies K z l 6 K a + l and thus

2'" 6

NZ,< N,,, < 2'",

whence Nu+,= 2'". 'THEOREM 7: The hypothesis (H) is equivalent in the system .Zo [TR] to the theorem: (T) For every infinite set X , the set 2x can be represented as the union U M~ of a strictly increasing sequence of sets M~ equipollent €
with X .

PROOF.(H)+ (T). Let ? = K,; from (H) it follows that the set 2x W(o& Choosing for ME the is equipollent with W(w,+,) =

u

~y<€<"y+l

image of the set W(w,)we obtain the desired representation of 2x. (T) --f (H). Assume that we have a representation of 2x as described in (TI and let x'= K,. Then 2', = K,.Z and thus Z 2 K,+l and a 2 my+,. Let S =

u M E . We shall show that 2x

= S. Indeed,

s"=

K,+1

E-y+l

by the assumption that ME c M,,, ME # M,, and that i@ = K, for 5 < q < a . If YE^^ then there exists q < a such that Y E M , , . Because 2,,= K,, it follows that the set M,, cannot contain the entire union S and thus there exists 5 < my+, such that ME-M,, # 0; hence 7 < ( and q < LO,+,. Therefore 2x c S; the opposite inclusion is obvious. From the equation 2x = S we infer that 2Y ' = S = K,+,. Q.E.D.

-

6. THE CONTINUUM HYPOTHESIS

333

In the next section we shall deduce from the hypothesis (H) several theorems in the theory of ordered sets. ' = K1has many The hypothesis (H) and even its particular case 2O consequences in various areas of mathematics and particularly in the theory of real functions. We shall give only one characteristic example#'THEOREM8 '): There exists a non-countable subset 2 of the set of real numbers c' such that the intersection of Z with every nowheredense subset of & is countable. PROOF.Let R be the family of closed nowhere-dense subsets of t (see p. 32). Since = c (see the exercise on p. 196), there exists a sequence F of type SZ whose set of terms coincides with the family R. Let E, = FEF,,. There exist uncountably many indices 5 such

u

W C

that EE# 0; otherwise there would exist an ordinal o! < 9 such that OF,,=U F , , for a < p < Q ,,
and then we would have

W B

u Fq = € sinceevery one-element set belongs

to R. But then the set € would be the countable union of nowheredense sets, in contradiction to one of the basic theorems of topology due to Baire2). Let S denote the family of non-empty sets Et. By the axiom of choice there exists a set 2 containing exactly one point from each of these sets. Therefore z'> No.Let H be a nowhere-dense subset of E. Because the closure of H is also nowhere-dense, it too belongs to R and thus = FE for some 5 < $2. Since Z n F E No,it follows that Z n H < No. # We conclude this section with the proof of a theorem which implies as an immediate consequence that the axiom of choice follows from the hypothesis (C) (see p. 329).

-

a

-<

THEOREM 9 3): If ni is an infiite cardinal and if neither between m and 2"' nor between 2"' and 2'* lies any cardinal, then m is an aleph. l)

N. Lusin, Comptes Rendus de I'Acad6mie des Sciences, Paris 158 (1914)

1259. 2,

3,

See. K. Kuratowski, Topology I, p. 414. See the paper of B. Specker cited on p. 295.

334

IX. INACCESSIBLE CARDINALS

PROOF.Let us abbreviate the formula

A 1(m < F < 2") as H(ttt).

<

F

Using Corollary 10, p. 296 we obtain nt nr+ 1 < 2m,whence nt = m + 1 by H(n1). From m < m + m < 2m-2= 2"" it follows now m < 2m < 2". H(m) implies that either IU = 2nt or 2m = 2". But the second equation is impossible by Theorem 9, p. 295, because m2 2 2m. We prove next that H(m) implies m2 = m. For nt < nr2 2m. 2" - 2m+m= 2". The equation nt2 = 2" is impossible by Theorem 9, p. 295. Thus by H(m) we have that m2 = m.

<

By Theorem 7, p. 278 2N0n)<22m and 2" 2 m + ~ ( m )< 2(2m+tt(m))= 22'". 2 ~ ( m )

< < 22" .22" = 22m.2 - 22m+l = 22",

< 2". From < nt*K(m) < 2'".K(nq < 2".2" = 2"+l = 2"

thus by H(2m), 2"' = 2"+K(m) and K(m) nt

< n+K.C(n1)

it follows by H ( m ) that m+K(m) = nt.K(nt), which implies (see Lemma 4, p. 293) that nr is an aleph. COROLLARY: The hypothesis (C) implies the axiom of choice').

57. qasets

In this section we shall show how the continuuin hypothesis can be used to solve the following problem in the theory of linear orderings'): given a cardinal in, does there exist a linearly ordered set H of power m such that every linearly ordered set of power m is

<

l) This corollary was proved by Tarski and Lindenbaum. See their paper Communication sur Ies recherches de la thdorie des ensembles, Comptes rendus de la SociktC des Sciences et des Lettres de Varsovie, Classe 111 19 (1926) 299-330. Our proof is due to Specker. Another proof was given by W. Sierpinski, L'hypothPse gdndraZisde du continu et I'axiome du choix, Fundamenta Mathematicae 34 (1947) 1-5. ') This problem was posed and solved by F. Hausdorff. See his Grundziige der Mengenlehre (Leipzig 1914), p. 181. The construction given here is due to W. Sierpinski, Sur m e propridtk des ensembles ordonnds, Fundamenta Mathematicae 36 (1949) 56-67.

335

7. qe-SETS

similar to a subset of H? For the case m = KOwe have already discussed this problem and solved it affirmatively (Chapter VI, p. 219).

DEFINITION: A linearly ordered set H (of arbitrary power) is said to be an qC-set if H 0 and if for every two subsets A, B of power < H, such that

+

( a €A ) A (b E B)+(a

(1)

< b)

there exist u , o , w E H such that (aEA)A(bEB)-+(u
The qo-sets are simply sets which are densely ordered and have no first and no last element. ‘THEOREM 1: If H is an q,-set and X i s a linearly ordered set ofpower < K,, then X is similar to a subset of H .

<

PROOF.Let = Kayo! E and let z be a one-to-one sequence of type o, whose set of terms is equal to X ; moreover, let be a oneto-one sequence (of type a,,)whose set of terms is the set H. We shall define by transfinite recursion a sequence of elements of H such that the set of terms of the sequence is similar to X. The construction is almost identical to the construction used in the analogous case of sets .of type q (p. 219). For e < w, let ye = min C

A {(xC z

xIp(q))

q<e

A

(x<+ R x I ~ ~ ~ J I I

[(re3 x z q )

9

where + x and + H denote the cclessthan” relation in the set X and in the set H , respectively. The function min is to be understood in such a way that min A(() = 0 in the case where l A ( [ ) for all (. c We shall prove by induction that yQ# yq for q < e < w,. In fact, let

c = {e < o a : qA <e

(ye

+

yq)}

and assume that e < o, and W(p)c C . To show that e E C it suffices to show that there exists an element a € H such that for all q < e, af

Xq(q)

and

(a 3 H ~ q p ( q J

(re

f

3x rq) -

336

IX. INACCESSIBLE CARDINALS

For this purpose let

A

= ( ~ q ( q ) : (7<

e) * (rq +xre>l,

< e) A (re + x r q ) } of power < 3 < 6.< Kcand satisfy (l), thus

B = (xp(q) : (V

The sets A and B are there exist u , v , w satisfying (2). If B = 0 then let a = w , if A = 0 let a = u , and if A # 0 # B let a = v . Since pl,, = 0, it follows from the inequality obtained above that re ~ X T , , which completes the proof of the theorem. We shall construct an rc-set, assuming that w, is a regular ordinal (see Exercise 3, p. 339). Let H, be the lexicographically ordered set of those sequences pl E (0, l}w(""for which there exists a number x < wcsuch that plX = 1 and p l o = O for D > X . Xq(e) + ~ ~ q ( q )

THEOREM 2: If a, is a regular ordinal then

€€l

is an qyset.

PROOF.Let A and B be subsets of HE of powers < HE and let u and be one-to-one sequences of the types a,,and w, respectively (p < 5 , Y < 5) such that the set of terms of the sequence u is A and the set of terms of p is B. Assume that A and B satisfy Londition (1). Denote by < the relation of lexicographical ordering. For every e < w,, the term cc, is itse1f.a sequence of type we whose terms are either 0 or 1 and there exists an ordinal x = x ( e ) < we such that Q ~ , ~ = ( ~1 ) and aQe= 0 for cr > x ( e ) . We shall call the for the sequence up and we shall employ similar ordinal x terminology for the sequences p, e < w v y where by A(u) we denote the critical ordinal of the sequence Po. The regularity of u), implies that the sequence of critical ordinals for the sequences u,, e < w,,is not cofinal with w,. Thus there exists an ordinal C < wc such that x ( e ) < C for all e < w,,.Let ply = 0 for y # 5 and let plS = 1. Then pl E HE and pl < cr, for all e < LO,,. Similarly there exists an ordinal C' < w E such that A(u) < 5' for all D < 0,. If yy = 1 for y c' and y y = 0 for y > C', then y E Ht and Po < y for all D < my.

<

337

7. Vt-SEW

To prove the theorem it remains to construct a sequence ~ E H , such that a, < 6 < Pu for all e < w,,and cs < w,. First we construct by transfinite induction a sequence 6* of type wt which does not necessarily belong to He but is such that

< 6* < Bu Let z be an ordinal <

e < a,,,0 < w,.

for

a,

OJ& and let y E (0, For y < z we put F ( y , y ) = 1 if there exists an ordinal e < wp such that yIy = aely = 1 and F ( y , y ) = 0 if there is no such e. and From the theorem on transfinite induction it follows that there exists a sequence 6* of type we such that

6:=F(6*ly,y) thus

(6; = 1) 3

VA

e-/,

for [(Me8

y < q ;

== 63)A (aey ==

B
Assume that a, > 6*; then there exists an ordinal y < we such that a,, = 1,s; = 0 and = 62 for 6 < y . By the definition of the sequence 6* it follows that 8; = 1, which is impossible. Thus a, < 6* for every e < LO,,. In turn, assume that 6* > p,,; thus there exists an ordinal y < w, such that 6; = 1, Buy = 0 and i@ = boafor 6 < y . From the definition of the sequence 6* it follows that for some e < w,, a,ly = 6*ly

and

aey= 1.

Hence a, > BU contrary to our assumptions. We now modify the sequence 6* so as to obtain the desired sequence 6. We examine two cases: Case I: For every yo < we there exist y > yo such that 6; = 1. In this case 6* $ HE and thus the strict inequality a, < 6* < ,flu holds for arbitrary e < cop and c < 0,. Let yo be an ordinal such yo and 6, = 0 for y > y o . Clearly that SFo= 1. Let 6, = 6; for y for every yo the sequence 6 belongs to He. We shall show that we can choose yo in such a way that the sequence 6 satisfies the desired conditions. Froin 6* < Let yo > max(5, it follows that there exist ordinals 6 such that 6; #Bud and the least such ordinal 6,, satisfies

<

r).

338

IX. INACCESSIBLE CARDINALS

the equations 6z0= 0 and & 8 0 = 1. Thus do < C' (since = 0 for S > C'); hence 660< ,tIa,ao and a8= pa, for S < a0. Therefore 6 < p,,. For e < opthere exists an ordinal S o @ ) such that = 0 and 6$o(e) = 1 and for all y < &(el, E e , y = 6:. Choose yo > So(@) for all e < cop. The sequence 6 obtained from 6* by the modification just described satisfies the condition olQ < 6 for e < cop. Case 11. There exists yo < wE such that 6; = 0 for y > yo. Let y1 > max (yo, 5 ,5') and modify the sequence 6* so that OY= 6; for y # y1 and 6;l = 1. It is easy to check that the sequence obtained in this way belongs to HE and satisfies the condition ctQ < 6 < 8,. 'THEOREM3: Ha+, = 2'"; i f 5 is a limit ordinal, then Et = 2'".

-

a<€

PROOF.Assume that 5 = u + l and yo < o,.The set Zyoof those sequences 9 E (0, l}wY'"t)for which qyo= 1 and q y = 0 for y > yo has a power < 2'". Since HE= Zyo,it follows that

u

-

Y0-p

<

Ht < 2'"0, = 2'"K,+1 2'"2'" = 2'a. Since Zmahas power 2'" and Zm,c HE,we conclude that Et = 2'"". The proof of the second part of the theorem is similar. 4: If m = Kt is strongly inaccessible then there exists 'COROLLARY an qt-set of power m. PROOF.For such t , 2*a = H ~ .

C

a<€

COROLLARY 5: If 2'" = then there exists an qa+l-set of power This corollary follows immediately from Theorems 2 and 3. It can be shown that if 2*" > Kafl then no q,+l-set has power On the other hand, for ordinals 5 where cot is not regular, every q,-set has a power > K,. This fact follows from the theorem (proved by Hausdorff) which states that every qa+l-set contains a subset of power 2'" '). l) A simple proof can be found in the paper of W. Sierpiriski cited on p. 334. Interesting algebraic applications of the r,-sets are given by P. Erdos, L. Gillman and M. Henriksen, An isomorphism theorem for real closed jields. Annals of Mathematics 61 (1955) 542-554. Applications t o logic are given by S. Kochen, Ultraproducts in the theory of models, Annals of Mathematics 74 (1961) 221-261.

7. r]c-SETS

339

Exercises are isomorphic. wausdorff] 1. Every two q p e t s of power Hint : Use an argument similar to that used in the proof of Theorem 2, p. 217. 2. If W is an q p e t and if H' is a subset of H,dense in H, then H' is also an ?pet. 3. If cut is a singular ordinal, then every q p e t is also an qg+pet. 4. If R = r] and I is a non-principal prime ideal in the family 2N, then the reduced direct product RN/Z iz of type ql. [Kochen] Hint: Letf. and g , (n,m e N ) be elements of RN such that (i: fn(i) > gm(i))EZ for arbitrary n , m. 'The essential step in the proof depends upon the construction of a sequence h E RN such that the sets {i: fn(8-j > h(i)} and ( i : h(i) > g,(i)) belong to I. If fn(i)
h(i) = 4 [fK(j)(i)+gL(j)(i)l-

If the sequences fn and g,,, do not satisfy the inequalities above, then they can be modified in such a way as to obtain functions belonging to the same classes mod Z and satisfying the desired inequalities. For example, monotonicity can be obtained by letting f:(i) = f o ( i ) , fi+,(i)= max(f;(i),f,(i)) and similarly forg,.

CHAPTER X

INTRODUCTION TO THE THEORY OF ANALYTIC AND PROJECTIVE SETS

In this chapter we shall deal with the so-called operation (A) and the projection operation. We shall also treat the family of sets resulting from the application of these operations to sets belonging to a given fixed family, and, in particular, to the family of closed subsets of the space NN.The results we shall obtain constitute an introduction to what is called descriptive set theory.

5 1. The operation ( A ) We define this operation, first introduced by Suslin in his investigation of Borel sets'), as follows: Let H be a function defined on the set of finite sequences of natural numbers which associates with each such sequence a certain set. At times we shall refer to such a function as a defining system. For a given defining system H and for an arbitrary sequence v e N N we set

where vln is, as usual, the sequence of n terms plo, ql, ..., pn-l obtained from y by restricting the domain of v to the set of numbers less than n. I) I t was M. Suslin (1894-1919) who first gave an example of a projective, nonBorel set. To construct this example he introduced an operation on families of sets t o be called later the operation (A). It is interesting that he conceived the idea of this operation after finding an e r o r in a paper of Lebesgue. More information about this topic is provided by W. Sierpinski, Les ensembles project@ et analytiques, MBmorial des Sciences Mathkmatiques, fasc. CXII (Paris 1950), p. 28.

1. THE OPERATION

(A)

34 1

The union of all P, is called the result of the operation ( A ) performed on the defining system H

A(H) =

u P, u n =

v

PENN

Hrpln.

n

Examples 1. Let F be an infinite sequence of sets. We define a defining system by letting

for eE N n f l , and for the empty sequence e = 0 by letting Ho = Fo. Then A ( H ) = U F,,. n

2. Let H:

= Fn for

e E N"+l and H;

A ( H ~= )

= Fo.Then

n F,,. n

The examples above show that the operation (A) is a generalization of the operations of countable union and intersection. The operation ( A ) is monotone in the sense that if HL c H: for every finite sequence e of natural numbers, then A(H') c A(H"). We call a defining system H regular if for any finite sequence e and for arbitrary nE N the formula He c He,,

holds. In other words, if e' is an initial segment of e, then He c He,. If H is an arbitrary defining system, then the defining system H' given by the formula

HL=

nH , , ~

for

eE N"

jCn

is regular and, as is easily shown, A ( H ) = A(H). It follows that in most circumstances we may consider the operation (A) as being performed only on regular defining systems. By m;e we shall denote the sequence whose first term is m and whose following terms are the terms of the sequence e. More generally,

342

X. ANALYTIC AND PROJHXTW? SETS

for e E N" and f e N " we shall denote by e;f the sequence of (n+m) terms whose first n terms coincide with the terms of e, and whose last m terms coincide with those off, in their original order. LEMMA: r f the defining system H is regular, then (1)

U n HppIn = U U nHm:yln,

(2)

U n K := ~ UU ~ f l~Ho;(kyjn).

9

m

n

v

k

r p n

n

v

n

PROOF.If X E n Hql,, then by letting yn = y: we have yln'

=

n

yo;y'(n, so that XE

n

Hppo:yln

=U U f l Hm:ylnm v n

Conversely, if for a Iked rn and every n E N the formula x E H,:,,,* holds, then by letting yo = m,yn, = y, we obtain X E Hqln,for every n. By the regularity of H it also follows that x E Hpp!o.Thus x E n H , , , n

and (1) holds. Equation (2) follows from (1) by substituting in (1) for the defining system H the system H' where

H; = He. and where f is a finite sequence. The following theorems are straightforward and deal with the relationship between the operation (A) and the operations of forming Cartesian products and projections. THEOREM 1: If H and Q are defining systems and Qe c Y for every finite sequence e E N", then

if Hec X and

A(H)xA(Q) =A(@,

where the defining system U is given by

ue = H & ( e ) l ~ ( n ) X Q L j ( e ) l L ( n )

for

e E N"

(see pp. 97-99 for the definitions of the functions K,L, K', L').

PROOF.From the distributivity of union and intersection over car-

1. THE OPERATION

343

(A)

tesian product we obtain

A ( H ) x A < Q >=

(unHqpJx(U n

=

U f7 (Hpp x PY

We replace the double union

n by 0.In the nrn

v

c p n

QVlm)

m

Qtplm).

nm

u by u and the double intersection w

8

expression HIlnwe replace

pl

by K*(8)and n by

P

K ( p ) ; in Qvlm, p by I,*(@ and m by L ( p ) (see p. 123-124). Then we obtain A ( H ) x A ( Q ) = A ( U ) because K*(8)IK(n)= K'(@Jn)\K(n),and similarly for the function L.

THEOREM 2: The set A ( H ) is the projection onto the X-axis of the set M = n M,, in the space X x NN,where n

M n

= {<x,41):

x E Hpin}.

In fact,

We shall now examine the iteration of the operation (A). Assume that with every finite sequence e is associated a defining system G(=). Let

u nG%;, M = A ( H ) = u n H~,,,, = u n u n~ 8 p . v m v n H,

=A

( G= ~

~n

v r n

OTHEOREM 3: There exists a deJining system S such that M and f o r every e the set S , is one of the sets G$T;).

=A(S)

PROOF.From (ll), p. 124 it follows that (3)

For an arbitrary finite sequence eE N", let where t is the largest natural number such that J(m, t ) < n. Note

344

X. ANALYTTC AND PROJECTIVE SETS

that the sequence e(m)is empty if for every i, J(m, t ) > n. Denote by d(e) the length of the sequence e and let (K'(e)fbdte)))

Se = GLt(e)(

(4)

( (e)))lL(dce))

(see p. 99 for the definition of K' and L'). We shall show that for every 6 E N ( K * ( Q )Imp)) (5) Sqp = GL*(s)(K(P)),L(~)In fact, letting e = 61p in (4)and noticing that d(61p) = p we see that it suffices to prove the equations (6)

K'(filp)lK(p) = K*@)IK(P)Y

(7)

L'(6lp)(K(P))IL(p ) = L*(@)(=(P')lL(p).

Equation (6) follows from the fact that K(6lp) is a finite sequence of length p whose terms are K(&), so that both the left and right side of (6) denote the sequence of length p(p) whose terms are K(6,).(We appeal, of course, to the fact that K ( p ) b p , see p. 97.) For the proof of (7)we observe that L'(6Jp)(K(p)) is a finite sequence whose terms are L(6,(K(p),l)) and whose length is t, where t is the largest number such that J ( K ( p ) , t ) S p . Since J ( K ( p ) , L(p)) =p , it follows that t = L ( p ) because the function J(x, y) is increasing. Thus the sequence L'(6Jp)(K(P)) is identical with L*(6)jff(P))IL(p).Since L ( p ) < p , it follows that both sides of equation (7) denote the same sequence. Thus equation ( 5 ) is proved. Comparing equations (3) and (5) we conclude that M = A (S). Q.E.D. 02. The family A(R)

Let R be an arbitrary family of sets. By A (R)we denote the family of all sets of the form A ( H ) where H is an arbitrary defining system and He€ R for every finite sequence e. The sets belonging to A (R) are called A-sets with respect to R. The family A ( R ) satisfies the formulas (see p. 127):

0) R c A (R), (ii) R' c R" + A(R') c A(R").

2. THE FAMILY

A(R)

345

For if X E R and if the defining system H satisfies the equation He = X for every e, then A ( H ) = X and thus X E A ( R ) . Formula (ii) is obvious. We shall prove now that formula (iii), p. 127 also holds. 'THEOREM1: A ( A ( R ) ) = A ( R ) .

PROOF.The inclusion A (R) c A ( A ( R ) )follows from (i) and (ii). To prove the opposite inclusion, assume that M EA(A.(R)), i.e. that M = A ( H ) where H e € A ( R ) for every finite sequence. Therefore for R every such e there exists a defining system G such that G f ~ for every finite sequence f and W e )= A (G). By the axiom of choice we associate with every e one defining system G ( e ) .Then M =A(H), He = A(G("1) for every e, GP) E R for every e and f. By Theorem 3, p. 343 it follows that there exists a defining system S such that S,E R for all e E N " and M = A (S). Thus M E A ( R ) .Q.E.D. Since the operations of countable union and intersection are special cases of the operation ( A ) (see Examples 1 and 2, p. 341), we obtain the following consequences of Theorem 1. THEOREM 2: For an arbitrary family R, the family A(R) is both a-additive and a-multiplicative.

THEOREM3: The family A ( R ) contains the family B(R) (see p. 129). For B ( R ) is the smallest cr-additive and a-multiplicative family containing R. THEOREM 4: A ( R ) is the smallest family S closed with respect to the operation (A) (that is, the set resulting from applying the operation (A) to a dejining system H, where H , E S for all e, is itselfa member of S). PROOF.The family A ( R ) contains R and by Theorem 1 is closed with respect to the operation (A). On the other hand, if S also contains R and is closed with respect to the operation (A), then R c S and A ( R ) c A (S) = S. In Section 4 we shall discuss the case where R is the family of closed sets in the space N N and we shall establish certain criteria for

346

X. ANALYTIC AND PROJECTIVE SETS

a set belonging to A ( R ) to be Borel. In this section we shall prove a basic theorem applicable to arbitrary families R, from which the above mentioned criteria follow easily.

DEFINITION: Let R be a family of subsets of the set X and let A, B c X. We say that the sets A and B are R-separable if there exist sets P , Q E R such that A c P, B c Q and P n Q = O .

'LEMMA:If A

=

u A, and B u B, and if every pair A,, =

n

B, is

n

R-separable, then the sets A and B are Rdu-separable.

PROOF.Let A,, c P,,,, B, c Q,,,,, where P,,,,ER, Q,,, E R and P,,,n Qn,, = 0. It follows that Ac and that

u nPn,m n

and

Bc

m

u nPn,mERdu, u n n

m

m

n

U f-7 Qn,m r n n

Qn,rnE%u.

We now have

un n

Pn,m

m

n

un m

Qn,m

=

u n(Pn,m n

Qq,p)

n,p m,q

n

c

U ( P n , p n Qn

p)

= 0,

n,p

proving the lemma. Note that the axiom of choice is used in this proof to associate with each pair A, and Bm the sets Pn,, and e n , , .

'THEOREM 5 '): If the defining systems H and Q are regular, i f He€ R and Q, E R and i f for all sequences Q], w E N N there exists an n such that HVln n Qvin = 0, then the sets A ( H ) and A ( Q ) are B(R)-separable.

PROOF.Assume that the sets A and B are not B(R)-separable. Because B(R)du= B ( R ) and A =

uun P

P

Hp:qln, ~

B

=

u U fl q

v

Qq;Vlrn

m

') See W. Sierpiliski, Le thtor8me d'unicitt de M. Lush pour les espaces abstraits, Fundamenta Mathematicae 21 (1933) 250-275.

2. THE FAMILY

A(R)

347

(see (l), p. 342), we can use the lemma and infer that there exist numbers po and qo such that the sets HPo;91n and U Qno;v,,,,

un ~

,

n

v r n

n

are not B(R)-separable. We shall define by induction sequences @ and k the sets

U fl H e l k ; q l n

(8)

such that for every

U U Qoikqln

and

~n

Q

v

n

are not B(R)-separable. For this purpose we define for arbitrary k i t e sequences f and g of natural numbers, the least number p such that the sets

u fl

Qp;L(p):q[n

~n

P(f9

and

g) =

u fl H/;K(p):p,n are not B(R)-separable, v

n

or 0 if such a p does not exist. Then the sequences e and cr defined inductively by

eo =PO, @&+I = Kc(@lk,

QO

9

= qo,

Ok+l

=

Lc(@lk,

satisfy the desired condition (8). Clearly, for k == 0 the sets (8) are not B(R)-separable, by definition of the numbers po and qw Assume that for some given k the sets (8) are not B(R)-separable. By (2) these sets can be represented as unions

U U f l Hpik;rn;cpln

and

m p r n

U U nQulk;k:gln* h

~

n

By the lemma there exist numbers m and h such that the two sets U f l H p l k ; m ; g / n and U Q s l t ; k ; p l n are not B(R)du-separable- One 9

~

n

9

n n

such pair of numbers is the pair &+I and ak+1. Since B(R)du = B(R), it follows that the sets obtained from (8) by replacing k by k+l are not B (R)-separable. By the regularity of the defining system H it follows that H Q l k ; q , l n c Hplk,and thus

u n v n

Helk:q,ln

= H~~~ ER

348

X. ANALYTIC AND PROJECTIVE SETS

and similarly

U n Qaik:qln 9

C Qulk

€2-

n

Since the left-hand sides of these inclusions denote sets which are not B(R)-separable, it follows that Hplkn Qo,t # 0 for all k, in contradiction to the hypothesis of the theorem.

8 3. Hausdorff operations In this section we shall consider operations on sequences of sets. These operations will be generalizations of the operation (Aj. First of all we note that an operation defined for arbitrary sequences of sets is not in general a function (see p. 72). Thus, for instance, the operation of forming the union of a sequence of sets is an operation defined by the propositional function

similarly the operation ( A ) is an operation defined by the propositional function A b X = A (=H,,n)l. X

v ~n

Hausdorff operations are also defined by propositional functions with the additional property that each such operation depends upon some fixed set B c NN,called the base of the operation. The propositional function A rx = A (X E 47("1)1 X

v

PEB n

defines the Hausdorff operation of base B; and the set X satisfying this formula is denoted by HB(F). By definition, (1)

H~(F) =

u nF ~ ( ~ ) .)1

rpzB n

l) HausdorR operations (also called in literature the &-operations) were defined independently by F. Hausdorff und A. N . Kolmogorov in 1927. See L. Kantorovitch and B. Livenson, Memoir on Analytical Operations and Projective sets, Fundamenta Mathematicae lS(1932) 214-279. Newer bibliography can be found in the paper: Ju. S. Oczan, Teorija operacij nad mnoiestvami, Uspehi Mat. Nauk 10, No. 3 (1955) 71-128 (Russian).

349

3. HAWSDORFF OPERATIONS

Examples 1. If B contains only a single element, the sequence (0,1, 2, ...),

then HE(&')= f Fn. I n

2. If B is the set of all sequences p) E N N whose range is infinite, then HB(F) = Lim sup F,. Indeed, if X E H , ( F ) , then there exists a p ) € B such that X E F ~ (for , ) all n; this x belongs to Fk for infinitely many values of k, because the range of p) is infinite. Conversely, suppose that x ~ L i msup F,,; then there exist infinitely many numbers k such that x E Fk. If p) is an infinite sequence whose set of terms coincides with those numbers k , then X E fI F,,) c HB(F). n

3. If B is the set of all constant sequences, then HB(F) =

u F,,. n

Now we prove a theorem which characterizes Hausdorff operations. 1: Every Hausdorff operation @ = HB has the properties: THEOREM (0 @ ( F ) =

u F,,; n

(ii) ( a . @ ( F ) ) A (b$@(G^))+ V [(aEE,)A (b $ GdIn

Conversely, every operation satisfying conditions (i) and (ii) is a Hausdorff operation. PROOF:(i) follows directly from the definition of a Hausdorff operation. If a E H B ( F ) and b $ H , ( G ) , then there exists ~ , , E Bsuch that A (a E Fpo(n)),and on the other hand for every p E B there exists an n n

such that b $ G H ~ )In . particular, if p = p0, there is an no such that b q? G v O ( ~Thus ~ ) . condition (ii) is satisfied by taking n to be p)o(no). Let @ be an operation (or equivalently, a propositional function) and assume that @ satisfies (i) and (ii) for arbitrary sequences of sets F and G. Let Dk= (p): (p),, = k ) }; then D is a sequence of sets

v n

contained in N N and @ ( D ) cNN.We shall show that @ = HB where B = @(D). For assume that there exists an a such that either (2)

a E HB(F)

a 4 @(J?,

3 50

X. ANALYTIC AND P R O ~ c l l v ESETS

or (3) In case (2) there exists v such that 9 E B and aE

(4)

nF ~ ( ~ ) . P

Since cp E @(D) and a $ @(F), it follows by (ii) that for some n ~

€

and 0 ~a $ F n .

Thus, for some p , cp(p) = n and a $ FqCp)contradicting (4). In case (3) we have first by (i) that a E Fn and then by the defi-

u n

nition of the operation HB

Let Z = (n: U E F , } and let yo be an infinite sequence with range and by ( 5 ) vo6 B and po# @(D). Beequal to 2. Thus A (a EF+,~(,,)) n

cause at the same time U E @ ( F ) ,we conclude applying (ii) that for some p , po$ Dp and a E Fp. The first of these formulas together with the definition of Dp implies that p $2,the second that p EZ,which is a contradiction. Thus both (2) and (3) lead to a contradiction, and so for all F, HB(F) = @ ( F ) and HE 3 @. Q.E.D. We shall show that in a certain sense the operation ( A ) is equivalent to a Hausdorff oFeration. Let e E Np and f E N4. We shall call the sequence e an immediate extension of the sequence f if p = q f l and if elq =f (equivalently, e results from f by the addition of a single term). Let gc be a one-to-one mapping of N onto the set of all finite sequences of natural numbers such that ac(0) = 0 and let

Bo = {cp:(q(O)

= 0) A

An ac(p(n+ 1))

is an immediate extension

of &J(n))} For every defining system Q let Qi = Q0c(,,). In this way to each defining system there corresponds in a one-to-one manner a sequence of sets.

351

3. HAUSDORFF OPERATIONS

THEOREM 2: For every dejining system Q we have

A(Q) = HB,(Q’). PROOF.From the definition of the operation ( A ) it follows that A(Q)

VI /2 ( X E QpIn) n

V A (x

QA(plln)).

p l n

For every the sequence y defined by (yn = o(9,ln) belongs to B,, because y ( 0 ) = 0 and c ~ ~ ( y , ,is + ~an ) immediate extension of ac((yn). In fact, the first of these sequences is equal to v,l(n+1) and the second to plln. It follows that

x E A(@

+

xE

U f7 Q&,n, = x E HBJQ’). tyd,

n

Assume now that x E HBo(Q‘),that is, that for some sequence y E Bo we have X E Q&, that is X E QoC(r(n)).From the definition of

n

n n

n

the set Bo it follows that for every n the sequence oc(y(n))has exactly n terms. Denote by q ~ the ~ last - ~term of the sequence oe(y(n)).It follows without difficulty that oc((y(n))= Vln, thus x E fI Qpln, which n

implies that x e A ( Q ) . Q.E.D. Exercise 1. Find the set B such that HB(F)= Liminf F. .

54. Analytic sets

LetX be a topological space (see p. 26) and P the family of closed subsets of X.

DEFINITION. The family A ( F ) is called the family of analytic subsets of the space X. From the theorems in $ 2 we obtain a number of theorems about analytic sets, for instance: THEOREM 1: The family A ( F ) is a-additive and a-multiplicative. THEOREM 2: B(F) c A ( F ) ; that is, every Bore1 subset of X is analytic.

352

X. ANALYTIC

AND PROJECTIVE SETS

THEOREM 3: The family A ( F ) is closed under the operation (A). In particular, the result of applying the operation ( A ) to a defining system H where He is BoreI for every e is an analytic set. Theorems 1 and 2 result from Theorems 2 and 3, p. 345, Theorem 3 from Theorem 1, p. 343. THEOREM 4: The Cartesian product of two analytic sets in the spaces X and Y respectively is an analytic set in the space X X Y. PROOF.By Theorem 1, p. 342 this product is the result of performing the operation ( A ) on the defining system U where, for every e, Ue is the Cartesian product of a set closed in X and of a set closed in Y . Thus U, is closed in the space X x Y . Q.E.D.

THEOREM 5 : Every analytic set in the space X is the projection onto the X axis of a set closed in the space XX NN.

PROOF,By Theorem 2, p. 343 it suffices to show that the set M n = (<x,Y ) : XEHqln) is closed in XX NN,provided that the sets He are closed in X. Assume then that (xo,qo)$ Mn,that is, xo$ Hqoln. Thus there is a neighborhood U of the point xo such that u 4 HqoI,,for every u E U. The set (p: pin = po)n}= Y is a neighborhood of the point po in the space NN.Thus U x V is a neighborhood of the point (xo, vo>in the space X x NN.For (x,p) E U x V we have x $ H,,,,, that is, ( x , p) ef M,,. Hence the com-

plement of M, is an open set. Q.E.D. Deeper theorems concerning analytic sets require additional assumptions regarding the space X . We shall not consider the general theory and limit ourselves to the consideration of the case where X = ( N N J k (k = 1,2, ...); the basic properties of analytic sets are already apparent in this simple case'). Note that it is possible to limit the discussion to the case where k = 1: I) The theory of analytic sets for complete separable spaces (that is, for continuous images of the space NN) is given by K. Kuratowski, Topology, pp. 478-514. 386-421. For more general topological spaces the theory of those sets is given by Z. Frolik, A contribution to the descriptive theory af sets and spaces, Proceedings of the symposium on General Topology and its relations to Modem Analysis and Algebra (Praha 1961), pp. 157-173.

4. ANALYTIC SETS

353

theorems proved for the space N N are automatically translatable into are homeomorphic: theorems for the space ( N N ) k ,since N N and there exists a one-to-one mappingp of the space N N onto the space (NNjksuch that both p and p o are continuous. Such a mapping is given by the formula p(p) = (p), ..., p-')), where q(j)(n) = q(kn + j ) for n E N and j < k. . We shall prove that p is a homeomorphism in the case k = 2. The function p is one-to-one. Assume that p # y and P ( Y ) =
x Np(l)lk.

The function p a is continuous. For if p(p) = ( q ( O ) , $'I) and if N,,, is a neighborhood of v, then p'(N,,J contains a neighborhood N , ( o ) , ~ x N&),k of the point p(p), where k > n/2. Because the function p is bicontinuous and one-to-one, it follows that x + p ( x ) maps the family of open sets in the space N N onto the family of open sets in the space ( N N ) k ;similarly for the family of closed sets and the family of analytic sets. We prove now several theorems concerning the relationship between analytic sets and continuous functions in the space NN. THEOREM 6 : I f f : N N + N N is continnous a n d i f Z is a set closed in NN, then the set fl(Z) is analytic. l) See p. 138 for the definition of the set Newhere e is a finite sequence of natural numbers.

3 54

X. ANALYTIC AND PROlJXllVE SETS

PROOF:Let Qe =f'(Ne n Z) = the closure of the set f'(N,n 2).

The defining system Q is regular and its values are closed sets. We shall show that

f'(z)= A(Q1. I f q E Z , thenp,ENqInnZforeveryn;thusf(q)Efl(NqInnZ)eQ,,,, which implies that f(v)E Q,," and f(p7) E A ( Q ) . Therefore f ' ( 2 )

n n

c A (Q).

Next assume that ~ E A ( Q ) i.e. , that there exists a q such that a~ We shall show that a = f ( y ) . If tl # f ( q ) ,then there exist dis-

nQqIn. n

joint neighborhoods Ne and Nd of the points tl and f ( a ) : if the first k-1 terms of the sequences a and f ( 9 ) are identical and differ in the kth term, then let e = a/@+ 1) and d =f(v)p)l(k+ 1). Since tl belongs to the closure of the set fl(NqInnZ), it follows that N e n f l ( N q l nn Z) # 0 and thus for every n there exists a point ') y nE NVln n Z such that $(yn)E N , for every n. By the continuity off the set f"(Nd) contains a neighborhood of the point q, for instance the neighborhood NVrq. Therefore ys~ f - ' ( N d ) and f ( y q ) ENd,so that f ( y q )$ N e . The contradiction thus obtained proves that a =f(v). It remains to show that FEZ.Otherwise a neighborhood Nqlmwould be disjoint from 2, contradicting the defnition of the point ym. Thus g, E 2 and a = f@) ~ f ' ( 2 ) . We prove the converse theorem. THEOREM 7: Every analytic set has the form f ' ( Z ) , where 2 is a set closed in N N and f is a continuous map of N N into NN2). l) The existence of the point rp, is a consequence of theinequalitv Nenf '(NqlnnZ) # 0; on the other hand, the existence of the sequence tp is a consequence of the axiom of choice. The use of this axiom in that place could be eliminated for it can be proved on the basis of the axioms ,Z alone that there exists a function which associates with each closed non-empty subset of NN one of its elements. ') It can be proved that each closed subset of NN is a continuous image of NN. Thus the family of analytic sets in the space NN coincides with the family of continuous images of this space. See K. Kuratowski. op. cit., p. 440.

4. ANALYTIC SETS

355

PROOF.By Theorem 5, p. 352 the analytic set A is the projection of the closed set W in the space N NX N N : A = g'(W), where the projection g is a continuous map of the space N Nx N N onto NN. Now apply the mapping p c (p. 353). The set 2 = p-I( W ) is closed in N N ;the composition f = g o p is a continuous map of the space N N into itself, and f '(2)= g'(p'(2)) = g'( W) = A . COROLLARY 8: If g : N t , + N N is a continuous function and A is an analytic set, then the set g'(A) is also analytic. PROOF.By Theorem 7 the set A has the form f ' ( 2 )where f is a continuous map of the space N; into itself and 2 is a closed set. Thus the set g'(A) is the image h'(2) where h = g of is a continuous map of NN into itself. Similarly one proves that i f g : (NN)' + (")" and i f g is continuous, then the image g'(A) of the analytic -set A in (NN)' is analytic in (NN)". Thus in particular the projection of an analytic set is analytic. We conclude that the convese of Theorem 5 holds for the space NN. We prove now an important theorem concerning separation: SEPARATION THEOREM.) Any two disjoint o T ~ E 9:~ (THE R ~ FiRST ~ analytic sets in the space N N are separable by Borel sets; that is, there exist Borel sets P and Q such lhat X c P, Y t Q and P n Q = 0').

PROOF.Let H be a regular defining system such that for every finite sequence e the set He is closed in NN.Let H: = NK,(e)n HLfCej. Clearly, A ( H ' ) c A ( H ) . We shall show that the opposite inclusion also holds. In fact, if y E A ( H ) , then there exists pl such that for all n y E Hp,,,;since p E NVI,,we obtain y E HL,,,, where 6 = J*(y, pl). Assume that the sets A = A ( H ) = A ( H ' ) and B = A ( Q ) == A(Q') are analytic in N< but are not separable by Borel sets. By Theorem 5, p. 346 there exist sequences e, ( T E N $such that the sets H:l,, and QL,, are separable for no n. Thus for every n there is a point E(")E Hir,, n Q:l,,. By the definition of the defining systems H' and Q' the point E(") is an element of NK,(sln) n NL+,,,). It follows that the first n terms ') This theorem (and its name) is due to N. Lusin. See his Lepns sur les ensembles umlyiiques (Paris 1930). In the proof of Theorem 9 we appeal to Theorem 5, p. 346, thus, indirectly, to the axiom of choice.

X. ANALYTIC AND PROJECTIVE SETS

356

of the sequences t'"),e* = K*(e) and b* = L*(a) are identical. Since n was arbitrary, we conclude that e* = a*. We shall show that e* E A nB. For this purpose we choose an arbitrary number no E N and an arbitrary neighborhood P of the point e*. This neighborhood contains a neighborhood of the form NQ.,,, = NKlCpln, where n is an arbitrary number greater than no. Thus P contains the point 5'") which is an element of H,& and thus an element of He,,,c HQ,no.Hence I'n Helno # 0, that is, e* E Hplno = IfglnO since He is a closed set for every e . Since no was arbitrary, it follows that e* E H p l n = A.

n n

In an entirely similar manner we can show that o* E By and because e* = o* we have proved that every two analytic sets which are not separable by Borel sets are not disjoint. Q.E.D. OTHEOREM 10: Every analytic subset of N N whose complement is analytic is Borel'). For the Borel sets separating X and N N - X are identical with X and NN-X. Clearly analogous theorems also hold for the space (NN)k. The separation theorem and Theorem 10 proved here for the space ( N N ) khold for more general spaces as well, for instance, for arbitrary compact spaces and also for complete separable spaces z, although not for arbitrary complete spaces and not even for metric locally separable spaces 3). Exercises 1. Using Theorem 5, p. 346 prove the separation theorem for compact spaces. 2. I f f is a one-to-one continuous mapping from the space NN into itself, then the image fl(A) of a Borel set A is itself a Borel set. [Suslin] Hint: Show that the complement of fl(A) is analytic. Use the equivalence V € f ' ( A ) =V" =f(ar))+(ccEA)]. U l) This is a theorem of M. Suslin; see Comptes rendus de l'Acadt5mie des Sciences, Paris 164 (1917), p. 89. The converse is also true, because every Borel set is analytic (see p. 351) and in the space (NN)k the complement of a Borel set is itself Borel. 2, See, for example, K. Kuratowski, Topology I, p. 458. ') W. Sierpiliski, Sur un espace complei qui n'admet pas le th&o&me de Souslin, Fundamenta Mathematicae 34 (1947) 66-68.

4. ANALYTIC SETS

357

3. If B is analytic in NN and F a sequence of analytic sets in NN,then the Hausdorff operation with base B applied to the sequence F gives an analytic set. 4. Let A be a set of points of the plane (x, y > where 0 < x < 1 and 0 < y < 1. We define as a base for the topology on A the family of all open segments parallel to the X-axis. Show that the continuous image of a closed set in this space is not always analytic. [Sierpidski] Hint: Let M be a set of power continuum lying in the interval (0,l) of the X-axis and not analytic, and y - a one-to-one map of the interval (0,l) of the Y-axis onto M. Take as f the function f ( x , y ) = ( ~ ( y )O>. ,

0 5. Projective sets These sets were first introduced by N. Lusin as a generalization of the analytic sets '). In this section we shall consider the basic definitions of the theory of projective sets. We shall formulate them under general assumptions. As a result the relationship between analytic and projective sets will not be apparent in the beginning, but on the other hand, the significance of projective sets for the mathematical theory of definability will become evident'). L e t S b e a s e t #O. F o r X c S q a n d Y c S ' l e t

XO Y=

{<XI,

.*.,xq,Y1, - - - , Y r > : < X I ,

* . . , x q > ~ X v ( Y -l *, - , Y r ) ~ y } ,

X O Y = { < X I , . - * , X q , Y 1 , . * - , Y r ) : <X1? . - . , x q > E X A < Y 1 , . - * Y Y P ) E Y ) . We call these sets the free union and the free intersection. (We use the notation (ul, ...,uq) for elements of Sq, for q = 1 we identify u1 with : <-%(1),

*.*,xn(q))EX}

and we say that X, arises from X by a permutation of coordinates. For 1 < i < j < q and X c Sq let IdijX=

((XI,

...,xq-J: ( X I ,

.*-,~j-l,xi,xj+1 * ,. . , x q - 1 ) ~ X } ;

N. Lusin, Lecons sur les ensembfes annlytiques, COILBore1 (Paris 1930). The connection between the theory of projective sets and the notion of definability was established by K. Kuratowski and A. Tarski in the paper Les opPrations logiques et les ensemblesprojectifs,Fundamenta Mathematicae 17(1931)240-248. I)

z,

358

X. ANALYTIC AND PROJECTIVE SETS

we say that Idil X arises from X by identi$cation of the i-th and j-th coordinates. Finally for 1 r < q we denote by P:(X) the set

<

V

( ( ~ 1 *, * * , x r ) :

<XI,* * * , x r , Y 1*, * - , Y ~ - ~ ) E x } ;

YI....>Yq-'

P:(X) is the projection of X onto S'. We note the following elementary laws concerning these operations.

(i) [PW)lYr = PX-Gf)7 where n' is the permutation of the set (1 , ...,q } which is identical with n on the set {1,2, ... r } and is equal to the identity permutation on the set { r f l , ..., q}.

(ii) E(Idij(X)) = Idij(P'f(X)) for 1 < i < j

< r < q.

(iii) I f X c S and Y c S , then Id12(X0 Y ) = X u Y

More generally, S f X c

Sq

and

Idlz(X 0 Y ) = X n Y

and Y c Sq, then

Idl,q+lIdz,q+z... Idq,2q(XO Y)= X U Y , Idl,q+iIdZ,q+z... Id,,zq(X O Y ) = X n Y .

(iv) Z J X c Sq, then

(P-X), (v) I f X c Sq and 1

=


Idij(S"X)

sq--x,.

q, then

= Sq-'-1dij(X).

(vi) ZJ X c Sq and Y c S', then (S4-X) 0 (S'-Y> (P-;y)

@

= P+'--(X

O Y),

(S'- Y ) = S4+'--(X 0 Y ) .

(vii) I f Y is a sequence of subsets of S' and X c Sq, then

X

oU Yn = U ( X o Yn)y n n

X@

n Yn= n ( X O yn), n

n

U Yn o X n

=

U (Yn o X ) , n

fn l (yn0x1 = n (ynO X ) . n

(viii) Under the same assumptions as in (vii),

359

5. PROJECTIVE SETS

xo U Yn = U (XOYn), U YnOX= U (YnOX)y n

n

n

n

xo n yn= n ( X O Y n ) , fl Yn O X = f l ( Y n O x ) . n

n

n

n

We omit the proofs of these propositions which are not difficult. We give several more definitions and then proceed to the definition of the class of projective sets. Let KOc 2Sqand denote by Koqthe intersection KOn 2". We

u

q> 0

shall assume that KO satisfies the following conditions, where q y r > 0 are arbitrary. (1)

O E K ~ ~and

SeKol.

Koq is a field of subsets of

(2)

Sq.

( 3 ) rfXeKOqand Y e K O , ,then X Q Y E K ~ , a~n+d X ~O Y E K ~ , ~ + ~ .

If X EKoqand n is

(4)

a permutation of the numbers 1,2,

...,q,

then

-GE KO¶. (5)

If 1 < i <j <

q and XeKOq,then Id, X E K ~ . ~ - ~ .

We.call a family KOsatisfying (1)-(5) a base. For an arbitrary family L c 2' we denote by P(L) the family

u

q>o

of all projections of sets belonging to L,i.e. P(L)= ( Y :

VV[(1 <~<~)A(xEL~~S~)A(Y=P~~(X))I}. X qr

We let

C(L) = { Y :

V4 Vx [ ( X E Ln 2s')

A

(Y = S ~ - X > I;)

that is, C(L) is the family of complements of sets belonging to L . Applying to the base KO the operations P and C in succession we obtain classes of projective sets with respect to the base KO;

4 = ~ ( K o )KZ , = C(K1)y

y

K2n 1 1

= P(Kz~),K2n+2 = C(K*n+l),

e . 0

The familyKnis the n-thprojective class with respect to KOand the union = K , is the family of all projective sets with respect to KO.We

u Kn n

define Kn,q= Kn n 2'

Q

.

360

X. ANALYTIC AND PROJECTIVE SETS

THEOREM 1: For all n E Nand for every base KOthe class K,, satisJies conditions (l), (3), (4), ( 5 ) .

PROOF.Instead writing “the family K,, satisfies condition (i)” we

>

shall write (in). Clearly (lo) A (30) A (4,) A (50). Assume that n 0 and that (1,) A (3,) A (4.) A (5,). We shall prove the analogous formula for n+l. Case 1 : n even. Because 0 c S2 and 0 E K,, by (I,,) we obtain 0 E P (K,) = K,,, since holds. 0 = PI(0). Similarly, S = P:(S2) implies S EK,,,,. Thus (4n+1) and (5,+1) follow from (i) and (ii). To prove (3,,+1) assume that X = P&(X*) and Y = PF1(Y*), where 1 q1 < q and 1 < rl < r and where X* EK,,,and Y*EK,,,,. Thus

<

<xi

3

.-. 7

xgl, ~ 1 ,

V

Y e

9

0 . .

..,U q - q l ,

Vl,.

Vrl>

..,V r - r l

[(XI

EX@ Y 7

9

xql

9

u1

7

*..

7

uq-ql)

EX*

v ( Y ~*,. . , Y r l , v l , -..7vr-r1>~Y*l. Applying (I,,), (4.) and (5,) we conclude that there exist sets X’and Y’ belonging to Kn,q+rsuch that (XI,

.-.,Xqi, YI,

.-. vr-rl>EX‘ ..., xqi, u1, ..- +,,)EX*,

Y r , , ~ 1 , > uq-q19 01,

-

=(XI,

... xql

<XI 9

7

9

9

... Y r , , ~ 1 , 7

*. *

9

Uq-ql

-

9

9

Y

01

=(y17

9

*--

,vr-rl) E Y’

. . - , y r l , v l ,* * . vr-rl)EY**

The union X’u Y’ = 2 belongs to K,,q+r;for the free union of these sets belongs to K,,,Z(q+r), and we obtain the ordinary union from this free union by iterated application of the operation Id (see (iii))7 which by (5,) does not lead outside of the family K,,.Thus (XI, **-yxq1,Yl> *-*,Yrl)EX@

V

4,. * * I ug-ql.

[(XI>

Y=

- - - 7 x q 1 , ~ 1 7-

. . , ~ r ~ r .**~uq--ql, ~ 1 , ~ 1

v1. -**. V+r1

that is, X 0 Y = (2) E K,,,q + r . The proof that X g Y E & , + , is similar.

---,vr-r1) 7 €21;

361

5. PROJECTIVE SETS

Case 2: n odd. follows directly from (In). ( 3 n + 1 ) follows from (3,) by (vi), (4,+1) and ( 5 n + l ) follow from (4,) and (5,) by (iv) and (v).

2: I f q > 0, then each of the families Kn,q is a lattice of sets; THEOREM 4 the families Km n 2' and Kn.4 n Kn+l,q are Jields of sets. PROOF.The free union and free intersection of two sets belonging to K,, n 2' also belong to the same class. By applying the operation Id we obtain the ordinary union and intersection and the operation Id Q does not lead outside of the class K,,. Thus K,, n 2' = Kn,4 is a lattice of sets. The family Km n 2" is a field of sets, because X E K , n 2" 4 implies S'--XE K,+, n 2' for n odd, S'-XE K,,--ln 2' for positive 4 even n and S'--XE KOn 2' for n = 0. Finally the family Kn,qn K,,+,,, is a field of sets, because by the first part of the theorem it is a lattice and by the definition of K,, it satisfies the condition X E K,,,qn Kn+,,q S4--XEKn,q n Kn+l,qTHEOREM 3 : The following inclusion relations hold among the families K,, (where the sign stands in place of the inclusion sign c): -+

K1

-+

K3 + ... KZn-1

Kz

4

K4

-+

... Kzn

-+

-+

KZn+l ...

KZn+2 4+1

PROOF.Let XEKOn2"; thus XO S E KOn2s'+' and X = PI (XOS ) E K l . Hence K o cKl and by taking complements we obtain Ko=C(Ko)c C(KJ = Kz. Arguing by induction we assume that KZnc KZn+z and Kzn-1 c Kzn+l. From this assumption it follows that

Kzn+l= P(Kzn) c P(Kzn+z)== KZn+3

7

and thus

Kzn+z = C(Kzn+l)c c(KZn+3)== KZn+4This argument establishes the inclusions in the upper and lower lines of the diagram. The inclusion KZn c Kzn+lfollows from Kzn+l = P(K2n) 3 Kzn; applying the operation C we obtain C(K,,,)c C(KZ,,+,),i.e. Kzn-1 c Kzn+Z.

362

X. ANALYTIC AND PROJECTIVE SETS

We shall now explain the significance of projective sets for the theory of definability of sets. Let @ be a propositional function of q >, 1 free variables ranging over the set S. The graph of @ in S is the set 2, = {(Xl,

... xq) E SQ:@(XI, ...,x,)]. y

We say that @ defines the set 2., The major problems in the theory of projective sets consist of investigating the projective class of sets definable by various propositional functions. We say that a propositional function @ is projective with respect to the base KOif Z , E K , , and we say that @ is a propositional function of class n with respect to KO if 2, E K, . THEOREM 4 l):

If @,

and O2are propositional functions of class n,

then

(a) Qil v G2 and 0, A Q2 are propositional functions of the same class; (b) 1@ is a propositional function of class n+l if n is odd, of class n- 1 if n is even, and of class 0 if n = 0; (c) @ is a propositional function of class n f 1 if n is even, and of

v x

class n if n is odd; (d) // @ is a propositional function of class n if n # 0 is even, of X

class n f 3 if n is odd, and of class 2 if n = 0. PROOF.(b) holds because Z l a is the complement of 2., (c) holds because the graph of @ is the projection of the graph x

of @. (d) follows from (b) and (c) by de Morgan’s laws. It remains to prove (a). Assume at first that the propositional functions and CD2 have no free variables in common. Then the graph of v a2is Z,, @ Z , and the graph of Q1A Q2 is Z,, @Z,,, so both and G2 have free variables in common, graphs belong to K,,. If A Q2 from the free then we can obtain the graphs of Ql v CD2 and union and from the free product of the sets Z,, and Z,, by identification and permutation of coordinates. I) Tbeorem 4 is a scheme: for each propositional function we obtain a separate theorem.

363

5. PROJECTIVE SETS

Theorem 4 is important because it allows us to estimate by a simple inspection the projective class of a propositional function. We give examples of bases. The most important example is the family of all Borel sets of finite dimension. Let S be a topological space; then S4 for positive q E N is also a topological space with the Tychonoff topology. Let B(S4) be the family of Borel sets of the space Sq and let B

=B(S)=

U B(S4). 4

THEOREM 5: If every open set in S is a Borel set in S, then B is a basis '). PROOF.Condition (1) clearly holds. To prove (2) it suffices to show that the complement of a Borel set is again a Borel set. Thus let R be the family of those sets 2 c S4 whose complement is a Borel set. By the hypothesis of the theorem it follows that the family of closed subsets is contained in R. De Morgan's laws show that the family R is both u-additive and u-multiplicative, and thus R contains the family of Borel sets in Sq. To prove (3) we let, for X c Sq: U(x> = { Y c S': X @ YEB(S4+')}, V ( x ) = {Y c S': X@Y€B(S4+')}. It may easily be shown that the free union and the free product of closed sets is closed; hence X EF

+ (F c

U ( X ) )A (F c V ( X ) ) ,

where F denotes the family of closed sets in the space Sq. Equations (vii) and (viii) imply that the families V(X) and V ( X ) are o-additive and b-multiplicative, thus

X E F + (B(Sr)c U ( X ) )A (B(S')

c

V(X)).

Now we denote by U the family ( X c S4: B(S') c U(X)}and by V l) The assumption of Theorem 5 is satisfied, in particular, for the space NN, where every open set is the union of all neighborhoods of the form N , included in it, consequently a union of a countable number of closed sets. The same holds true for the space ( N N ) k .

364

X. ANALYTIC AND PROJECTIVE SETS

the family {X c S'J: B(S")c V ( X ) ) . Then F c U and I;c V. We shall show that the families U and V are a-additive and a-multiplicative. We assume that X i e U for i E N; that is, B(S') c U(Xi) or, in other words, Xi0 Y E B ( S ~ +for ' ) every Bore1 set Y c S'. From (vii) and (viii), p. 358 it follows that Xi) 0 Y E B(S4+') and Xi) 8 Y E B(Sq+')

(u I

for every set Y E B(S'), thus

(n

u Xi

i

EU

and

nXi E U.Hence the family I

i

U is both +additive and a-multiplicative. Similarly, we can show that the family V is a-additive and o-multiplicative. Since both families contain the family F of closed sets, they also contain the family B(S4). This means that for arbitrary sets X and Y such that X E B ( S ~and ) Y EB(S"),X@Y eB(Sq+') andX@YEB(S*+').Thus condition (3) holds. The proofs of (4) and ( 5 ) run similarly: denoting by W' and W" the families consisting of sets X e B ( S 4 ) such that X , E B ( S ~ )and IdiJEB(S4"), we show that these families contain F and are both a-additive and c-multiplicative. Q.E.D. We shall call the sets projective with respect to B simply projective classes, without reference to the base B. Ordinarily we shall denote the projective class zero by B(S), the (2n- 1)st class by P,(S)and the 2nth class by C,(S). The following corollaries result from Theorem 5. COROLLARY 6: The Cartesianproduct of two subsets of S belonging to any of the classes B(S),P,(S), C,(S) (n 2 1) belongs to the same class. In fact, the Cartesian product is here identical with the free product. COROLLARY 7: If ~ E ( S ~and ) ~ 'if tke graph of f (i.e. the set W = { (x,y)E S P + q : f (x)= y } ) belongs to the class P,(S),where n 2 1,then

Y E P , ( S ) -tf-l(Y)~p,(S)

and

YEC,(S)

-P

f-'(Y)EC,(S).

PROOF.From the equivalences

x Ef-i(Y) = f ( x )

EY

= Y

[ ( f ( x ) = y> A

(u E Y)]

it follows that the set f - l ( Y ) is the projection of the set ( ( x , y ) :

36 5

5. PROJECTlVE SETS

((x,y ) E W ) A ( y E Y ) ) . This set is in turn obtained f r o a the sets W and Y by the operations of free product and of identification of arguments and thus is an element of the class Pn(S).Since n 2 1, it follows that f - ’ ( Y ) belongs to Pn(S). The proof of the second part of the corollary is similar and is based upon the equivalence

O THEOREM: 8 If S = ( N N ) k ,then the class P l ( S ) is identical with the class of analytic sets.

In fact, in this case the projection of a Bore1 set is analytic and conversely (see p. 355). REMARK:A theorem analogous to Theorem 8 holds not only for spaces ( N N ) kbut more generally for complete separable spaces ’ ) . We give now a second example of a base. We shall call the set X c Nq elementary if there exist polynomials f and g of q variables and with natural numbers as coefficients such that (XI,

...) x q ) € X = [ f ( X 1 , ...

)

xq) = g ( x 1 ,

Let 04 be the least field of sets contained in all elementary sets. THEOREM 9: The family D =

u

Dq

... Nq

)

x,)]. and containing

is a base.

4

We omit the proof of this theorem. The projective sets with respect to D constitute the family of arithmetically dejinable sets; they have numerous applications in mathematical logic 3. We conclude this section by giving two examples of projective sets. # 1. We shall denote by G the family of open sets in the space under consideration. Let F be a closed set lying in the plane (or more generally I) A detailed discussion of projective sets for such spaces can be found in the monograph of K. Kuratowski, Topology, pp. 453-478. See S. C. Kleene, Hierarchies of number theoretic predicates, Bulletin of the American Mathematical Society 61 (1955) 193-213.

366

X. ANALYTIC AND PROJECTIVE SETS

F c C").We say that a point p E F is linearly accessible if there exists a segment p q which except for the point p has no points in common with F. Denoting by A the set of linearly accessible points of F and by Ip-ql the distance between p and 4 we have

=F n

+ IP-qj)

{P: 'd A ( 4 fP)A[(IP-rI+Ir-q/ q

r

v(r = p ) v l ( r 4 .

The set A is projective because each of the propositional functions

T ( p ,4 )

(4 = p ) ,

d(P,4, r) -[Ip--rl A(r) =(reF )

+ 1-41

=

IP--411

and

is projective (where, of course, py4 and r are variables for the elements of €'). From the definition of A we shall evaluate its class; we shall show that A is a set of the first projective class. Each of the sets {
{

and

491

{I:

is closed. The set { { p , y , r ) : @ ( p ,q, r ) } , where

@(Py q , r ) = ( q # P ) h [ ( I P - - r l + I r - 4 ] f I P - q 4 / ) v ( r = P ) v l ( r E F ~ 1 , is a Ga set since it is the intersection of an open set with the union of two open sets and one closed set; for { : @(P, 4 , I) = > {) n({

" { " (
The set ( ( p , q ) : A @ ( p , q , r ) } is Gal), and finally the set r

') This follows from the remark that a projection of a closed and bounded set in the space 6. is a closed set, and consequently, a projection of any closed set is an F,-set. For other spaces (for example NN)this theorem - as we know from 5 4 -is false.

367

5. PROJECTIVE SETS

v Y ( P ,4)L

A = {P:

where

4

W P ,4)= ,AW P ,4, r),

is of the first projective class, as the projection of a Csset '). 2. Let a be a double sequence of natural numbers (i.e. a E N N x N ) such that the set {(m, n, p ) : a,,,,, = p } is arithmetically definable. Denote by A the set { m : lim inf am,,= m}. We shall show that n+ m

+

the set A is arithmetically definable. For

Writing

v [x+y = z] instead of x < y

we may rewrite the formula

Y

above as

mEAr/\VAV[(n+u=q)v(p+v=a,nn)l P

E

q

nu*

A Vq A V [(w + amn) v (n+u p n uvw

= 4)v @+o = w)I*

Since each of the propositional functions w # a,,,,,, q+u = n, p + v = w is projective with respect to D, the set A is also projective with respect to D. Using Theorem 4 it is easy to see that if the propositional function w = a,,,,, is of the class KOwith respect to D, then the set A belongs to the class K4 with respect to D .

8 6. Universal functions Let S be a set and KO a base, that is, a family contained in the union 2s4satisfying conditions (1)-(5) on p. 359. As on p. 359

u

q>o

we denote by K,, the nth projective class with respect to KO and we let K,,,4= 2" K,,. 1: For an arbitrary family R of subsets of DEFINITION

u S4,f is

qcN

l) The arguments above d o not imply that our estimate is sharp, i.e. that we cannot lower the class in this estimate. However, in our case the estimate is sharp, because there exists a closed set F c C3 for which the set A is non-Bore1 (this was proved by 0. Nikodym, Fundamenta Mathematicae 7 (1925) 250 and by P. Urysohn, Proceedings K. Akademie van Wetenschappen 28 (1925) 984).

X. ANALYTIC A N D PROJ'@CTWE SETS

368

a universalfunction for R if the domain off is S and the range o f f is R.

DEFINITION 2: A sequence F of functions is a normal sequence of universal functions for R if (1) Fq is a universal function for the family R n 2# (q E N, q > 0); (2) for q > 0 the set P, = { ( x , s ) : xeFq(s)} belongs to R n 239+1. THEOREM I: If F is a normal sequence of universal functions for the n-th projective class K,,, then the set Mq

= {<xo, * - - ,

xq-2,

s>: (XO,

xq-2,

.'.Y

belongs to Kn+l,q-Kn,qif n is odd and to

s> $w91

Kn-l,q-Kn,q

if n is even.

PROOF.If Mq E Kn,q, then for some soE S

...,xq-2, xq-l> E Mq <xo, ...,+ l > for arbitrary x,, ...,X,-~ES,By the definition of <xo,

3

0 0 , ".,xq-2,Xq-l)

$F&q-J

E iF,(so)

Mq we have

= (xo, ...,xq- 2,Xq-l>

EFq(S0).

Letting = so we obtain a contradiction. Thus Mq$&,q. The set M q arises from by identifying the qth and (q+ 1)st coordinates and then by complementation. The first operation leads from a set belonging to Kn,el to a set belonging to K,,q (see Theorem 1, p. 3601, the second operation leads from a set belonging to K,,,q to a set belonging Kn+l,qor to Kn-l,q depending upon whether n is odd or even (see p. 362). Thus Theorem 1 is proved.

rq

COROLLARY 2: r f for every n 2 1 there exists a normal sequence of universal functions for the class K,,,then all inclusions in the diagram (see p. 361) Ki -+K3 + ... -+ KZn-1-+KZn+1-+...

KO/"

I/I

are strict. Moreover,

\r

/I

6. UNIVERSAL FUNCTIONS

369

PROOF.From Theorem 1 it follows that Kzn+t-KZn+z # 0 # K2n+2-K2n+t (3) for every n 2 0. If KZnbl= K2n+l,then by the diagram above K2n+2

c Kzn-t = K 2 n + l

contradicting (3); similarly, if KZn= K2n+2,then Kzn+t

c Kzn = KZn+2

also contradicting (3). Moreover, K2n-1= K2n+2implies K2,,+*c K2n+l, and KZn= K2,+l implies KZnfl= KZn+2. Analogously we show that KO# Kt and KO# K2. Corollary 2 shows that if for every n 2 1 there is a normal sequence of universal Functions for K,, then for arbitrary n >, 1 there exist sets belonging exactly to the nth projective class. Thus in this case the classification of projective sets cannot be reduced to a smaller number of classes. We shall prove that the hypothesis of Corollary 2 is satisfied when S = NN, n > 0 and the base KO is the family B = B(NN)OF Bore1 sets. The construction of the desired sequence of universal functions proceeds in several steps. Firt we let Gq be the family of open sets of the space (NN)qand let G = Gq.

u

q>o

LEMMA 3: There exists a normal sequence of universal functions for the family G. PROOF.Every open set X E G , is the union of the neighborhoods it contains, that is, of the Cartesian products

where e is a finite sequence of natural numbers. Using the one-to-one function 0' introduced on page 98 to map N onto the set of finite sequences of natural numbers, we can represent every neighborhood in the form N0qaj)where ai E N f o r j < q. Let C4 = -c;-l (see p. 98) map N

n

j<4

onto N q . Then Cq(n) is a sequence of q terms C,(n,O), Cq(n, l), ...,

3 70

X. ANALYTIC AND PROJECTIVE SETS

[ J n , q - 1). Thus an arbitrary neighborhood in (NNJqcan be represented in the form where n E N . Let The function H a is universal for the family Ga. In fact, for every p the set Hq(v) is open because it is a union of neighborhoods. If X EGq and p is the sequence of all natural numbers m such that O ( m ) c X, then X = Hq(v). Thus the range of Hq is Gq. It remains to show that the set rq= {(x, v): x ~ H , ( p j }is open. r,,,, where Tq,,= ((x, 9): x E O(p(m))}; This set is the union

u

msN

it suffices to show that each set that (x, V) = (970, ..., pq-l, p) ~ r (yo,

rq,m is open. For this purpose assume q , m i.e-

...)~ q - i ) E o ( ~ ( m ) = ) Il~oCcq(P(m),i), j
or Let 0' =

vjENoc
1<4

(vo, ...,Q ) ~ v} - ~ in , the space (NN)q+'.We shall show that v') E 0', then p'i(m+ 1) = pj(mf1) In fact, if (v;, Sri; E Not cq(v(m),j)

for j

0' c rq,m. and

< 4-

It follows that p'(m) = q(m) and thus V; E N o c cq (v,(m),j) for j < This shows that (pi, ..., pi-^) E O(cp'(m)), whence Q.E.D. ( ~ i.,.., ~ i - 1 ,p') E rqm , LEMMA 4 : There exists a normal sequence of universal functions for the family PI(") of analytic sets. PROOF.Let 4q(v)

=:

("Y

= ("1'

v (x, Y > $ Hq+1(v)1 {x: v ((XY YY v)

n {x:

n

V,+l)I.

371

6. UNIVERSAL FUNCTIONS

Since the set P,,, is open, it follows that Flq(q)as the projection of a closed set is analytic. If X is an analytic subset of the space (NN)', then X is the projection (say with respect to the qth axis) of a closed set Y in the space (NN)qtl(p. 352). Thus, for some 9, Y = (NN)q+l -Hq+l(v) and it follows that X = Flq(q). Finally the set { ( x , q>: X E Flq(v)} is analytic, because

V

xEFiJ(P) = x E ( N N Y A

Y<.[

YY(P)

$rq+&

Y,

hence this set is the projection of a set closed in (NN)q+I. Lemma 4 shows that for the family of projective sets the hypothesis of Corollary 2 hoMs in the case n = 1. In order to extend this result to higher projective classes we shall make use of a mapping which will allow us to replace a finite number of projections by one projection. Let h be a fixed positive integer; for j < h, and for arbitrary v E N N , let (see p. 353) @ = P [ p ( h n + j ) ] . It is easy to check that if q > 0, nEN

then the function &,h:

<%Y

..'Y9 ) q - l ~ 9) *
vq-l,Q)(o), **.,Q)('-'))

maps the space (NN)'Jf'onto the space (NN)q+".

LEMMA 5: If n 2 1 and B belongs to one of theprqiective classes Pn(NN) or Cn(NN)and i f B c (NN)q+h,then the set &;\(B) belongs to the same projective class and is a subset of ( N N ) q f l .

PROOF.The invariance of the projective class follows from Corollary 7, p. 364 and from the remark that the set W = {(x,~): f,,,(x) = y ) is closed. In order to establish the validity of this remark we assume that (X,Y)+W

and

x=
Y

=
and

.yq-l'fio, .-.

Y

2qk-l>.

From the inequality y # fqr(x)it follows that either (Pi # yi or ( P ~ ./L) fij for at least one i < q1j < h. If for instance pi(n) # yi(n)then, taking as a neighborhood U of the point (x,y) all those points (X',Y') =

(74,...,Q);-1,p),yo, ...Yy;-l,%, ...,4-1) I

)

for which the equations qj(n) = P?;(n)and yj(n)= y;(n) hold f o r j < q,

372

X. ANALYTIC AND PROJECTIVE SETS

we infer that the sets U and W are disjoint. Similarly, one shows that if v(J)# 6,, then there exists a neighborhood of the point (x,r) which is disjoint from W.Thus the complement of W is open and W itself is closed.

n

6: For each of the projective classes P n ( N N )and C n ( N N ) , THEOREM 1, there exists a normal sequence F@),F'(") of universal functions.

PROOF. The existence of the sequence F ( ' ) was proved in Lemma 4. Thus it suffices to prove that (I) the existence of F(") implies the existence of F'(") and (11) the existence of F'(") implies the existence of F("+I). (I).Letting Fi(")(y)= (NN)4-F$')(y) we obtain, as can easily be checked, the desired sequence of universal functions. (19. For this case, let

{ V (x, y> E F;Y&~)}.

~F+l)(p1) = ( N N ) ~n X :

v

The set {{x, y ) : x E F $ + ' ) ( ~ ) }is the projection of a set of the class Cn(NN),because x E F t + l ) ( Y )= [(x, y) E F;y&)].

vv

Therefore this set belongs to the class Pn+l(NN). For every y , F p l ) ( y )E Pn+l(NN).It remains to show that every X of the class P,+l(NN) can be represented in the form F:+')(y) for some yo. If X c ( N N ) qthen X arises from a set B E C,,(NN)by finitely many projections. If h is the number of projections, then B c (NN)4+h. Since permutation of coordinates does not change the projective class of X we may assume that

This condition is equivalent to

that is,

373

6. UNIVERSAL FUNCTIONS

The set f,sk(B) belongs to Pn+l(NN)(Lemma 5); it follows that for some plo, f4;i(B) = Fiti(y0),which implies that

x EX= // [(x,q ) E Fiy](qO)] 3 x E F$"")(plo).

Q.E.D.

P

Theorem 6 implies, in particular, that in the space N N for every n there are sets which belong to the nth projective class but not to any lower class. We conclude this section by giving examples of families for which there exists no normal sequence of universal functions. THEOREM 7: There does not exist a normal sequence of universal functions either for the family B of Borel sets in the space N N or for the family K, of all projective subsets of NN. PROOF. Assume that @ is a universal function for the family Bn2"" (or for K m n 2"w). It suffices to show that the set {{y, pl): y ~ @ ( p ) } is not Bore1 (or projective). Supposing that it were Borel (projective) then the set {q: q E @(pl)}, and thus the set {T: q $ @(y)}, would also be Borel (projective), since the operations of identification of coordinates and of complementation do not lead outside of the classes B and K,. From the assumption of the existence of a normal sequence of universal functions it then follows that for some plo, { q : q $ @(q)}=@(ploj, that is, pl # sP(pl) = pl E @(plo). Substituting plo for p we obtain a contradiction. Q.E.D. In 0 11 we shall use the following theorem, which can be proved without difficulty by applying the methods obtained above. O

THEOREM 8: For arbitrary n and for k 2 I the classes B ( N ~ ) ~ ( N ~Pn(NN) } ~ , n (NNjk

and

c,(NN)

n

are a-additive and @-multiplicative.

PROOF.The theorem holds for the zero projective class, because this class coincides with the family of Borel sets. If the theorem holds for the (2n+ 1)st class (i.e. for Pn(NN)), then it holds for the (2n+2)nd class (i.e. for Cn(NN)), because sets belonging to this class are just complements of sets belonging to the (2n+l)st class. Thus it suffices to prove the theorem for the class Pn+l(NN)assuming that

3 74

X. ANALYTIC AND PROJECTIVE SETS

it holds for the class C,,(NN). (If n = 0 then we replace C,,(NN)by the class of Borel sets.) Therefore, we assume that &EP,,+~(") and X i c ( N N ) kfor i E N . By Lemma 5 it follows that there exists a set Yi c (NN)k+' such that YiE Cn(NN),and xeXis

v [ ( x , ~ ) E Y ~for]

iEN.

B

Thus XEUXi-V i

which implies that

[ ( x y 6 ) ~ Uyi], f

8

u Xi EP,,,,( N N ) ,because by the inductive assumpi

tion

UI YiECn(Nh)*

To prove that tain X E

ni X i E P " + ~ ( Nwe~ )apply (10) from p.

123 and ob-

n xi= p, v ((X,6)Ey i ) = v A ((x, w ) ) ~Y*), i

i t 9

B

m

where I?(~) is defined as on p. 99. It remains to show that the set {(x,6): ( X , ~ ( ~ ) ) E Y belongs , , , } to class C , , ( N N ) . But this set is equal to the set f - ' ( Y , x N N ) , where f is the function defined by the equation f ( x ,6) = ( x , O(m)). Because the graph of this function is Borel, applying Corollary 7, p. 364 we obtain the desired result. Exercises 1. Prove that the classes Pn(NN) are closed under the operation (A) for n > 1. 2. Prove that there exists a normal sequence of universal functions for the families F d , Ga, Fd,Gee, etc. in the spaces NN, (NN)Z,... and derive as a corollary a statement about the inclusion relations which hold between these families.

3. The pair (U,V), where U and V are subsets of the space (NN)Z,is a universal pair for a family K of subsets of NN if and only if for arbitrary sets A, B E K there exists V E N N such that A = {y: and B = {y:(p, y ) V~> .Prove that if F i s a normal sequence of universal functions for the n-th projective class Pn(NN) orC,(V", thenthepair(U, V)where U = ((9,y ) : y ~ F ~ ( K * ( c p )and ) } V = {: y E Fl (L* (9)))is universal for the family P,,(NN) n NN or C,,(NN)n NN and that U and V both belong t o P,,(NN)or to Cn(NN).[Lusinl

375

7. SIEVES

57. Sieves In $9 7-11 we shall use the theory of ordinal numbers to investigate the projective sets of the space ( A V ~ )We ~ . shall denote this space by X , where k is an arbitrary fixed natural number (without loss of generality we may let k = 1; see p. 353). The applications which we have in mind make use of a special ordering of the set d = N"+l, that is, of the set of a11 non-empty

u n

finite sequences of natural numbers. 0 will denote the one-to-one mapping of d u (0) (the set'of all finite sequences of natural numbers) onto N which satisfies the condition a(0) = 0 (see p. 98). For arbitrary sequences e E N", f~ Nm(n > 0, m > O), let

e 42 f -

V (f=elk),

kim

e -5 f-[(e=f)v(e <1f)v(e<2f)l.

The relation e + 2 f holds exactly when f is a proper initial segment of e. The meaning of the relation 31is illustrated by the following examples: 41<3, lo>,

41<5>,

<4,1,5>41<4,2,8,2).

the set d into type q. 3 -is obvious. Antisymmetry. Assume that e 3 f and f 3 e. If neither e is a proper initial segment off nor f is a proper initialsegment of e, then either e =f, or there exists a least k < min(n, m) such that ek #&. In the latter case it follows by the assumption that e 3f and f 4 e; but e 3 f contradicts the inequality fk < ek and f 3 e contradicts fk > ek. Thus the only remaining possibility is e =f. Connectedness. Assume e # f and e non <J Iff is a proper initial segment of e, then e 4 2f and thus e 3 5 Otherwise there exists a least k < min(n, m) such that ek Zfk; and by the assumption that e non < A we have ek > h and thus f -3e. Transitivity. Clearly e S2f F Z g-+ e F2g.

THEOREM 1: The relation PROOF.The reflexivity of

3 orders -

376

X. ANALYTIC A N D PROJECTWE SETS

If e + l f ~ 2 and g k is the least number such that ek #A, then ek >fk and ek > gk because f is a proper initial segment of g. Moreover, ep =fp = gp for p < k . Thus e Flf F 2 g e >lg. If e g2f + l g and k is the least number such that # gk, then f k > gk and fp = gp for p < k. If the length of e is greater than k, then it follows that ek =fk > gk and ep = g p for p < k , and thus e + l g. On the other hand, if the length of e is less than k, then e F2g, because e j =J; = g j for all terms of the sequence e. Thus e F1f >2 g +e>g. Finally assume that e > f F g and let k and h satisfy the conditions

-,

ek>fk,

P
qgh7 If k g k and p < k -, ep = gp;if k > h, then eh < gh and q < h + e4 = gq. Thus e F1g -, e F l g . The four cases above exhaust all possibilities, and thus the transitivity of the relation 4 is proved. The set J has neith2 thejirst nor the last element. For if n > 0 and e = (eo,e l , ... ,en-l), then > e > <eo,el,...,en- 1,O). (eo+l,el: ... The set d is dense. Assume first that e F2f and let e = (fo , ... and f = (fo, ...,fk-l,fk, ...,fm>, where k m. Then the sequence (fo,fi, ... ,fk-l ,A+ 1 ) lies between e and f. Assume next that e F1f and that k is the least number such that e, #fk. Then ek >f k . If the length off is > k+ 1, then

<

e >l(eO~ . . - 7 e k - 1 , f k 9 f k + l + 1 )

=

(hi * . - 7 f k - 1 7 f k , f k + 1 + 1 )

>If;

if, on the other hand, the length off is k+ 1 , then 7fR-17 ek, 0) >If* ek-1, ek7 0) = ( f 0 7 e > 2 (e07 The set J is, therefore, densely ordered without first or last element, thus the order type of d is '11 (see p. 219). # W e can prove Theorem 1 differently. Let Ro be the set of rational numbers (2m+ 1)/2" contained in the open interval (0,1). Y

n

2-ck,where

Every number r E Ro can be represented in the form k-1

377

7. SIEVES

n > 0 and 0 < c1 < c2 ... < c, . Associating the number r with the sequence e(r) = ( ~ ~ - 1c2-cl-1, , ..., C , - - C , - ~ - ~ ) we obtain a oneto-one mapping of Ro onto d such that rl > r2 =e(rl) -$ e(r2).# THEOREM 2 : I f g , e N N ,then ?In

> yl(n+l)

f o r n>O.

PROOF.The sequence cpln is a proper segment of the sequence ?In+ 1. THEOREM 3 : Let e(")E Nkn for n E N and k, > 0. Assume that e(") + e("+') for n E N . Then there exists an increasing sequence of natural numbers p , such that k , >, n+ 1 and such that the sequence e(P,)/(n+ 1) is a proper initial segment of the sequence e(Pn+l)/(n+2)for every n E N . PROOF.If e?) S1e("+'), then e$?, > ep+'); on the other hand, if e(")F2e("+'), then e p ) = e$"+').Thus the sequence e$') is non-increasing. Because there exists no infinite decreasing sequence of natural numbers, it follows that, from a value of n on all the terms e p ) are equal. Let po be the first number such that e p ) = e$'O)

for

n >po.

Then kpo>, 1 and e(")/l= e@o)ll for n > p o . Assume that s > 0 and that the numbers p j for j conditions PO < P I

n >pj

-+

< ...
(kn >,j+1)

A

< s satisfy the

k , >,j+L

(e(n'I(j+l) = e"j'l(j+l)).

We consider now the sequence of numbers el") for n >,P,-~. Since e(")ls= e(ps-l)Is, it follows that for n >ps-l the sequence e(") has length >, s and that the equations e:") = ejn+') hold for all j < s. If e(Ps-l) F2e(@,then the sequence e(") has at least s f l terms. If e(PS-')

F1e(") then there exists h < min(kPs-l, k,) such that e p - ' )

>

>

> ep). Thus h s and again k , s+ 1. It follows that all the sequences e(:) for n >,pS-' have length at least s+l and that the term e p ) is defined. the numbers el") form a non-increasing sequence. Thus For n there exists a least number p s > pS-' such that e?) = eLPS)for all

378

X. ANALYTIC AND PROJECTIVE SETS

n >p,. It follows that

k, >, s+ 1

and

n > ps -+ (kn2 s+ 1 ) A (e(")l(s+ 1 ) = e"S)/(s+ 1 ) ) .

The sequence of numbers pn defined in this way by induction satisfies the assertion of Theorem 3. COROLLARY 4: r f the hypotheses of Theorem 3 are satisfied, then there exists an infinite sequence q~ E N Nsuch that for some infinite increasing sequence of natural numbers p n pln = e("n)I(n+l)

for

nEN.

PROOF.We may take for q~ the sequence defined by the formula pn = e p ) , where p n is the nth term of the sequence defined in Theorem 3. We introduce the notion of sieve'). A function L ~ ( 2 ~ that ) ~ is, , a function with domain J such that L, is a subset of X for every e E d, is called a sieve. The sieve L is called open (closed, Borel, analytic, etc.) if for every e E J the set L, is open (closed, Borel, etc.). For any sieve L and for x EX we let IL(x) = {eE 6:X E I,,}.

Thus for every x the set IL(x) is a subset of 6. Denote by R ( L ) the set of x E X for which there exists an infinite sequence e(")of elements of I'(x) such that e(")> e("+')for n E N . Then x E R = {the set IL(x) is not well-ordered}. We say that the set R(L)is sifted through the sieve L. We can picture the function L as a subset of the plane, where J is the vertical axis, X the horizontal axis and L, as a horizontal segment with ordinate e. For el 3 e2 we can interpret e, as lying higher than el on the vertical axis. The set IL(x) then consists of those e that the vertical line through x intersects L,. The sifted set R(L) consists of points x such that the intersection of the vertical line through x and I) The notion of sieve is due to N. Lusin; see his paper, Sur les ensemblesanalytiques, Fundamenta Mathematicae 10 (1927) 1-95. In a particular case this notion was considered by Lebesgue in 1905. See H. Lebesgue, Sur les fonctions reprisentables analytiquement, Journal de Math., 1905, pp. 213-214.

3 79

7. SIEVES

the sets L, is not a well-ordered set (the intersection contains an infinite decreasing sequence). 'THEOREM5: The set sifted through an analytic sieve is analytic. PROOF.By definition, x e R ( L ) if and only if there exists an infinite sequence of elements of d such that e(")t e("+l)

Letting

vn= a(e(")) Fn # 07

and

x E Le(n) for all n E N .

we have 0'

(FnJ F be( P n + d

and

x E -b('pn)

for all n E N . Thus Clearly the set QL of those pairs (v, x ) for which P), # 0 is Borel. ( ~analytic. ~) The set QY of those pairs ( 9 1 , ~ ) for which X E L , C is In fact, (9, x> E

QL

E

'v v [ ( x E L , ) (vn A

q:N esJ

= ql A

( 4 4 =q)],

and thus

Q Y = U U [ ( ~ " X L e ) n ( { ~P ):n = q } X X ) n Z e q ] , qEN e c d

where Z,, = 0 or Z,, = N N X X depending on whether a(e) # q, or a(e) = q. Since the set N N x L, is analytic, it follows that the set Q; is analytic. Similarly one proves that the set Qr' which consists of those pairs (tp,x) for which uc(pn)> U~(P),,+~)is a Borel set. Thus x ER(L)= 'd [ E 'p

n (Qh QYnQ3], n

and the set R(L) is therefore the projection of an analytic set and, as such, is itself analytic (see p. 355). THEOREM 6: Every analytic set can be obtained by applying the function R to a closed sieve. PROOF.Let M

=

u f-1 H,,,, where H is a regular defining system ~n

such that for every e E ~3 the set He is closed. For simplicity we may assume that H, = X .

X. ANALYTIC AND PROJECTIVE SETS

380

Let L , = He for e E d ; the function L is then a closed sieve, We shall show that M = R(L). First we suppose that x E M , i.e. that there exists a sequence 9 E N N such that x E (7 H,,, = L,,, . Then yln E IL(x) for all n > 0 and thus n

n n

the set IL(x) is not well-ordered, because for every n > 0, qln pl(n-t-1). Hence X E R ( L )and M c R(L). Next we suppose that x ER(L), i.e. that there exists a sequence of elements of d such that e(")E IL(x), x E Le(n)= H,(n) and e(")+ e("+') for all n E N . By Corollary 4, p. 378 there exists a sequence such that for some infinite sequence of natural numbers p n , we have e@n'[n= yln. From the regulariLy of H it follows that X E HJP,,)c HJP,,)~, = HPIn and thus X E H q I n .Q.E.D.

un (

~

n

Theorem 6 does not imply that there is a one-to-one correspondence between analytic sets and closed sieves: in fact different sieves can determine the same sets. Let V be a segment of d such that = 11 and let f be an order-preserving map of d onto V . For any sieve L, let L: = L,,,,, if eEV, if e$V. L: = 0

v

It follows from definition that (e E v) -+ [(e E 1,. ( X I )

= ( f WE I d x ) ) ] . Thus for arbitrary e E d we have f(e) E ILL.(")= e E ZL(x); that is, IL,(x) Clearly the sets IL(x) and I&) are similar, thus R(L) R(L') although the sieves L and L' are in general distinct. Now let ZI be an element of d greater than all'the elements of V and let L" differ from L' only in that L:' = X . Clearly ILt,(x)= IL,(x)u { v } , and thus R(L") = R(L'). We have proved the following theorem:

=f'(I,(x)). ==

7 : For every analytic set M there exists a closed sieve L THEOREM such that M = R(L) and such that the set IL(x), for every x , has a largest element. Similarly, replacing w by a sequence of type w all of whose terms are greater than those of V we obtain the following

7. SIEVES

381

THEOREM 8: Fof every analytic set M there exists a closed sieve such that M = R ( L ) and for no X E X the set IL(x) has a largest element.

5 8.

Constituents

We shall use sieves to analyze sets of the class C l , that is, sets which are complements of analytic sets. ! we let For a given sieve L and a given ordinal a < 2

CJL) = ( x : IL(x)= a}. We call this set the u-th constituent determined by the sieve L. THEOREM 1: X-R(L)

=

u C,(L).

a
For X E X - R ( L ) if and only if there exists an ordinal M which is the order type of the set IL(x);it follows that a < Q and thus XEC,(L). We shall show that under certain assumptions all constituents are Borel sets. For this purpose we need several auxiliary theorems. For arbitrary sieves L and L’, let T

= T ( L ,L’) =

{ ( x ,x’): the set IL(x)is similar to a part of IL.(x’)},

T* = T*(L,L’) = { ( x ,x‘): there is an e in I,, (x’) such that I&) is similar to a part of the segment O[e) of IL,(x’)}.

LEMMA 2: If the sieves L and L’ are Borel, then the sets T and T* are analytic. PROOF.The condition (1)

( x , x’) E T

is equivalent to the existence of a one-to-one order preserving function which maps the set IL(x) into IL.(x‘). Because the sets IL(x)and I,,[x’) are countable, we may restate the condition above in the following form: there exist two infinite sequences e(O),e(l),...,

f(O),f(l),...

of elements of d such that IL(x) is the set of terms of the first sequence, ZLt(x’)contains the set of terms of the second sequence and

382

X. ANALYTIC A N D PROJECTNE SETS

where A m , = { ( y , W ,X ,

XI):

vm

B m n = ((9,P, X , x'):

= n},

ym

= n}

and Z,,,, = 0

or

Zpq,,= N N xN N xX X X

depending on whether the sentence

(."(P)

3

= ( o C ( q ) .uC(s)) i

is false or true. Since the sets A,,, Bmn and ZPqrSare Borel, it follows that H5 is also Borel. The sets H3 and H4are Borel. We shall prove only that H3 is Borel, because the proof for H4 is similar. From the definition of the set IL( x ) we have (rp, y , XY x'> E H3

=A n

v{ q

(rpn

= 4)A ( x E LoC(q)) A C(X

@ J%(*))

"v P

(rpp

= 41

1

383

8. CONSTITUENTS

thus

and it follows that H3 is Borel since the class of Borel sets is closed with respect to the operations of union, intersection, complementation (see pp. 363-364) and Cartesian product. Condition (1) is thus equivalent to

v [(cp, PV

p, x , XI> E H2 n H3 n H4 n &I,

and thus the set T is the projection of a Borel set and as such is analytic. The proof that T* is analytic is similar except that we replace condition (4) by

The set H: of quadruples satisfying this condition is Borel. Since T* is a projection of the set H2n H3 n H: n H,, it is analytic. O THEOREM 3 : If L is a Borel sieve, then every constituent C,(L) belongs to the class PI n C,; that is, every constituent is analytic and has analytic complement.

PROOF.We may assume that the constituent C,(L) = C, is non-empty. Let x,EC, and let T = T ( L , L ) . Clearly, . (XEC,) ~((X,XO)fT)h(<XO,X)ET). Conversely, suppose that <x, xo)E T and (xo,x} E T. The first condition implies that the set IL(x) is similar to a subset of IL(xo). This set is of type a, thus Zr,(x) is well-ordered and its type is a. Similarly, the second condition implies that the type of ZL(x) is >a. Thus the type of I L ( x )is exactly a. In this way we have shown that

<

Ca = {X: ((X,xo)ET) A ((Xo, x > E T ) } ,

which proves that the constituent C, is analytic. The complement of C, consists of those x which satisfy one of the

384

X. ANALYTIC AND PROJECTIVE SETS

following conditions: the set IL(x) is not well-ordered; the set ZL(x) is well-ordered and its type is < a; the set I,(x) is well-ordered and its type is > a. The set of x satisfying the first condition is analytic because it is equal to R(L). The set OF x satisfying the second condition is equal to the set { x : ( x , x0) ET*},and the set of x satisfying the third condition is equal to [X--R(L)] n { x : (xo,X) E T* } .Thus X--C, = R ( L ) u {x: (x, x J E T * } u {x:

x )ET * ),

(~0,

and the set X-C, is analytic. COROLLARY 4: Every constituent is Borel. PROOF.The Family Pl n C,is identical with the family Bo (see p. 356). O COROLLARY 5: Every set M E C , is the union of K1 Borel sets. For every such set is equal to X--R(L), where L is a Borel sieve. Thus M is the union of K1 constituents, all of which are Borel. O

Exercises 1. Deduce from Corollary 4 that every set belonging t o Pz is the union of analytic sets and thus the union of N, Borel sets. 2. Let H be a defining system such that the sets He are Borel. Let H:

= He,

and

H," =

H,"" = H , " n

I,

u H& k

nH f for limit ordinals 3,

b
(the expression e ; k denotes the sequence obtained from e by adding k as the last term). Prove that

x- U nHVln = U E', 9

n

where

E'=

a
U H," e-N

(union indexed by sequences of one term)'). [Sierpihski] ') This statement shows that it is possible t o decompose an analytic complement into the union of X, Borel sets without using the notion of sieve. See W. Sierpihski, Sur une propriktk des ensembles (A), Fundamenta Mathematicae 8 (1926) 362.

3 85

9. UNIVERSAL SIEVE AND THE FUNCTION I

$9. Universal sieve and the function t With each point of the set

v

of the space N N we associate the order type t (v)

. {ac(vo),uc(vJ, *.*,aC(vn), --.l.-{O) ordered by the relation

c

J

5.

By Theorem 1, p. 375 and by the universality of the type q (p. 219), for every countable order type z there exists a such that t(p))= z. For if z is the order type of a non-empty set E c J and if eo,el ... is an infinite sequence (possibly with repetitions) whose terms constitute the set E, then it suffices to take for 9 the sequence 97, = a(e,). If E = 0, then it suffices to take for cp the sequence all of whose terms are 0. By Q we shall denote the set of all q~ for which t ( q ) is an ordinal number.

THEOREM 1: The set NN- Q is analytic; there exists a Bore1 sieve U (all of whose values are Fb-sets) such that N N - Q = R(U) and such that for every a < Q the constituent C,(U) consists of those q~ for which t ( v ) = a. PROOF.For e in d let U,= {v: // (aC(qn)= e)}. The set U, belongs n

to the class F,, because ~1 E ue

V

[(vn

= 4) A (4 #

0)A ( U c ( q )

= e)],

nq

and the set {v: v, = q} = A,, is closed. The set I u ( y ) consists of all non-empty sequences of the form ac(qn) and thus I,@) = t(v), which proves the theorem. We call the sieve U universal; this name is justified by the fact that among the sets I u ( v ) occur ordered sets of all possible countable order types. In particular, the constituents C,CU) are non-empty for all a < Q'). Using the set Q and the constituents C,(U)we can introduce a ge~

l) Neglecting non-essential differences, the set Q and the sieve U were defined already by Lebesgue. See his paper cited on p. 378.

386

X. ANALYTIC A N D PROJECFIVB SETS

ometrization of the cardinal numbers < 9 '). This geometrization relies upon the fact that every relation R ( q Byy, ...) among ordinal numbers < 9 can be replaced by a relation R' among elements of N N which holds for the points y , y , 8 , ... exactly when R ( t ( v ) , t ( y ) ,...). We shall not examine the details of this geometrization here and shall only define two operations which we shall use later in 0 11. For y E N N and q EN- (0)we denote by e ( y , q) the sequence defined by the following formulas: v n if oc(vn)-3 oC(q)or y n = 0, ~n = 0 if ac(yn>> oc(q); in addition we let E(Q),O)= v.

{

THEOREM 2: For every 91 E N N and q E N - { 0 } the set lv(e(yyq)) consists of those elements of the set Iv(y) which are + o"(q); moreover, for every 4 the set {(q, y>: y = ~ ( yq)} , is Borel. The first part of the theorem follows from the fact that I J v ) consists of all finite non-empty sequences of the form bC(yn)and that I U ( 4 y , q ) )consists of those UC(Q)n) which are jdo(@. The second part of the theorem follows from the formula

[Y

E(QI,

A (<(yn =

4)1

vn) A

[(qn = 0)v ( ~ c ( ~ n3 ) oC(q))])

v [ ( n = 0) A (0"(4) 5 8(qn))])

An V ((91. p

= P) A

({(Yn

= P> A

[(P ==0)v ( ~ Y P 3 ) OC(q))l>

v [(vn = 0)A (uc(q) F~o(P))J)).

THEOREM 3 : If q # 0, then the set { y : oo(q) is the last element of the set l u ( y ) } is Borel.

PROOF.This set is equal to (91 : V n

A { ( v n = 4 ) A [ ( v m = 0)v (aC(ym) 5aa(q))])1* m

I) This msthod of investigation of ordinal numbem is due to K. Kuratowski. Detailed information on geometrization of ordinal numbers can be found in his Topology, pp. 503-508.

9. UNIVERSAL SIEVE AND THE FUNCTION

t

387

We prove now two important theorems about C,-sets. The first of them (Theorem 4) will show that in a certain sense the set Q is universal. The second (Theorem 7) will give necessary and sufficient conditions for a set to be Borel. THEOREM 4: For every set M c X belonging to the class C, there exists a function f E: ( N N ) Xwhose graph { ( x , y ) : g~ =f ( x ) } is Borel and such that M =f - ' ( Q ) . In other words, every set of the class Cl is the Borel inverse image of the set Q. PROOF.Let eo, e l , ... be a sequence such that d is the set of ist terms. Assume that X - M = R ( L ) where L is a closed sieve and let f ( x ) be the sequence y defined by the formula =

Q I ~

{

0 a(e,)

if if

en $ ZL ( x ) (i.e. x $ Len), enEIL(x)(i.e. xELen).

The graph off is Borel, because

1

A { [ ( x L,JA (Vn = 011v [ ( x ~ ~ e ,A, )(yn = en))] ( f ( x )= and the sets Lenare Borel by assumption. Let x EX and let y =f ( x ) . By definition it follows that Iu ( y ) = ZL(x), and hence t ( y ) = IL(x).Thus t ( y ) is an ordinal number if and only if the set IL(x) is well-ordered. Hence @ E M )E ( x . X - R ( L ) )

e {t(p')is an ordinal number}

= ( y e Q>= ( f ( 4E Q), which implies that M = f - ' ( Q ) .

COROLLARY 5 : The set Q is not Borel I). PROOF.Let M EC, be a non-Bore1 set in the space N N (see p. 373). By Theorem 4, M =f - ' ( Q ) and thus

l) This is a theorem of Lusin and Sierpihki; see their paper: Sur un ensemble non mesurable (B), Journal de Mathdmatique, 1923, p. 68.

388

X. ANALYTIC AND PROJECTIVE SETS

If Q were Borel, then the formulas above would imply that M belongs to the class PI n C,,and would be Borel since in the space NN the intersection PIn C,coincides with the class of all Borel sets (P. 356). O COROLLARY 6: The notion of well-ordering for subsets of N is not definable in terms of the operations of the propositional calculus and quanti3cation limited to elements of N * ) ,

PROOF.Assume that {R is a well-ordering relation for a subset of N ) = @(R), where @ is a propositional function constructed from expressions of the form x R y , x= y , by means o f the operations of the propositional calculus and quantification over elements of N. For Q~E", let

R, = (<m, n> : (vm z 0)A (pin # 0)A (uWn) 4 .C<~m))}. The set {pl: mR,n} is Borel. It follows that, for every propositional function @ of the kind under consideration, the set {q: @(I?,)) is Borel. Thus the set (q: R, is a well-ordering} s Borel, which is impossible since

(R, is a well-ordering} = (the-set IU(q)is well ordered) EVPQ.

'THEOREM7: Let M be an arbitrary subset of the space X. The following three conditions are equivalent: (i) M EPIand for every Borel sieve L, M = R(L) implies that there exists a, < l2 such that all constituents C,(L)for u > a, are empty. (ii) There exist a Borel sieve L and and C,(L)= 0 for u > c10.

go

< 52 such that M = RCW

(iii) M EPIn C,. I) This is a theorem of A. Tarski, Grundzige des Systemenkalkiirs Il, Fundamenta Mathematicae 26 (1936) 301. The proof given here is due to K. Kuratowski, Les rypes d'ordres &$nissables et les ensembles boreliens, Fundamenta Mathematicae 29 (1937) 97-100.

9. UNIVERSAL SIEVE AND THE FUNCTION

t

389

PROOF.(i) --t (6). M EP1 implies that M = R(L) for some Borel sieve L. By (i) for every such L there exists a. < 9 such that all constituents with indices > a. are empty. (ii) * (iii). By (ii), X - M =

u C,(L) u C,(L). Every constit=

UCR

u
uent Ca(L)is Borel and thus the union of countably many C,(L)is also Borel. Thus the sets X--M and M belong to the class P, n Cl . (iii) + (i). Let M belong to P1n C, and let L be a Borel sieve such that M = R(L). Assume that there exist non-empty constituents C,(L) for arbitrarily large countable a. Let T = T ( U , L); we shall show that

For if x $ M and (pyx) E T ,then the set &(p) is similar to a subset of ZL(x) and thus is similar to a subset of a well-ordered set. Thus Iu(pl) is well ordered, that is, p E Q. Conversely, if ~ E and Q Iu(pl) = a, then by assumption - there exists an x which does not belong to M such that I,,@) > a, and thus x> E T . Formula-(l) has been proved. Since the complement of M is analytic and since the set T is also analytic (see Lemma 2, p. 381), it follows by (1) that Q is also analytic. Since Q E C ~we , have that Q E Pln Cl , which implies that Q is Borel. Thus the assumption that there exist non-empty constituents Ca(L) for a arbitrarily large leads to a contradiction. ( 9 9

O COROLLARY 8 I): A set M c X is Borel if and only if it satisfies one of the conditions (i), (ii) or (iii) of Theorem 7.

Exercises 1. Let M be Borel and a. an ordinal < 8.Construct a sieve L such that R(L) = M and C,,(L) # 0. 2. Prove chat if L is a Borel sieve in NN and E is an analytic set contained in l) This corollary is due to N. Lusin. See his paper Sur les ensembles anabtiques. Fundarnenta Mathematicae 10 (1927) 71.

390

X. ANALYTIC AND PROJECTIVE SETS

X--R(L), then there exists an ordinal a0 < sd such that E c

u

C&).

[Lusin]

=
Hint: Use an argument similar to that used in the proof of Corollary 8. 3. Prove that if L is aBorel sieve, then the sets ( x : IL(x) = 7)and ( x : ZL(X) = w*> are Borel l).

5 10. The reduction theorem and the second separation theorem Let K be an arbitrary family of sets. We say that K has the reduction property if for arbitrary hil and M2 belonging to K the union Ml u M , can be represented as the disjoint union Z1u 2, of two sets Z1 and 2, belonging to K and contained in Ml and Mz respectively: (1) Z l u Z , = M l u M z ,

ZlnZ,=O,

Z i c M i for i = 1 , 2 .

O THEOREM 1: The family Cl = Cl(NN)of complements of analytic sets has the reduction property ').

PROOF.Let X - M i = R(Li), i = 1,2, where the Borel sieves L1,Lz are chosen in such a way that for every XEXthe set I&) has a last has a last element element and such that for no X E Xthe set I&) [see p. 380-381). Let 2, = { x : <x, x> $ T(Lz,Li)} n MI9 2 2

= { x : (x, x> $ T ( d ,Lz)} n Mz-

The sets Zi belong to the class Cl (see Lemma 2, p. 381), and Zi c Mi for i = 1,2. Moreover, x E Z= ~ (x E Ml) A [the set IL2(x)is not similar to any part of Z"l(x)],

x E Z2= ( x E M,)

A

[the set IL1(x)is not similar to any part of I&)].

Clearly, the sets Z1 and Z, are disjoint. For XEZ,nZ, implies l ) See S . Hartman. Zur Geometrisierung der abzahlbaren Ordnungstypen, Fundamenta Mathematicae 29 (1937) 209-214. D. Scott has proved that for every order type T the set { x : Irr(x) = T > is Borel. See his paper Invariant Borel sets. Fund. Math. 56 (1964) 117-128. See also C. Ryll-Nardzewski, On Borel measurability of orbits, ibidem, 129-130. ,) See K. Kuratowski, Sur les thiorhes de se'paration duns la thiorie des ensembles, Fundamenta Mathernaticae 26 (1 936) 183-1 91.

391

10. THE REDUCITON THEOREM

that both of the sets ZLI(x) and I&) are well ordered and none is similar to any part of the other, which is a contradiction (see p. 231). It remains to be shown that Ml u Mz c Z1u 2,. Assume that x E M1 -Z1, then I&) is well ordered and IL,(x) is similar to a part of the set ZLI(x). It follows that x e M z and that the set IL,(x) is not similar to any part of the set I&), since by assumption the types of the sets ILI(x)and I&) are distinct. Thus x e Z z and Ml-Zl c 2,. Similarly, Mz-Zz c Z1 and hence Mi c Z1u Z 2 for i = 1,2. Q.E.D. The reduction property is closely associated with the second separation principle. We say that the second separation principle holds for the family K if for arbitrary sets Hl and H, belonging to K there exist sets D1 and Dz such that

Hi- HZ c Dz, Hz- Hi c D1, and X - D i e K , i = 1,2. DlnDz=O

'COROLLARY 2: The second separation principle holds for the family P1I). PROOF.Applying the reduction theorem to the sets X- Hi we obtain sets Di belonging to Cl such that (X- H l ) u ( X - Hz) = D1 u Dz and Di c X - Hi for i = 1,2. It follows that Hi n Di

Hz-Hl

=0

for i = 1,2, and thus

= Hz n [ ( X - H I ) u (X-H,)] = Hz n (Dl u

Dz) = Hz n D1 c D1

and similarly Hl-Hz c Dz. It was shown by Novikov that if X = N N (or, more generally, if X is a compleie separable space), then the reduction property holds for the family Pz and the second separation principle holds for Cz2).It is likely that the reduction property and the second separation principle for other projective classes are independent from the axioms of set theory. On the other hand, it has been shown that the reduction propI)

See N. Lusin, Lecons sur les ensembles analytiques (Paris 1930), p. 210. P. Novikov, Sur la sdparabilitd des ensembles projectifs de seconde classe, Eun-

damenta Mathematicae 25(1935) 459-466.

X. ANALYTIC AND PROJECTIVE SETS

392

erty for the classes P. (n 2 2) and the second separation principle for the classes C, (n 2 2) are not contradictory with the axioms '1. Exercises 1. Show that if the reduction property holds for the family K , then the second separation principle holds for the family C(K)consisting of all complements of sets belonging to K. 2. We say that the first separation principle holds for the family K. if for arbitrary disjoint sets M, and M2belonging to K there is a set E E K n C ( K )such that M, c E and E n M2 = 0. Show that if the family K has the reduction property, then the iirst separation principle holds for C(K). 3. Using Exercise 2 above and Theorem 1, p. 390 prove Theorem 6, p. 355. 4. Neither the first nor the second separation principle holds for the family C, in the space NN. Hint: Let be a universal pair for the family C,, where the sets U and V are complements of analytic sets (see Exercise 3, p. 374). By virtue of reduction property there exist sets U,c U and V, c V such that U u V = U,u V,. If the sets U,and V,could be separated by a Borel set W,then the function F(q) = { y : : y E F(p)) would be Borel. which is impossible (see p. 373). Note, moreover, that the second separation principle for C, implies the first.

0 11. The problem of projectivity for sets defined by transfinite induction Let H be a function which assigns to every set M c (NN)k= X a set H ( M ) c X. By the theorem on inductive definitions, for every Z c X there exists a sequence of type Q of subsets of X such that

Do = Z , Let E =

DE+'= H(D,),

DA=

u De for L a limit ordinal.

€
u D,. The problem which we shall deal with in this section

edl

concerns the projectivity of the set E2). Applying the general idea of geometrization of ordinal numbers, we let D' {: (F E Q)A (xeDt(pJ}L-

I) J. W . Addison, Separation principles in the hierarchies of classical and effective descriptive set theory, Fundamenta Mathematicae 46(1959) 123-135. 2, This problem (in a more general setting) was solved by K. Kuratowski, Les ensembles projectifs et l'induction transfinie, Fundamenta Mathematicae 27 (1 936)

269-276.

11. THE PROBLEM OF PROJECTIWIY

Clearly x E E =

393

v [(v, x ) ED'];thus, in order to evaluate the projec(P

tive class of the set E, it suffices to consider the same problem for the set D'. We shall use the following notation. If A c NNx Y where Y is an arbitrary space, then A'" will denote the set {y: (v,y> E A } we call it the crossection of A through q~.We shall denote the projective classes P n ( N Nn ) ( N N ) kand Cn(NN)n (NN)kby Pn,k and Cn,k, or by Pa and C,, if k is fixed and if no confusion can occur. DEFINITION: A function HE"')'' is projective of class P, (or C,,) if for every set A EPn,k+Z (or A e C n , k + 2 ) the set {(F, y, x } : x EA(')(*)} belongs to the class Pn,k+z (or to the class Cn,k+Z). In the sequel we shall restrict our attention to functions of class P,,, but all arguments can be shown to hold for functions of class C,,.

LEMMA1: If H is a function of class P,, (n 2 l), then

( M cX)n ( M E P,,) + H(M)EP,,. PROOF.If M E P,, then the set A class Pn,k+Z, and thus the set

= N Nx

N Nx M belongs to the

B = { ( ~ , Y , X ) : XEH(A(")(~))} = { ( ~ , , w , x )X: E H ( M ) } belongs to the class Pn,k+P. It follows that H ( M ) is the projection of the set B: xEH(M)=

v [(FYy,x>"q;

I ,

thus, H(M)€Pn since n 2 1. LEMMA 2: lTfZePn,kand H i s aprojective function of class P,,( n > l), then D , E P , ,for ~ all E < 9.

PROOF.Let K = (5 < 9:DcE p,,} and let W(a)c K; by the induction principle it suffices to show that ~ E KThis . is clear for a = 0 because Do= 2 E P,,. If a is a limit ordinal, then 0, is the countable union of sets belonging to the class P,, and thus D,EP,, (see p. 373). Finally, if a = p+ 1 then Ds E P,, by hypothesis and thus D, = H(Dp)E P,, by Lemma 1.

394

X. ANALYTIC AND PROJECTIVE SETS

LEMMA3: Under the same hypotheses as in Lemma 2, the union U(Ct(U)x 0 6 ) belongs to the class P n , k + l . €
PROOF.Each constituent Ce(U) is Borel and the class P. is closed with respect to the operations of countable union and of Cartesian product. In the following lemma we use the notation Kq = (p): #(q) is the last element of the set Iu(p))}. The set Kq is Borel as was proved in Theorem 3 on p. 386. We shall use T to denote the set T(U, V ) (see p. 381). For p), W EQ the condition t ( y )< t (y) is equivalent to (9, y ) E T . LEMMA4: If Z E P , , H is a function of class P,,(n >, 1) and if x)ED', then there exists a set A c N" x X having the following proper ties: (p),

(1)


(2) (3)

~ E C , ( U+ ) A(V)= Z ; ( Y E Q ) A ( ( ~ , ~ ) ) E T ) A ( ~ E K ~ ) A ( ~ # O+ ) (A(V)=H(A(e(V*Q))));

(4)

(YE

Q ) A ( ( Y , p)) E T ) A // (Y, #Kq) + (A(v)= 4#0

u A(%

4)));

4#0

(5)

A € P n , k+l * Conversely, if there exists a set A satisfying (1)-(5) where p) E Q, then (9,x>ED'. PROOF.The first part of the lemma follows from the fact that the set

satisfies conditions (1)-(5). To prove the second part we assume that A satisfies (1)-(5) and that Y E Q.Let CI = t(p)).We shall show by induction on ,B that if ,B a, then t ( Y ) = p +A@') = D 8. (6)

<

Let Y be the set consisting of all countable ordinals > a and of ordinals 6 a which satisfy (6). Assume that W ( p ) c Y. It suffices to show that BEY. Clearly we may also assume that ,B < a. If ,B = 0

395

11. THE PROBLEM OF PROJECTNITY

then p satisfies (6), since t ( y j = 0 implies that peC0(V) and thus by (2), A(') = 2 = Do: = y f l < a , and thus (y,p} If = y + l and t ( y ) = p then-) E T , y E Q and there exists a q # 0 such that @(q) is the last element From (3) it then follows that A(') = of the set Zu(y), that is, H(A(8(V*@)), and because (seep. 386) t(a(y, q)) = y E Y, we have A(t(v.4)) = D, and A(V)= H(D,) = D,+l = DB. If 4, is a limit ordinal and tcy) = ,4, then I,(y) < a, thus p E Q and (y, p>E T; moreover, y #Kq for all natural numbers q # 0. From (4) it follows that A(v) = On the other hand, for every

u

qfO

q, t(E(y,q)) < i ( y ) = j3, and every ordinal y < p can be represented in the form t(E(y, q)) where q is a natural number # 0. Since i ( c ( y , q)) E Y, we have

Thus formula (6) holds. Substituting in (6) /I= t(q) and y = q, we obtain A ( @ = Dt(cp), and thus by (1)

E Ct(q)(V>X

Dt(p)

c

D'.

Q.E.D.

Now let F be a universal function for the class the set = {<6, x): x E F@)j

P,J,k+l such

that

w

belongs to class P , , k + 2 . The existence of such a function was proved on p. 372. Notice that W(4)= F(6) according to the definition. Every set A E P,,,k+l can be represented in the form F(6), that is in the form W(')and W(')E P,,,k+l for every 6. Hence substituting A = W(')in Lemma 4 we obtain the following statement: 5 : If Z E Pn,kand H is a function of class P,, , then ihe formula LEMMA ( y , x) ED' is equivalent to the exisience of a point 6 E N N such ihat for all y:

(7) (8)

((6,Q1, X> E w)A (v E Q); y € C o ( V )+ (W(Q(V)= 2);

The lemma above allows us to solve the problem of projectivity of the set D'. Namely, by Lemma 5 we have the equivalence

are the propositional functions (7)-(10). where Let Hi = {(q, x, 8 , y ) : Qi} for i = 1,2,3,4. Then (1 1)

(v, x> ED' =

v A [(q, x,@,y> *

LEMMA 6: If n

Y

E H l n H 2 n f4nH41.

,

1 then the sets Hi belong to class C,,+l,if3 for

i = 1,2,3,4.

PROOF. The set HI can be obtained from the set [ W x N N ]n [NNx x NNx Q xx] by permutation and identification of coordinates. Therefore H1 belongs to class Pn,k+3 since W E P , , ~ and + ~ QEC,,,c P.,,. Thus HZE cn+lk+3 , * The set H2 can be defined by the equivalence

(9% x, 8,Y>E H2

= {(y 4 CO(u)) v

A, [(Y E W(')('))= (Y E z>) Y

=A {(Y 4 co cv>> v [ K @ Y Y u>E w >A (u E a 1 Y 9

v

Y,Y> f# W )A 0,$311

v

= 1 {[YE CO(v)l A [(<.a, Y,Y>4 m v (Y 4211 Y

A

I((@,

y , Y> E

w v (YEZII}.

The propositional function Q written inside the brackets { } defines a set which is the intersection of a Bore1 set (the first component of 8 in square brackets [ I), a set of class Cn(the second component in [ I), and a set of class Pn (the third component in [ 3). Thus 0 belongs

397

1I . THE PROBLEM OF PROJEECITVITY

to class

Prefixingwith the existential quantifier

v the propositional Y

function 0 we do not change the projective class of the set defined by 0, since the class P,+lis closed with respect to projections. Thus the propositional function 1 0 determines a set of class C,,, .

v Y

The argument for the sets H3 and H4 is similar. We can define H3 equivalently by (12)

(VY x, 6,Y> E H3 z5

A {@,@,y) v A [(yEW(9)(lp))~(yEH(W(9)(s(lP.,’))]} Y AY 4+0 A { @,(qyY) [((o,yyJJ> E w)A ( Y E H(W‘”‘(e‘v’4))))]

qzo E

V

V

[(@,y,y)$ w)A (y 4 H(W(’)(8(V,q)’))]},

where 0, is the propositional function

1[(Y E Qj A ((Y, V) E T )* (Y EK,)]. Clearly 0, determines a set of class Pzsince Q belongs to C,, T is analytic, and K, is Borel. The propositional function (6, y,y ) E W defines a set of class P,, since WEP,,. On the other hand, the propositional function Y E H( W(s)(e(cp*q))) also defines a set of class Pn by the assumption that H is a function of class P,,. Thus the propositional function ((6,y,u) E w)A (y E H( W((s)@(lp’q))))

determines a set of class P,,. It can be shown similarly that the propositional function written in the last square brackets on the right-hand side of equivalence (12) determines a set of class C,,. It follows, similarly as for the set H,, that H3E Cn+,,k+3. The proof that H4E Ckfl, kf3 follows in an analogous manner from the equivalence

398

where

X. ANALYTIC AND PROJECTIVE SETS

O* is the propositional function

Q)A ((Y, V )

(Y $ Q* From formula (11) and Lemma 6 we obtain the desired result: (YJ

E

A

THEOREM1: If Z E P , , , and ~ H is a projective function of class p,, (n 2 11, then the sets D' and E are projective of class Pn+z.

PROOF.From (11) it follows that

v 1v ( ( ~ , x Y s , Y ) ~ H l n H , n H 3 n H 4 ) . The complement of the set H I nHz H3 H4 belongs to Pn+l.Application of the quantifier v corresponds to the operation of projection (y,x)EDr=

6

I

n

n

P

and thus yields a set of the same class, Pn+l; application of negation yields a propositional function which determines a set of class C,,+l and the projection corresponding to the application of the quantifier

v 6

gives a set of class Pn+2. From the above it follows by the equivalence x E B =

v ((v, x )

E D')

Q

that the theorem holds for the set E. It can be shown by an analogous argument that if Z E C , , , ~ and if H is a projective function of class C,, ( n 2 11, then the sets D' and E both belong to Pn+2. In general, deciding whether the function H is projective and evaluating its class presents no difficulty if the formula x E H ( M ) is defined by an equivalence of the form XEH(M)

= q x ,M ) ,

where @ is a propositional function whose bound variables range over the sets N and NN.For example, if all quantifiers occurring in Q, bind only variables which range over the set of natural numbers, then H is a Bore1 function.

LIST OF IMPORTANT SYMBOLS conjunction (logical product) of the sentenm p and q 1 disjunction (logical sum) of p and q 1 implication “if p then q” 2 equivalence of p and q 2 negation of the sentencep 2 “xisaset”5 “ x is an element of y” 5 “x is not an element of y” 5 “ x is the relational type of y” 5 sumofthesetsAand3 6 difference of the sets A and B 6 intersection of the sets A and 3 7 symmetric difference of A and 3 7,14 inclusion relation 7 empty set 9 “the sets A and B are congruent modulo the ideal I” 17 space (universe) 18 complement of the set A 18 closure of the set A 26,28 interior of the set A 28 boundary of the set A 31 derivative of the set A 32 sum of elements of a Boolean algebra 33 difference of elements of a Boolean algebra 33 symmetric difference of elements of a Boolean algebra 33 product of elements of a Boolean algebra 33 zero element of a Boolean algebra 33

LIST OF IMPORTANT SYMBOLS

value of the Boolean polynomialf 34 “Boolean polynomial f is immediately transformable into the polynomial g” 35 order relation in a Boolean algebra 37 unit of a Boolean algebra 37 pseudo-difference of elements of a Brouwerian lattice 43 pseudo-complement of an element of a Brouwerian lattice 43 universal quantifier 45, 46 existential quantifier 45, 46 class of propositional functions of general set theory 50 class of propositional functions of general set theory extended by the primitive terms P,Q,... 51 union of sets belonging to the family A 52 power set of the set A 52 the set of x which belong to A and for which @(x) 54 image of the set A obtained by the transformation @ 55 infinite system of axioms of set theory 55 system of axioms which does not contain the axiom of choice 55 unordered pair of the elements a and b 59 ordered pair of the elements a and b 59 intersection of sets belonging to the family A 61 Cartesian product of the sets X and Y 63 the set of real number 63 “a is in the relation R to b” or: “the relation R holds between a and b” 64 left domain of a relation 64

RC

right domain (range, counter-domain) of a relation 64 domain of the functionf 69 range of the functionf 69 field of the relation R 64 inverse of the relation R 65

40 1

LIST OF IMPORTANT SYMBOLS

R oS x/R C/R Yx

composition of the relations R and S 65 equivalence class containing the element x 68 quotient class of the set C with respect to the relation R 68 set of all mappings of X into Y 70 ‘Yrnaps X into Y” 70 value of the function f at x 70 restriction f of g for which &Cf) = A 72 function defined by the propositional function ... x

... 73

image of the set X under the relation R 74 set of values of the function f on the set X 74 inverse image of the set Y under the relation R 75 inverse image of the set Y under the function f 75 factor Boolean algebra 79

Vh reT

least upper bound of a set 82

/\A

greatest lower bound of a set 82

f€T

0 OA

]

1l A ) ( A , R,, R 1 , ... ,Rk-l> < A ,R )

zero element of an ordered set 82 unit element of an ordered set 83 relational system of characteristic b0, p l , ...,pk-]) 84 relational system of characteristic (2) 84 isomorphism of the relational systems (A, R ) and ( B , S) 84

successor of the set X 88 nth iteration of the function a 93 “the set X has n elements” 100 complete graph with the field X 107 union of the sets Ft belonging to the family W being the range of F 1 11 intersection of the sets belonging to the family W being the range of F 111 limit superior of the sequence Fo, 4 , ... 121

402

LIST OF IMPORTANT SYMBOLS

Lim inf F,, n=m

Lim F,, n=

m

Rs Rd

R.3

limit inferior of the sequence Fa, F,,

... 121

limit of the sequence Fa, Fl , ... 122 least family of sets containing the family R and closed under the union of sets 127 least family containing the family R and closed under the intersection of sets 127 family of sets of the form

u Ha, where W ERN 129

family of sets of the form

nn W,,where H ERN

n

129

least u-additive and a-multiplicative family containing R 129 Cartesian product of sets 131 Cartesian power of the set Y 132 Cantor set 137 the generalized Cantor set 137 equivalence relation modulo the ideal I142 inverse limit of the inverse system U = (X,T,F, f) 147

cut in an ordered set 159 order relation between cuts 159 family of all cuts of an ordered set 159 “the set A is equipollent to the set B” 169 cardinal number or power of the set A 174 cardinal number of infinite countable sets 174 “the sets X and Yare equivalent by finite decomposition” 191 power of the continuum 193 order type of the relational system 206 the element x precedes y under the relation R 209 initial segment defined by the element x under the ordering relation R 209

403

LIST OF IMPORTANT SYMBOLS

order types of sets

I ;:: 217

inverse of the order type a 222 limit of the Rsequence ~ ( y for ) y

Lim {CJ} &
< t 235

“the propositional function @ defines an operation” 238 “the operation defined by the propositional function @ is increasing” 238 limit of the operation defined by the propositional function @ 238 “the operation defined by the propositional function @ is continuous” 238 “there exists exactly one y” 238 “the least ordinal number p which satisfies the condition ... p ...” 238 “ I is the inductive sequence for @ of type a + 1” 244 formalization of a definition by means of transfinite induction 244 family of sets of rank at most a 246 ath power of the ordinal number n 246 natural sum of the ordinal numbers a and p 260 natural product of the ordinal numbers a and p 260 ordinal number in the sense of von Neumann 269 cardinal number of the ordinal E 274 smallest uncountable ordinal number 274 power (cardinal number) of the ordinal number B 274 Hartog’s aleph function 278 initial ordinals of the powers a and Nt 279 set of all initial ordinals w < 91 280 index of the initial ordinal 280 initial ordinal whose index is equal to a 281 the least number E such that the limit ordinal with wc 282

a

is cofinal

power of an initial number whose index is equal to

a

282

tth term of the exponential hierarchy of cardinals 297

404

LIST OF IMPORTANT SYMBOLS

N, is strongly inaccessible 326

sequence whose first term is m and whose following terms are the terms of the sequence e 341 sequence of (n+m) terms whose first n terms coincide with the terms of the sequence e, and whose last m terms coincide with those o f f in their original order 342 family of A-sets with respect to the family R 344 family of analytic subsets of a topological space 351 free union of the sets X and Y 357 free intersection of the sets X and Y 357 the set which arises from the set X by permutation of coordinates 357 the set which arises from the set X by identification of the ith andjth coordinates 358 projection of the set X onto the set Sr 358 family of all projections of sets belonging to the family L 359 family of complements of sets belonging to the family L 359 set of all non-empty finite sequences of natural numbers 375 sieve 373 ath constituent determined by the sieve L 381

AUTHOR INDEX Addison, J. W. 392 Bachmann, H. 289, 331 Baire, R. 242,333 Banach, S. 188, 190, 315 Bar-Hillel, Y. 58 Bendixson, I. 277 Bernays, P. 57, 58, 89 Bernstein, F. 190, 191, 192,267,289 Birkhoff, G. 41,269 Bolzano,B. v, 169 BooIe, G. 10, 33 Bore], E. vi, 53, 130, 190, 241 Bourbaki,N. 80 Brouwer, L. E. J. 42 Bruns, G. 160 Burali-Forti, C. 234 Cantor, G. v, vi, 61, 68, 86, 169, 173, 179, 182, 188, 190, 191, 192, 195, 206, 219, 228, 229, 231, 232, 239, 246, 274, 308, 328 Cauchy, A. 67, 130 Cohen, P. J. vi Davis, A. 224 Davis, R. L. 87 Dedekind, R. v, 103, 160,212 Dirichlet, L. 102, 104 Du Bois Reymond, P. v Ehrenfeucht, A. 325 Erdos, P. 110,215,338

Fichtenholz, G. 303 Fraenkel, A. A. vi, 53, 57, 58, 89, 262 Fralssk, R. 219 Frayne, T. 142 Frege, G. 61 Frolik, A. 352 Gillman, L. 338 Godel, K. vi, 58 Gratzer, G. 167 Hajnal, A. 215 Halmos, R. 33 Hame1,G. 267 Hanf,W. 320 Hardy, G. H. 182 Hartman, S. 390 Hartogs, F. 278,292 Hausdorff, F. 13, 18, 53, 287, 303, 310, 334, 338, 339, 348 Henkin,L. 156 Henriksen, M. 338 Hessenberg, G. 229,259,283 Hoborski, A. 252 Huntington, E. V. 41 Iseki, K. 32 Jacobstahl, E. 261 Jhnsson, B. 167,219 Kantorovitch, L. 303,348

406

AUTHOR INDEX

Keisler, J. H. 320, 323, 325, 326 Kleene, S. C. 365 Kochen, S. 338, 339 Kolmogorov, A. N. 131,348 Konig, D. 104,107 Konig, J. 190, 203

Lebesgue, H. 53,340,378,385 LeSniewski, S. 297 Levi,B. 186 Levy, A. 187,326-328 Lindenbaum, A. 186,191,279,297,334 Livenson, B. 348 Lusin, N. 333, 335, 357, 374, 378, 389-391

LoS, J. 142, 325, 326

Mahlo, P. 314 Marczewski, E. 22, 122, 126 Mc Kinsey, J. C. C. 26,42,44 Mc Naughton, R. 58 Mc Neille, H. M. 160 Mirimanov, D. 57 Monk,D. 326 Morel, A. C. 142 de Morgan, A. 4, 12,47,113,152 Morley, M. 219 Mycielski, J. 325

Rado, R. 110 Ramsey, F. P. 107, 109, 110 Rubin, H. 269, 295 Rubin, J. 269,295 Russell, B. 61, 133, 246 Ryll-Nardzewski, C. 390 Schonflies, A. 215 Schroder, E. 190 Scott, D. 142,326, 390 Sierpidski, W. 53, 131, 187, 198, 224, 252,269,287, 313, 334, 338, 340, 346, 356, 357, 384,387 Sikorski, R. 33, 156, 287 Skolem, Th. 26, 57, 110 Specker, E. 295,296, 333, 334 Stone, A. H. 31 Stone, M. H. 13, 167 Suslin, M. 340, 356, Tarski, A. vi, 26, 42, 44, 115, 129, 153, 155, 167, 186, 188, 191, 193, 288, 293, 295, 297, 303, 311, 313, 319, 320, 323, 325, 326, 334, 357, 388 Teichmiiller, 0. 269 Tennenbaum, S. 220 Tychonoff, A. N. 135 Ulam, S. 318,319 Urysohn, P. 367

Oczan, Yu. S. 348

de la Vallk Poussin, Ch. 122 Vaught, R. L. 219,269 Vitali, G. 69 Wang, H. 58 Whitehead, A. N. 133,246 Wiener, N. 59

Peano, G. 69,89

Zermelo, E. v, vi, 56, 57, 186, 261, 311 Zorn, M. 263

von Neumann, J. 57,269 Nikodym, 0. 367 Novikov, P. 391

SUBJECT INDEX abscissa of a point 63 absorption of a cardinal number by another cardinal number 193 abstract measure theory 315 accumulation point 28, 32 additivity of the operation of forming images and inverse images 117 alephs 282 a-sequence 235,237 algebra, atomic Boolean 152 Boolean 33 Brouwerian 42 factor Boolean 79 algebras, cylindrical 156 analytic set 351 analytically representable function of class a 242 antecedent of an implication 1 anti-lexicographical ordering 226 antinomies 4, 61 antinomy of Russell 61 arithmetically definable set 365 A-sets 344 associative law 171 associative law for the addition of cardinal numbers 183 for the multiplication of cardinal numbers 184 for the product of cardinal numbers 202 associative laws for operations on sets 10

atom of Boolean algebra 152 atomic Boolean algebra 152 atomic formulas of the class R? 51 axiom, multiplicative 133 axiom of choice 53 of difference 6 of existence 6 of extensionality 5, 51 of infinity 52 of pairs 52 of power set 52 of regularity 56 of relational systems 86 of replacement for a propositional function 54 of subsets for a propositional function 54 of Tarski 326 of the empty set 51 of unions 6, 52 axioms of Boolean algebra 33 of lattice theory 41 axiom system of set theory of Godel-Bernays 58 of von Neumann 57 of Zermelo 57 of Zermelo-Fraenkel 57 Bake space 137 base, 359 closed, of a topological space 119 Hamel 268

open, of a topological space 120

408

SUBJECT INDEX

base for an ideal 307 of a Hausdorff operation 348 Bernstein formula 289 binary operation 64 Boolean algebra (ring), 33, 134 atomic 152 distributive 152 factor 79 Boolean polynomial 34 Boolean ring (algebra) 33, 134 Bore1 sets of type a 241 boundary of a set 31 boundary set 32 branch, 104 infinite 104 branch of length n 104 Brouwerian algebra 42 Brouwerian lattice 42 cancellation laws 187 canonical mapping 79 Cantor-Bernstein (Schroder-Bernstein) theorem 190 Cantor set 137 cardinal number 174 cardinal C 193 cardinal, hyper-inaccessible 314, 315 inaccessible 314 measurable 315 strongly inaccessible 311 weakly inaccessible 310 Cartesian power of a set 132 Cartesian product of sets 62, 131 of relations 133 of spaces 138 of operations 133 Cauchysequence 67 characteristic function of a set 122 characteristic of a relational system 84

choice function 74 class of projective sets with respect to a base 359 closed base of a topological space 119 closed ordered set 262 closed set 27, 119 closed subbase of a topological space 120 closure 26, 135 cofinal ordinals 235 cofinal sets 81 coinitial sets 81 commutative diagram 72 commutative law 171 for the addition of cardinal numbers 183 for the product of cardinal numbers 202 commutative laws for operations on sets 10 commutative ring 16 compact topological space 139 complement ofagraph 107 of an element of a Boolean algebra 37 of a set 18 complete graph 107 complete lattice 83 complete ordering 83 components of a conjunction 1 composition of relations 65 conjunction 1 consequent of an implication 1 constituent 20 constituent determined by a sieve 381 continuous a-sequence 237 continuous function 77, 125 continuous set 212 continuously ordered set 212 continuum hypothesis, 328 generalized 328 convergent sequence 122

SUBJECT I W E X

coordinate axes (as subsets of product) 63 coset of a function 75, 186 countable (denumerable) set 174 cover of a set 81 critical ordinal of an a-sequence 237 cross-section of a set 393 cut 159 cylindrical algebras 156 Dedekind infinite set 103 defining system 340 definition by induction 92 dense in itself set 32 densely ordered set 210,211 denumerable (countable set) 174 derivative of a set 32 derivatives of order a 241 descriptive set-theory 340 difference of elements of Boolean algebra 33 of ordinals 248 of sets 6 digits of an expansion of ordinal numbers 256 direct predecessor of an elements 209 direct product of Boolean algebras 134 of sets (functions, relations) reduced modulo an ideal 143 direct successor of an element 209 directed set 81 Dirichlet’s drawer principle 102 Dirichlet’s principle for infinite set 104 discrete topology 137 disjoint sets 9 disjunction 1 distributive Boolean algebra 152 distributive lattice 41 distributive laws for operations on sets 10 distributive law for the product of cardinal numbers 202

409

domain of a function 69 of a relation 64 drawer principle of Dirichlet 102 element, extendable, of a set 188 maximal (minimal), of ordered set 82 unit, of ordered set 83 zero, of ordered set 82 element lying in the cut of a set 212 of a set 5 effective proof 47 elementary propositional function 145 elementary set 365 embedding preserving suprema (infima) 158 empty set 9 epsilon-ordinal 254 equipollence 169 equivalence of sentences 2 equivalence classes of a relation 68 equivalence relation 66 qE-set 335 Euclidean algorithm for ordinal numbers 250 even ordinals 242 existential quantifier 46 expansion of an ordinal number for the base, 256 exponents of 256 exponentiation of alephs 287 exponential hierarchy of cardinal numbers 297 extendable element of a set 188 extension of an element 188 of a function 72 of a relational system 157 factor Boolean algebra 79

410

S

family of sets of rank at most a 246 family of sets, 51 closed under a function 126 independent 304 inductive 268 monotone 81 a-additive 128 a-multiplicative 128 field ofsets 34 of agraph 107 of a relation 64 of a relational system 84 filter 162 finalsegment 209 finite inteisection property 139 finite sequence 91 finite set 100 first distributive law 3 first element of linearly ordered set 207 first monotonic law for addition 247 for ordinal multiplication 248 first separation principle 392 firstseparation theorem 355 first term of an ordered pair 59 free intersection 357 freeunion 357 function 69 function, analytically representable, of class a 242 characteristic, of a set 122 choice 74 continuous 77,125 domain of 69 elementary propositional 145 extensionof 72 Hartogs’ aleph 278 increasing 230 monotone, of subsets 19i non-trivial 317

m INDEX

function, nth iteration of 93 one-to-one 70 open propositional 141 projective 393 propositional 45 propositional, projective with respect t o a base 362 propositional, of class n with respect to abase 362 propositional, with domain limited to aset 45 propositional, of general set theory extended by primitive terms 51 universal 367 function consistent with a relation 78 defined on a set 70 defined almost everywhere 317 induced by a relation 78 with the domain limited to a set 45 functions of two and more variables 48, 73

gap 212 generalized associative laws 114 generalized associative law (for cardinal numbers) 198 generalized Cantor set 137 generalized commutative laws 115 generalized commutative law (for cardinalnumbers) 198 generalized continuum hypothesis (H) 328 generalized distributive laws 115 generalized distributive law for multiplication with respect to addition 199 generalized Hausdorff formula 289 generalized logical conjunction and disjunction 46 geometrization of the theory of cardinal numbers 386

41 I

SUBJECT INDEX

Godel-Bemays axiom system of set theory 58 graph 107 of propositional function 362 greatest lower bound 82

Hamel basis 268 Hartogs’ aleph function 278 HausdorB operations 348 HausdoriT recursion formula 287 hereditary set 230 homeomorphism 78 hyper-inaccessible cardinals 314, 3 15 hypothesis (C) 329

ideal, 17 m-additive 317 prime 163 principal, generated by an element CI 163 ideal of a distributive lattice 163 identification of coordinates 358 image of a set 55 of a set under relation 74 immediate extension of a sequence 350 immediate transformability of polynomials 35 implication 1 inaccessible cardinal 3 14 inclusion relation 7 increasing function 230 independent sets 22, 267 index of an initial ordinal 280 induction, 89 transfinite 229 inductive definition 92 inductive family of sets 268 inductive set 89 inequalities between cardinal numbers 185

infimum of elements 150 infinite branch 104 infinite set 100 initial ordinal 279 281 initial ordinal initial segment 209 initial vertex of a branch 104 interior of a set 28 intersection of sets 7 of sets belonging to a family 61 interval 209 invariance under an isomorphism 84 inverse image of a set under relation 75 inverse limit of an inverse system 147 inverse of an order type 222 of a relation 65 inverse system 147 isolated point 178 isomorphic embedding of a relational system 157 isomorphic relational systems 84 w.1

Konig’s infinity lemma 104 Konig’s theorem 203 lattice, 41 complete 83 distributive 152 modular 43 lattice of sets generated by a family closed under operation 127 law, associative 171 associative, for the addition of cardinal numbers 183 associative, for the multiplication of cardinal numbers 184 associative, for the product of cardinal numbers 202 cancellation 187

412

SUBJECT INDEX

law, commutative 171 commutative, for addition of cardinal numbers 183 commutative, for multiplication of cardinal numbers 184 commutative, for the product of cardinal numbers 202 distributive for operations on sets 10 distributive for cardinal numbers 202 first distributive 3 first monotonic, for addition 247 first monotonic, for ordinal multiplication 248 generalized associative 114, 198 generalized commutative 115, 198 generalized distributive 11 5 generalized distributive, for multiplication with respect to addition 199 second monotonic, for addition 241 second monotonic, for ordinal multiplication 249 law of associativity of conjunction 3 of associativity of disjunction 3 of commutativity of conjunction 3 of commutativity of disjunction 3 of contradiction 4 of contraposition 4 of double complementation 18 of double negation 4 of excluded middle 4 of exponents for the Cartesian product 172 of hypothetical syllogism 4 of trichotomy 292 laws, associative, for operations on sets 10 commutative 10 distributive 10 logical (tautology) 2 monotonic, for ordinal subtraction 248 de Morgan’s 4, 12,47, 152

laws of absorption 3 of subtraction 11 of tautology 3 , l l least upper bound 82 left distributivity of ordinal multiplication with respect to ordinal subtraction 249 lexicographical ordering 224 left domain of a relation 64 Uvy’s scheme 327 limit of a sequence 122 of A-sequence 235 limit ordinal 235 limit inferior 121 limit superior 121 linear order relations 81 linear (total, complete, simple) ordering 206 linearly accessible point 366 logical conjunction, generalized 46 logical disjunction, generalized 46 logical law (tautology) 2 logical product 1 logical sum 1 lower section of a cut 159

m-additive ideal 317 Mahlo classification of inaccessible cardinals 314 maximal element o f a s e t 262 of aq ordered set 82 mapping (transformation) 70 maximum principle 262,269 m-compact topological space 323 mean-value theorem 191 measurable cardinal 315 method of transfinite induction 230 modular lattice 43 monotone family 81

SUBIII(=T

monotonic function on subsets 192 monotonic laws for ordinal subtraction 248 de Morgan’s laws 4, 12,47, 152 minimal element of an ordered set 82 minimal extension of an ordered set 160 multiplicative axiom 133 multiplicativity of the operation of forming inverse images 117

natural addition 259 natural interpretation of axioms 27 natural multiplication 259 natural numbers 88 natural product 260 natural sum 260 negation 2 neighborhood 135 von Neumann’s axiom system of set theory 57 von Neumann’s ordinals 262 non-trivial function 317 normal form of sets 20 normal sequence of universal functions 368 nowhere dense set 32 nth iteration of a function 93 nth projective class with respect to a base 359 numbers of the second class 274

odd ordinal 242 one-to-one function 70 one-to-one sequence with n terms 100 open base 120 open propositional function 141 openset 27 open (closed, Borel, analytic) sieve 378

INDEX

413

open subbase of a topological space 120 operation (A) 340 order relation 80 in Boolean algebra 37 in a lattice 41 order of a function 77 of a value of function 77 order type 206 vj 217 1 219 w 216 ordered pair 59 ordered union of linearly ordered sets 213 ordering, anti-lexicographical (by the principle of last differences) 226 lexicographical (by the principle of first differences) 224 ordinal exponentiation 252 ordinal number (ordinal) 232 ordinal, epsilon- 254 expansion of 256 exponent 253 even 242 initial 279 initial, ma 281 limit 235 odd 242 power of 253 regular initial 308 singular initial 308 ordinal number in the sense of von Neumann 269 ordinals, natural addition of 259 natural multiplication of 259 natural product of 260 natural sum of 260 quotient of 250 weakly inaccessible 309 ordinate of a point 63

414

SUBJECT INDEX

partial order relation 80 partition 81 pair, ordered 59 universal, for a family 374 unordered 59 partition of a set 67 permutation of a set 71 of coordinates 351 Peano axioms 89 perfect set 267 point, accumulation 32 isolated 178 linearly accessible 366 point of wtesian product 63 points of the space 26 polynomial immediately transformable into polynomial 35 power set 52 power ofaset 174 of an ordinal 253 of a cardinal number 184 of the continuum 193 prime ideal 144 prime ideal of a distributive lattice 163 principal ideal generated by a 163 principle, first separation 392 maximum 262,269 second separation 391 Zorn maximal 263 principle of duality 41 of induction for ordinals 275 of transfinite induction 229 principal ordinals of multiplication 261 problem of elimination 23 product, cartesian, see cartesian product direct, of Boolean algebras 134

product, direct, of groups, rings, etc. 134 logical, of sentences 1 natural 260 product of cardinal numbers 183,202 of order types 223 of ordinal numbers 246 projection of relation 64 projective classes 364 projective set 357 proof by induction 89 proof, effective 47 proper cut 212 proper extremum of a function 178 proper subset 8 property, finite intersection 139 reduction 390 property invariant under isomorphism 84 propositional functions 45 propositional function of two variables 48 with domain limited to set 45 projective with respect to a base 362 proto-order relation 80 projection 358 pseudo-complement 43 pseudo-difference 43 quantifiers 46 quasi-order relation 80 quotient class of sets with respect to a relation 68 quotient of ordinals 250 Ramsey’s theorem 107, 109 range of a relation 64 of a function 69 reduction property 390 refinement of a cover 81

SUBJECT INDEX

regular closed set 38 regular initial ordinal 308 regular system 341 relation 64 relation, binary 64 equivalence 66 inclusion 7 inverse 65 linear order 81 order 80 order, in a Boolean algebra 37 order, in a lattice 41 partial order 80 proto-order 80 quasi-order 80 transitive 8 relation of lying between, for elements of 209 of preceding, for elements of a set of preceding, for intervals of a set “less than”, for cardinal numbers relational system 84 relational system embedded into a tional system 157 relational type 5, 87 remainder of an ordinal number 250,259 o f a s e t 235 result of the operation (A) 341 right domain of a relation 64 ring 16 R-separable set 346 Russell’s antinomy 61

a set 209 210 185 rela-

scattered set 21 1 Schroder-Bernstein (Cantor-Bernstein) theorem 190 second distributive law 3 second monotonic law for addition 247

415

second monotonic law for ordinal multiplication 249 second separation principle 391 second term of an ordered pair 59 sequence, Cauchy 67 convergent 122 finite 91 immediate extension of 350 infinite 91 A- 135 normal, of universal functions 368 one-to-one, with n terms 100 transfinite, of type a (or: a-sequence) 235 a-sequence, 235 increasing 235 continuous 237 set, 4 A- 344 analytic 351 arithmetically definable 365 boundary 32 Cantor 137 closed 27, 119 closed ordered 262 continuous 212 continuously ordered 212 countable 174 Dedekind’infinite 103 dense in itself 32 densely ordered 210 densely ordered in linearly ordered set 21 1 directed 81 elementary 365 empty 9 rly- 335 finite 100 fmite, in the direction of the kth axis 286 generalized Cantor 137 hereditary 230

416

SUBJECT INDEX

set, infinite 100 independent 267 inductive 89 nowhere dense 32 open 27, 119 ordered 80 perfect 267 power 52 projective 357 quasi-ordered 80 regular closed 38 scattered 211 sifted through a sieve 378 successor of 88 well-ordered 228 set of representatives 69 sets, Borel 130 Borel, of type a 241 cofinal 81 coinitial 81 congruent modulo ideal 17 disjoint 9 disjoint modulo ideal 17 equipollent 169 equivalent by finite decomposition 191 independent 22 reduced modulo ideal 143 R-separable 340 similar 83 sets of the same cardinal number (of the same power) 173 sieve, 378 universal 385 sifted set 378 similar sets 83 similarity of relations 208 singular initial ordinal 308 u-additive family of sets 128 a-additive zero-one measure 317 a-multiplicative family of sets 128 space (universe) 18

space, 26 Baire 137 compact topological 139 ttt-compact topological 323 Stone 167 topological 26, 119 strongly inaccessible cardinal 311 subgraph 107 subset 7 subsystem of a relational system 157 successor of a set 88 sum, logical, of sentences 1 natural 260 sum of cardinal numbers 181,197 of elements of a Boolean algebra 33 of order types 222 of series of cardinal numbers 198 of sets 6, 52 supremum of elements 150 Suslin problem 220 symmetric difference 6 of sets 7 of elements of a Boolean algebra 33

Tarski recursion formula 288 tautology (logical law) 2, 11 theorem, Cantor-Bendixson 277 Cantor-Bernstein (Schroder-Bernstein) 190 first separation 355 Konig’s 203 mean-value 191 Ramsey’s 107, 109 Tychonoff’s 139 well-ordering (Zermelo’s) 261 theorem on definition by transfinite induction 239 on diagonalization 179

SUBJECT INDEX

theorem on the existence of pair 59 on the existence of union 59 on the existence of unordered triples, quadruples, etc. 59 topological space 26, 119 topology, 26 discrete 137 Tychonoff 135 transfinite sequence of type a 235 transformation (mapping) 70 transformability of Boolean polynomials 35 transitive relations 8 Tychonoff theorem 139 Tychonoff topology 135 type of a relational system 86

union of sets 52 unit element 83 of the ring 16 of Boolean algebra 37 universal function for a family 367 universal pair for a family 374

417

universal quantifier 46 universal sieve 385 universe (space) 18 unordered pair 59 upper section of a cut 159

value of a Boolean polynomial 34 of a function 70 vertices of a field of a graph 107 VN ordinal 269

weakly inaccessible cardinal 310 weakly inaccessible ordinal 309 well-ordered set 228 well-ordering theorem (Zermelo’s theorem) 261

Zorn maximal principle 263 zero element 82 Zermelo set theory 57 Zermelo-Fraenkel set theory 57