Equivalents of the Axiom of Choice II

EQUIVALENTS OFTHE AXIOM OF CHOICE, 11

Herman RUBIN Professor of Mathematics and Statistics Purdue University West Lafayette Indiana U.S.A.

and

Jean E. RUBIN Professor of Mathematics Purdue University West Lafayette Indiana U.S. A.

1985

NORTH-HOLLAND AMSTERDAM NEW YORK OXFORD

ELSEVIER SCIENCE PUBLISHERS B.V., 1985 All rights reserved. No part of this publication may be reproduced, stored in a retrievalsystem, or transmitted, in any form or by any means, electronic, mechanicat, photocopying, recording or otherwise, without the prior permission of the copyright owner.

ISBN: 0 444 87708 8

Published by:

Elsevier Science Publishers B.V. P.O. Box 1991 1000 6 2 Amsterdam The Netherlands Sole distributors forthe U.S.A. and Canada: Elsevier Science Publishing Company, Inc. 52Vanderbilt Avenue NewY0rk.N.Y. 10017 U.S.A.

Llbrary of Congress Cataloging in Publication Data

Rubin, Herman. Equivalents of the axiom of choice, 11. (Studies in logic and the foundations of mathematics ; V. 116) Bibliography: p. Includes indexes. 1 . Axiom of choice. I. Rubin, Jean E. 11. Title. 111. Series. QA248.R8 1985 511.3'22 84-28692 ISBN 0-444-87706-8

PRINTED IN THE NETHERLANDS

In dedication to o u r children

Arthur Leonard Rubin Leonore Anne Rubin Findsen i n h o p e of t h e future: and t o

Alfred Tarksi i n r e m e m b r a n c e of t h e past.

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PREFACE (1963 Edition)

In 1955 at Stanford University Professor Patrick Suppes gave a course on axiomatic set theory, during which time Professor Herman Rubin was asked to give a series of lectures on the axiom of choice. During the course of these lectures it was noted that,' while there was much in the literature on the axiom of choice, the material was available only in many diverse journals and books. It was suggested that we collect this material to make it more readily available. It seemed like a simple enough project to begin with; then it grew and grew some more and now has blossomed forth into a book. The book consists of a selection of the forms of the axiom of choice which appeared in the literature together with additional forms which were obtained in the process of writing the book. It would have been a hopeless task to try to include all of the forms of the axiom of choice which appeared in the literature, so we chose the forms which in our opinion were either used often in practice, unusual, relatively unknown, or particularly weak or strong. We hope that we have included all of the interesting equivalents of the axiom of choice. We assume a knowledge of logic and elementary set theory (von Neumann-Bernays-Godel set theory), but we do include a list of definitions of set theoretical symbols and terms in the section entitled "Preliminary Definitions and Theorems". In Part I we discuss propositions which are equivalent to the usual form of the axiom of choice. These equivalents will be referred to as set forms. In Part I1 we discuss stronger forms - essentially forms which are obtained from the set forms by changing the word "set" to "class". These latter forms are called class forms. The set forms of the axiom of choice are V

vi

PREFACE (1963 Edition)

the forms which are most often used in practice. In preparing this,monographfor publication we first prepared a draft and sent it to several people for their comments and corrections. We are very grateful to the people who did reply. We believe that the quality and usefulness of the book was greatly improved by their comments. In particular we should like to thank Professor Alfred Tarski for his many useful comments and corrections. Others whom we should like to thank are Professoxs E . W. Beth, A. Levy, D. Scott, and R. Vaught. Our typists, Ann Breen and Barbara Johnson, also deserve credit for bearing with us under strain, and we should like to thank the Mathematics and Statistics departments at Michigan State University for their cooperation. HERMAN RUBIN JEAN E. RUBIN September 1961 East Lansing, Michigan

PREFACE (1985 Edition)

In the twenty years since "Equivalents" (Rubin & Rubin 1963) has been published, there have been so many new developments with respect to the axiom of choice that we thought it was time for a new book to describe these developments. Many new equivalents have been discovered since 1963. The section on algebraic equivalents has been greatly expanded to include statements about vector spaces, groups and other algebras. There are so many forms in logic, analysis and topology that a new section was added to include these forms. Another new section was added for those forms which are equivalent to the axiom of choice only under the axioms of extensionality and foundation, In addition, new forms were added to the other sections. We have also rewritten all the material using modern notation in a style which, we hope, is easier to read. We should especially like to thank the following people for their help: our son Arthur who proof read Part I and made significant contributions to Part 11; Norbert Brunner who, while visiting Purdue during the 1983-84 academic year, made many suggestions for improving the manuscript; Paul Howard who proof read part of the manuscript and made important contributions; Wilfrid Hodges who sent us a list of some of the new algebraic equivalents of the axiom of choice: Andreas Blass who corrected some errors and sent us his unpublished proof that the "vector space theorem'' implies the axiom of choice; and our typist, Judy Snider, who did an admirable job of typing and cutting and pasting. HERMAN RUBIN June 1984 JEAN E. RUBIN West Lafayette, Indiana vii

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TABLE OF CONTENTS

.................. v . . . . . . . . . . . . . . . . . . vii .................... . . . xi . . . . . . . . . . . xv .................... 1 . .......... . . . . . . . . . . . . . . .. .. .. . . . . . . . . . . . . . . 21 . . . . . . . . . . . . . . . . . . 31 . .. . . . . . . . . . . . . . . . . . .. .. .. 7933 .. . . . . . . . . . . . . . . . 137 . . . . . . 163 PART I1 .Class Forms . . . . . . . . . . . . . . . . . . . 185 1 . The Well-Ordering Theorem . . . . . . . . . . . . . 187 2 . The Axiom of Choice . . . . . . . . . . . . . . . . 191 3 . Maximal Principles . . . . . . . . . . . . . . . . . 2 0 3 PREFACE (1963 Edition) PREFACE (1985 Edition) INTRODUCTION PRELIMINARY DEFINITIONS AND THEOREMS PART I .SetForms 1 The Well-Ordering Theorem 2 The Axiom of Choice 3 The Law of the Trichotomy 4 Maximal Principles 5 Forms Equivalent to the Axiom of Choice Under the Axioms of Extensionality and Foundation 6 Algebraic Forms 7 Cardinal Number Forms 8 Forms from Topology, Analysis and Logic

1

i

. . . . . . . . . . . . . . . . . . . 247 . . . . . . . . . . . . . . . . . 271 . . . . . . . 279 . . . . . . . . . . . . . . . . . . . . . . . . 289 . . . . . . . . . . . . . . . . . . . . . 305 . . . . . . . . . . . . . . . . . . . . . 311 . . . . . . . . . . . . . . . . . 315

List of the Set Forms List of the Class Forms List of Forms Related to the Axiom of Choice Bibliography Index of Authors Index of Symbols Index of Technical Terms

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INTRODUCTION

Equivalents of the axiom of choice appear frequently in almost all branches of mathematics in a large variety of different forms. In this monograph we should like to describe some of these equivalents. By the "Axiom of Choice" we mean a statement similar to the following statement given by Russell [1906] and Zermelo [19081.

AC: If x is a pair-wise disjoint family of non-empty sets, then there is a set C which consists of one and only one element from each set in x. The reluctance that some mathematicians have about using the axiom of choice is due to its non-constructive nature. There is no rule for constructing the choice set, C. Since, in our opinion, it is more desireable to have a constructive proof rather than a non-constructive proof, the axiom of choice and other non-constructive principles should be avoided whenever possible. At times, of course, non-constructive principles are unavoidable and the least the mathematician can do at these times is to declare their use in his/her proof. The set theory in which we prove of the axiom of choice is called NBGO without the axiom of foundation). It Bernays and G6del and it is described entitled "Preliminary Definitions and

most of the equivalences (NBG with atoms but is due to von Neumann, in detail in the section Theorems".

The relative consistency and independence of the axiom of choice with NBG (and NBGO) has now been completely determined. xi

xii

INTRODUCTION

(NBG is consistent if and only if NBGO is consistent). Kurt Godel [1938,1939,1940] has shown that if NBG is consistent so is NBG + AC; Abraham Fraenkel [1922] proved that if NBGO is consistent so is NBGO + 1 AC (see also Fraenkel [19371, Lindenbaum & Mostowski [1938] and Mostowski [1939]); and Paul Cohen [1963,19641 proved that if NBG is consistent so is NBG + 7 AC (see also Cohen [1966] and Shoenfield [1971]) . In Part the axiom of from the set "class". It is a class.

I1 of this monograph we discuss "class" forms of choice, CAC. Generally, these forms are obtained forms in Part I by replacing the word "set" by is clear that CAC implies AC because each set

Using a Fraenkel-Mostowski model in which the class of atoms is a proper class it is easy to see that if NBGO is consistent, so is NBGO + AC + 7 CAC. (This result is credited to E. Specker. See Bernays [1958] p.196.) Later Easton [1964] proved that if NBG is consistent, so is NBG + AC + lCAC. (See Felgner [1976a]). Thus AC does not imply CAC in NBG. However, it has been shown by a number of people that NBG + CAC is a conservative extension of NBG + AC. That is, even though AC does not imply CAC, if a statement about sets, a , is provable in NBG + CAC, Q is also provable in NBG + AC. (See Cohen [19661 ,p.77 , Felgner [1971,1976al, Gaifman [1968], Grishin [1972], Jensen, Kripke and Solovay.) The relative strength of the axiom of choice and other well-known propositions has now been demonstrated using Cohen's forcing techniques. For example, Cohen 119631 has shown that the axiom of choice does not imply the continuum hypothesis in NBG (assuming, of course, that NBG is consistent). It was known much earlier that the generalized continuum hypothesis does imply the axiom of choice in NBGO. (Lindenbaum & Tarski [1926], Sierpinski [1947]). Halpern & Levy [1971] proved that the Boolean prime ideal theorem, BPI, does not imply the axiom of choice in NBG. (Halpern [1961,1964] proved that BPI does not imply AC in NBGO.) Sageev [19751 proved that the 2k = k principle, (that is, for all infinite cardinals k, 2k = k) does not imply AC in NBG. (Halpern & Howard [1974]

INTRODUCTION

xiii

proved that the 2k = k principle does not imply AC in NBGO.) Clearly, AC implies each of BPI and the 2k = k principle. In closing, we'll mention a few of the unsolved problems. It is well-known that the axiom of choice is used in the proof of each of the following propositions: Every subgroup of a free group is free. (NielsonSchreier Theorem.) (Lauchli 119621 proved that its proof is dependent on the axiom of choice and Howard [19831 proved that it implies the axiom of choice for finite sets.) For all sets x

and

Y, F ( x ) < P ( y 1

For all infinite cardinals

2

k, k = k

For each set x, and partition (Partition Principle).

P

of

or 3

P(Y) < $"XI.

. x, P 6 x.

However, it is not known if any of the above propositions imply in NBGO or NBG. Other unsolved problems are mentioned in the text.

AC

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PRELIMINARY DEFINITIONS AND THEOREMS

We assume an elementary knowledge of logic and set theory. We shall use the following logical symbols: 1 A V

-+

(3x1 (VX)

for for for for for for for for

not and or if then if and only if (iff) identity there exists an X for all X.

...

...

The system of axioms for set theory which we adopt is NBGO (von Neumann-Bernays-Godel set theory with atoms and without the axiom of foundation). (See von Neumann [19281, Bernays [19371 and Godel [1940].) Thus, our undefined terms are: class, atom, and the element relation, E . The convention is made that capital letters, XI Y,..., denote variables whose range is all classes and all atoms. AXIOMS Al:

Characterization of Atoms

DEFINITION 0.1 : (a) X is a Set, set(x)

iff

Class(X) xv

A

(3Y)(Class(Y) A X E Y).

xvi

PRELIMINARY DEFINITIONS AND THEOREMS

(b) X

is a proper class, Pr(X)

iff

Class(X)

Set(X).

Axiom of Extensionality Class(X)

A

Class(Y)

[(Vu) (u E X

+

It follows from A2 it is denoted by PI. A4:

7

denote variables whose

Lower case letters, x, y, z , . . . , range is all sets and atoms. A2:

A

that the

X

. u E Y) of

A3

-+

X = Y].

is unique and

Class Construction Schema

If @(u) is a formula in the language of NBGO which has no bound class variables and in which X is not a free variable then (3x1 [Class(XI

(Vu)(u E

A

x

-

Q

(u)1 1

.

(Godel [1940] has shown that in NBGO, A4 can be replaced by a finite number of axioms.) It follows from A2 that the class X of A4 is unique and it is denoted by

x DEFINITION 0.2:

Let

X

= iu: O(u) 1 .

and

Y

be classes.

(a) An element is a set or an atom. (b) Subclass:

X 5 Y

(Vu)(u E X

iff

(c) Proper Subclass: X c Y (d) Power Class: (e) Union: (1) (2)

(f)

ux x u

iff

+

X c Y

p(X) = {u:u _c XI.

= cu: (3v

E X) (u E v) I .

Y = cu:u E

Intersection: (1) nx = {u: (VV E

x

v u E Yl.

x) (U

E

V) I

,

u E Y). A

X # Y.

PRELIMINARY DEFINITIONS AND THEOREMS

x n

(2)

Y = {u:u E

Class Difference:

X

Y

Complement: - X = {u:u also be denoted by V See 0.4(a).) Disjoint:

X

Y

and

Pair-wise Disjoint: of distinct sets in

4

-

x

A

=

{u:u E X

u E

YI A

X

4

u

Y].

XI. (The complement of X will X where V is the universe.

are disjoint if X

xvii

X

n

Y = 8.

is pair-wise disjoint if each pair are disjoint.

Power Set Axiom Set (x) * Set

)

.

Union Axiom Set(X) * Set(UX). Pairing Axiom (3x1 (Set(x) A (Vu) [u E x It follows from and it is denoted by

A2

x

-

(u = v

V

that the set x

u

=

w) I .

of

A7

is unique

= {v,w).

DEFINITION 0.3 : Unordered Pair: v and w. Unit Set:

{v,w)

{u} = {u,u}

is called the unordered pair of is called a unit set.

Ordered Pair: (u,v) = {{u),{u,v)} pair of u and v. Relation:

is called the ordered

A (binary) relation is a class of ordered pairs.

Function : Func(F) [F is a relationh (Vu,v,w) ((u,v),(u,w)

-

Domain: Range:

EF-+v=w)].

$(X) = {u:(~v)(u,v) E XI. .!#,(X) = {v:(~u)(u,v)

E XI.

R is a relation we frequently write xRy instead of (x,y) E R; and if F is a function, we write F(x) = y or

xviii

PRELIMINARY DEFINITIONS AND THEOREMS

Fx = y A8:

instead of

xFy

or

(x,y) E F.

Axiom of Replacement (Func(F) A Set (3 (F)) ) * Set (2(F)).

A9:

Axiom of Infinity (3x1 [Set(x)

0 E x

A

A

(Vu)(u E x

A

Set(u)

+

u

u

{ul E x)].

Axioms Al-Ag are the axioms of NBGO. A10:

Axiom of Foundation (Class(x)

A

x # 0)

+

(3u E x)(Set(u)

A

u

n x

=

0).

Axioms A7 and A10 imply there donotexist any atoms. Al-AlO are the axioms of NBG. Another form of the axiom of foundation which is sometimes used is the following: A'10:

Axiom of Foundation, alternate form IClass(X)

A

X # 0

A

( V u E X)Set(u)l

+

13u f

x) (u n x = 0)

Of course, A10 i s equivalent to All0 if there do not exist any atoms. As is well-known, the set form of A10, (that is, replace "Class(X)" by "Set(X)" in A101 is equivalent to the class form in NBGO. For suppose the set form of A10 holds and let X be a class. Suppose u E X I u is a set and X n u # fl. Let z = X r l u. Then the transitive closure of z, TC(z), is a set , TC(z) = z U ( U z ) U (UUzIU

... .

Let w = TC(z) n X. Then w # 0 is a set so by the set form of the axiom of foundation, there is a set v E w such that v n w = 0. Clearly v E X. If v n X # 0 then there exists an a E v n X. By definition of TC(z), a E v E TC(z) implies a E TC(z). Thus a E v n w, which is a contradiction. All proofs, unless otherwise stated, are in are independent of the existence of atoms.

NBGO

and

PRELIMINARY DEFINITIONS AND THEOREMS

xix

DEFINITION 0.4:

v

(a) Universe:

=

iu:u = ul [(vu) (u E v)].

(b) Unordered n-tuple: (xltx2r * * txn-1) u {xn}-

-

(cI

Ordered n-tuple:

=

...rxn) =

(xlrx2r

(d) If F i s a function and the language of NBGO, (F(X1t...rXn):~(~lr...rX

(x1r(x2r...rxn)).

@(X~,...~X n)

i s a formula in

n ) I = Iu:(3Xl...x n )(U = F(X1r.a-rX n ) O(Xlr...,x n 1 ) ) .

A

(e) Cartesian or Direct Product: Class(X) (f) Image:

A

Class(Y)

-t

x

(h) Inverse:

RI\X = ((urv):u

R-l = {(x,y):(y,x)

( i ) Indexed Classes:

Suppose

u xi

=

t~

{xi:

i E I).

r) xi

=

n {xi:

i E I}.

iE1

Y

{(u,v):u

=

E

x

A

v E Y}.

R"X = {v: (3u E X) ((ulv) E K) 1.

(9) Restriction:

iE1

x

(vi E I) (xi = Y)

-t

X xi

iE1

E X

A

(u,v) E R}.

E R).

(Vi E I)Set(Xi).

= Y '

.

We may also take the union, intersection, and Cartesian product of proper classes by using the following device: Let R be a relation such that for each a set. (R"Xi could be a proper class.)

i E 1, Xi

is

PRELIMINARY DEFINITIONS AND THEOREMS

xx

x

R"Xi = {f:Func(f)

A

$(f) = I

A

(vi t I) (f(i) E R"Xi)I.

A

$(F) = X

A

%(P) _C Y.

if1

(j) Functional Notation: F:X

+

Y

iff

Func(F)

is 1-1 or an injection iff E F u =

F

(VUrVrW) ((UrW)r(VrW)

F:X < Y

X < Y F:X

F:X X

iff iff

Func(F) (3F)(F:X

Onto> Y iff surjection. )

<*

Y

iff

<* Y iff X

F:X

=

fl

iff

A

F

V).

is an injection.

< Y).

F:X

-+

Y

onto

F:Y

k Y iff F:X and X and Y

X =Y

+

V

A

%(F) = Y. (F is called a

x.

(3F)(F:X

<*

Y).


A %(F) = Y. (F is called a bijection are called equipollent or equinumerous.)

(lF)(F:X = Y).

(k) Class-Valued Functions: F is a class-valued function if whenever X E $(F) and (X,Z) E F then either Z = fl or Z is a unit set. Then F[Xl = { y : ( x , { y } ) E F). (F[X] can either be fl, a set or a proper class. Using this definition we can distinguish the two cases when F[Xl = 0, (1) if X fZ $(F), then F"{x) = 0 ; ( 2 ) if (xrO)E F, then F"{x) = I01.1

(m)

An

n-ary Relation is a class of ordered n-tuples.

We shall make the convention that whenever we write:

xXi, etc., we shall assume that iE1

X

and

Y

are classes and

PRELIMINARY DEFINITIONS AND THEOREMS that for each DEFINITION 0.5:

i E I, Xi Let

R

xxi

is a class. be a relation.

(a) R

is a reflexive on

X

if

(Vu E X) (u R u).

R

is irreflexive on

X

if

(Vu E X) (1u R u).

(b) R R (c) R

is symmetric on

X

if

is anti-symmetric on is transitive on

X

X if

(Vu,v E X) (URV + vRu). if

(Vu,v,w E X) (uRv

(d) R is a partial ordering on symmetric and transitive on (e) R

is connected on

X

if

(f) R is a linear ordering on partial ordering on X. (9) u

if

(Vu,v E X) (uRv A vRu

X if X.

R

if

R

vRw

-+

uRw).

is reflexive, anti-

(Vu,v E X) (uRv X

A

u = v).

-+

V

vRu

V

u = v).

is a connected

is said to be an R-first or R-smallest element of u E X and (Vv E X) ( u f v * uRv).

(h) u is said to be an R-last or R-largest element of u E X and (Vv E X) (u # v + vRu).

X

X

(i) u is said to be an R-minimal element of and (Vv E X) (vRu + uRv).

X

if

u E X

(j) u

X

if

u E X

and

is said to be an R-maximal element of (Vv 6 X) (URV * vRu).

if

(k) If Y _c X, u is said to be an R-lower [R-upper] bound of Y in X if u E X and (Vv E Y)(u # v + uRv) vRu)l. [(Vv E Y)(u # v -+

(a)

If Y c X, u is said to be an R-greatest lower bound or infimum [R-least upper bound or supremum] of Y in X if u is an R-largest [R-smallest] lower [upper] bound of Y in X.

xxii

PRELIMINARY DEFINITIONS AND THEOREMS

(m) R is a well ordering on X if R is a partial ordering on X and every non-empty subset of X has an R-first element. (n) (X,R) is a partially (linearly, w s ) ordered class if X is a class and R is a partial (linear, well) ordering on X. (0)

If (X,R) and (Y,S) are partially ordered classes, (X,R) is isomorphic to ( Y , S ) , (X,R) ;(Y,S) iff there is a function F such that (1) F:X

IJ

Y,

and

(VU,v E X) (uRv

(2)

*

F(u)SF(v)).

is an initial R-segment of Y

(p) X 5 Y

if

(Vu E X) (Vv E Y) (vRu + v E X). u E Y then generated by u

(9) If

X is the initial R-segment of if X = {v E Y: vRu}.

Y

(If R is transitive, an initial R-segment of Y generated by u is an initial R-segment of Y.) DEFINITION 0.6:

Ordinal Numbers

(a) A class X

is transitive

(b)

x

(c) X

sY

iff

x E

Y v

is an Ordinal

(1) (X,s) (2)

x

iff (VU,V)(UEVAVEX+UEX).

= Y.

iff

is a well ordered class and

(Vu E X) (u is the initial €-segment of by

(d) X

is an Ordinal Number iff X a set.

(e) X

B

X

generated

U)

is an Grdinal and

is called a successor class iff E X A (Vu)(Set(u) A u

E

x

+

u

u

{u) E XI.

X

is

PRELIMINARY DEFINITIONS AND THEOREMS w =

n {x: x is a successor set}

(w

xxiii

is the set of

natural numbers). A set x is finite iff (3n E w ) (x fi: n). Otherwise x is called infinite. A set x is called transfinite if w < x. and

0 = B

On

is the class of all ordinal numbers.

Lower case greek letters, a , 6, y,..., variables whose range is On. a

will be used to denote

iff a E B.

8

iffa
is a successor ordinal iff

a is a limit ordinal iff

( 3 8 ) ( a = f3 U ( 8 1 ) .

a # fl

and

is not a

a

successor ordinal. a 2

w

where

is an initial ordinal

iff ( V B

(Initial ordinals are denoted by wo = w.)

a)

wa

B.<

a.

for

a

E On,

We shall assume all the well-known properties of ordinal numbers, some of which are listed below. (All these properties are provable in NBGO.) THEOREM 0.7: (a) Every ordinal is transitive. (b) On

is an ordinal.

(c) A set of ordinal numbers is an ordinal number is transitive. (d) Recursion Theorem: F with domain On

If

g:V

+

V

iff

it

then there is a function

s u c h that for all a E On, F(a) = G(F"a).

xxiv

PRELIMINARY DEFINITIONS AND THEOREMS

(el Transfinite Induction:

If X 5 On

( V a E On) ( a _C X * a E X)

then

(h)

( V a , B ) ( a E B V a = B V B E a).

(i)

(Va

# 0) ( a is a limit ordinal

such that

X = On.

iff a = U a ) .

(j) If two well ordered sets are isomorphic there is a unique

isomorphism mapping one onto the other. (k) For each well ordered set there is a unique ordinal number which is isomorphic to it.

THEOREM 0.8: (a) If

vo

=

Axiom of Foundation

0,

(va+l = F’(v~))

(Va)

ordinal, Va =

and if

a

is a limit

U VB then B
V = (b)

U Va aEOn

iff A10.

If Vo = A = class of atoms, ( V a ) (Va+l =

?(v,))

,

then

a is a limit ordinal then

Va =

U

VB

and if

B
v = DEFINITION 0.9:

u E

U Va aEOn

iff A’10.

Rank

U Va and Set(u) then the of u I P(u), aEOn is the smallest ordinal a such that u 5 Va. -(Each Va is transitive so u E V a and S e t ( u ) implies

(a) If

u _CVa.)

PRELIMINARY DEFINITIONS AND THEOREMS (b) If

u E

U

aeon

and Atom(u)

Va

then

xxv

p ( u ) = 0.

THEOREM 0.10: u E

U Va, p ( u ) is the smallest ordinal number a €On such that for all v E u, p(v) < a . If

DEFINITION 0.11:

a

Hartogs Function (Hartogs [1915]).

For each set

x, r(x) = { a : a

<

XI.

THEOREM 0.12: (a) For each set

x, r(x)

(b) For each infinite set ordinal.

is a set. x, T(x)

is an initial

(c) For each set x, T(x) is the smallest ordinal such that it is not the case that a < x.

a

It has been shown independently by Gauntt 119671 and Levy [1969] that the notion of cardinality is undefinable in NBG". However, it is convenient for us to use cardinal notation even though all statements we make about cardinals can be replaced by statements about sets. Thus, we shall use the following notation.

If x

is a set, 1x1

is the cardinal number of

x.

We assume cardinal numbers have the usual properties. set(lx1)

set(x)

+

If

and

x

y

are sets,

PRELIMINARY DEFINITIONS AND THEOREMS

xxvi

(It is well-known that cardinal numbers are definable in NBG. Using the axiom of choice, the cardinal number of a set x can be defined as the initial ordinal equipollent to x. (von Neumann [1928]). Alternatively, the cardinal number of a set x can be defined as the set of all sets of smallest rank which are equipollent to x (Scott [19551)). we shall use lower case greek letters, K , p , v , . . . , to denote cardinal number variables. We shall also make the convention that when we write 1x1, we assume x is a set. DEFINITION 0.13:

1x1 =

(r(x)( =

K +

K*.

( V a l (Ha = I w a l ) . An aleph is the cardinal number

of an infinite well

ordered set. A cardinal 1x1 =

is transfinite, if

K

K

and

Addition:

/ y (= p

Kg

i K.

then

If x l l y = 8 ,

K

+

p =

Subtraction: If there is a unique ~ = p + y then ~ - p = y . Multiplication: Division: K

= u*y

~ * =p

Ix

x

yI

K’

= Ixyl.

2K = (P(x)I. If for each

i

y

such that

.

If there is a unique then K + = y.

Exponentiation:

lx U yI.

y

such that

PRELIMINARY DEFINITIONS AND THEOREMS

xxvii

(The sum and product of an infinite number of cardinals is in general not definable without the axiom of choice.) We shall use some of the following well-known properties of cardinal numbers. THEOREM 0.14:

For each aleph, N, there existsa unique ordinal that H = K,. If

K,

P

2

then

2

K

+

1.1 s

such

K-V.

K

For each pair of alephs,

ct

K',

and

(1) K + N' = H * N ' = max(&,K'). (2) K2 = K. If K 1 r K2i Pl and p 2 are transfinite cardinals such that at least one is an aleph, then

If K and aleph then

p

are transfinite cardinals and N?;

K

+

V + K s

K

v K s u

K

is an

and

If K is an infinite cardinal then K * is an aleph and it is the smallest aleph which is not s K .

. 2K 22 . KL

For all cardinals

K,

K*

s 22

For all cardinals

K,

K*

s

Even though we are using NBGO as our axiom system, it follows from the work of Mostowski that NBGO is a conservative extension of ZFO, Zermelo-Fraenkel set theory without the axiom of foundation. That is, if @ is a formula in the

xxviii

PRELIMINARY DEFINITIONS AND THEOREMS

language of ZFO, provable in ZFO. in NBG" than in

then 0 is provable in NBGO iff it is Thus nothing more about sets is provable ZFO.

PART I, SET FORMS 51.

THE WELL-ORDERING THEOREM

Before 1904, when Zermelo [1904] published his proof that the axiom of choice implies the well-ordering theorem, the wellordering theorem,was considered as self-evident. Cantor and others used it without hesitation. WO I-WO 3 are three classical forms of the well-ordering theorem. WO 4(m) and WO 5 are due to Azriel Levy [1962] and WO 6 is a generalization of wo 5. WO 1: Every set can be well-ordered.

WO 2:

Every set is equipollent to an ordinal number.

WO 3: number. Let

m

Every set is equipollent to a subset of an ordinal be a natural number, m

2

1.

WO 4(m): For every set x there exists an ordinal number a and a function f defined on 01 such that f(0) < m for every B < a and IJ f(B) = x. B
WO 5: WO 4 (m)

.

There exists a natural number m

2

1

such that

WO 6: For every set x there exists a natural number m z 1, an ordinal number a , and a function f defined on such that f(B) < m for every 0 < 01 and U f(B) = x.

a

B
WO 4(m) could be restated in the following way: Every set is the union of a well-ordered set of finite sets each of which has not more than m elements. Similarly for WO 5 and WO 6.

1

PART I, SET FORMS

2

Later we shall see in section 5 that if we omit the bound m on the finite set, WO 4(m) remains equivalent to the wellordering theorem, but then the axiom of foundation is required in the proof. WO 3 and WO 3 WO 1. To prove It is clear that WO 2 WO 2 see Godel [1940]. (The proof is given in that Wo 1 GBdel [1940] p. 2 7 , theorem 7.7. This proof is quite similar to the proof of Theorem 2 . 8 which we shall give in detail.) WO 4(1), The following equivalences are also clear: WO 1 Wo 4(m) WO 4(n) if m s n, and WO 4(m) WO 5 WO 6. Therefore, to prove that the above six forms are all equivalent it remains to prove the following theorem. (The proof is a modification of a proof due to Levy [19621.) -+

-+

-+

-

-+

-+

THEOREM 1.1.

WO 6

-+

-+

WO 1.

PROOF. Let x be an arbitrary set. For each set define a set of natural numbers Ny as follows: (i)

m

E Ny

if

y,

(3f)( 3 a ) [f is a function, a E On, b(f) = a,

LJ f(B) = y, and ( V B ) [ B < a f(B) < mll. B
(WO 6

Next, we shall prove the main result: (ii) If y x y 5 y and m is a natural number such that implies m 1 E Ny. m > 1, then m E N Y

-

Suppose y and m satisfy the hypothesis of (ii) and m E Ny. Then there exists a function f and an ordinal number a satisfying (i). Define (iii) uBy6 = (f(B)

x

f(Y))

n

f(61,

where

B , Y, 6 < a.

is a set of ordered pairs, so that it is a relation NOW, UBy6 and we may talk about its domain and range. Moreover, we have

THE WELL-ORDERING THEOREM

51.

3

For each B < a with f(B) # 0 , let A@ and p B be the lexicographically < - first ordinal numbers y and 6 such that $ ( u B y 6 )# 0 and $(uBys) < m. (That is, first find ordinal numbers y and 6 which satisfy the conditions. Then let X B be the < - smallest such y which satisfies the let p B be the < - smallest conditions. Then given X B , which satisfies the conditions.) Now define:

and

wB = f(B)

-

vB.

Next we define a function g

6

as follows:

3(g) = a + a, if

B < a

if

a

s-

(The sum of two Bachmann [1955] proving that g satisfies (i)

.

then

8, and

g(B) = v

B

-

B f a z y < a

then

g(B)

=

wy

.

ordinal numbers is defined, for example in p. 4 5 . ) We shall prove that m - 1 E Ny by is a function and a + a an ordinal which That is, we have to prove U g(8) = y and B
if 8 < a + a then g(B) < m - 1. The first equality follows from set theoretic considerations and from the fact that U v B 5 U f(B) = y, If B < a, we have v B < m and %
-

-

number in { X : A

# 0

+

< a,

3 ( u h y 6 ) = ml}

PART I, SET FORMS

4

v = '

and

Y

If

y <

If

a s y

a

then and

g(y) = vy y

-

a

z

h c

-

v

g(y) = wx

.

w

= f(y)

Y

Y '

. 01

then

Again we shall show that g is a function and a + a an ordinal number which satisfies (i) with m replaced by m - 1. As in case 1, we have U g ( y ) = y. If y < a, y
vy < 1 < m

-

1.

If a s y and

y

-

c1

g X I

g(y) =

wh

=

f(X) v x ; if f ( X ) # 0, vA = 1 and g ( y ) 6 m - 1, while if f(X) = g ( y ) = @ < m - 1. Hence, if h < a + a, then g (A) < m - 1. This completes the proof of (ii)

a,

.

It follows from (ii) and mathematical induction that if then y can be well-ordered. NOW, to complete the proof of the theorem we must show that our original set x can be well-ordered. Therefore, we prove: y x y _c y

(iv) For every set x x u ( Y x Y ) _c y. Construct

y

there exists a set

as follows:

y

such that

51. zo = x

Let

THE WELL-ORDERING

z

and

~ -+zn ~U ( z n

5

THEOREM

2,).

X

Let

y =

m

U zn. n= 0

m < n

If

x _c y .

zm

then

_C zn

and

zm x zm _c z n .

Clearly,

Also, m

u

Y X Y '

n,m=O

,"

zn

zm

m

C -

x z

'max(m,n

n m=O

max (m,n)

m

C -

x U (y x y)

Hence,

Therefore,

5

'max (m, n +1

y.

i s a s u b s e t of a s e t which c a n b e w e l l -

x

ordered, so t h a t

U

n,m=O

x

c a n be w e l l - o r d e r e d a l s o , q . e . d .

When A l f r e d T a r s k i was s t u d y i n g t h e n o t i o n of f i n i t e (see T a r s k i [1924a,1938a]), h e d i s c o v e r e d s e v e r a l d e f i n i t i o n s o f f i n i t e which w e r e e q u i v a l e n t t o t h e axiom of c h o i c e . The one w e g i v e h e r e i s e q u i v a l e n t t o t h e w e l l o r d e r i n g theorem. (We give several others i n 57). I t i s e a s y t o p r o v e i n NBGO t h a t (*)

If

w e l l orders

i s f i n i t e and

x.

R

R

The c o n v e r s e of x

If

(**)

such t h a t

x

w e l l orders x then ( * ) is e q u i v a l e n t t o

R

-1

is i n f i n i t e then t h e r e e x i s t s a r e l a t i o n

w e l l orders

x

I t i s e a s y t o see t h a t

but (**)

R-l

R

does not.

is equivalent t o t h e w e l l

o r d e r i n g t h e o r e m b e c a u s e e a c h f i n i t e set c a n be w e l l o r d e r e d and e a c h i n f i n i t e o r d i n a l i s w e l l o r d e r e d by

s-

b u t n o t by

Thus, w e h a v e t h e f o l l o w i n g e q u i v a l e n t t o t h e w e l l o r d e r i n g theorem : WO 7 :

For e v e r y s e t

ordering

R

of

x , R-'

x,

x

i s f i n i t e i f f f o r each w e l l

also w e l l orders

x.

2.

6

PART I, SET FORMS

The final form in this section, WO 8 , was discovered through conversations with Norbert Brunner and he noticed a simple proof that WO 8 implies WO 1. We say f is a choice function on a set x of non-empty sets if a(f) = x and for all u E x, f(u) E u. It is well known that if there is a {fl} then x can be well ordered. choice function on f " x ) (See Theorem 2 . 9 ) . Therefore, it was interesting to observe that the following statement is equivalent to WO 1.

-

WO 8: Every set on which there is a choice function, can be well ordered.

WO 1 because if Clearly WO 1 -+ WO 8. Conversely, WO 8 x is any non-empty set then there is a choice function on Thus, WO 8 implies x can be well ordered. {{u}:u E XI. -+

52.

THE AXIOM OF CHOICE

Apparently, the first specific reference to the axiom of choice was given in a paper by G. Peano [1890]. In proving an existence theorem for ordinary differential equations, he ran across a situation in which such a statement is needed. Beppo Levi [19021, while discussing the statement that the union of a disjoint set t of non-empty sets has a cardinal number greater than or equal to the cardinal number of t, (The partition principle) remarked that its proof depended on the possibility of selecting a single member from each element of t. It is not known whether the partition principle implies the axiom of choice. Others, including Cantor, had used the principle earlier, but did not mention it specifically. We give here nineteen forms of the axiom of choice. AC 1: If s is a set of non-empty sets, there is a function f such that for every x E s , f(x) E x. AC 2 : If t is a pairwise disjoint set of non-empty sets, there is a set c which consists of one and only one element from each set in t. AC 3: For every function f there is a function g that for every x, if x E a(f) and f(x) # a, then

such

g(x) E f(x). AC 4: that $(f)

For every relation r there is a function f = a(r) and f _c r. (See AC 20 in section 5 . )

such

5: For every function f there is a function q such that a(g) = %(f) and for every x E $(g), f(g(x)) = x. (If f is 1-1 then g is its inverse.) AC

7

PART I, SET FORMS

8

AC 6: The Cartesian product of a set of non-empty sets is non-empty. AC 7: The Cartesian product of a set of non-empty sets of the same cardinality is non-empty. AC 8 : If s is a set of pairs of equipollent sets then there is a function which associates with each pair a bijection mapping one onto the other. AC 9: If s is a set of sets of the same cardinality then there is a function which associates with each pair a bijection mapping one onto the other. Let

n

1:

2

be a natural number.

AC 10(n): If s is a set of infinite sets, there is a function f such that for each x E s , f(x) is a decomposition of x into a set of disjoint sets of size between 2 and n. AC 11: AC 10(n).

There exists a natural number

n

1:

2

such that

AC 12: If s is a set of infinite sets there is a natural number n 2 2 and a function f such that for each x E s I f(x) is a decomposition of x into a set of disjoint sets of size between 2 and n. Each of AC 10(n), AC 11 and AC 12 remains equivalent to the axiom of choice if we require that the size of each of the sets in the partition be exactly n. Let

m

5

1

be a natural number.

AC 13(m): If s is a set of non-empty sets there is a function f such that for each x E s , f(x) is a non-empty subset of x with f(x) < m. AC 14: AC 13(m).

There is a natural number m

1:

1

such that

AC 15: If s is a set of non-empty sets then there is a natural number m 2 1 and a function f such that for each x E s I f(x) is a non-empty subset of x with f(x) < m.

92.

Let n l < k < n .

and

k

THE AXIOM OF CHOICE

9

be natural numbers with

n > 2

and

AC 16(n,k): If s is an infinite set then there is a set tn of n-element subsets of s such that each k-element subset of s is a subset of exactly one element of tn.

-

{ P I ) , F is the set AC 17: If s is a set, t = p ( s ) of a l l functions from t to s and g is a function from F to t then there is an f E F such that f(g(f)) E g(f). B

AC 18: For any non-empty set A, and for every function whose domain is A and Ba is a non-empty set for each

a € A,

AC 18 is equivalent to the axiom of choice if XaIb is = b. As we shall see later a set or a proper class or X a,b in section 5, if we allow B to be a relation so that for each a E A, Ba is a class, the resulting form is equivalent to the axiom of choice in NBG. (See AC 21). AC 19: For any non-empty set A, each of whose elements is non-empty,

u

n u b = 17 f(a) a€A bEa f€C(A) a€A where

C(A)

I

is the set of all choice functions on

Dual forms of AC 18 and AC 19 interchanging union and intersection. equivalent to the axiom of choice.

A.

can be obtained by The dual forms are a so

In 1904, Zermelo [1904] stated a principle of choice similar to AC 1 and proved that it implied the well-order ng theorem. (Zermelo took s to be the set of all non-empty subsets of a given set and he stated in his paper that the proof of the theorem came out of conversations with Erhardt Schmidt.) AC 3 and AC 6 are essentially restatements of AC 1. Bertrand Russell [1906] gave a principle analogous to AC 2. He announced this principle as a possible substitute

10

PART I, SET FORMS

for Zermelo's but he believed that it was weaker. (Russell actually stated his principle in the form of AC 6 with disjoint sets. However, by "product" Russell meant the following: x € Sa if x _c Sa and x n Sa is a unit a€ A aEA set for each a E A. This definition of product is due to Whitehead [1902]. Hence Russell's form is the same as AC 2. See 0.4 (i) in the section entitled Preliminary Definitions and Theorems for our definition of product.) Zermelo, in [1908a] stated, and in [1908b] proved, that Russell's and his formulations of the axiom of choice are equivalent. The name "axiom of choice" is due to Zermelo [19041, and the name "multiplicative axiom" or "multiplicative principle'' is due to Russell [1906]. AC 4 and AC 5 were given by Bernays [1941]. (Bernays actually gives CAC 4 and CAC 5 which are stronger than AC 4 and AC 5. CAC 4 is given as an axiom and CAC 5 is proved equivalent to it.) AC 7 is an interesting variation of AC 6 which is based on some topological results due to Ward [1962]. (See TOP 2 in Section 8.) AC 8 and 9 follow from work of Pincus [1974]. AC lO(2) is due to Sierpinski [19651 and AC 10(n) for n > 2 , AC 11 and 12 are generalizations which follow from results of A . Levy 119621. AC 13(m) and AC 14 are due to Levy [1962] and AC 15 is a generalization of AC 14. (If the bound m on the finite sets is omitted in AC 10(m) and AC 13(m), the resulting statements remain equivalent to the axiom of choice, but the axiom of foundation is required in the proof. See MC 1 and MC 3 in section 5.) Sierpinski [19401 proved that the axiom of choice implies AC 16(3,2) and Sobocinski [1964] proved the converse. Frascella [1965,1966] and H. Rubin and J. Rubin [1967] independently proved that AC 16(n,k) for n > 2 and 1 < k < n is equivalent to the axiom of choice. Tarski I19511 proved that AC 17 is equivalent to the axiom of choice. AC 18 is due independently to C o l l i n s [19541 and Farah 119551; and AC 19 is a generalization of AC 18.

x

We shall first show that AC 1 - AC 7 are all equivalent. AC 1 obviously implies AC 2 , and it is clear

THE AXIOM OF CHOICE

52.

from the definition of product that equivalent. THEOREM 2.1:

AC 2

-+

AC 1

11 and

AC 6

are

AC 1.

PROOF. Let s be a set of non-empty sets. Define x x: x E s } . Then the choice set of AC 2 is the required choice function on s . t =

{{XI

THEOREM 2 . 2 :

AC 1

-+

AC 3.

PROOF: Let f be an arbitrary function. Let s = p(f) and let F be the choice function on s . Define a function g so that for each x E a(f), g ( x ) = F(f(x)). Then g is the required function. THEOREM 2 . 3 :

AC 3

AC 1.

i .

PROOF: Let s be a set of non-empty sets. Let f be a 1-1 function such that g(f) = s . Define a function F such that for each x E s , F(x) = g(f-l (x)), where g is defined by AC 3 . Then F is the required choice function. THEOREM 2 . 4 : r g

=

AC 4

+

AC 5.

PROOF: Let f be an arbitrary function and let {(x,y):(y,x) E f } . AC 4 implies that there is a function such that a ( g ) = and g _c r. Clearly, for every

a(r)

= 9(9) I f (g(x)) = x. x E l(f)

THEOREM 2.5: PROOF: function h

AC 5

+

AC 4 .

Let r be an arbitrary relation and define a as follows: h

=

{((x,y),x):(x,y)

E rl

.

AC 5 implies that there is a function g such that and for every x E $ ( g ) , h(g(x)) = x. NOW, g(x) is an ordered pair, so we define f(x) to be the second coordinate of g(x), for each x E a ( g ) = a(r). Clearly, a ( f ) = $(r) , f is a function, and f _c r. a ( g ) = %(h)

THEOREM 2.6: AC 4 -f AC 3. PROOF: Let f be an arbitrary function. relation r as follows:

Define a

PART I, SET FORMS

12

AC 4 implies that there is a function $ ( g ) = a(r)

and

THEOREM 2.7: PROOF: function h

g _c r. AC 3

g -f

g such that is the required function.

AC 4 .

Let r be an arbitrary relation. as follows:

Define a

For each x E $(r), h(x) = {y:(x,y) E rl. AC 3 implies that there is a function f such that if x E a(h) and h(x) # $2, then f (x) E h(x). f is the required function. It is clear that THEOREM 2.8:

AC 6

AC 7

-f

+

AC 7.

AC 6.

PROOF: Suppose s is a set and for each i E s , xi is w a non-empty set. Let y = ( C, xi) Define F:y X xi y iEs for i E s such that for f E y and u E xi, F(f,u) = g, where g(0) = u and g(n + 1) = f(n). Then F is a 1-1 function mapping y x xi into y. Since y 4 y x xi, it follows that for each i E s, y x xi FZ y . Thus, AC 7 implies ( y x xi) # $2. But if (y x xi) # JJ then that iEs iEs xi # 0, so AC 6 holds.

.

-f

x

x

x

iEs

Next, equivalent because WO the f of x E s its

we shall show that each of AC 1 - AC 7 is to the well-ordering theorem. WO 1 implies AC 1 1 implies U s can be well-ordered. Then define AC 1 to be the function which assigns to each first element.

THEOREM 2.9:

AC 1

+

WO 2.

(The intuitive idea of the proof is as follows: Let x be a non-empty set and f a choice function on the set of all non-empty subsets of x. We well-order x as follows: Let f(x) = a be the first element of x; f(x {a}) = b be the second element of x; f(x {a,b)) = c be the third element of x; etc. The proof given here is a modification of Zermelo's I19041 proof using ordinal numbers instead of

-

-

52.

THE AXIOM OF CHOICE

13

well-ordered sets.) PROOF: Let x be a non-empty set. Let f be a choice function on the set of all non-empty subsets of x and define f(B) = u where u 4 x. We define a function G as follows: For all ordinal numbers a, G ( a ) = f(x G"a). (It follows by transfinite induction that G is defined for all a E On.)

-

is 1-1 on g ( G ) f l x. If a < 8 , then G ( a ) # G ( B ) since G ( B ) = f(x

(1) G-l G(B)

E x,

and

G(a) E G"8.

-

and G"f3)

G(a),

f x

-

G"B

(2) There is an ordinal number a such that G " a = x. For there must be some ordinal number 8 such that G ( B ) 4 x. If this were not so, by (1) G would be a 1-1 mapping of On into x, which contradicts x being a set. Let a be the smallest ordinal number such that a E { B : G ( 8 ) 4 XI. Then by the definition of a, G " a _c x, and by the definition of G , x G " a = ,@.

-

From (1) we see that an ordinal number satisfying (2) yields the desired result, q.e.d. Next, we shall show AC 8 is equivalent to the axiom of choice. It is clear that AC 1 + AC 8. THEOREM 2.10:

AC a

+

AC 1.

PROOF: Let s be any set of non-empty sets. Then for each x E s , x x w (x x w ) U { P I . Thus, AC 8 implies that there is a function f such that for each x E s , fx is a onto x x w. Define a bijection from (x x w ) U {B} function g on s such that for each x E s , g(x) is the first coordinate of fX(p). The function g is the required choice function. To prove that AC 9 Theorems 2.8 and 2.10. THEOREM 2.11:

+

AC 1, we use techniques given in

AC 9 * AC 1.

PROOF: Let s be a set of non-empty sets. Let y = ( U s ) w . Then we prove as in Theorem 2.8 that for each

PART I, SET FORMS

14

x E s , x x y = y. A l s o , it is easy to see that for each x x y = (x x y) u 191). Let t

=

{x

x

u

y: x E s )

I(x

x

y)

u I@]:

x E

XES,

.

s3

The set t satisfies the hypothesis of AC 9. Let f be a function such that for each x E s , fx is a bijection from (x x y) U { g ] onto x x y. Define a choice function g on s such that for each x E s , g(x) is the first coordinate of

fx (91)

. Thus, AC 1

-

since it: is clear that AC 1 * AC 9 .

AC 9 ,

The following implications are also clear: WO 1

AC 10(n)

AC 11

AC 12, AC 1 +,AC 13(1), if m 5 n, and AC 13(n) AC 14 AC 15. We shall show that AC 10(n) AC 13 (n - 1), AC 11 + AC 14 and AC 12 * AC 15 + WO 6, which is sufficient to show that each of AC 10(n) - 15 is equivalent to the axiom of choice. -+

AC 13(m)

-+

-+

-+

AC 13(n)

-+

-+

-+

THEOREM 2.12: AC 10(n) * AC 13(n

-

If n 1).

is a natural number,

n

2

2

then

Suppose s is a set of non-empty sets. Let t = {(x x w ) U {a): x E sl where a 4 x x w f o r any x E s . The set t satisfies the hypothesis of AC 10(n) so there is a function f such that for each x E s , f(x) is a decomposition of (x x w ) U {a} into disjoint sets of cardinality between 2 and n. There is exactly one such set u, which has a as an element. Then the set of first {a] is a non-empty subset of x which coordinates of u has at most n - 1 elements, q.e.d. PROOF:

-

The same proof gives us that AC 11 * AC 15.

-+

AC 1 4

and

AC 12

THEOREM 2.13:

AC 15 + WO 6.

(The proof is similar to 2.9.) Let x be a set and let s be the set of all nonempty subsets of x. By AC 15, there i s a natural number m and a function y such that for each y E s , g(y) # 91, g(y) 5 y and g(y) < m. PROOF:

92.

THE AXIOM OF CHOICE

15

Define g ( @ ) = u where u $ x. Define a function G as follows: For all ordinal numbers a, G ( a ) = g(x IJ G ( B ) ) . $
-

(2) There is an ordinal a such that U G " a = x. Then define f = G l a , and we have m is the natural number, a is the ordinal number and f is the function which satisfies WO 6. Next, we shall show that if + WO 4(n - k).

n > 2 and

1 < k < n

then

WO 2 * AC 16(n,k)

that

THEOREM 2.14: If n and k n > 2 and 1 < k n then

are natural numbers such WO 2 AC 16(n,k). -f

PROOF: Suppose s is an infinite set. WO 2 implies that there is an initial ordinal wa such that s = w a . Let K be the set of all k-element subsets of s and N, the set of all n-element subsets of s . Then K w N = w a . Suppose and N = {ng: $ < w a ) . K = CkB: B < w a }

Nowl we shall construct a subset t

5 N which has the

desired property. Define a sequence of sets, T y l by transfinite induction as follows: To = @. Suppose T has Y the property that for all B < y < w a , k B is a subset of exactly one element of T and for all B such that Y Y 2 B < w C 1 l k B is a subset of at most one element of Ty. If ky is a subset of exactly one element of Ty define T = T If ky is not a subset of any element of T y l let Y' n ' + b e the smallest element x E N such that k c x, but Y Y for all $ < y, k g x and for all $ > y if there is a y E N such that ks _c y then k B $ x. (An example of such . ~ u the ~ - ~ui's ~ are an x is x = k U ~ ~ ~ , u ~ , . . where Y (UTy U k,,).) NOW, define distinct elements of s T = T lJin;). If y is a limit ordinal define T = TB Yfl Y B
-

"

PART I, SET FORMS

16

THEOREM 2 . 1 5 : If n and k are natural numbers such that n > 2 and 1 < k < n then AC 16(nfk) WO 4(n-k). -+

PROOF: Suppose x is an infinite set, and n and k satisfy the hypothesis. (There is no l o s s of generality in proving WO 4(n-k) for an infinite set x.) Let y be a well-ordered set such that y n x = and y w where

w

=

{u _c x: u

sz

n

- kl.

(For example, let y = I'(P(x)), where r is Hartogs' function; see 0.11.) Apply AC 16(n,k) to x U y. Let tn be a set of n-tuples of elements of x U y such that each k-tuple of elements of x U y is a subset of exactly one element of tn. For each u E x, let s

U

=

{V

E v

E tn: u

A

k

-

1 < v

n y}.

Then there is a v E su such that k 4 v y. For suppose not. Let a be any (k-1)-element subset of y. Then a U {u} is a subset of exactly one element in su. Moreover, each element of su contains exactly one subset a U {u} where a is a (k-1)-element subset of y. Consequently, sU

{ a _CY: a

F=

k

- 13

= K.

Since y can be well-ordered, K FS y. We claim K 4 w. function F such that for each a E K, F(a) = va (a U where va is the unique element of su with a U { u } 5 is a 1-1 function. This contradicts the definition of for y 4 w. Therefore, there is a v f su such that k
The {u}),

vat yf

Now let

M = {v E tn:k

4

v n y}

and

We have shownthat for each u E x, M n su # 8 . Furthermore, since each element of L has at least k elements, each

52.

THE AXIOM OF CHOICE

17

element of L is contained in exactly one element of M. Since y can be well-ordered and L is a set of finite subsets of y, L can be well-ordered. For each a € L, let G(a) = ua a, where ua is the unique element of M such Then G yields a well-ordering of a collection that a c u - a' of subsets of x, each of which has at most n - k elements and whose union is x. Thus WO 4(n - k) holds.

-

We note that AC 16(n,k) can be generalized similarly to the generalizations of AC 10(n) (AC 11 and AC 12) and AC 13(m) (AC 14 and AC 15), but we shall not state the generalization here. We shall show next that AC 1 * AC 17. THEOREM 2.16:

AC 1

+.

AC 17.

PROOF: Let s , t = y ( s ) hypothesis of AC 17. Then if then

f

E F , g(f) E t

THEOREM 2.17:

and

AC 17

- {a}, f

F and g satisfy the is a choice function on t,

f(g(f)) E g(f). -+

AC 1.

PROOF: Suppose AC 1 is false. Then there is a set s such that if t = y ( s ) ( $ 1 , and F is the set of all functions mapping t into s then for all f E F there is a u E t such that f(u) f u. Suppose a s. For each ordinal a and for each f E F, define

-

f(s hf(a)

-

- hl;a),

if

a

otherwise

f(s

h;a)

E s

-

h;a

=

Just as in the proof of Theorem 2.9 (for G ) we can prove that hf-l is 1 - 1 on g(hf) n s and therefore, there is an a such that hf(a) = a. Let a. be the smallest such a. If hjaO = s then s can be well-ordered, which implies t has a choice function, contradicting our assumption. Thus, hjao c s. Define g(f) = s h"a f 0' Then g is a function with domain F, range contained in t and f (g(f)) g (f) for all f E F. This contradicts AC 17, q.e.d.

-

PART I, SET FORMS

18

To complete the proofs of equivalences in this section we shall prove AC 1 * AC 18 -+ AC 19 AC 1. -+

THEOREM 2.18: PROOF : (i)

AC 1 * AC 18.

Suppose

n

U XaIb. aEA b€Ba

u E

a E A

Then, for each Let

Ca Let

f

for a l l

Conversely, f E Ba such a €A for each a E A, that u E XaIb.

x

b E Ba

{b E Ba: u E Xa,b}

be a choice function on E 'a,f(a)

and

=

there is a

{C,:

a E A,

such that

u E XaIb.

.

a E A}.

Then

so

f E

x

Ba aEA

suppose (ii) holds. Then there is an that for all a E A, u E Xa,f(a). Therefore, there is a b E Ba, namely, b Consequently, (i) is true, q.e.d.

=

f(a), such

The axiom of choice was not used in the second part of the proof of 2.18. Since AC 19 is a special case of AC 18, it remains to be shown that AC 19 -+ AC 1. THEOREM 2.19:

AC 19

-+

AC 1.

PROOF: Let A be a set of non-empty sets. For each a E A, let ua = {c U { u } : c E a}, and let B = {u,: a E A } . Then u E U b for each ua E B. Therefore, bEua

n

(J

uaEB bEu,

b # fl

;

AC 19 implies

n

U f(ua) # 9, where fEC(B) uaEB

C(B) is the set of choice functions on B. Consequently, C(B) # fl and a choice function on B determines a choice function on A.

52.

THE AXIOM OF CHOICE

19

In closing, in this section, we should like to say a few words about the principle of dependent choice. DC(a): If R is a relation between subsets and elements of a set x such that (Vy _C x) ( y < ci (3u E xj(y,u) E R ) then there is a function f mapping c1 into x such that ( V B < a)(f"Brf(B)) E R +

The first mention of the principle of dependent choices was in Bernays [1942] and rediscovered independently by Tarski [1948]. The statement in these papers has the following form: DC: If R # 9 is a relation such that Z(R) _c $(R) then there is a function f with domain w such that for all n E w (f(n),f(n+l)) E R.

-

.

It is easy to see that DC DC(w) The forms DC(a) were given by Levy [1964] and Theorems2.20 and 2.21 are also due to Levy [19641. (See also Felgner 119711 pp. 146-166.) THEOREM 2.20:

WO 1

-+

(Va)UC(a).

PROOF: Let a be an ordinal and let x and R satisfy the hypothesis of DC(a). WO 1 implies x can be well-ordered. Define a function f on a so that for all 6 < a , f(B) is the first element a E x such (f"6,a) E R. It follows from the hypothesis that f is defined on a and f is clearly the desired function.

THEOREM 2.21:

(Vci)DC(a) + WO 3.

PROOF: Let x be any set and let r be Hartogs' function. Then T ( x ) = {a:a < x} is the ordinal number, A , such that for all a < A, a < x but X x. Define a relation R _C p(x) x x such that

If x < X then WO 3 holds, so suppose it is not the case that x < A. Then for all y 5 x such that y < A there is a u E x y with (y,u) E R.

-

20

PART I, SET FORMS

Thus x and R satisfy the hypotheses of DC(X). Therefore, there is a function f such that for all B < A , (f"B,f(R)) 6 R. The function f is 1-1 because by the definition of R, f(R) 4 f"B. Thus if f(a) = f(R) with c1 < R we would have f(B) = f(a) 6 f"R. This implies X < x, which contradicts the definition of A . Consequently, x < A , q.e.d.

THE LAW OF THE TRICHOTOMY

53.

The equivalence of the axiom of choice and the trichotomy was given by Hartogs [1915]. A s in the case of the wellordering theorem, the trichotomy was considered self-evident and was used without hesitation before 1915. T1:

For all sets

x

and

y, x < y , x = y

or

y < x.

It is clear from the properties of ordinal numbers that WO 2 implies T1. To prove the implication the other way we use Hartog's function r , where r(x) = {a: u < XI. (See 0.11 and 0.12). THEOREM 3.1:

T1

+

WO 3.

PROOF: By T1, either x < T(x) or r ( x ) 4 x. The first inequality implies WO 3 since r ( x ) is an ordinal number. The second is impossible because it implies T ( x ) E T ( x ) . ( N o ordinal number can be an element of itself.)

We shall next consider a proposition which is similar to T1. It was stated by Lindenbaum in Lindenbaum and Tarski [1926] and later proved by Sierpinski [1948]. T2: For every two non-empty sets, there is a mapping of one onto the other. Clearly T1 implies T2. Before proving that T2 implies the axiom of choice we prove the following lemma. LEMMA 3.2: For any two sets x and y, if there is a mapping of x onto y then there is a 1-1 function which maps Y ( y ) into B ( x ) . g(t)

PROOF: L e t f = f-l"t. Then

map x onto y. For g is a function from

t f p(y), define P(y) into Y ( x ) .

PART I, SET FORMS

22

To show that g is 1-1, we note that if s, t E y(y), Therefore, if g(s) g(t) = f-1"s f-l"t = f-1"( s t) f-l" (t ; which s) = g ( s ) = g(t), f-l"(s t) = fl and implies t _c s and s _C t; so that s = t.

-

-

THEOREM 3 . 3 :

T2

N

+

.

-

WO 3.

Consider the non-empty sets x and T2, either there is a mapping of x onto r ( P ( x ) a mapping of r ( Y ( x ) ) onto x. If the first mapping exists, then 3 . 2 implies there is a 1-1 function which maps $ ( r ( f ) ( x ) ) ) into y ( x ) . But since r @(x) ) < $ ( r ( y ( x )) 1 , this implies that r ( y ( x ) ) < p ( x ) , which is impossible. Hence, there must be a mapping f of r(P(x)) onto x. For each t E x, define g(t) to be the smallest element of f-1"{t3. We see that g is a 1-1 function from x into r ( Y ( x ) ) . Hence x T ( ? ( x ) ), which proves WO 3. PROOF:

There are two additional forms of the trichotomy due to Tarski [19641, T3(n), and Kruse 119621, T4. For convenience, we shall state these forms in terms of cardinal numbers rather than sets. (All statements about cardinal numbers which we use may be replaced by statements about sets.) In what follows, K , p , v and these letters with subscripts, represent cardinal numbers. (See Definition 0.13 and the discussion which precedes it.) For example, T1 stated in terms of cardinal numbers would be: For all cardinals If

n

I

2

K

and

p,

K

< p,

K

or p <

= p,

K.

is a natural number,

T3(n): If x is a set of cardinal numbers such that 1x1 = n then there exists K , p E x such that K < 1.1 or p <

K.

T3(2) is T1 and it is clear that T1 T3(n) natural numbers n L 2 . We shall show that T3(n) each natural number n 2 2, -+

T3(n)

THEOREM 3 . 4 : WO 3 . -+

For each natural number

n

I

-+

2,

for all WO 3 for

THE LAW OF THE TRICHOTOMY

3.

23

Suppose x is an infinite set such that Ix = K and K 2 KO. (There is no loss of generality in assuming K z KO, for, if not, replace x by x U w. If x U w can be we1 ordered then so can x.) Let n ;r 2 be a natural number such that T3(n) holds. For each m E w, define P,(K) so that pjn(K) PO(^) = 0, P1(~)= K , and for all m z 1, Pm+l(~)= 2 Let K, be any aleph such that K, 4 Pn(lc). (For example, let PROOF:

N,

=

lr(pn(K))l.)

Let

Then y is a set of cardinal numbers such that IyI L n. Suppose there is a j and k, with 0 i j < k < n such that

Ka+k

(1)

5

Ha+,

Or

K,+k

Pn-j ( K )

5

-

The first inequality is impossible because j < k and the second is also impossible because it implies N, i P n ( ~ ) which contradicts the definition of Ka. Thus, IyI = n. Applying T3(n) to y , there exists natural numbers and 0 i j, k < n such that

j , k, j # k, (2)

Na+k + Pn-k ( K ) < Ka+j + Pn-j ( K )

*

If j < k then we obtain the inequalities (1) which have been shown to imply a contradiction. Therefore, we must have k < j. Let p = Pn-j(~), then from ( 2 ) we obtain K,+j

+ U

’ H,+k 2

+ Pn-k(K)

Pn-k(Kc) ’n-(j-l)

(K)

Thus it follows from 0.14(d) that K i q.e.d.

2’

= 2’. i

and this implies

PART I, SET FORMS

24

The form of the trichotomy due to Kruse [1962] is the following : T4:

either

For all cardinals p , if K~ i; K * or K~ i; K ~ .

p i

K

~

,K~

i;

2'

then

It is clear that T1 + T4. Moreover, once we prove that T4 implies the axiom of choice, we will have another proof that the generalized continuum hypothesis (GCH) implies the axiom of choice, because it is clear that GCH + T4. Before proving that T4 implies AC we acquire some preliminary results.

DEFINITION 3 . 5 : the set of all 1-1

For any set x, we define K(x) to be functions mapping an ordinal into x.

K(X) = if: (3a E 0n)f:a

< XI.

If 1x1 = K , let IK(x) I = K t We shall call K Kruse's function. It is easy to prove that K has the following properties

.

LEMMA 3.6:

For each set

x,

X x: (3y c x) ((y,R) is a well-ordered set) I .

5x

(a) K(x)

k

(b) K(x)

< ?'(x

(c) K(x)

< ?'(8(x)).

{R

x

XI.

(d) x < K(x).

(e) If

1x1

LEMMA 3.7: (a)

Kt*pt

(b) K

5

is an aleph then For all cardinals (K

S

Kg

+

K~

-t p )

+

.t

K+

=

K+

K(x) K

k

and

P(x).

u,

THE LAW OF THE TRICHOTOMY

53.

(b) If Kt

+

Kt

KO

2

K

= Kt*2 = Kt-1'.

(c) Suppose

K

LEMMA 3.8: then

K

<

+

=

K

Then using 3.7 (a) we obtain

K.

If

ho

(K

K)'

=

.

K t

is an aleph and

K

i K,

+

+

i K

K+

K

= 2K i K .

K~

PROOF: by 3.6 (d) obtain,

then K + 1 = K . Moreover, Therefore, using part (a) we see that

= Kt*Kt i

(K'12

25

K

Suppose K and K satisfy the hypothesis. Then < K ~ ,thus using 3.7 (b) and hypotheses, we K

+

K

f;

Kt

+

K t =

Kt

i K

+

Therefore, there exists p i K such that This implies u i K , (0.14(e)) so K + p K + K = K. It f o ~ ~ o wfrom s 3.7 (c) that K~ i K + N, there exist cardinals K~ f; K that K' = K~ + u . Thus ,

K . K

+

K

=

K

+

p

.

Consequently, ( K + ) ~ = K ~ . Since and p i K such

= K.

from which it follows that either K~ i K or p s K . (See 0.14(d)). Suppose 1-1 i K , then K t = K~ + 1-1 f; K + K = K . which contradicts 3.6 (a). Thus, K~ s K which implies K t is an aleph. Therefore, by 3.6 (e) K~ = 2 K s N, q.e.d. To prove T4 implies the well-ordering theorem we shall prove two things : (1) For every cardinal

that

K,

there is an aleph, K , such

2 K i 2'. (2)

If

N

is an aleph, so is

2K.

Then since K < 2K, it follows from (1) that K < 2', and from ( 2 ) that K is the cardinal number of a well-ordered set.

N

THEOREM 3.9: such that 2 K

T4 * For each cardinal i 2'.

K

there is an aleph

PART I, SET FORMS

26

K

5

PROOF: We may assume, without l o s s of generality that KO. Let K = ( 2 K ) * , where if 1x1 = K , K = Ir(y(X))I.

have ,

K

K

5 2K

+ K,

+ zN

K

5

We

.

zK+’

T 4 implies that K + 2 ’ s z K + x

o r z K + K s K + 2 N.

If the first inequality holds then 2 ‘ s 2 K + K , which by 0 . 1 4 (d), implies 2’ s’ 2K, contradicting the definition of K. Similarly, the second inequality implies 2 K s 2,’ which is what we wanted to prove. THEOREM 3.10: PROOF:

(*I

T4

-+

For every aleph,

K, 2’

is an aleph.

First we shall show that T4 implies

(Va

E On) Ha+1

i 2

.

N,

.

If Ka+l 4 K then K is not an aleph and K * = K a+l. If we could find a cardinal p such that p < K * < 2’, then since p i K,, we would obtain the Let us now construct this cardinal p. contradiction K * i 2 Ka

Let

K

= 2

N,

.

Therefore, it follows from 3 . 6 (1)

2 K

<

Kt

S

2K

= 2K

(b) and (d) that

.

Moreover, we have by 0.14 ( 9 ) and (h), m

Therefore,

By (1) and ( 3 ) , both so T 4 implies

K~

and

K t S K + K *

K

or

+

K*

are between

K + K * S K

t

.

K

and

2K

53.

THE LAW OF THE TRICHOTOMY

27

t If the first inequality holds it follows from 3.8 that K < K = 2K i K * which implies that K is an aleph contradicting the hypothesis. If the second inequality holds then K * i K t . Let x be a set so that 1x1 = K . Then there is a set w 5 K ( x ) such that r ( x ) = w . Each element of w is a function from an ordinal number onto a subset of x. Let y = U %(f) _c x. fEW

Since w can be well-ordered, y can also be well-ordered. Let u = 1y/ i K . Then, w _c K(y) and 1~ < u * i K * . Moreover, since K * = /wi, K * 5 p t and since is an aleph it follows from 3.6 (e) that ut = 2’. Thus, p is the required cardinal because p Y* s 2’.

’

Now we shall show that for each ordinal a, 2 Ha aleph. Let KR = (2 ) * . Then, K a < K R and K R (*), we have

+ and

2

8, )J

+ Kg

2 K R = 2”.

tiR and

are between

+

t:&

or

2

H6

2Ka

k

2

is an Ka . By

Therefore, both

. +

Kg+l

T4 implies Ng i

t:

R+ 1

.

The first inequality is impossible because of the definition of K g and the second inequality implies 2” s K g + l , q.e.d. THEOREM 3.11: PROOF:

T4

+

WO 3.

3.9 and 3.10.

Next, in this section on the trichotomy, we shall add three forms due to Sobocinski. T5 and T6 are in Sobocinski [1961al, and T7 in [1961bl. T5: or

p i

For all cardinals

K

and

u , if

1~ < Z K

K,

and alephs

fc, if

then

K

5

)J

K.

T6: For all cardinals K S K or k s ~ .

K < 2K

then

28

PART I , SET FORMS

For all cardinals KIl and if for all v1 and K~ < p implies w1 = w 2 , then K~ i K~ T7:

K2,

and

p, if

W2,

K1

+

or

K~

i K ~ .

V1

=

No i

K2

+

V2

K1,

=

Clearly, T1 implies each of T5, T 6 , and T7. It is clear that T 5 implies T 6 . We shall show T 6 implies WO 3 and that T 7 implies T1. THEOREM 3 . 1 2 :

T6

-f

WO 3 .

PROOF: Let K 2 K O be a cardinal. properties of cardinals (0.14(i)) that

If equality holds, then Otherwise, T 6 implies K*

K*

K

or

22K

i

It follows from

which implies

22K i

K*

K

is an aleph.

.

The second alternative implies K is an aleph. If the first K alternative holds, repeat the process. If K * < 22 apply T 6 to obtain K*

or

i 2K

i K*

.

The second alternative implies K is an aleph. alternative holds repeat the process, obtaining

The first alternative is impossible. is an aleph, q.e.d. THEOREM 3 . 1 3 :

PROOF: K +K

u = 2

’.

Let

T7 K~~

K~

The second implies

V1

K

T1. 2

Ho

be arbitrary cardinals.

Then it follows from 0.14 (f) that

K1

and

-+

If the first

= 2 K1+K2 + v 1 = 2 K l + K 2

Let

53.

THE LAW OF THE TRICHOTOMY

29

Thus, the hypothesis of T7 is satisfied, so K~ and K~ comparable. (The proof given here is simpler than the one given in Sobocinski [1961bl.)

are

The final forms of the trichotomy which we shall consider here are due to Alas [1968].

P<,(Y)

Let

p,,(x)

T8:

For all sets

4

be the set of all finite subsets of

x

and

y,

y,w(x)

x

and

y,

x 6 p,,(y)

$

p,w(y)

x.

or

Ip
T9:

For all sets

or

P<JY) < x. T10: Y <

For all sets

x

and

y, x < p , , ( y )

or

Y<,(X).

T1 implies each of T8, T9 and TlO, and each of T8 and T9 implies T10. We shall show T10 + WO 3. THEOREM 3.14:

T10

+

WO 3.

PROOF: Let x be an infinite set and let y = T(p<,(x)). Then since y can be well-ordered, y,,(y) F;: y. 1'10 implies x < y or y 6 p,,(x). The second alternative is impossible and the first implies WO 3.

This Page Intentionally Left Blank

54.

MAXIMAL PRINCIPLES

As mathematics developed further there also developed a need for another non-constructive proposition; a principle, which Kuratowski, Hausdorff, Zorn, and others, used to replace transfinite induction and the well-ordering theorem. It appears,

at first glance, unrelated to the axiom of choice, but actually is equivalent to it. This principle and principles similar to it are often referred to as f m m s of Zorn's Lemma. (According to Zorn (see P. Campbell [1978]), Artin and Chevalley first referred to the principle as Zorn's Lemma in 1933.) We prefer to call these forms maximal principles since it is a more descriptive title. Also, we discuss forms which are far afield from Zorn's original form, but are still maximal principles. The history of maximal principles is quite tangled. (See Campbell [1978] for a more detailed description.) The earliest reference to a maximal principle in the literature is in Hausdorff [1907] in which a special case of M 6 is derived. Later in Hausdorff 119091, M 6 itself is proved. In his book, Hausdorff [1914], M 5 is derived from the well-ordering theorem. In 1927, in the second edition of his book, Hausdorff derives a form similar to M 4 from the axiom of choice. (In 1910, Janiszewski [1910], Mazurkiewicz [19101 and Zoretti [1910, 19121 published a special case of M 4 in the form of a theorem in topology. Brouwer [1911] derived a more general topological theorem from the well-ordering theorem. Reference was made in Hausdorff [1927] to Janiszewski [1910, 1911, 19121 and Brouwer [1910], but not to Brouwer [1911] which was the second half of Brouwer [1910].) P.

31

32

PART I, SET FORMS

Kuratowski [1922] derived minimal principles equivalent to M 3 and M 4 from the well-ordering theorem. (In our forms M 1 - M 4 there is an obvious duality between minimal and maximal principles. We shall not discuss the minimal principles here.) Kuratowski used a minimal principle to prove a theorem in analysis. (Some of the preliminary work in Kuratowski’s paper is due to Hessenberg [1909] and Kuratowski referred to the work of Brouwer [19111.) R. L. Moore [19321 derived a minimal principle equivalent to M 3 from the well-ordering theorem. (Moore lists Kuratowski [1922] in his bibliography.) It wasn‘t until 1935 that Zorn published his paper. He was the first one to use a maximal principle in algebra. He stated without proof that M 3 is equivalent to the axiom of choice (to be proven in another paper which was never published). According to Zorn himself (as stated in Campbell [1978] p. 83) he saw a copy of Moore‘s book [1932] but did not see the maximal principle. He also knew about Kuratowski [1922] but had “formed an erroneous impression about the thrust of the article”. Also, Zorn credits Artin for noticing that M 3 implies the axiom of choice. 0. Teichmuller [1939] and J. W. Tukey [1940] independently

gave M 7 and proved it equivalent to the axiom of choice. Tukey also gave M 1, a generalization of M 3 . Teichmbller was unaware of Zorn’s work while Tukey refers to Zorn. N. Bourbaki [1939] gave a proposition similar to M 1 and states, without proof, that it is equivalent to the well-ordering theorem. (He stated a lemma similar to Kuratowski’s lemma (see 4.17) to indicate how one would prove that M 1 implies the well-ordering theorem. M 7 is also stated by Bourbaki, but no proofs are given.) M 2 was given by Kneser [1950] and Szele [1950]. There are two different forms of the maximal principles. M 1 - M 4 postulate the existence of a maximal element, and the remaining forms postulate the existence of a maximal set. M 1: If R is a partial ordering relation on a non-empty set x and if every subset of x which is linearly ordered by R has an R-upper bound, then x has an R-maximal element.

MAXIMAL PRINCIPLES

54.

33

M 2: If R is a partial ordering relation on a nonempty set x and if every subset of x which is well-ordered by R has an R-upper bound, then x has an R-maximal element.

DEFINITION 4.1: A nest is a class which is linearly ordered by inclusion, 5. M 3 : If every non-empty nest which is a subset of a nonempty set x has its union an element of x, then x has a c-maximal element. M 4: If every non-empty well-ordered nest which is a subset of a non-empty set x has its union an element of x, -_ then x has a 5-maximal element.

Clearly M 2 -+ M 1 M 3 and M 2 -+ M 4 + M 3. There are many other maximal principles we could have chosen just by changing the ordering relation or the conditions on the upper bound. We chose M 1 since it is a common form and often used in practice and M 4 for historical reasons. We shall discuss other variations of M 1 - M 4 and also of M 5 and M 6 later. +

M 5 : If R is a partial ordering relation on x, then there exists a 5-maximal subset of x which is linearly ordered by R.

of

M 6: For every set x which is a nest.

x, there exists a 5-maximal subset

Next, we define a property of finite character. By a property we mean a formula in the language of NBG which has at least one free variable. DEFINITION 4.2: P(X) is said to be a property of finite character if P(B) and for each class X,

P (X)

-

For each finite x _c X, P (XI.

Equivalently, if we let Q = {x: x Then

P(XI

is finite

A

P(x) 1

is a property of finite character iff

PART I , S E T FORMS

34

P(X)

+

(Vx)(x _c X

A

x

is finite * x E Q)

.

For example, if P ( X ) is any one of the following then it is a property of finite character: 1.

X

is partially ordered by

2.

X

is linearly ordered by

3.

u f

4.

The elements of

R. R.

x. X

are pairwise disjoint.

On the other hand, the following are not properties of finite character:

between

5

can be linearly ordered. and 2 ) .

6.

X

is well-ordered by

7.

u E

8.

X

5.

X

(Note the difference

R.

x. is infinite.

is directed by R , where R is a directed relation on X iff R i s a partial ordering on X and for all u , v E X there is a w E X such that u R w and v R w. 9.

X

(Note that if each finite subset of X is directed by R then is directed by R, (in fact X is linearly ordered by R ) but if X is directed it does not follow that each finite subset of X is directed. On the other hand, if X is wellordered by R then each finite subset of X is well-ordered by R , but the converse does not necessarily hold.)

X

M 7: For every set x and every property P of finite character, there exists a 5-maximal subset of x which has the property P . Clearly M 7 M 5 M 6. In order to prove these seven forms are equivalent, it remains to be shown that M 3 -+ M 7 and M 6 * M 2 . It turns out that it is easier to prove the latter implication in two steps, M 6 * M 5 and M 5 -+ M 2. -+

-f

THEOREM 4 . 3 :

M 3

* M 7.

14.

MAXIMAL PRINCIPLES

35

PROOF: Let x be a set and P be a property of finite We shall show that character. Let y = it: t 5 x and P(t)). if n _c y, n # 8, and n is a nest then U n E y. Clearly U n 5 x. Let u be a finite subset of U n . Then since n is a nest, u is a finite subset of some element of n. Since every element of n has the property P, it follows that P(u). Therefore, we obtain P(Un) by the definition of finite character. Fusthermore, the empty set has property P. Thus y satisfies the hypothesis of M 3 , so y has a maximal element. M 6

THEOREM 4.4:

+

M 5.

PROOF: Let R be a partial ordering relation on x. Let be the set of all subsets of x which are linearly ordered by R. By M 6, there is a maximal subset n of y which is a nest. Let m = U n . We wish to show that m is a maximal element of y . Clearly m _c x. Let u, v E m, u # v; then (3w)(u, v E w and w E n). R linearly orders w, and hence either u R v or v R u. Consequently R linearly orders m. Suppose m is not maximal; then for some z E y, we have m c z . We observe that n U { z ) is a nest since every element of n is a subset of z . Since n is maximal, we must have { z ) _c n. That is, z E n, whence z 5 m, which contradicts our assumption about z. Consequently, m is maximal, and therefore M 5 holds. y

THEOREM 4.5:

M 5

-+

M 2.

PROOF: Let R be partial ordering relation on x. Let y be the set of all subsets of x which are well-ordered by R. Define a relation S on y as follows: S =

{(t,u):t

E y

A

u E y

A

t

is an initial R-segment of

ul.

is clearly a partial ordering relation on y; therefore, n of y which is linearly ordered by S. Let m = U n . We shall show that m has an R-upper bound, and that such an element is an R-maximal element of x. S

by M 5 there is a maximal subset

36

PART I, SET FORMS

First let us show that m E y. Clearly m 5 x. Let z be any non-empty subset of m. Then for some t € n, z n t # Pr. Let u be the R-first element of z n t. It remains to be shown that u is R-first in z. Now let v E z , v # u . If v E t, we have u R v and not v R u by the well-ordering property. If v t, v E w for some w E n. By the definition of S and the construction of n, t is an initial R-segment of w. Hence u R v and not v R u . Therefore, u is the unique R-first element of z so that R is a well-ordering on z and m E y. By an argument similar to that of the preceding theorem, m is maximal in y. Consequently, m cannot have a strict R-upper bound. For if b were a strict R-upper bound for m, we would have m' = m U {b} E y which contradicts the maximality of m. Since m E y, it follows from the hypothesis of M 2 that m has an R-upper bound. Let b be such a bound. Suppose z E x and b R z. If not z R b, z would be a strict R-upper bound for m, which is impossible. Therefore, b is an R-maximal element of x, q.e.d. There are some additional maximal principles which we shall discuss briefly. They are quite similar to the forms M 1 - M 7 and first appeared in the literature in Kuratowski [1922]. We shall call them MI1 M'7. The hypotheses for these forms are the same as that of the corresponding unprimed forms, but the conclusions are stronger. The conclusions for M I 1 - MI4 state that there exists a maximal element larger than any given element: the conclusions for MI5 - MI7 state that there exists a maximal subset containing any given subset with the required property. We shall state only two of these additional forms in a formal way: it will be clear what the other forms are.

-

M'l: If R is a partial ordering relation on x and if every subset of x which is linearly ordered by R has an R-upper bound and if y E x , then there is an R-maximal element z in x such that y R z .

MAXIMAL PRINCIPLES

94.

37

M'7: For every set x and every property P of finite character and for every subset y of x such that P(y), there is a 5-maximal subset z of x such that y 5 z and

P(Z).

It is clear that M'n M n for n = 1,...,7. prove the implications the other way. -+

THEOREM 4.6: PROOF: Let bounds for {y}. obtain M'n. THEOREM 4.7:

M n

+

M'n

for

Next we

n = 1,2,3,4.

y E x and let w be the set of all upper Then apply M n to the set w and thereby M 7

+

M'n

for

n

=

5,6,7.

The others are PROOF: We give the proof for M ' 5 . analogous. Let y be a subset of x which is linearly ordered by R. "Linearly ordered" is a property of finite character. Call it P. We define a new property Q as follows. For any subset w of x I Q(w1 if and only P(w Uy) . Q is clearly a property of finite character. By M 7 there is a maximal subset z of x which has the property Q. Therefore z U y is a maximal subset of x which contains y and is linearly ordered by R. We shall next l o o k at some of the ways the ordering relation R in M 1 and M 5 can be changed and still obtain a maximal principle which is equivalent to the axiom of choice. The results are due to Harper and Rubin [1976]. Consider the following statements: M l(Q,U): Every non-empty set which is Q-ordered by R in which each U-ordered subset has an R-upper bound, has an R-maximal element. A s possibilities for

Q

and

U

we shall take the

following: A:

arbitrary (any binary relation)

TR:

transitive

AS:

antisymmetric

PART I, SET FORMS

38

C:

connected

P: partial ordering (reflexive, antisymmetric and transitive)

L:

linear ordering (connected partial ordering)

W:

well-ordering

D: directed (upwards), where X is directed by R if R is a partial ordering on X and each finite subset of X has an R-upper bound. (It is sufficient that each two element subset has an R-upper bound.) RA: ramified, where X is ramified by R if R partial ordering on X and each initial R-segment of linearly ordered by R. F: forest, where partial ordering on X well-ordered by R.

T: tree, where forest under R and

X X

is a X is

X is a forest under R if R and every initial R-segment of is a tree under R if X has an R-first element.

is a X is

is a

Each forest is a union of pairwise disjoint trees. In what follows in this section, the maximal principles which hold for forests also hold for trees and conversely. We shall just state the results for forests. Thus, for example, M 1(P,L)

is M 1

and

M l(P,W)

is

M 2.

If Q1 and Q2 are properties of relations we write Q1 -t Q2 if every relation with the property Q1 also has the P because every linear property Q2. Thus, for example, L ordering is a partial ordering. -f

The following facts hold:

4.

MAXIMAL PRINCIPLES

39

(3)

W

\

L

bD AS

&

\

&

_3

TR

C TR

We shall show first that axiom of choice if (*)

Q

(4)

M l(TR,W)

(5)

M l(Q,U)

= TR,P,RA, or F -+

+

M

and

l(Q,U),

M l(Q,U)

is equivalent to the

U = C,AS & C, TR & C,L,D,

for all

M l(F,U), for all

Q

and

Q

and

(6) The statements M l(F,U), for all equivalent.

U

U

in

U

in

in

(*)

(*)

(*)

or

W.

above. above.

are

Thus, to prove that each pair of statements M l(Q,U) with and U in ( * ) are equivalent, it is sufficient to prove M l(F,C) -+ M 1(TR,W). (The proof is similar to one given by Felgner [ 19671 ) Q

.

THEOREM 4 . 8 :

M 1(F,C)

-+

M l(TR,W).

PROOF: Let R be a transitive relation on x and suppose each R-well-ordered subset of x has an R-upper bound. Let y = {u: u 5 x and R well-orders u}. Define a relation S on y such that for u, v E y , u S v iff u is an R-initial segment of v. Then ( y , S ) is a

PART I, SET FORMS

40

forest. Let z be a connected subset of y. Then U z is an R-upper bound of z in y. Thus, M l(F,C) implies that y has an R-maximal element, w. Since w is well-ordered by R, w has an R-upper bound. This R-upper bound is an R-maximal element of x. Now we have proved the following result: THEOREM 4.9: Each pair of statements in the following set are equivalent: {M

l(QrU):

Q

= TR,P,FA, or F and U = C,AS&C,TR&C,L,D or Wl.

The statement M l(D,W) is also equivalent to M l(TR,W). Clearly, M l(TR,W) -+ M l(D,W). To prove the converse we shall show that M 1 ( D , W ) + M l(F,W). Then it follows from 4 . 9 that M 1(D,W) * M l(TR,W).

THEOREM 4.10:

M l(D,W)

+

M l(F,W).

PROOF: Let x be a forest under a relation each well-ordered subset has an R-upper bound. Case 1.

Suppose

x

in which

R

can be well-ordered by a relation

S.

In this case we use S to construct a subset y 5 x which is well-ordered by R such that an R-upper bound for y is an R-maximal element of x. The construction is by transfinite recursion as follows: For each ordinal a, define

f (a) =

I

The S-first element v E x that ( V u E f"a)u R v, a,

- f"a such

if no such element exists, where

a

L

x.

Let a,, be the smallest a such that f(a) = a. Let y = f"cxo. Just as in 2.9 we show f(ao is a bijection. Moreover, it is clear from the construction that ( a o , 5 ) & (y,R), so y is well-ordered by R. Also, by construction, an R-upper bound of y is an R-maximal element of x. Case 2 .

x

cannot be well-ordered.

94.

MAXIMAL PRINCIPLES

41

Define y = If: ( 3 a E on) (3u

5 x)f:

a

M

u).

We define a relation S on y so that for f, g E y, f S g iff g(f) c g(g) v f = g. Since x cannot be well-ordered, S is directed on y. Moreover, if z _c y is well-ordered by S, then z has an S-upper bound constructed as follows: Let u = U g(f). Well-order u by T defined by: if a, b E u , fEz a # b, and fa and fb are the S -first elements of z for which a E fa and b E fb, a T b if fa S fb, and either E f,'(b). T well-orders u so there is fa # fb or f,'(a) a function g and an ordinal a so that g: c1 w u. The function g is then an S-upper bound of z . Thus, (y,S) satisfies the hypothesis of M 1(D,W) so y has an S-maximal element, f. It is easy to see that %(f) = x, contradicting the fact that x cannot be well-ordered, q.e.d. Thus, M 1(D,W) is equivalent to each of the statements given in 4.9. We shall show later, that each of these statements is equivalent to some form of the axiom of choice. (All of these statements also have equivalent primed forms. See p 36.) Next, we consider variations of M 5. Wallace [19441 and Gottschalk [1952] had forms similar to M 5 (A,AS) and M 5(A,C). Hoft & Howard [1973J have a form similar to M 5(C,RA). The remaining forms are due to Harper & Rubin [19761. M 5(Q,U): Every set which is Q-ordered by a 5-maximal subset which is U-ordered by R.

R

contains

For example, M 5(P,L) is M 5. THEOREM 4.11: are equivalent:

Every pair of the following statements

M 5(A,U), where

U = A S , C, A S K , TR&C, P I L, D, or RA.

M 5(AS,U), where

U

M 5(TR,U), where

U = AS, C, A S K , TR&C, P I L, D, or RA.

=

C, A S K , TR&C, L, D, or RA.

PART I , S E T FORMS

42

M 5 ( C , U ) , where

U = AS,

M 5 ( P , U ) , where

U = C , A S K , T R & C , L , D, or RA.

M 5 ( D , U ) , where

U = C , A S & C , TR&C, L , or RA.

M 5 ( R A , U ) , where M 5 ( F , U ) , where

A S K , P I L, D, or RA.

U = C , A S K , T R & C , L , or D.

U = C , A S K , TR&C, L , D, or W.

By using a proof similar to the proof of 4 . 3 , it is easy to show that M 3 implies each of the above 5 0 statements. PROOF:

The following facts are easily verified: (1) M 5 ( A S , C )

-

M 5(AS,AS&C), M 5(AS,TR)

M 5(AS,TR&C) (2)

M 5(TR,C)

M 5(C,AS)

M 5(C,D)

M 5(Q,AS&C)

M 5(Q,C)

if (5)

M 5(RA,D)

(6)

M 5(F,C)

-

M 5 (TR,L).

-

M 5(CIAS&C), M 5(C,TR)

M 5(C,P)

(4)

-

-

Q = P, D

-

M 5(AS,P),

-

M 5(TR,P)

M 5(AS,L).

M 5(TR,TR&C) , M 5(TR,AS)

M 5 (TR,AS&C)

(3)

-

-

or

-

M S(C,TR&C),

M 5(C,L)

M 5(Q,TR&C)

RA.

-

.+

M 5(C,RA).

M 5(Q,L),

M 5 ( R A , L ) , M S(D,RA) * M 5 ( D , L ) .

M 5(F,AS&C)

M 5(F,D)

.+M

-

M 5(F,TR&C)

-

M 5(F,L)

-f

M 5(Q,,U)

(7) M 5 ( A , U )

for all

+

U.

-

5(F,W).

A l s o , using the fact that if M 5(Q2,U)

,

Q1 + Q2 then we obtain the following implications:

M 5(QfU),

for all

Q

and

U.

14.

MAXIMAL PRINCIPLES

43

-

Also a relation R is connected on x iff 5 = (x X x) R is antisymmetric on x, and R U Irx is a linear ordering on x iff (x x x) * (R I) is a linear ordering on x. Using these facts it is easy to prove (9) and (10) below.

-

M S(AS,C)

(9)

-

M 5(C,AS), M S(AS,AS&C) M 5(C,L).

M 5(AS,L)

(10) M 5(A,AS)

-

-

M 5(C,AS&C),

M 5(A,C).

It follows from the preceding results that to complete the proof of the theorem it is sufficient to prove that (a) M 5(D,L) + M 5, (b) M 5(F,L) + M 2, and (c) M 5(TR,AS) AC 2. (We shall show later that the axiom of choice implies a maximal principle.) -f

(a) M 5(D,L)

by

-+

M 5.

PROOF: Suppose x R. Define y = iu

5 x:

u

is a set which is partially ordered

is linearly ordered by

RI.

Define a relation S on y so that if u, v E y, u S v iff Then S is a directed relation on y. (j3 is an S-upper bound for every subset of y.) Let z be a c-maximal S-linearly ordered subset of y. Then it is easy is a 5-maximal R-linearly ordered subset of to see that U z

v 5 u.

X.

PROOF: Suppose x is partially ordered by R-well-ordered subset has an R-upper bound. Let y = {u f x: u

is well-ordered by

R

and each

RI.

Define S on y so that for u, v E y, u S v iff u is an R-initial segment of v. Then y is a forest under S and the union of a 5-maximal S-linearly ordered subset of y is an R-well-ordered subset of x with an R-maximal element. (c) M 5(TR,AS)

+

AC 2.

PART I, SET FORMS

44

PROOF: Let x be a set of non-empty pairwise disjoint sets. Define a relation R on U x as follows: For all a, b E U x , where a E u E x and b E v E x, a R b iff u = v. Then R is transitive and a 5-maximal R-antisymmetric subset of U x is a choice set for x, q.e.d. All of these maximal principles have primed forms: Mt5(QIU): In every Q-ordered set, every U-ordered subset can be extended to a 5-maximal U-ordered set. For all of the above choices for U, except U = W, MI6 * M'5(Q,U) for all Q. Also, M'S(Q,U) + M 5(Q,U). Thus, Theorem 4.11 holds for the primed forms. In Theorem 4.12, below, Paul Howard sent us the proof of (a). In part (b), we prove that a primed form of M 5(Q,U) implies the axiom of choice, while it is unknown whether the corresponding unprimed form does. THEOREM 4.12: (a) M 5(A,TR) * AC 2. (b) M'5(A,TR

PROOF: empty sets.

+

Reflexive) * AC 2 .

(a) Let x be a set of pairwise disjoint nonDefine R on U x as follows:

a, b E Ux

-+

(a R b

-

(31E x)(a,b E u

A

a # b)).

M 5(A,TR) implies that x has a 5-maximal subset y that R is transitive on y. For each u E x, u y y, we would have a R singleton because if a,b E u b R a but not a R a. It follows from the maximality that u n y # 8.

such is a b and of y

(b) Let x be a set of pairwise disjoint non-empty sets. Suppose x has at least two elements. Define a relation R on U x such that if a,b E U x , where a E u E x, b E v E x then a R b iff u # v V a = b . Let u, v E x, u # v, and a E u, b E v. Then ia,b} is an Any 5-maximal R-transitive R-transitive subset of U x

.

54. subset of

Ux

MAXIMAL PRINCIPLES

which contains

{a,bl

45

is a choice set for

x.

A result due to Gottschalk [19521 can also be written in if we add to our definition of types of the form M 5 ( Q , U ) relations, UN, universal: a relation R is universal on x if x X x 5 R. Gottschalk's form was M 5(A,UN). It is clear that M 7 -+ M 5(A,UN) because the property of being universal is a property of finite character. THEOREM 4 . 1 3 :

M 5(A,UN)

-+

M 5(A,C).

PROOF: If R is any relation then R U R - l U I is connected. An R U R-l U I -maximal universal set is an R-maximal connected set. The form M 5(A,UN) can be generalized to n-ary relations. M 5 (A,UN(n): For any set there is a 5-maximal subset y

x and n-ary relation R 5 x such that yn c - R.

M 5(A,UN(n)) is equivalent to the axiom of choice if n 2 2 because M 7 M 5(A,UN(n)) -+ M 5(A,UN(n-1)) and since M 5(AfUN(2)) is M 5(A,UN), by 4.13, M 5(A,UN) M 5(A,C). -+

-+

It is also clear that the following two variations of M 5(A,UN(n)) are equivalent to the axiom of choice. (1) There is a natural number n

2

2

such that

M 5 (A,UN(n)).

(2) For every set x there is a natural number n 2 2 such that for each n-ary relation R there is a 5-maximal subset y 5 x such that yn _c R.

M 5(A,UN(n))

+

(1) -+ (2)

-+

M 5(A,UN).

(We shall say more about the relation complement in section 5 . )

UN

and its

We shall next show that these maximal principles are equivalent to the other forms of the axiom of choice we have given. Since there is so much in the literature about the equivalence of a maximal principle and the axiom of choice or the well-ordering theorem we shall give several of these proofs. The first two are well known.

46

PART I, SET FORMS THEOREM 4.14:

M 3

AC 1.

-+

PROOF: Let S be a collection of non-empty sets, and let C be the collection of all choice functions on subsets of S. It is clear that every nest of functions in C has its union in C, and therefore, by M 3 , C has a maximal element f. Clearly, f is the required choice function. A similar proof could be given that we shall prove THEOREM 4.15:

M 1

-+

M 3

-+

WO 1.

Instead

WO 1.

PROOF: Let x be an arbitrary set, and let P be the set of all ordered pairs (y,w) such that y 5 x, w 5 y x y and w is a well-ordering on y. Let us define a relation R as follows: If (y,w) and (y',~') are elements of P, then (y,w)R(y',w') if Y 5 Y', w = w' n (Y x Y), and y x (y' y) _c w'. That is, w' is an extension of w, w = w'ry, and y is an initial w'-segment of y'. If s is an R-linearly ordered subset of P, then an R-upper bound for s is ( U a ( s ) , U % ( s ) ) . (We leave the proof that ( U a ( s ) , U % ( s ) ) E P and is an R-upper bound for s , as an exercise.) Therefore, by M 1, P has an R-maximal element (z,v). Suppose m E x z. We define z ' = z U {m) and and we observe that (z,v)R(z',v') and v' = v U ( z ' x {rn)), not (z',v')R(z,v). This contradicts the maximality of (z,v), so z = x and hence v well-orders x, q.e.d.

-

We now proceed to the slightly more difficult task of proving that the axiom of choice implies a maximal principle. We shall first prove that the well-ordering theorem implies a maximal principle and then give several proofs that the axiom of choice implies a maximal principle. The proof of the next theorem was given in Hausdorff [1914]. THEOREM 4.16:

WO 2

-+

M 5.

The intuitive idea of the proof is as follows: Let x be a set and let R be a partial ordering relation on x. By WO 2, there is an ordinal number c1 such that c1 is equipollent to x. We construct a maximal R-linearly ordered

14.

MAXIMAL PRINCIPLES

47

subset of x by the following method: Let the first element of x belong. If the second element of x stands in the relation R to the first element, the second element is included, otherwise not. We continue by including those elements which are R-connected to the elements already included. Formally, let F be a 1-1 function from Define the function G as follows:

It is clear that the range of ordered subset of x.

G

a

onto

x.

is a maximal R-linearly

The proof of Theorem 4 . 1 7 is due to Kuratowski 119221. Similar proofs were given by Hausdorff [19271, Bourbaki [19501, Kneser [1950] and Szele 119501. THEOREM 4.17:

AC 3

-f

M'4.

Let us first prove the following lemma, which is independent of the axiom of choice: LEMMA: Let a E x and suppose that every non-empty wellordered nest of elements of x has its union in x. Furthermore, let f be a function on x to x, such that for all y E x, y _c f(y). Then there is a z E x such that a _c z and z = f(z). To prove this let f be a function on x to that for all y E x, y _c f(y). Define a function transfinite induction as follows:

x G

such by

G ( 0 ) = a,

G(a

+ 1)

G(a) =

=

U

f ( G ( a ) ), G(B),

if

a

is a limit ordinal.

B
We observe that if a < 8 , G ( a ) _c G ( B ) . Since every non-empty well-ordered nest of elements of x has its union in x, G is a function from On into x. Since x is a set, the function G cannot be 1-1. Hence for some ordinals

PART I, SET FORMS

48

Therefore, G ( a ) = G ( a + l), and thus f ( G ( a ) ) . Since G ( 0 ) = a, a _c G ( a ) and this completes the proof of the lemma. CY

<

6,

G(a) = G(6).

G(a) =

Now let us use this lemma to prove the theorem. Let x be a non-empty set with the property that every well-ordered non-empty nest of elements of x has its union in x, and let a E x. Define a function h on x as follows: For all y E x, h(y) = { z : z E x and y c z } . By AC 3, there exists a function f such that B(f) = a(h) and for all y E x, if h(y) # pI then f(y) E h(y). If h(y) = fl define f(y) = y. NOW, f satisfies the hypothesis of the lemma - f is a function from x to x such that for all y E x, y _c f(y). Therefore, there is a z E x such that a 5 z and f ( z ) = z . Such an element z is a maximal element of x. The proof of Theorem 4.18 is due to Frink [19521 and is similar to the proof given by Zermelo [1908]. THEOREM 4.18:

AC 3

+

MIS.

PROOF: Let x be an arbitrary set, R a partial ordering relation on x, and y an R-linearly ordered subset of x. Let L = { z : z _c x and R is a linear ordering on z and y _c z ) . Suppose M I 5 does not hold. Then there is no maximal element of L. By AC 3, there is a function f on L such that for each z € L, f(z) E x z and z U {f(z)l E L. (Formally we obtain this function f as follows: For each z E L define hl(z) = {w:w E L and z c w}. Because we are assuming that L has no maximal element, hl(z) # @. AC 3 implies there exists a function h2 such that a(h,) = a(h2) and h2(z) E hl(z) for all z E L. Now define z. Since h3(z) # pI for all z E L, again hj(z) = h2(z) by AC 3, there exists a function f such that a(f) = a(h3) = L and for all z E L, f(z) E h3(z).) Let us denote z U ( f ( z ) l by z', which we shall call the successor of z .

-

-

Let

K 5 L.

We shall call

K

(a) Y E K , (b) z ' E K

whenever

z E K,

complete if

§4.

(c) if

N

MAXIMAL PRINCIPLES

is a non-empty nest and

N _C K ,

49 then

UN E K.

Let % be the set of all complete sets. Since L E X, is non-empty. Since the intersection of any non-empty set of complete sets is complete, J = n;f is the smallest complete set. Let us now demonstrate that J is a nest. If z E J, z is normal if for every u E J, z _c u or u _C z . We shall prove that every element of J is normal, thereby proving that J is a nest. Let z be a normal element of J. (J has a normal element since y E J is normal.) Define K ( z ) to be the set of all u E J such that u _c z or z ' c u. Clearly y E K ( z ) since y 5 z . Suppose u E K ( z ) . If z ' _c u, then z ' _c u' and u' E K ( z ) . If u _C z and u' z then z _c u' since z is normal. But we have u _c z _c u', which implies u = z or u ' = z . The latter equality is impossible since we have assumed u' $ z . Consequently u = z , so z ' _c u ' and It i s a l s o clear that if N _c K ( z ) is a nest then u' E K ( z ) . UNEK(z), because if there is an n E N such that z ' _c n then z ' _C UN, otherwise UN_cz. Therefore, K ( z ) is a complete subset of J. Since J is the smallest complete set it follows that K ( z ) = J. Now we are able to show that the set of normal elements form a complete set. First, y is normal. Next, suppose z is normal, then J = K ( z ) . Therefore, for all u E J either u _c z, which implies u _c z ' , or z ' _c u. Consequently z ' is normal. Finally, suppose N is a nest of normal elements, (every subset of normal elements is a nest). Suppose u E J. If every element of N is a subset of u then U N _c u . If there is an n E N such that u _c n then u _c UN. Hence, U N is normal. Therefore, the set of normal elements of J forms a complete set which implies all elements of J are normal and thus J is a nest. is a nest and a complete class, by (c) U J E J, (UJ)' E J, which implies (UJ)' _c UJ, which is a contradiction, q.e.d. Since

J

so that by (b)

PART I, SET FORMS

50

The proof of the following theorem is due to Weston [1957]. AC 1

THEOREM 4.19:

+

M 1.

PROOF: Let x be an arbitrary non-empty set and R a partial ordering relation on x. For each R-linearly ordered subset y of x, let 9 be the set of all R-upper bounds of y. Let f be the choice function on the set of y in x all non-empty sets of the type 2. If y is an R-linearly ordered subset of x, y is said to be an f-chain if y # fl and if for all z c y with 2 n y # g, f ( 2 ) is an R-minimal element of 2 n y. (At least one f-chain exists since {f(i)} is an f-chain.)

-

The following properties hold for f-chains: (a) If y is an f-chain, z 5 yI and 2 n y = 0 then z = y. (In fact all linearly ordered sets satisfy this condition.) h

e

$ # Pr then y'

(b) If y is an f-chain and is also an f-chain. To see this observe that if either

z c y'

(i) z c y

and

inyffl

or

(ii) z 5 y

and

2

or

(iii) z $ y.

n

and

= y U {f(?)}

^z n y' # fl

then

y = fl

Since y is an f-chain (i) implies that f(z^) is an R-minimal element of 2 n y'. If (ii) holds then 2 n y' = {f (i)I , but (a) implies 9 = 2. Therefore, f(2) is an R-minimal element y and of 2 n y'. And (iii) is impossible because if z z c y', then f ( 9 ) E z which implies 2 n y' = 8. Since (i), (ii), and (iii) are the only possibilities it follows that y' is an f-chain. (c) For any two f-chains, one is an initial R-segment of the other. Let

y

and

z

be f-chains and suppose

u = 1s: s E z

n

y

and

s R t].

t E z

- y.

Let

MAXIMAL PRINCIPLES

94.

We shall show that

u

=

y.

Clearly

u _c y.

51

Let us assume

(i) u c y. It is also clear that have

u

5

Since

z.

t E z

and

t

u,

we

(ii) u c z . Also,

i n z, (iii) G n z t E

and therefore

+ 8.

Since z is an f-chain, it follows from (ii) and (iii) that f(6) is an R-minimal element of 6 n z . In particular, since tciinz, (iv) f(G) R t. NOW, if f(6) E y then since f(<) E z , by (iv) we would have f(6) E u, which is impossible. Therefore, (v) f ( G ) i y . If 6 n y # a, since y is an f-chain, (i) implies f (6) E 6 n y _c y, which contradicts (v). Therefore, (vi) G n y

=

la.

By (i) and (vi) it follows from (a) that

(vii) u = y. A

A

-

u, then it follows from (vii) that there NOW, suppose r E y is an s E u such that r R s (if not r E 6 but r & contradicting (vii)). Since s E u, s R t, so that the transitivity of R implies r R t. NOW, we have r E y and z. NOW, r R t, so that since r & u we must have r E y if we define a set

-

v

=

{s: s E y

n

z

and

s R r],

then by an argument analogous to the argument used to derive (vi), we obtain (viii)

0

n

But since r R t

z =

and

a. t E z, t E

n

z

which contradicts

PART I, SET FORMS

52

(viii). Therefore, u = y of z .

and

y

is an initial R-segment

Since the set of all f-chains is a nest, its union is linearly ordered by R. Let us next show that the union is an f-chain. Let c = Uly: y is an f-chain). Suppose y c c and $ n c # @. Then there is f-chain z such that ? n z # 0 and an f-chain w such that y 5 w. We wish to show first that y _c z . Since z and w are both f-chains it follows from (c) that either w _c z or z is an initial R-segment of w. In the former case, y c z . In the latter case, if y z then there exists an element s E y such that s E w 2 and s E 2. This implies that for every t E $, s R t, which implies $ z = @, which contradicts the choice of z . Hence, y c z . (y # z since ? n z # 8 . )

-

NOW, we see that there exists an f-chain z such that y c z and $ n z # @, therefore f(?) is an R-minimal element of ? n z _c ? n c. It remains to be shown that f ( 9 ) is an R-minimal element of ? n c. Suppose not,then there is a t E 9 (1 c such that t f? z and t R f (?). Also, there is an f-chain w such that t E w. Since z and w are both z , it follows from (c) that z is f-chains and since t E w an initial R-segment of w, so that for every s E z , s R t. This implies, in particular, f(?) R t. Consequently, since R is anti-symmetric, t = f($), which is a contradiction. Hence, f($) is an R-minimal element of n c and c is an f-chain. Moreover, G = @ because, if 2 # 8, then we could which is also an f-chain by (b), construct c' = c U {f(;)} and c c c' which contradicts the choice of c.

-

Since c is linearly ordered by M 1 implies that c has an R-maximal is an R-maximal element of x, because n # m such that m R n then n E G , q.e.d

R the hypothesis of element m. Clearly m if there exists an which is a contradiction,

Of the forms we have already given of the maximal principle, only two do not involve a relation or a property and state that there is a maximal element or subset with

54.

MAXIMAL PRINCIPLES

53

respect to a natural relation among sets (M 3 and M 6). We shall concern ourselves only with the maximal subset form M 6. The question can then be raised as to whether the corresponding forms with other natural relations among sets are of equal strength. Let x and y be two arbitrary sets. These sets y, are disjoint unions of the appropriate two of x n y, x y x. (x = (x y ) u (x n y) and y = (y x) U (x n y)). The immediate symmetric set-theoretical relations derivable from these are whether x n y is the empty set, and how many of x y and y x are the empty set. If x and y are distinct and non-empty, it can readily be seen that exactly one of the relations x n y = 9, one of x y and y N x is empty (x 5 y or y 5 x), and none of the three sets are empty (x y and y $ x and x n y # 9 ) , must hold. Accordingly we introduce the following notation:

-

-

-

-

-

-

-

DEFINITION 4.20:

Let

and

X

be any two classes.

Y

Then (i) X D Y

if

x n Y = 8

X E Y

if

X

(iii) X K Y

if

X _ c Y V Y z X

x ii Y

if

x

CY A Y G X

(v) x J Y

if

x

g

if Y

X C Y V Y C X V X ~ Y = @ V X U Y = V .

(ii)

(iv)

(vi) X

~

~

Y

Y

#

~

~

Y

I

g

x

~

x

n

!

f

#

~

~

x

U

Y

Note that the properties b, Z, and 5 are the complements of the properties D , K, and J respectively, and as we have previously observed, exactly one of D , K, and J must hold if the classes are non-empty. Therefore, = K or J , 3 = D or K , and = D or J. We shall also use the following consistent ambiguity: We shall say that a class has a given one of the above properties if every distinct pair of its elements has the property. If a class X has the property D , we shall write DIX] and similarly for the others. Let us now introduce the following notation:

#

V

PART I, SET FORMS

54

M 8(A): property A .

Every set has a 5-maximal subset with the

G. Kurepa [1952] has shown that M 8(A)

is equivalent to the axiom of choice if A is 5 , J, or 5 . R. Vaught [1952] proved that M 8 ( D ) is equivalent to the axiom of choice. M 8(K) is M 6. J. D. Halpern [19611 in his Ph.D. thesis constructed a Fraenkel-Mostowski model in which M 8 ( K ) is true but the axiom of choice is false. (In section 5 we shall AC. The result is due to Felgner and show NBG F M 8(K) Jech [1973].) On the other hand Kurepa I19521 has shown that the following statement is equivalent to the axiom of choice:

.

M 9 (K): Every set can be linearly ordered, and every set has a 5-maximal subset with the property K. Clearly M 7 implies each of the forms M 8 (A), A = D, D, K, J, 5, and M 9 ( K ) , since the properties considered are all properties of finite character. (While "can be linearly ordered" is not a property of finite character, M 7 implies every set can be well-ordered which, in turn, implies every set can be linearly ordered.) THEOREM 4.21: s E

M 8(;)

+

M 8(D).

PROOF: Let x be an arbitrary set. x we define a set Ns as follows:

Ns = {u: u

= Is)

For each element

or (3t)[t E x , u = {s,t), and s D t]).

Now if s # t, then NsD Nt if and only if s D t. M 6 ( B ) implies that {Ns: s E x) has a maximal subset w which has the property 5. We also have that Ns = Nt if and only if s = t; therefore, {s: Ns E w) is well-defined and is clearly a maximal subset of x which has the property D. (The proof shows that is a canonical symmetric relation. The same is true for D.) THEOREM 4.22:

M 8 (J)

-+

M 8 (6).

PROOF: Let x be an arbitrary set. For each s E x, define su = s U ~(s,u)l, where u is a set such that (Vtj( V s ) [ ( s E x and t E x) -t (s,u) f tl. Therefore, if s # t

54.

MAXIMAL PRINCIPLES

55

--

sUK tU, which implies suJ tU suE tU. If s # t, so that sufi tU s E t. M 8 ( J ) implies that sU n tU = s n t, Isu: s E x} has a maximal subset which has the property D.

then

THEOREM 4.23:

M 8(D)

-t

AC 2 .

PROOF: Let s be a collection of non-empty disjoint sets. The elements Let x = ~~(O,v),(l,w)~: w E x and v E w). {(O,v),(l,w)~ and ~(O,v'),(l,w')~ of x are disjoint if and only if w # w'. By M 8 ( D ) , there exists a maximal subset y of x which has the property D. Then the choice set for the collection s can be taken to be {v:(3w){(O,v),(l,w)~ E y), q.e.d. THEOREM 4.24:

M 8(5) * M 8(K).

Let x be an arbitrary set and let u Ux. For E x, define su = s U { u } . Then for all s , t E x, su tU # g , SO that suz tU sU K tU. A l s o Su K tU s K t since u Ux. Therefore, if Isu: s E x) has a maximal subset which has the property 5 then x has a maximal subset which has the property K , q.e.d. PROOF:

each

-

s

THEOREM 4 . 2 5 : Let

M 9(K)

-t

AC 2.

PROOF: Let s be a family of non-empty disjoint sets. If x E u E s , define R be a linear ordering of U s . xR = { z : z E u

Since all

-

R

and

is antisymmetric, if x # y

x, y E Us.

z R

XI.

then

xR # yR

for

Also

(a) xRE yR if and only if x and y belong to different elements of s , because if x and y belong to different elements of s then xR n yR = p which implies xR K yR. Conversely, suppose x and y belong to the same element of 8 . Since R is connected on U s , either x R y or y R x. But x R y implies x E yR which implies xR _C yR, and y R x implies y E xR which implies yR _c xR. Therefore, if x and y belong to the same element of s then xR K yR.

PART I, SET FORMS

56

Hence, it follows from (a) that a maximal subset of yields a choice set {xR:x E U s 1 which has the property for s , q.e.d. We have now shown that the following implications hold: M 7

+

M 8 ( J ) * M 8(D)

M 7 * M 8(?)

+

+

M 8(D)

+

M 8 ( K ) = M 6, M 7

AC 2, -+

M 9(E) * AC 2.

We have shown previously that M 6 * M 7

AC 2 * M 7.

and

It has been shown by C. C. Chang 119603 and Azriel Levy M 8(D) can be generalized by extending the property D to Dml and, in an analogous way, we were able to generalize M 8(D). (Chang actually obtained a stronger form, M'8(Dm) but Levy by using AC 13(m), was able to weaken it by eliminating the prime. We give here the weaker form, M 8 (D,) ) [1962] that

.

DEFINITION 4.26: If m is a natural number, m 2 2, a class X is said to have the property Dm (6,) if every subset of X which has m distinct elements has an empty (non-empty) intersection. (D2 = D and E2 = 6.) Now we claim that M 8(D,) and M 8 ( E ) are equivalent mto the axiom of choice. Clearly M 7 -+ M 8(Dm) and the proof is similar to the proof that that M 8(6,) + M 8(Dm) M 8 ( E ) * M 8(D), 4.21. THEOREM 4.27:

M 8(Dm) * AC 13(m

- 1).

PROOF: Let x be a set of non-empty sets. We can assume without l o s s of generality that the elements of x are disjoint, since otherwise we can replace each element y of

x by

{(y,t):

t E Y}.

Let u = {{{O,t},(l,y)~:

t E Y

Let v be a maximal subset of For each y E x, define

u

f(y) = It: t(O,t),(l Then

f(y) _C y

since v 5 u;

f(y

and

Y

E

XI.

which has the property y)l

<m

Dm.

E vl.

-

1 since v

has the

54.

MAXIMAL PRINCIPLES

property Dm; and f(y) # j3 q.e.d. (See 4.23).

since y #

57

and

v

is maximal,

Just as in the case of Levy's forms of the well-ordering theorem and the axiom of choice (WO 4 ( m ) and AC 7(m)) we are able to obtain weaker forms of M 8(Dm). M 10:

There exists a natural number

m

2

2

such that

M 8(Dm).

M 11: For every set x there exists a natural number m 2 2 such that x has a 5-maximal subset which has the property Dm' M 11

Clearly, we have M 8(Dm) * M 10 * M 11. Also, -+ AC 15; the proof is analogous to the proof of 4.27.

There is some difficulty involved in trying to obtain similar results for Dm. Because of the definition of Dm, (definition 4.2 6) the statement: For every set x, there exists a natural number m such that x has a 5-maximal subset which has the property S:

holds independently of the axiom of choice. x be an arbitrary set.

2,

2

m'

To see this let

Case 1. Every finite subset of x has a non-empty intersection. Then x itself is the required maximal set for every rn 2 2 . Case 2. There exists a finite subset y of x with n distinct elements which has an empty intersection. Then y is a maximal subset of x which has the property fin+l, so that m = n + 1. To eliminate this rather trivial case we modify the definition of Dm slightly as follows: DEFINITION 4.28: If m is a natural number, m 2 2 , a class X is said to have the property D!m if every subset of X which does not have more than m distinct elements has a non-empty intersection.

58

PART I, SET FORMS

Therefore, if a class has the property E!, every subset of m elements has a non-empty intersection, every subset of m - 1 elements has a non-empty intersection, etc, so that if Em is replaced by D!m in S , the proof of case 2 is not valid. In fact, we are able to prove that the following propositions are equivalent to the axiom of choice. M 12: M 8(E!m).

There exists a natural number m

2

2

such that

(It is rather easy to show that M 12 with E!m replaced by Dm is also equivalent to the axiom of choice. The proof that it implies M 10 is analogous to the proof of 4 . 2 0 . ) M 13: For every set x, there exists a natural number 2 such that x has a 5-maximal subset which has the property Elrn.

m

2

It is clear that M 7 M 12 -+ M 13. We shall prove M 13 + M 11. The proof is a generalization of the proof of 4.21. -+

THEOREM 4.29:

M 13

-+

M 11.

PROOF: Let us first observe that if every element of x is disjoint from every element of y, then a subset w of x U y has the property Dm if and only if w n x and w rl y both have the property Dm. Consequently, to prove that x has a maximal subset with the property Dm it is sufficient to prove the fact for u = (Iwl where

w

x

x)

u

E(a,a): a E w )

I

is the set of natural numbers.

In what follows the range of the variables m, n, etc., is the set of natural numbers greater than 1. The set

u

satisfies the following property:

ii) There is a 1-1 function 4 from that for each s E u, there is an a E w s

n

$(a) =

0.

into u such that

w

such

54.

MAXIMAL PRINCIPLES

59

Note that (i) enables us to constructively enlarge any subset v of u with less than n elements to an n-element subset Q(n,v) with nJl(n,v) = B . Define (ii) zn and for each

It: t c u, t

=

s E u,

and

k

n,

nt

and

=

f3 l ,

n,

(iii) fn(s) = Ih:h E

x

m

zm

and

s E h(n)l.

Let (iv) y =

u

n We observe that

f;lu

.

(v) If v _c y then n v # B if and only if for each (a) v fl u:f < n or (b) v n f"u n = n and nf;l"v = 8.

n,

m

(v) is clearly true if h E

n

v

-

v = 9.

(Vw)[w E v

+

Otherwise, if

h E

x

m= 2

zm

h E wl

(Vn)( V s ) [ s E fil"v

-f

h E fn(s)l

.+(Vn) Is fil"v * s E h(n) 1 .+(Vn)[fil"v _c h(n) 1 . (Vs)

E

Now fil"v w v n f;u since fn is 1-1, and if n # m then f"u By (ii) and (iii) h(n) k n so that fn1"v < n. n n fgu = f3. 1" If fn v n then f-ll'v = h(n) and consequently has an n empty intersection. Conversely, if for each n (a) or (b) is satisfied, we can define h by

-

Then

h E nv.

PART I, SET FORMS

60

let v be a maximal subset of y with the property - NOW, D!,. This is equivalent to v n f"u being a maximal subset n of u:f with the property D!m for every n. Also, it has at least m elements. follows from (i), that v n f;u But if w _c f"u has at least m elements, w has the m property D!m if and only if fil"w has the property Dm. -1 Thus, fm v is a maximal subset of u with the property Dm. The theorem follows easily. 11

The concept

6!,

can be extended further.

DEFINITION 4.30: A class X has the finite intersection property (f.i.p.) if each finite intersection of its elements is non-empty. M 14: f.i.p.

Every set contains a 5-maximal subset with the

We shall show M 14 is equivalent to the axiom of choice. (Klimovsky [1956] also has a proof that M 14 implies the axiom of choice). Since the f.i.p. property is a property of finite character, M 7 M 14. -f

THEOREM 4 . 3 1 :

M 14

-f

M 4.

Suppose x is a set in which each 5 well-ordered nest has a 5-upper bound. Let PROOF:

y = { u _c x: and for each

u E y,

u

is a 5-well-ordered nest

let

A~ = {v E y: v

is an initial 5-segment of

ul.

Then AU n Av # iff either u is an initial 5-segment of v or v is an initial 5-segment of u. Let w = {AU:u E yj. M 14 implies that w has a 5-maximal subset t with the f.i.p. Then U t is a 5-well-ordered nest in x. A 5-upper bound of U t is a 5-maximal element of x, q.e.d. In most cases for statements equivalent to the axiom o f choice in NBGO, if the word "set" is changed to "class", the resulting statement is stronger. M 14 is a notable exception.

94.

MAXIMAL PRINCIPLES

61

The following statement, which we shall call M 14S, is also equivalent to the axiom of choice in NBGO. M 145: the f.i.p.

Every class contains a 5-maximal subclass with

Clearly M 14s

-+

M 14.

THEOREM 4.32:

M 14

-+

M 145.

PROOF: Let X be a class of non-empty sets. (There is no loss of generality in assuming 0 B x.) Let x E x and let Y = { x n y: y E X I . Since Y _c y ( x ) , Y c-maximal subset W

is a set. M 14 implies that Y has a with the f.i.p. property. Then let

We claim that Z is a 5-maximal subclass of X with the f.i.p. It is clear that Z has the f.i.p. because if u is a finite subset of Z then for all a E u, a n x E W. Since W has the f.i.p.,

0 #

f? ( a n x ) = x n n a - + n u = n a # 0 . aEu a ~u a Eu

Suppose Z is not 5-maximal. Then there is a z E X such that Z U { z ) has the f.i.p. Since every element of W is a non-empty subset of x and W is a 5-maximal subset of Y with the f.i.p., x E W. Consequently, x E Z. Therefore, if u is any finite subset of Z then x (lz (nu) # 0. This implies that x n z E W, which in turn implies that z E Z, q.e.d. M 14 also has a primed form which is equivalent to the axiom of choice. If y s x and y has the f.i.p. then there is a c-maximal set z with the f.i.p. such that y 5 z 5 x. M'14:

Next, we turn to some more generalizations of the form M 8 ( A ) and thereby answer some of the questions asked in

PART I, SET FORMS

62

Kurepa [ 1 9 5 2 1 . In order to simplify the notation we define a property K* as follows: DEFINITION 4 . 3 3 : A class X is said to have the property K* if X can be linearly ordered and X has the property K. We shall show that M 8(D V V J V 5 V K V K*) is equivalent to the axiom of choice. (The proof is due to A. Rubin and H. Rubin.) Thus it follows that if A is a disjunction of a subset of the six properties D, 5, J, 5 , K, or K* then M 8 ( A ) is equivalent to the axiom of choice because M 7 M 8(A) M 8(D V V J V 5 V K v K*) -+

.

-+

(While K* is not a property of finite character, K* is the conjunction of and X can be linearly ordered. i? a property of finite character and M 7 -+ WO 1 -+ X can be linearly ordered,) D

Let B stand for the disjunction We shall prove: THEOREM 4 . 3 4 :

M 8(B)

-f

V

5

V

J V

f

V

K

is

V

K*.

AC 2 .

PROOF: Let S be an infinite set of pairwise disjoint non-empty sets, no set of which is an element of another or of itself. (There is no loss of generality because, for example, a we could replace each element x E S by x x {a] where is an infinite ordinal such that c1 f U S . ) We define the following sets: C

is the set of all choice functions on subsets of

T = { u,x}: u E x E s } P

is the set of all linear orderings on

M = {{p}

X

xu: p E P

A U

E

x E S

A Xu

t v E x: v p u } l lJ I { q l

where

q

If

x

=

s),

is a fixed element such that r E T and

f E C,

US.

q flP.

S.

54.

If

f E C

H(f,m)

and

=

Y = {N(r,f)

where

a # 0

g E C

any

MAXIMAL PRINCIPLES

m E MI

{(f',m'): x

f' c f

V

(f' = f

A

m ' _c m)]

(a U H(g,m)): r E T I f, g E C, m E MI,

is a fixed element such that

and

63

m E M.

An element of

Y

a @ UH(g,m)

for

will be designated

by

Yrfgm' The idea of the proof is to show that a 5-maximal subset of Y with one of the six properties D, 5, J, 5 , K or K* induces either (a) a 5-maximal nest in C (The union of such or (b) a 5-maximal subset a nest is a choice function on S . ) of T whose elements are pairwise disjoint (This set also induces a choice function on S.) or (c) a 5-maximal subset {{q) x S I with the property z. (If of M

-

-

xu, {p') X xAI E M X S), {p} x xu 5 {p') X xA1 iff p = p', x = x' and u p u'. Thus a 5-maximal subset has the property that if { p ) x xu, N 5 M satisfying {p) X y, E N then x # y. Therefore, {u:(3xE S){pj X x u € N] is a choice set for S.)

{PI

X

We make the following observations: (1) If Y can be linearly ordered, this induces a linear ordering on U S. (2)

Yrlflglml

I

N(rllfl) n N(r2,f2) =

Yr2f2g2m2 E Y

a.

This is true because

are disjoint iff

PART I, SET FORMS

64 A direct product B

# Q

because (3)

K

A

x

B = Q

iff

In our case

a # 8.

Yr f g m

1 1 1 1

Yr2f2g2m2 iff

r1

Case 1: Y has a 5-maximal subset or with the property J.

It follows from ( 3 ) that if is a 5-maximal nest contained in J

A = fl V B = Q.

=

r2, fl = f2

Z

yrfgm E 2, C.

Case 2: Y has a 5-maximal subset or with the property 5 .

Z

and

with the property {g E C:yrfgm E z ) with the property

If rl # r2, N(rllfl) n N(r2,f2) # B iff rl n r2 = Q. Thus, it follows from (2) and ( 3 ) that if f, g E C and m E M, {r E T:Yrfgm E Z ] is a 5-maximal subset of T whose elements are pairwise disjoint. Case 3 .

Y

has a 5-maximal subset

2

with the propertyD.

If rl = r2, N(rllfl) n N(r2,f2) = iff fl K f2 and fl # f2. Let r E T so that there are f, g E C and m E M with yrfgm E 2. (It follows from ( 2 ) that such an r E T exists.) Then, {f E C:yrfgm E Z ) is a 5-maximal nest in C. Case 4:

Y

has a 5-maximal subset

Z

with the property

K*.

By (11, P # Q. Therefore, there exists a p E P and u E x E S such that {pj x xu E M I and for all p E P I u E x E S, ({PI x xu)E({q) x S ) . Let g E C be such that there exist r E TI f E C and m E M with yrfgm E 2. (It follows from ( 3 ) that such a g E C exists.) Then {m E M {{q} x S 3 : yrgfm E Z ) is a 5-maximal subset of M {{q] x S ) with the property E, q.e.d.

-

MAXIMAL PRINCIPLES

54.

65

All the M 8(A) forms have primed forms which are also equivalent to the axiom of choice. The following propositions are put in here since they deal with the properties D, E, J, j , K and E l and they were suggested by some problems proposed by Kurepa [19521. They are not the same type of maximal principles which were considered previously. Let us introduce the following schemata.

M 15(A:B): the property A property B.

Any set which contains a 5-maximal subset with contains a _c-maximal subset with the

M 16(A:B): Every set can be linearly ordered and any set which contains a 5-maximal subset with the property A contains a 5-maximal subset with the property B.

Under the following conditions M 15(A:B) is equivalent to the axiom of choice: A # B, A = D, 6 , J, 5, K;

or 2, and B

=

D,

J, j , or K.

It follows from J. D. Halpern's Ph.D. thesis 119611 that M 15(A,i?) is not equivalent to the axiom of choice in NBGO if A = D, D, J, 5 , or K (See remarks following M E(A)). However, M 16(A:B) is equivalent to the axiom of choice for all A and B, A # B, such that A = D, E, J, 5, K, or K and B = D, 5, J, 5 , K, or E. Clearly, M 16(A:B) M 9(E) M 16 (A:i?).

-f

M 15(A:B);

M 8(.B) * M 15(A:B);

and

-f

and

THEOREM 4.35: B = K , 5 , or

M 15(A:B)

.+

M 8(B),

for

A = K, J , or

-

D

D.

PROOF: Let x be an arbitrary set and let a be a nonempty set such that for all u E x, a n u = Let y = x U {a]. Then { a } is a 5-maximal subset of y with the property K, J, and D. If M is a 5-maximal subset of y with the property K , 5 or D then M {a} is a 5-maximal subset of x with the same property.

a.

-

COROLLARY 4.36:

M 15(5:i?)

-f

M 8(i?).

PART I , SET FORMS

66

PROOF: 4.35.

Let b

Let

z

E y}.

y

be d e f i n e d a s i n t h e proof of

i?

-DCOROLLARY .

4.37:

5.

L

b

-

A 5-maximal

s u b s e t of

M 16(A:K)

-+

Uy.

Let

i s a 2-maximal s u b s e t

{a U { b ] )

Then

with t h e property

i n d u c e s a 5-maximal

or

and

be an e l e m e n t s u c h t h a t

z = {u U {b): u

of

x, a,

z

s u b s e t of

K

x.

for

M 9(K),

5,

A = K, J,

THEOREM 4 . 3 8 :

(a) B = K

M 15(A:B)

-

(b)

M 15(D:j)

+

M 8(K).

(c)

M 15(6:J)

+

M 8(i?).

PROOF: . L e t

a @ Ux. of

y

for

M 8(B),

-+

E,

or

A = D

and

K.

or

be a s e t and

x

y = {u U {a]: u E

Let

with t h e p ro p e rt y

a 5-maximal s u b s e t of

Iu: u U { a ) E M I

y

D

a

be an e l e m e n t such t h a t

XI.

Then, 5-maximal s u b s e t s

a r e one e l e m e n t s e t s .

with t h e property

i s a _c-maximal s u b s e t of

K

x

If

is

M

or

then

w i t h t h e same

property. The proof i s s i m i l a r i f a 5-maximal s u b s e t of

y

i s r e p l a c e d by

"D"

because

is

y

itself.

and

K

are

with t h e property

5

"F"

This proves ( a ) .

P a r t ( b ) follows because

i d e n t i c a l on

To p r o v e ( c ) w e n o t e t h a t a 5-maximal s u b s e t

of

y

y.

with t h e p ro p e rt y

K.

with t h e property

J

i n d u c e s a 5-maxima1 s u b s e t of

COROLLARY 4.39:

M 16(A:c)

+

M 9(i?),

(b) M 1 6 ( 5 : J )

+

M 9(i?).

(a)

THEOREM 4 . 4 0 :

PROOF:

Let

z = y U {a).

properties

Then

i?

M 15(A:K)

x, a, {a)

and

J.

5-maximal c h a i n i n

x.

and

for

-+

y

A = D

M 8(K),

or

for

x

-

D.

A =

or

be d e f i n e d a s i n 4.38.

J.

Let

i s a 5-maximal s u b s e t of z w i t h t h e A 5-maximal c h a i n i n z induces a

MAXIMAL PRINCIPLES

54.

67

THEOREM 4.41: B = J

(a) M 1 5 c A : B ) or D.

-f

M 8(D),

(b) M 1 5 ( E : B ) * M 8 ( D ) , PROOF: a such Let ua ua n va Define v E x,

Let

for A = D,

5, K,

for

or

B = D

or

k,

and

5.

x

be an arbitrary set and choose an element that for a l l u , v E x, if u # v then ( u , a ) f v. = u U C(u,a)}. Then, if u , v E x and u # v, = u n v. Let y = x x {a}: then ua n y = t(u,a)}. z = {u I and w = z U {y}. Then, for all u, a’ u E X if u # v, ua K va. Therefore, uaEy

and

u ~ J v ~ . + u ~ ~ v ~ . + u ~ v ,

ua J y

and

ua

The set

{y]

5 va

-

ua D va

.+

u D v

is a 5-maximal subset of

J.

w

. with the

property D and If M is a 5-maximal subset of w with the property 5 then {u E x: ua E M {y}} is a 5-maximal subset of x with the property D. Moreover, J and D are identical on w. Thus, we obtain M 15 ( A : B ) * M 8 ( 5 ) , for A = D o r 5 and B = 6 or J.

-

-

To obtain the result when A = K or K, we note that c-maximal subsets of z with the property K are one element sets and a 5-maximal subset of z with the property K is z itself. If M is a 5-maximal subset of z with the property 6 or J then {u E x: ua E M} is a 5-maximal subset of x with the property D. This completes the proof of (a) and the proof of (b) is similar. THEOREM 4 . 4 2 : B = i5 or K.

M 1 5 ( A : B ) * WO 1,

for

A =

J

or

-

J,

and

PROOF: Let x be an arbitrary set. Let 2 be the set of all well-orderings on subsets of x. For each R E 2 define TR and let

= {S

E

2:

S = R V S

= {TR: R

E $,}.

is an extension of The set

R}

8 has the property that

PART I, S E T FORMS

68

each pair of its elements either has the relation K or D. Therefore, 5-maximal sets in & with the property J are one element sets, and & itself is a 5-maximal set with the property 3. M 1 5 ( A : E ) , for A = J or 5 , implies that there is a 5-maximal subset M of 5 which has the property D. Since 5 and K are identical on &, this implies that M is a 5-maximal nest in 5 . Let

u

s =

R.

TREM

Since S is a union of comparable well-orderings of subsets of x, it is a well-ordering of a subset of x. Suppose u E x .$(s). Let

-

s*

=

s u (a(s, x

{ul)

u

t(U,U)I,

and let M* = M U I S * ] . Since S is an extension of R for each R E M and S * is an extension of S, M* is a nest and M c M*. This contradicts the maximality of M. Thus, a ( S ) = x and S well-orders x. THEOREM 4 . 4 3 : (a)

M 15(?:D)

-+

M 8(D).

(b) M 1 5 ( 5 : J ) * M 8 ( J ) . PROOF: Let x be a set, let and define y as follows:

In

be the set of integers

For notational convenience, let ui = u x {i) x {1,2} and let v . = (Ux) x { j E In: j 5 i} x {l}. Each of the following is easy to verify for all i , j E In: ( 1 ) Ui A u i

5 v 3. ’

(2)

vi

(3)

ui J v

(4)

ui D v .

1’ 3’

.u

A

u’, for

for

i

5

j,

for

i

5;

j,

for

i > j,

A

=

D

or

J,

(5)

ui D u!

3'

54.

MAXIMAL P R I N C I P L E S

for

i # j.

69

The set {vi:i E In) is both a 5-maximal subset of y with the property J (by (2) and (4)) and also a 5-maximal subset of y with the property D (by (2) and ( 3 ) ) . Therefore, the hypotheses of both M 15(3:D) and M 1 5 ( D : J ) are satisfied. M 1 5 ( 5 : D ) implies that y has a 5-maximal subset z1 with the property D. It follows from (2) that at most one vi can be an element of zl. If no vi is an element of z1 then {u:ui E zl} is a 5-maximal subset of x with the property D for each i E In (by (1)). I f vi E z, then for each j > i, { u : u j E z,} is a 5-maximal subset of x with the property D (by (1) and ( 3 ) ) . This proves part (a).

To that y follows vi E z2, then i

prove part (b), we use the fact that M 1 5 ( 5 : J ) implies has a 5-maximal subset z 2 with the property J. It from ( 2 ) , ( 3 ) and ( 5 ) that there is exactly one and it also follows from (5) that if ui, u!J 22 = j . Thus, there is a j E In, j s i such that if then k = j . Then by (1) { u : u . E 2 , ) is a Uk '2 3 c-maximal subset of x with the property J, q.e.d. This completes the discussion of M 1 5 ( A : B ) and M 16(A:B) where A # B and A , B = D, D l J, 5, K, or We shall show in section 5 that in NBG M 8 ( K ) AC.

E.

-f

The next form to be considered in this section is an obvious extension of the idea of constructing maximal subsets with respect to "natural" relations. The relation to be considered is the €-relation. M 17: Every set x contains a 5-maximal subset y such that for every s , t E y, s # t, either s E t or t E s.

It is clear that M 7 + M 17 since M 17 deals with a property of finite character. We shall prove that M 17 + WO 1. THEOREM 4 . 4 4 :

M 17

-f

WO 1.

PROOF: Let x be an arbitrary set and let {W: (3y)[ y _c x and W is a well-ordering of define a relation R on @ as follows:

&

=

y] 1 .

We

70

PART I, SET FORMS If

W, W' E 2)

then

is an extension of W. For all function f on %r as follows:

W'

(i) f(W) = f(U) (ii) f(W) E f(U)

We claim:

W E

we define a

.+W

-

= U and W R U.

It is clear that if W = U then f(W) = f(U). converse it is sufficient to prove that (iii) for all

W ' E %r

{{WI}

.

n {f(W'):W' R W and W'

We leave the proof of

To prove the

E

%r)

=

fl

(iii) as an exercise.

To prove (ii) we note that if W R U then clearly f(W) E f(U). Suppose f(W) E f(U); then either f(W) = {U) or W R U, but the first alternative is impossible. M 17 implies that % ( f ) contains a maximal subset Q such that for every S, T E Q, if S # T then either S E T or T f S. Let 2 = {U:f(U) E Q ) . Since f is a 1-1 mapping of U onto $ ( f ) preserving order (by (i) and (ii)) it follows that ! ,? is a maximal subset of @ such that for all U, W E 21, either U R W or W R U.

We claim that U z is a well-ordering on x. Since Ug is a union of R-comparable well-orderings of subsets of x, u i l is clearly a well-ordering on a subset of x. Suppose there exists an s E x such that s p a ( U 2 ) . Then for all

u

E

il,

s fi

i(u). Let

v

=

u 2L u ( B ( U 2 1 )

x

CS})

u

t(S,S)I,

A*

= 21 U {V}. Since for each U E Z*, U R V, and the maximality of 2 is contradicted. Hence, Ug is a well-ordering on x .

a n d let

:d c d * ,

M 18, the next maximal principle in this section, is put in primarily as a curiosity. Blair and Tomber [1960] discuss

71

MAXIMAL PRINCIPLES

94.

two forms of the axiom of choice for finite sets. It was pointed out to them by H. Rubin that if a slight change was made in the wording, one would obtain forms of the general axiom of choice. M 18 is a combination of these two forms. M 18: Suppose x is partially ordered by t E x, let + t = { s E x : t R S A t # s ~ .

Suppose every R-linearly ordered subset of bound. Then there is a t E x such that element.

x

R.

For each

has an R-upper has no R-first

We shall show M 18 is equivalent to M 1. THEOREM 4.45:

M 1

-+

M 18.

PROOF: If x satisfies the hypothesis of M 18. Then it also satisfies the hypothesis of M 1. Consequently, x has -+ -+ an R-maximal element, t. Since t = 8 , t has no R-minimal element. THEOREM 4 . 4 6 :

M 18

+

M 1.

PROOF: Suppose M 1 is false. Then there is a set x a partial ordering relation R such that every R-linearly ordered subset of x has an R-upper bound, but x has no R-maximal element. Define a relation S on x x w as follows: If (u,m), (vln) E x x w then (u,m)S(v,n)

iff

(u # v

A

u R v) v (u = v

A

m

5

and

n).

satisfies the hypothesis of M 18. For suppose Then (x x w,S) u is an S-linearly ordered subset of x x w. Then is an R-linearly ordered subset of x. Every R-linearly ordered subset of x has a strict R-upper bound. If not,then an R-upper bound would be an R-maximal element of x. Let b E x be a strict R-upper bound of a ( u ) . Then (b,O) is an S-upper bound of u.

a(~)

-

Suppose (ulm) E x x w. Then we claim (u,m+l) is an R-first element of (u,m), because (u,m)S(u,m+l) and if (u,m)S(v,n) and (u,m) # (v,n), then (u,m+l)S(v,n). This contradicts M 18, q.e.d.

PART I, SET FORMS

72

We end this section with fix-point theorems which are equivalent to maximal principles. Abian [1983] gave M 19(P,W) in an abstract. The other forms and all the proofs are our own. M 19(Q,U): Every set x such that x is Q-ordered by R and every U-ordered subset has an R-upper bound,has the property that every function f from x to x such that for all u E x, u R f(u), has a fixed point.

We shall show M 19(Q,U) * M 1 (Q,U). It is clear that M l(Q,U) -+ M 19(Q,U) because an R-maximal element must be a fixed point. THEOREM 4.47:

M 19(Q,U)

+

M l(Q,U).

PROOF: Suppose x is a set and R a relation which satisfy the hypotheses of M l(Q,U). Suppose x has no and define a relation R-maximal element. Let y = x x w on y so that (u,m)S(v,n)

iff

(u # v A u R v)

V

(u = v A m

s

S

n).

It is clear that when one considers the possible values for Q and U given on pp. 37-8, that if R is a Q-ordering on x then S is a Q-ordering on y. Also, since x has no R-maximal element, if each U-ordered subset of x has an R-upper bound then each U-ordered subset of y has an S-upper bound. Let f be a function defined on y such that for all (ulm) E y, f(u,m) = (u,m+l). Then for all (u,m) E y, (u,m)Sf(u,m), but f has no fixed point. This contradicts M 19(Q,U). Thus, x must have an R-maximal element, q.e.d. COROLLARY 4.48: M 19(Q,U) is equivalent to the axiom of choice if Q = TR, P I RA or F and U = C, AS & C, TR & C, L, D, or W and also if Q = D and U = W. PROOF: Theorem 4.47 and preceding remarks, and Theorems 4.9 and 4.10.

85.

FORMS EQUIVALENT TO THE AXIOM OF CHOICE UNDER THE

AXIOMS O F EXTENSIONALITY AND FOUNDATION

As we state in the Introduction, all proofs in this book, unless otherwise specified, are in NBGO. Thus, the proofs are valid whether or not there are atoms and whether or not the axiom of foundation holds. H. Rubin [19601 discovered two forms of the Axiom choice, PW and LW below, in which he assumed there were no atoms (or the class of atoms was a set and could be well-ordered) and the axiom of foundation to prove they implied the axiom of choice. Also, since both PW and LW are true in the Fraenkel-Mostowski model given for example in Halpern [1962] and the axiom of choice is false in most Fraenkel-Mostowski models, it was clear the proof of equivalence did not hold in NBGO. Later Felgner & Jech 119731, Howard & Rubin I19771 and others discovered other statements in the same category. This section is devoted to such statements. Our proofs below which require the stronger assumptions will be prefaced by "NBG IThus, we shall assume in these proofs that there are no atoms and that the axiom of foundation holds. (In most instances, the conditions on atoms can be weakened.)

".

PW: The power set (the set of all subsets) of a well-ordered set can be well-ordered. LW:

Every linearly ordered set can be well-ordered.

INJ: If x < A.

x

is a non-empty set and

13

A

is a proper class then

PART I, SET FORMS

74

PROJ: If x is a non-empty set and A can be mapped onto x.

A

is a proper class then

MC 1: If s is a set of non-empty sets then there is a function f such that for each x E s, f(x) is a non-empty finite subset of x. MC 2: sets.

Every set is the union of a well-ordered set of finite

If s is a set of infinite sets, then there is a function f such that for each x E s, f(x) is a partition of x into pairwise disjoint finite sets, each of which has at least 2 elements.

MC 3:

DEFINITION 5.1: If R is a partial ordering relation on X then Y 5 X is called an R-antichain iff for all u, v E Y, if u # v then 1 uRv A vRu. For example, a set with the property can t icha in -

.

A:

-

K

(see 4.20) is a

Every partially ordered set contains a 5-maximal antichain.

MC 1 is usually called the multiple choice axiom and A. Levy [1962] proved that it does not imply the axiom of choice in NBGO. MC 2 and MC 3 are variations of MC 1. (See WO 4 ( m ) and AC 10(m).) INJ and PROJ are forms similar to a form given by von Neumann [1928]. (They are called the injection and projection principles in Felgner & Jech 119731.) The statement A appears in Kurepa (19521. If R is taken (As is well known, there is no to be 5, then A is M S ( i ? ) . loss of generality in assuming that a partial order is 5, because if R is any partial ordering relation on a set x c c and for each u E x, u = {v E x: vRu1, then if y = {u:u E x), (XtR)

(Yis).) Felgner

&

INJ

Jech [1973] prove the following implications: +

PROJ

+

AC

+

MC 1 * A

-t

LW

+

PW.

(They also proved that none of the arrows are reversible in NBGO). H. Rubin [1960] proved PW implies AC in NBG and

55,

Rubin

&

EQUIVALENCES UNDER EXTENSIONALITY AND FOUNDATION Rubin [ 1 9 6 3 ] prove that

AC

-+

INJ

15

in NBG.

The following implications are clear: INJ PROJ, AC 1 MC 1, MC 2 MC 1. The proofs that MC 1 MC 2 and MC 1 -+ MC 3 are similar to the proof of 2.9. The proof that MC 1 is similar to the proof of 2 . 1 2 . MC 3 +

-+

-+

-+

-+

THEOREM 5.2:

PROJ

WO 3.

-t

PROOF: Let x be a non-empty set. PROJ implies that there is a function F mapping On, the class of ordinal - 1 { s j is a non-empty numbers, onto x. For each s E x, F class of ordinal numbers. Let as be the first element of F-l" {s]. Then x w { a s : s E XI. 11

THEOREM 5.3:

MC 1

A.

-+

PROOF: Suppose x is partially ordered by R. MC 1 implies there is a function f such that for each non-empty subset u 5 x, f(u) is a non-empty finite subset of u. Define a function g such that €or all u such that BZU_cX, g(u) = {a: a

is an R-minimal element of

f (u)1.

Then g(u) is finite, non-empty and an R-antichain. Now, using transfinite induction, we construct a 5-maximal antichain, A _c x. A = (J A, , where aEOn Aa = g(1a E x: a is R-ipcomparable with all b E U A B ] ) . B
THEOREM 5.4:

y = t(u,a): S

-+

LW.

Suppose x

PROOF:

Define

A

on

y

p/

is linearly ordered by

# u _c x

A

R.

Let

a E u3.

so that

(u,a)S(v,b)

iff

u = v

A

aRb.

implies that y has a 5-maximal S-antichain, B. If (u,a), (u,b) E B then a = b, so B is a choice function on the non-empty subsets of x. Then it follows that x can be well-ordered. (See 2.9.) A

PART I, SET FORMS

76

THEOREM 5.5:

LW

-t

PW.

PROOF: Suppose x can be well-ordered by R. We shall show p ( x ) can be linearly ordered. Define a relation S on p(x) such that if u, v E p ( x ) , iff u = v V (u # v A the R-first element of (u u v) (u n v) is an element of u).

u S v Then P(x)

-

(p(x),S) is a linearly ordered set so can be well-ordered, q.e.d. NBG k A C 3

THEOREM 5.6:

A

WO 2

+

LW

implies

INJ.

PROOF: Let x be a non-empty set and A a proper class. For each s E A let p ( s ) be the rank of s , (See 0.9) and A is a proper class and the class of let B = { p ( s ) : s E A). all sets with a given rank is a set so B is a proper class of ordinal numbers. Every proper class of ordinal numbers is equipollent to On. Let F be a bijection mapping On onto B. By WO 2 f such that that for each such that for 1-1 function

there is an ordinal number a and a function f:x = a. By AC 3 there is a function g such B E a , g(B) E o-”’{F(B)I. Define a function h each u E x, h(u) = g(f(u)). Then, h is a which maps x into A.

(Actually, 5.6 is provable in NBGO + the class of atoms is a set.) THEOREM 5.7:

NBG k P W

+.

+

axiom of foundation

WO 1.

PROOF: Let s be a set. The axiom of foundation implies that each set has a rank so suppose p ( s ) = a . We shall prove by transfinite induction that there exist a function F such that for each B < a, FB well-orders z ( B ) = { x : p ( x ) = 8). For all u E s , p ( u ) < a . Thus, the above function F can be used to well-order s. For every

6, let

x

=

VB =

r(v a)

=

‘(r is Hartogs’ function.)

{x:p(x) < 8)

(Y:Y < va l

and let

55.

EQUIVALENCES UNDER EXTENSIONALITY AND FOUNDATION

77

X

is an ordinal number which doesn't depend upon 5. Moreover, X can be well-ordered so PW implies that there is a relation R which well-orders ? ( A ) . The induction hypotheses is that for all y < 5, where F is defined so that F well-orders Z ( Y ) . Y Y NOW, we well-order V B in the natural way. That is, define S B on V B so that if x E z(y,) , y E z Y 2 ) then x S y iff y1 < y 2 V (yl = y 2 A x F y). There exists a Y1 unique ordinal 6 < A and a unique function G5 such that GB:(VB,SB) z ( 6 , g ) . Suppose x, y E z ( B ) . Then x, Y E 9 ( V 5 ) . Define F B on z ( 5 ) so that if x, y E z ( B ) ,

5 <

a,

x

F5 y

-

G"x R G" y. B 5

Since G"x, G"y c 6s A, R is defined, so F5 is defined, B 5 and since R is a well-ordering relation, so is F5. Moreover , the definition of R does not depend upon B , therefore, F is a function, q.e.d. (Note: it is easy to give an incorrect proof of this theorem if one is not careful. For example, we might try to well-order V B for each 5 using transfinite induction as follows: Vo = 8, so clearly Vo can be well-ordered. Suppose V5 can be well-ordered. Then PW implies that can be well-ordered. Suppose B is a limit 'f3+1 = $'(Vg) ordinal and for all y < 5, Vy can be well-ordered. Then "can be well-ordered", but the problem is that it = requires some form of the axiom of choice to "construct" that well-ordering of VB when 6 is a limit ordinal. In.the correct proof of the theorem we construct the well-ordering using the relation R.)

YB

It follows from 4 . 3 5 , 4 . 3 6 , and 4 . 3 8 that each of M 1 5 ( A : i ? ) , for A = D, J, j or K, imply M 8 ( i ? ) , each is equivalent to AC in NBG.

so

We continue with some forms of the axiom of choice which are due to P. Howard 6r J. E. Rubin [1977]. Let ACp and Q P MCQ be defined as follows:

PART I, SET FORMS

78

P

ACQ: For each set x, if x is a P-ordered set of non-empty sets such that each u E x can be Q-ordered, then there is a function f such that for each u E x, f(u) E u. P

MCQ: For each set x, if x is a P-ordered set of non-empty sets such that each u E x can be Q-ordered, then there is a function f such that for each u E x, f(u) i s a non-empty, finite subset of u.

For P LO: DLO:

and

Q

we take the following:

linear ordering defined linear ordering.

For example, ACLo is the axiom of choice for a linearly ordered family of non-empty sets and MCDLO is the multiple choice axiom for a family of non-empty sets, each of which has a defined linear ordering. ORD and W-ORD

are due to J. Truss [1972].

ORD: If x is a set and f is a function such that for each u E x, f(u) is a non-empty set of linear orderings of u, then If(u): u E XI has a choice function. W-ORD: If x is a set and f is a function such that f o r each u E x, f ( u ) is a non-empty set of well-orderinqs of u, then {f(u): u E X I has a choice function. TCL is M 5(TR&C,L) of section 4 and appeared in P. E. Howard & J. E. Rubin [19771. TCL: Every transitive and connected set contains a 5-maximal linear ordered set.

-

--

Thus, TCL M 5(TR&C,L) directed.)

-

-

M 5(TR&C,AS) M 5(TR&C,AS&C) M S(TR&C,P) M 5(TR&C,D) * M 5(TR&C,RA). (Here, D means

We shall prove the following implications in NBGO:

55.

EQUIVALENCES UNDER EXTENSIONALITY AND FOUNDATION

-

LO --+LW eMCDLO

+M

C

~

~M

C

+

~

H

LO ACDLO

~ A ~ C

--+ ~

~

79

Pw

~

W-ORD FIGURE 5.1 It can be shown using results in Howard ti Rubin [ 1 9 7 7 ] and a model of R. J. Gauntt that none of the arrows in the above diagram are reversible in NBGO. However we have shown that PW -+ AC in NBG, so all the statments are equivalent in NBG. The following implications are easily verified: ACLo >-

MCLo

\1

/&LO

AC

ACLO A

THEOREM 5.8:

TCL

MCEE

1

C

-

+ ~ ~M

+MCiEO

C 4 ~ ~ M

C

~

~

~

ACLO .

PROOF: Suppose (x,R) is a linearly ordered set of pairwise disjoint non-empty sets. Define a relation S on U x such that if a, b E Ux, where a E u E x and b E v E x, aSb iff uRv. Then S is transitive and connected on Ux. A 5-maximal S-linearly ordered subset of U x is a choice set for x. Conversely, suppose set. Define a relation u S v

iff

(x,R) is a transitive and connected S on x so that

u = v V ( u R v ~ v R u )

is an equivalence relation on x. For each u E x let Since R is [u] = {v E x: u S v). Let y = {[u]: u E XI. transitive, we can define a relation T on y so that S

PART I, SET FORMS

80

[u]T[v] iff u R v. (If u ' E [ul and v' E [vl, then u ' R v' iff u R v.) The relation T linearly orders y. A choice set for y is a 5-maximal linearly ordered subset of x, q.e.d. THEOREM 5.9:

ACDLO

-

MCDLO.

PROOF: Clearly, ACDLO MCDLO. To prove the converse, let x be a set of non-empty pairwise disjoint sets, each of which is linearly ordered by R. Let € be a multiple choice function on x. Then {a:(]u E x) (a is the R-first element of f(u))l is a choice set for x. LO LO The proof that ACDLO MCDLO is similar to the proof of 5.9. LO THEOREM 5.10: LW ACDLO. -f

-

-

PROOF: Suppose (x,R) is a linearly ordered set of nonempty pairwise disjoint sets such that each set in x is linearly ordered by a relation S. Then U x can be linearly ordered, because if a, b E U x where a E u E x and b E v E x, define

a T b

iff

if

u # v I

a S b

if

u = v .

1

u R v

The relation T linearly orders Ux. Then LW implies U x can be well-ordered. Using this well-ordering a choice set for x can be defined. Conversely, suppose (x,R) Define y = {u x { s } : u 5 x A

is a linearly ordered set. s

well-orders

u}.

If (u,s) is a well-ordered set, let ord(u,s) be the ordinal number of ( u , s ) . If (u,,s,) and (uz,s2) are well-ordered sets with the same ordinal number then there is a unique isomorphism f mapping u1 onto u2, f:(ul,sl) ;(u2,s2).

EQUIVALENCES UNDER EXTENSIONALITY AND FOUNDATION

55.

u1

x

We define a linear ordering, TI on {s11 # u2 x { s 1 , then T (u,

{ s , f ) iff

x

or

(f:(uI,sl) ;( u r l s 2 ) and if

element of The set z = {(x

-

x

{b

f

ul: b # f(b))

u)

X

is}:

u cx

A

as follows, if

ord(ul,sl)

(ul

{s,})

y

a

then

s

81

< ord(u2,s2)

is the sl-first

a R f(a)).

well-orders u]

is a linearly ordered set (T induces a linear ordering on z ) of non-empty sets, each of which is linearly ordered by LO R', where (vl,s) R' (vz,s) iff v1 R v2. Thus, ACDLO implies there is a choice function f on z. Then, wellorder x so that the ath element of x, aa = the first U {ag]) x { s } ) , where 5 is the wellcoordinate of f((x

..

ordering of

U {ag}

B
B
THEOREM 5.11:

ORD

induced by

-

a.

ACLO.

PROOF: Suppose x is a set of non-empty sets, each of which can be linearly ordered. For each u E x, let f(u) be the set of all linear orderings of u which have a first element. (Clearly, if u can be linearly ordered then it can be linearly ordered so it has a first element.) If g is a choice function on If(u): u E XI, then we can define a choice function, h, on x such that for each u E x, h(u) is the first element in the ordering g(f(u)). Conversely, suppose x is a set and for each u E x, f(u) is a non-empty set of linear orderings on u. We shall show that each f(u) can be linearly ordered. Then, it follows that ACLO + ORD. LO LO LO Clearly, ACLO ACLO + ACDLO. By 5.10, ACDLO + LW, and -+

by 5.5, LW PW. Therefore, ACLO + LW * PW. Suppose u E x. Then, by hypothesis, u can be linearly ordered. Thus, LW implies u can be well-ordered. Therefore, u x u can also -+

PART I, SET FORMS

82

be well-ordered. PW implies f’(u x u) can be well-ordered, Since f(u) _c p(u x u), f(u) can be well-ordered, and therefore linearly ordered, q.e.d. Clearly ORD .+ W-ORD. The proof of the following theorem is due to J. Truss [1972].

THEOREM 5 . 1 2 :

W-ORD

PW.

-+

PROOF: Suppose a is any ordinal. We shall show that can be well-ordered. (This is sufficient since every set which can be well-ordered is equipollent to an ordinal.) Let y = a U {XI where x f? a . Let

~ ( c L )

z = Iy x {t}:

!a # t c

B(a)).

For y x It1 E z, let f(y x It]) be the set of all well-orderings, R, of y x {t) such that there is an s E t ( s _c a ) such that comes first,

(i (ii iii

is next, x

It)

is last,

where the ordering in each of (i) and (iii) is the natural ordering induced by on a . The set z and function f satisfy the hypotheses of W-ORD. A choice function g , on If (y x (t}): !a # t 5 p ( a ) 1 enables us to choose for each t E P ( a ) {!a) an s E t. (Namely, the s from the s x It1 which precedes the element Thus, we have a (x,t) in the well-ordering g(f(y x It))).) { g ) . This choice function allows choice function on p ( a ) us to well-order p ( a ) . (See 2 . 9 ) .

-

-

We have proved earlier, 5 . 5 , that LW + PW so this completes the proof of the implications given in Figure 5.1. The following weaker version of

ORD:

If x is a set such that for each u E x, u can be linearly ordered, then there is a function f such that for each u E x, f(u) linearly orders u

EQUIVALENCES UNDER EXTENSIONALITY AND FOUNDATION

95.

does not imply the axiom of choice. Halpern & Levy [ 1 9 6 7 ] . )

83

(This was proved by

There is also a weaker version of MC 1 which was shown by Halpern & Levy [ 1 9 6 7 1 not to imply the axiom of choice in NBG, namely the Kinna-Wagner principle (see Kuratowski 1 1 9 2 2 1 ) : KW 1: If x is a set of sets, each of which has at least 2 elements, then there exists a function f such that for each u E

X I

a

# f(u)

Kinna KW 2 :

&

= u.

Wagner [ 1 9 5 5 1 have shown that KW 1

For each set x, there is an ordinal

a

-

KW 2, where

such that

x < P(a). However, in each Fraenkel-Mostowski model PW is true. Therefore, in each Fraenkel-Mostowski model, KW 2 is equivalent to the axiom of choice and is therefore false. But trivially AC. we have KW 2 + PW

-

M. N. Bleicher [ 1 9 6 3 , 1 9 6 5 ] discusses some variations of the multiple choice axiom, MC 1, which are useful in the study of vector spaces. For m, n E w we define

I

(m,n) =

m if n = O n if m = 0 greatest c o m o n divisor of

and

m

n, otherwise

MC 4(n): If s is a set of non-empty sets then there is a function i such that for all x E s , f(x) is a non-empty finite subset of x and (If(x)(,n) = 1. Thus, according to the definition of (m,n), MC 4 ( 0 ) is AC 1 and MC 4(1) is MC 1. It is clear that for all n E w, AC 1 + MC 4(n) + MC 1. We have shown that AC. Therefore, NBG F M C 4(n) AC for each NBG F M C 1 n E w.

-

-

Since every finite set can be well ordered the wellordering theorem is equivalent to the statement: (*)

If

x

is infinite then

x

can be well ordered.

84

PART I, SET FORMS

The word "infinite" means "not equipollent to a natural number". Mathematicians have studied various other definitions of "infinite" or equivalently "not finite". (See, for example, Tarski [1924], Mostowski [1938], Levy [19641 and Truss [1974]). All alternate definitions considered are equivalent if the axiom of choice is assumed. Since the well-ordering o € a set implies the well-ordering of any subset, for any definition of infinite which has the properties that a superset of an infinite set is infinite and that there is an infinite set, ( * ) is equivalent to the well-ordering principle. Brunner I19821 has obtained some results along this line which we shall describe below: DEFINITION 5.13: i) A set x is said to be Dedekind-infinite (D-infinite) if it contains a countably infinite subset. ii) A set is said to be structured if it is the union of two disjoint infinite sets. Consider the following statements: INF 1: For every set x, if every infinite subset of D-infinite, then x can be well-ordered.

x

is

INF 2: For every set x , if every infinite subset of structured, then x can be well-ordered.

x

is

Clearly, WO 1 * INF 2 INF 1. Brunner [1982] proves that INF 1 does not imply INF 2 in NBG and that INF 2 does not imply WO 1 in NBGO. It follows from results of Levy [19581 that INF 2 * LW in NBGO. Thus, in NBG, INF 2 AC. +

-

THEOREM 5 . 1 4 :

INF 2

+

LW.

PROOF: It is sufficient to prove that every infinite linearly ordered set is structured. Suppose x is an infinite linearly ordered set. If x has an infinite number of finite initial segments or an infinite number of finite terminal segments then x has a countably infinite subset, so x is structured. Otherwise, there is some u E x such that both the initial segment and terminal segment generated by u are

EQUIVALENCES UNDER EXTENSIONALITY AND FOUNDATION

15.

infinite so, again, x is structured. us that INF 2 AC;:. )

85

(A similar proof gives

-+

U. Felgner [1974] proved that the statement:

The union of a well-ordered family of sets, each of which can be well-ordered, can be well-ordered, does not imply the axiom of choice in NBG. However, it is easy to see (J. E. Rubin [ 1 9 8 2 ] ) that if either of the "wellordered" in the above statement is replaced by "linearly ordered", or anything weaker than linearly ordered, the resulting statement is equivalent to the axiom of choice in NBG. Thus, we make the following definition:

The union of a P-ordered family of sets, each of which can be Q-ordered, can be well-ordered.

UE:

For LO: DLO: WO: DWO:

P

and

we take:

Q

linear ordering, defined linear ordering, well-ordering, defined well-ordering.

It is easy to see that the following implications are all provable in NBG': P UQ AC

+

AC

Q'

for all

- - uwo,

P

and

UL0

(because x

-

=

~{x)).

- - - .wo-

'LO "wo "DLO UDWO, (because x = UI{uI: u E XI). LO wo LO Lw UDLO UDLO UDWO*

-

uwo

Q.

86

PART I, SET FORMS

Since LW + AC AC in NBG.

in

NBG, all the above forms are equivalent to

In section 4 , we defined a universal, UN, relation as follows: R is UN on X iff X x X 5 R. We proved that M 5(A,UN)

-

AC,

where

M 5(A,UN): Given any relation R , every set contains a c-maximal subset on which R is UN. We say that a relation R is vacuous, VC, iff is UN. As defined previously in section 4 , TR means transitive; AS, anti-symmetric; and C, connected. Let ID mean irreflexively directed and IRA, irreflexively ramified. We shall show that each of M 5(Q,VC), for Q = TR, AS, AS & TR, ID, or IRA is equivalent to AC in NBG. Clearly, M 7 + M 5(Q,VC). THEOREM 5.15:

M 5(AS

&

TR,VC)

-

A.

PROOF: If x is irreflexively partially ordered by y _c x is an R-anti-chain iff y x y 5 R. COROLLARY 5.16:

M 5(AS,VC)

-+

A.

COROLLARY 5.17:

M 5(TR,VC)

+

A.

COROLLARY 5.18:

M 5(C,UN)

-+

R,

A.

.+

PROOF: M 5(C,UN) M 5(AS,VC) because a relation R is connected iff E is anti-symmetric, and R is UN iff E is VC. THEOREM 5.19: M 5(IRA,VC) ACDLO. PROOF: Let x be a family of pairwise disjoint sets such that for each u E x, RU linearly orders u. Define a relation R on U x such that -+

a R b

iff

(3u E x) (a,b E u

The relation R is IRA on of U x is a choice set for

Ux. x.

A

a Rub

A

a # b).

A 5-maximal VC

subset

EQUIVALENCES UNDER EXTENSIONALITY AND FOUNDATION

55.

THEOREM 5.20:

87

M 5 (ID,VC) + ACDLO.

PROOF: Let x be a family of pairwise disjoint sets such that for each u E x, RU is an irreflexive linear ordering of u. Let y = C(a,n):

Define a relation (a,n)R(b,m)

n E R

w A

on

iff n < m

V

y

(3u E x) (a E u)}

.

such that

[n = m

A

(3u E x) (arb E u

A

a RU b) I .

Clearly, R is ID on y. Thus, if z is a c-maximal VC subset of y, the set of all first coordinates of elements of z is a choice set f o r x. Our next statement, M 19, is obtained from M 8 (6) by

changing the word "set" to "class". (By 4.20, x y iff X n Y # 8 . ) With the aid of the axiom of foundation, it can be shown that M 19 is equivalent to the set form of the axiom of choice in NBG. M 19: D.

Every class has a 5-maximal subclass with the property

Clearly, M 19 THEOREM 5.21:

-+

M 8(6). NBG F W O 2

-+

M 19.

PROOF: Suppose X # 9 is a class and let x implies there is an ordinal a such that a w x. a bijection mapping a onto x. Also, On w a x F be a bijection of On onto a x On. For each define a(B) and b(B) by F(B) = (a(B),b(B)).

E X.

WO 2 Let f be On. Let 8 E On,

For each B E On, let S ( B ) be the set of all elements X which contains the a($)th element of x, have rank b ( B ) , and have a non-empty intersection with each element of In symbols, S(Y) for Y < B of

S ( B ) = Iy E

x:

VY

E 6) ( V Z E s A f (a(@))

In NBG,

S(B)

is a set for each

88

PART I, SET FORMS

-

then Y is a 5-maximal subclass of X with the property D. Clearly, Y _c X and by construction, Y has the property D. Suppose Y is not 5-maximal. Then there is a z E X Y such that for all y E Y, z n y # 0 . Since x € Y, z n x # 0 . Since f:u = x, there is a B u such that f(0) E z and since F: On u x On, there is an ordinal y such that F ( y ) = (R,p[z)). Consequently, z E S ( y ) 5 Y. This is a contradiction, q.e.d.

-

Moreover, if the word "set" is changed to "class" in each of M 8(Dm), M 12 and M 13, a similar proof shows that each is equivalent to the set form of the axiom of choice in NBG. However, as we saw earlier the class form of M 14 is equivalent to the set form of the axiom of choice in NBGO. Our next statement in this section is a class form of AC 4 . AC 20: For every relation R whose domain is a set (its range may be a proper class) there is a function f such that $ ( f ) = a(R) and f _c R. Clearly, AC 2 0 + AC 4. (It was shown in Felgner & Z'lannagan 119781 that this arrow is not reversible in NBGO.) THEOREM 5 . 2 2 :

NBG F A C 4

+

AC 20.

PROOF: Let R be an arbitrary relation whose domain is a set. Define a relation R ' as follows: R'

=

{(x,y)

E R:

( V z ) ((X~Z) E R

* P(Y)

5 P(z))}.

In NBG, the class of all elements with a given rank is a Set, so R' is a set. Applying AC 4 to R' gives us the desired function. AC 2 1 is a class form of AC 18. AC 21: For any non-empty whose domain is A,

set

A, and for each relation

B

55.

where

EQUIVALENCES UNDER EXTENSIONALITY AND FOUNDATION Ba

is defined to be

B"{a}

for each

89

a E A.

We shall prove that AC 2 0 * AC 21. The proof that AC 2 0 * AC 2 1 is similar to the proof of 2.18, except we use AC 2 0 to construct f rather than AC 1. THEOREM 5 . 2 3 :

AC 2 1 * AC 2 0 .

PROOF: Let R be a relation whose domain is the set A. (The range of R may be a proper class.) For each a E A, define BU[al = {{u,tl:(a,t) E R } . Then u E

n U b. a €A b€BU [a1

AC 2 1 implies that there exists an

f E

x

BU[a]

such that

aEA

for all a E A, u E f (a). (We are using AC 21 with X a r b = b.) NOW, from the definition of BuIaI, f(a) = Curt), where (art) E R. Define a function g with domain A such that for all a E A, g(a) = t iff f(a) = {u,tl. Then a(g) = a(R) and g _c R, q.e.d. Next, we give a class form of

DC(a)

(see section 2,

p . 19).

DCC(.a): If R is a relation between subsets and elements of a class X such that (Vy _c X) (y < a * (3u E X)(y,u) E R ) then there is a function f mapping a into x such that (VB < a)(f"Brf(B)) E R. Clearly, DCC(a) + DC(a) for all a E On. Felgner [1976al proves that it is not the case that DC(a) + DCC(a) in NBGa but that DC(a) + DCC(a) in NBG. Thus, it follows from 2.20 and 2 . 2 1 that in NBG, (Va)UCC(a) is equivalent to the set form of the axiom of choice. We give Felgner's proof of DC(a) * DCC(a) here. THEOREM 5 . 2 4 :

NBG

( V a ) (DC(a)

PROOF: Suppose X and R DCC(a). For each f3 < a, let

+

DCC(a)).

satisfy the hypotheses of

PART I, SET FORMS

90 z ( 6 ) = {y

5 X: y = 6

(vz

A

5 x) ( 2 = 6

-+

P(y)

P(Z))).

5

It follows from the axiom of foundation that if there is a ! and that z ( B ) is a Y c x such that y % B then z ( B ) # J set for all B < a . Therefore, v = U z ( B ) is a set. Next for each

y E z(0),

E R

w(y,B) = {a E X:(y,a) Then w(y,B)

B
let

(Vb E X) ((y,b)

A

is a set and

s

U

=

U

E R

-+

p(a)

i

p(b)) 1 .

is a

w(y,B)

B
subset of X. Let R' = R n s and R ' to get the desired result.

DC(ff) to

The last forms we shall discuss in this section are forms of the trichotomy due to Kruse 119631, each of which is equivalent to PW. For each cardinal statement:

K,

let

T(K) stand for the following

T(K): For all cardinals yl and p 2 , then either p l 5 p 2 or p 2 s ul.

if

K 5

ul,

y 2 i 2K

Thus, T(K) is the trichotomy for cardinals between K and 2 K . We show that each of the following statements are equivalent to PW. N

T 11:

{ a E On: T(K,)

A

T 12:

{a E On: T(KJ

A

i 2

T(2

Ha

)I

is unbounded.

is unbounded.

T 13: For all cardinals K and p , if there is an ordinal such that K , p 5 2 I a l then K 5 p or p s K .

Clearly

PW

implies each of

THEOREM 5.25: PROOF: Let number such that

T 11 H

+

be an aleph. K, d ,'2 N+,l

3. Therefore, T(Na) and 24 are comparable; that is,

and

T 13.

PW.

N < Hal

Thus, we have that both

T 11, T 12

Let a be an ordinal Ha T(Ka) and Ha+1 s 2

and '2

+ K,

implies that

are between and

'2

. Ha

+

c1

55.

EQUIVALENCES UNDER EXTENSIONALITY AND FOUNDATION

91

The first inequality is impossible by the definition of Ka and the second inequality implies 2& s Hali which implies PW. THEOREM 5.26: T 12 + PW. PROOF: Let H be any aleph and let a be an ordinal such that N Na, N, ZC ,'2 T(N,) and T(2Ha) Using the proof of ( * ) in Theorem 3.10, we see that T(2 kcr ) implies that Nail s Z K a . Now use 5.25 to obtain PW. THEOREM 5.27: PROOF:

T 13

+

Let B E On,

Let CL = max(wB,p). Then or p i implies that K 5 and K I; p implies PW.

PW.

let K, K.

K

= 2 Nc1

and let

p = (2N8 ) * .

Therefore, T 13 By the definition of p , p P;

p 2 2IaI.

K,

The cofinality index of a limit ordinal X I cof(X), is the smallest ordinal c1 such that there exists a 1-1, order preserving function f mapping a into A , with $(f) cofinal in A , that is, for each y E X there is a B E a such that y i f ( B ) . T 14:

{cr:T(Ka)

A

cof(wa+l)

2

8)

is unbounded for each 8 E On.

It is easy to show that PW implies T 14. For the proof that T 14 implies PW, the reader is referred to Kruse 11963, Theorem 5.3, p.2171.

This Page Intentionally Left Blank

56.

ALGEBRAIC FORMS

T h i s s e c t i o n i s devoted t o a l g e b r a i c s t a t e m e n t s which a r e e q u i v a l e n t t o t h e axiom of c h o i c e .

Many of t h e s e s t a t e m e n t s

were used i n a l g e b r a l o n g b e f o r e it was known t h e y were I t w a s n ' t u n t i l 1953 t h a t e q u i v a l e n t t o t h e axiom of c h o i c e . J . Schmidt [1953] and D .

S c o t t [1953] i n d e p e n d e n t l y proved

t h a t s e v e r a l of t h e s e s t a t e m e n t s imply t h e axiom of c h o i c e . S u b s e q u e n t l y , a wide v a r i e t y of s t a t e m e n t s a b o u t many t y p e s of a l g e b r a s w e r e proved e q u i v a l e n t t o t h e axiom o f c h o i c e . DEFINITION 6 . 1 :

Algebra

(1) 4 = (A,{Ok:k E K}) a l g e b r a i c system i f A

i s c a l l e d an ( a b s t r a c t ) a l g e b r a o r an

i s a s e t and f o r each

k E K , Ok

f i n i t a r y o p e r a t i o n d e f i n e d on A. ( A 0-ary o p e r a t i o n is a c o n s t a n t and f o r n > 0, an n-ary o p e r a t i o n i s a f u n c t i o n mapping t h e d i r e c t p r o d u c t of A w i t h i t s e l f n - t i m e s i n t o (2) B

3

= (B,iOkrB:k E K ) )

and, i f

A

bl,b 2 , . . . , b n

i s c a l l e d a sub-algebra of A is a n n-ary o p e r a t i o n t h e n f o r a l l

Ok

E B , Ok(bl,b2,

...,b n )

c a l l e d a proper subalgebra i f

(That is,

E B.

c l o s e d w i t h r e s p e c t to t h e o p e r a t i o n s

Ok

for

k

A.)

if

is

B

E

is a

K.)

is

3 # $.

When it w i l l n o t l e a d t o any c o n f u s i o n , w e s h a l l c a l l t h e

set "A"

of

t h e a l g e b r a and t h e s e t

"B"

of

$

t h e sub-

algebra. (The f o l l o w i n g d e f i n i t i o n of a l a t t i c e i s g i v e n i n Birkhoff

[19481 p.

19.)

DEFINITION 6 . 2 :

Lattice

(1) $ = ( L , { v , A ) )

i s c a l l e d a l a t t i c e i f t h e following

93

PART I , SET FORMS

94

conditions a r e s a t i s f i e d f o r every (a)

L

(b)

a v b E L

and

a A b E L ,

(c)

a V a = a

and

aA a = a ,

(d)

a v b = b V a

(e)

a

v (b v c)

(a

A

a v

(f) V

v

i s a s e t and

a , b,

b) (a

A A

and

and

c

and

in

L:

a r e binary operations,

A

a A b = b A a ,

= (a V b) V c

and

a

A

(b

A

c) =

c, b) = a

i s c a l l e d t h e join and

and A

a A ( a V b) = a ; the m e e t . ( I t should be c l e a r

from t h e c o n t e x t when t h e s e symbols a r e l a t t i c e o p e r a t i o n s and when t h e y a r e s e n t e n t i a l c o n n e c t i v e s . ) ( 2 ) 1 E L i s c a l l e d an i d e n t i t y element i f f o r a l l a E L , a ~ l = a . ( 3 ) 0 E L i s c a l l e d a z e r o element i f f o r a l l a E L , a v O = a .

(4) E c_ L a V b E E

i s c a l l e d an i d e a l i f whenever a E L

and whenever

c a l l e d a proper i d e a l i f

F C_ L

(5) A

(If

b E F

a E L

is

from 6 . 2 (7)

for a l l bound o f

If

a, b € L, a s b

(1) ( f ) t h a t

V a if a€A a E A then x

a

V b

va

= b

a s x

x =

A.

if

i

y.

F

is is c a l l e d an u l t r a f i l t e r

a V b = b

(It follows

i f and o n l y i f

for a l l (x

a , b E F,

b E F , a V b E F.

and

F # L. F i s a 5-maximal p r o p e r f i l t e r .

F

(6)

a E A

a A b = a).

and i f

a

5

y

i s c a l l e d t h e l e a s t upper

may be w r i t t e n a s VA.)

a €A I\a if x i a f o r a l l a E A and i f y s a f o r aEA a E A t h e n y i x. ( x i s c a l l e d t h e g r e a t e s t lower x =

all

E

i s a proper i d e a l

E

i s c a l l e d a f i l t e r i f whenever

and whenever

c a l l e d a proper f i l t e r i f if

b E E l a A b E E.

and

# L.

1 f E.)

then

a

E

a, b E E l

bound of

A.

I\ a a tA

=I\A.)

ALGEBRAIC FORMS

§6.

x

95

i s c a l l e d complete i f f o r e a c h nonempty s u b s e t A of L, a 6 L and A a E L. a€A a €A (9) c E L is c a l l e d compact i f whenever c 5 a a €A where A s L , t h e n t h e r e e x i s t s a f i n i t e s u b s e t B o f A such t h a t c 5 a. a€ B The l a t t i c e

(8)

v

v

v

(10)

E

c_

i s c a l l e d a prime i d e a l i f

L

a , b E L, i f

and f o r a l l

a

A

b E E

i s an i d e a l

E

then e i t h e r

a € E

or

b E E.

A l a t t i c e i s an a l g e b r a i c s y s t e m , b u t a s d e f i n e d a s u b a l g e b r a i s n o t n e c e s s a r i l y an i d e a l . However, i f t h e (Note:

l a t t i c e were d e f i n e d a s f o l l o w s a s u b a l g e b r a would be an i d e a l .

x

= (L,{V}

U {Oa: a E L})

o p e r a t i o n such t h a t

where

Oa(b) = a

A

i s a unary

Oa

b,

for a l l

a E L.)

Every l a t t i c e w i t h an i d e n t i t y e l e m e n t and a t

AL 1:

l e a s t one o t h e r e l e m e n t c o n t a i n s a 5-maximal p r o p e r i d e a l . AL'1:

Every p r o p e r i d e a l of a l a t t i c e w i t h an i d e n t i t y

element can be extended t o a 5-maximal p r o p e r i d e a l . ( I f i n AL 1 and AL'1 w e r e p l a c e " i d e n t i t y " by " z e r o " and "5-maximal p r o p e r i d e a l " by " u l t r a f i l t e r " , t h e r e s u l t i n g s t a t e m e n t s , DAL 1 and DAL'1, a r e e q u i v a l e n t . )

AL'2:

A

If

a E A, b u t

and

i s a n a l g e b r a i c system, a

a subalgebra,

B , t h e n t h e r e e x i s t s a c_-maximal

j!

s u b a l g e b r a which c o n t a i n s contain

B

B

as a s u b s e t , b u t d o e s n o t

a s an e l e m e n t .

a

D E F I N I T I O N 6.3:

if

C

class

C i s called a f i n i t a r y closure operator i s a monotone unary o p e r a t i o n s u c h t h a t f o r e v e r y

X:

C ( X ) = U { C ( y ): y C _ X

If

C(X)

5X

AL'3: operator,

then If

P

X

and

y

is finite).

i s c a l l e d C-closed.

x i s an a r b i t r a r y s e t , C a f i n i t a r y c l o s u r e a p r o p e r t y of f i n i t e c h a r a c t e r , and y a

PART I , SET FORMS

96

C-closed s u b s e t of

x

which h a s t h e p r o p e r t y

i s a 5-maximal C-closed s u b s e t o f a s u b s e t and h a s t h e p r o p e r t y

x

P I then there

which c o n t a i n s

y

as

P.

AL'1 was g i v e n by D. S c o t t [1953] and AL 2 w a s g i v e n by &'3 is a

Robert B l a i r i n 1957 i n an u n p u b l i s h e d p a p e r .

A L ' 2 and AL 3 a r e f a l s e i n

g e n e r a l i z a t i o n of A L ' 1 and AL'2.

t h e c o r r e s p o n d i n g forms w i t h o u t primes a s t h e e may e x i s t no subsets with t h e desired property.

Dzik [1981] a l s o p u b l i s h e d

some e q u i v a l e n t s s i m i l a r t o AL'3. W e s h a l l prove t h e f o l l o w i n g i m p l i c a t i o n s : M 3

+

AL'3

AL'2

+

+

AL'1

-+

A1 1

+

M 7.

I n s e c t i o n 4 i t was shown t h a t M 3 and M 7 are e q u i v a l e n t . The proof t h a t

M 3

+

i s s i m i l a r t o t h e p r o o f o f 4.3.

AL'3

A L ' 1 i s a s p e c i a l c a s e o f AL'2 and i t i s c l e a r t h a t AL'1

+

AL 1.

THEOREM 6.4:

PROOF:

Let

AL'3

4

=

AL'2.

-+

(A,COk: k E K})

be an a l g e b r a i c system.

a E C ( x ) i f and o n l y E x and a n o p e r a t i o n

I f x c_ A , d e f i n e C ( x ) a s f o l l o w s : i f t h e r e e x i s t elements al, a2,...,an Ok, k E K

a = Ok ( a l , a 2 , . . . , a n ) .

S i n c e Ok is a is a f i n i t a r y closure o p e r a t o r , and B is a s u b a l g e b r a o f 4 i f f B i s C-closed. Moreover, t h e p r o p e r t y o f e x c l u d i n g a n e l e m e n t from a set i s a p r o p e r t y o f f i n i t e c h a r a c t e r . Hence, AL'2 i s a s p e c i a l case such t h a t

k E K,

f i n i t a r y operation for a l l

C

o f AL'3. THEOREM 6 . 5 :

PROOF:

x

-+

M 7.

be a s e t and

L e t u s assume t h a t

character. P,

Let

AL 1

x

P

a p r o p e r t y of f i n i t e

d o e s n o t have t h e p r o p e r t y

f o r i f it d o e s t h e n it i s t h e maximal s u b s e t which h a s t h e

property

P.

L e t u s a l s o assume t h a t a t l e a s t one non-empty

s u b s e t o f x h a s t h e p r o p e r t y , o t h e r w i s e t h e empty s e t i s t h e r e q u i r e d maximal s e t . L e t y = ( t : t 5 x and P ( t ) } U { X I . If

s , t E y,

define

s

A

t = s

n

tand

ALGEBRAIC FORMS

56.

s v t =

1;

u

t

97

u

t)

otherwise

,

if

P(S

then ( y , { v , ~ ) ) i s a l a t t i c e . The i d e n t i t y e l e m e n t o f t h e l a t t i c e i s x. By AL 1, ( y , { V , A ] ) h a s a maximal p r o p e r ideal

uz

W e s h a l l show t h a t

z.

with t h e property

P.

Clearly,

i s a maximal s u b s e t o f

uz

c

x.

Next,

x f z

x

since

z i s a proper i d e a l . L e t t be a f i n i t e s u b s e t o f uz. Then t h e r e i s a f i n i t e number o f e l e m e n t s s1,s2, sn o f n z such t h a t t 5 si. S i n c e z i s a n i d e a l and x f z , n i= 1 n U si E z , which i m p l i e s P( U s i ) . C o n s e q u e n t l y , P ( t ) , i=1 i=l which i m p l i e s P ( U z ) . U z i s maximal s i n c e z i s maximal, q.e.d.

...,

u

I n t h e proof of 6.5 w e j u s t r e q u i r e d AL 1 t o h o l d f o r complete l a t t i c e s . A L ' 4 w a s g i v e n i n 1959 by E . W. Beth.

I t is a set

t h e o r e t i c a l i n t e r p r e t a t i o n of t h e s y n t a c t i c a l form t h a t

Beth

I t i s one of t h e most g e n e r a l forms o f t h e

a c t u a l l y states.

axiom of c h o i c e t h a t w e have come a c r o s s .

Before s t a t i n g it,

we give t h e following d e f i n i t i o n : DEFINITION 6 . 6 : (a)

Let

m

and

a n n-ary r e l a t i o n .

i f t h e r e e x i s t elements ( X 1 i X 2 t - - - r X n ) E R. (b)

Let

(w,m,n,u,v)

be n a t u r a l numbers and l e t

m < n , (x1,x2,...,xrn) E a n ( R ) ~ ~ + ~ , x ~ + ~ , . s.u.c h , xt h a t n

R:z.

R:X

i f for all

and n , f o r a l l u E Xm and f o r a l l w (w,m,n) E a 53( R ) t h e r e e x i s t s a v E Xn s u c h t h a t E R.

m

(Let

Xo =

($1

and

AL'4: I f x is an a r b i t r a r y set, t h e n t h e r e i s a 5-maximal s u b s e t z o f

and

be

R

m

be a quinary r e l a t i o n , then

R

n a t u r a l numbers such t h a t

n

Then f o r

X1 = {O)

x X.)

y 5 x, and R:y x such t h a t y 5 z

PART I , SET FORMS

98

AL'4 i s s i m i l a r t o t h e p r o o f of W e s h a l l show t h a t AL'3 i s a s p e c i a l c a s e of A L ' 4 .

The proof t h a t 4.3.

M 3

AL'4

THEOREM 6 . 7 :

Let

PROOF:

C

a finitary

y

a p r o p e r t y of f i n i t e c h a r a c t e r , and

P

a C-closed s u b s e t o f quinary r e l a t i o n

AL'3.

-+

be an a r b i t r a r y s e t ,

x

closure operator,

+

x

which h a s t h e p r o p e r t y

P.

Define a

a s follows:

R

( w , m , n , u , v ) E R i f f m and n a r e n a t u r a l numbers; m u E x ; v E x n ; and e i t h e r n = 0 , w = JJ, and P@,(u)), or

n = 1, w E C ( $ ( u ) ) , Then, w e c l a i m ,

C-closed.

and

R:z

v = (0,w).

i f and o n l y i f

and

P(z)

z

is

( W e l e a v e t h e d e t a i l s of t h e proof t o t h e r e a d e r . )

NOW, it i s c l e a r t h a t

AL'4

+

AL'3.

The f o l l o w i n g maximal p r i n c i p l e , AL 5 , i s due t o Mrowka [1955,1958].

(Mrowka [1955] showed t h a t t h e axiom o f c h o i c e

i m p l i e s AL 5 , b u t h i s proof o f t h e c o n v e r s e was i n c o r r e c t . T h i s w a s c o r r e c t e d i n Mrowka [ 1 9 5 8 ] . )

AL 5 d e a l s w i t h a

s p e c i a l t y p e of l a t t i c e , a Boolean a l g e b r a (see 6 . 2 ) . DEFINITION 6.8:

3

Boolean Algebra

i s s a i d t o be a Boolean a l g e b r a i f t h e

= (B,{V,A})

following condi t i o n s a r e s a t i s f i e d :

jj

(a)

a A ( b V c ) = (a A b) V ( a A c )

(b) (a v b)

(a

V c)

(c)

0

and

(d)

For e v e r y

A

such t h a t

b

is a l a t t i c e .

AL 5 :

0 L S, from S .

1

are i n a E B

a v b = 1

i s unique.

d e n o t e d by

for a l l

B.

b. E B ( I t c a n be shown t h a t

t h e r e e x i s t s an e l e m e n t

and a A b = 0. i s c a l l e d t h e complement o f

b

a V (b A c ) =

and

a , b , c E B.

a

and i s

-a.) If

B

i s a Boolean a l g e b r a and S B such t h a t 5-maximal p r o p e r i d e a l d i s j o i n t

then t h e r e e x i s t s a

56.

ALGEBRAIC FORMS

99

(See 6 . 2 ( 4 ) f o r t h e d e f i n i t i o n of a p r o p e r i d e a l . non-empty i d e a l if

a E E

i n a Boolean a l g e b r a

E

B

Every 0,

contains

for

A a = 0 E E.)

then - a

The proof t h a t AL'3 + AL 5 is analogous t o t h e proof o f 6 . 4 s i n c e a Boolean a l g e b r a can b e d e f i n e d a s an a l g e b r a i c system i n such a way t h a t an i d e a l i s a s u b a l g e b r a (see t h e remarks f o l l o w i n g 6 . 2 ) . AL 5

THEOREM 6 . 9 :

PROOF:

Let

M 7.

be an a r b i t r a r y s e t and

x

f i n i t e character.

-+

3

Let

= (B,{v,A})

a property of

P

be d e f i n e d a s f o l l o w s :

v = U, and A = n , (0 = 0 and 1 = x ) , t h e n 8 i s t h e Boolean a l g e b r a of t h e set of a l l s u b s e t s o f x. L e t

B = p(x), S = {y: y

P(g)

5x

and

1

P ( y ) }.

S _c B

Then

and

f o r e v e r y p r o p e r t y of f i n i t e c h a r a c t e r

0

S

since

Therefore,

P.

AL 5 i m p l i e s t h a t t h e r e e x i s t s a maximal p r o p e r i d e a l

d i s j o i n t from

It i s clear t h a t

S.

maximal s u b s e t o f

UE

E

i s t h e desired

x.

The Boolean P r i m e I d e a l Theorem ( B P I ) i s t h e s t a t e m e n t : BPI:

Every Boolean a l g e b r a c o n t a i n s a p r o p e r prime i d e a l .

I n a Boolean a l g e b r a , a p r o p e r i d e a l i s prime i f f it i s

-c-maximal.

W e o b t a i n AL 1 from

BPI

by changing "Boolean

a l g e b r a " t o " l a t t i c e w i t h a u n i t element and a t l e a s t one o t h e r e l e m e n t " , and w e o b t a i n

"8"

o r "{l}"

for

"S".

minor v a r i a t i o n s of B P I .

from

BPI

AL 5

by s u b s t i t u t i n g ,

Thus, AL 1 and AL 5 a p p e a r t o be However, i n 1 9 6 7 , Halpern & Levy

[1971] proved t h a t B P I does n o t imply t h e axiom o f c h o i c e i n NBG.

( E a r l i e r , Halpern [ 1 9 6 1 , 1 9 6 4 ] proved t h a t B P I does n o t

imply t h e axiom of c h o i c e i n N B G O . )

t o o b t a i n AL 1 and a t f i r s t glance.

AL 5

The changes made i n B P I

a r e n o t a s i n i n o r a s one may t h i n k

P. Howard [ 1 9 7 7 ] d i s c u s s e d v a r i a t i o n s of AL 5 o b t a i n e d by p l a c i n g c o n d i t i o n s on t h e s e t AL 5 ( V ) :

that

0 $? S

If

and

B S

S.

i s a Boolean a l g e b r a and i s closed with respect t o

S _c B V

such

( j o i n ) , then

PART I . SET FORMS

100

t h e r e e x i s t s a c_-maximal i d e a l d i s j o i n t from

S.

i s a Boolean a l g e b r a and S C_ B s u c h is closed with respect t o (complementat i o n ) , t h e n t h e r e e x i s t s a 5-maximal i d e a l d i s j o i n t from S . If

AL 5 ( - ) :

0

that

and

S

f!

-

B

S

It is c l e a r t h a t

AL 5 ( V )

THEOREM 6 . 1 0 :

PROOF:

AL 5

Suppose

-+

AL 5 ( V )

-+

AL 5 .

implies

AL 5(-)

Suppose

x

j J ( u x), where

i s ux.

Let

and l e t

S = S'

-+

iff

I

such t h a t

{a} E I.

maximality of

u E x

a Then S

n

Let

=

8.

V.

Thus,

( T h e r e i s no l o s s

c a n b e r e p l a c e d by A

is

u E x

But

UX:

0

is

@

u

x 4.)

and

ideal i n

B

t h e r e i s e x a c t l y one

disjoint a E u

{ a ) , {b} E I.

such t h a t

{a,b) E S, x

Suppose t h e r e i s

S.

contradicting the fact that

is

(3u E x ) ( a E u

and

{ a } E I ) }.

I n AL 5 , i f w e r e q u i r e t h a t

S be c l o s e d w i t h r e s p e c t (meet) t h e r e s u l t i n g s t a t e m e n t i s e q u i v a l e n t t o B P I . That i s t h e s t a t e m e n t :

to

A

SM: 0 g S

and

If

B S

1

T h e r e i s a t l e a s t o n e b e c a u s e of t h e

A choice set f o r

{a E

n,

is closed with respect t o

S

b e a 5-maximal

a , b E u, a # b ,

and

0.

U,

and t h e d e f i n i t i o n o f

I

{ a , b j E I. I =

x

is

Then I

Then f o r e a c h

S.

n

AC 2 .

u

V

- {O,l}.

complementation. from

and

i s a set of p a i r w i s e d i s j o i n t s e t s

of g e n e r a l i t y b e c a u s e e a c h B =

AL 5(-).

with respect t o

e a c h o f which h a s a t l e a s t f o u r e l e m e n t s . Let

-+

AL 5.

THEOREM 6 . 1 1 :

PROOF:

AL 5

i s a Boolean a l g e b r a , S c_ B

B

0 j? S. L e t 2 b e t h e c l o s u r e of S Then f o r any i d e a l I 5 B , I fl S = AL 5 ( V )

and

i s a Boolean a l g e b r a and is closed with respect t o

S _c B A,

such t h a t

then t h e r e i s a

56.

ALGEBRAIC FORMS

5-maximal ideal disjoint from BPI

-f

then

101

S.

To show that SM .+ BPI, take S = 111. To prove that SM, first note that BPI ..BPI', where BPI': If I is a proper ideal in a Boolean algebra B, I can be extended to a proper 5-maximal ideal.

(Clearly, BPI' + BPI. To prove the converse, show that if I is a proper ideal in B, then applying BPI to the factor algebra, B/I, induces a &-maximal ideal in B which contains I.) Let S 5 B be such that 0 jZ S and S is closed with respect to A . Let J = {a E B: (3b E S ) (a s b) 1 . The set J is a proper ideal in B. Use BPI' to extend J to a c-maximal proper ideal, I. Then I is a 5-maximal proper ideal disjoint from S.

-

On the other hand, suppose we restate

SM

as follows:

S: If B is a Boolean algebra and S 5 B such that S is closed with respect to A then there is a c_-maximal proper ideal I such that I n S 5 lo].

It is clear that S implies BPI, but it is not known whether BPI implies S. A corresponding change in the wording of AL 5, AL 5 ( V ) , or AL 5(-) does not change the strength of these statements. They remain equivalent to the axiom of choice. Andreas Blass raised the question whether the following weakened from of AL 1 implies the axiom of choice. AL 1D: Every distributive lattice with an identity element and at least one other element contains a 5-maximal proper ideal. The dual to AL lD, DAL lD, is clearly eauivalent. DAL 1D: Every distributive lattice with a zero element and at least one other element contains an ultrafilter. DAL 1D. We shall prove Clearly. AL 1 -+ AL 1D DAL 1D * M 14. (The proof is due to Klimovsky 119581.) CI

102

PART I , SET FORMS

THEOREM 6 . 1 2 :

PROOF:

DAL I D

M 14.

b e a non-empty s e t and l e t

x

Let

+

B = fp(U x ) .

$ ’ = (B, { U , n,-}) i s a Boolean a l g e b r a , where U i s set u n i o n , n i s s e t i n t e r s e c t i o n , is the relative complement (-u = Ux u ) , 0 = 0 and 1 = Ux. A l s o x C_ B.

Then

-

-

be t h e c l o s u r e of

Let

2.

If

x

has t h e €.i.p.

x

empty) t h e n

x

with respect t o

U

and

II i n

( e v e r y f i n i t e i n t e r s e c t i o n i s non-

i t s e l f i s t h e 5-maximal s u b s e t .

If not then

x s o ( x , { U , n } ) i s a d i s t r i b u t i v e l a t t i c e which s a t i s f i e s t h e h y p o t h e s i s of DAL 1 D . Thus, t h e r e i s a n u l t r a f i l t e r u 5 x. Then, it i s e a s y t o see t h a t u fl x i s a c_-maximal 0 E

s u b s e t of

x

with t h e f . i . p . ,

q.e.d.

I t f o l l o w s from t h e work of Rubin & S c o t t [1954] t h a t i f

t h e words “c_-maximal p r o p e r ” i n AL 1 D a r e r e p l a c e d by “ p r i m e “ , t h e r e s u l t i n g s t a t e m e n t i s e q u i v a l e n t t o BPI. W e s h a l l c o n s i d e r n e x t s e v e n more a l g e b r a i c e q u i v a l e n t s of t h e axiom of c h o i c e . AL 6 :

2

If

i s a complete l a t t i c e , a , b ,

= (L,{V,A})

i s compact, c s b and c ,d a t h e n t h e r e m E L s u c h t h a t a 5 m i b , c d m and if t h e r e i s a d E L s u c h t h a t m d s b then c i d (In o t h e r w o r d s , m i s a maximal e l e m e n t i n t h e i n t e r v a l between a and b which i s n o t g r e a t e r t h a n o r e q u a l t o c ) .

c E L, a s b, c

exists an e l e m e n t

B e f o r e s t a t i n g t h e r e m a i n i n g forms i n t h i s s e c t i o n w e r e q u i r e some f u r t h e r d e f i n i t i o n s . DEFINITION 6.13:

Let

4

=

(A,{Ok:k E K))

be a n a l g e b r a i c

s y s t e m (see 6 . 1 ) . (1)

whenever

A subalgebra

B =

nc)

B

is s a i d t o be m e e t - i r r e d u c i b l e i f i s a set of s u b a l g e b r a s t h e n

where

B E L ‘ .

0 5 A

(2)

either

a E 0

i s s a i d to g e n e r a t e A i f f o r e a c h a E A o r t h e r e e x i s t s a n a t u r a l number n 2 1, k E K

and

b l ,b 2 , . . , b n E A

and

a = 0

k blrb2t-

-

*

such t h a t tbn) *

B

generates

{bl,b2,...,bn}

56.

ALGEBRAIC FORMS

103

(3) B _ c A is said to be algebraically dependent if there exists a natural number n 2 1, k E K, b E B and {b) generates {al,a2,...,an} al,a2,...,an E A such that B and b = Ok(al,a2,...,an). Otherwise, B is said to be algebraicallv independent.

-

(4) B _c A is said to be a basis for A if B is algebraically independent and generates A. (A basis is a -c-maximal algebraically independent subset of A.)

(5) An element x E A is said to absolutely dispensible if for all sets S _c A, whenever A is the smallest subalgebra containing S as a subset then A is the smallest {x} as a subset. subalgebra containing S

-

AL 7 - AL 12 all refer to an algebraic system subalgebras in that system.

and

AL 7: Every subalgebra is the intersection of all meetirreducible subalgebras containing it as a subset. AL 8: Every proper subalgebra has a proper meetirreducible subalgebra containing it as a subset.

AL 9: The set of absolutely dispensible elements is equal to the intersection of the set of all 5-maximal proper subalgebras. (We make the convention that n 0 is the whole algebra. ) For every element in a basis there exists a =-maximal subalgebra which does not contain it as an element. AL 10:

AL 11: If A has a finite basis and B c A is a subalgebra then there exists a 5-maximal proper subalgebra which contains B as a subset. AL 12:

Every algebra has a basis.

Diener [1956] proved AL 6 and AL 8 equivalent of the axiom of choice. AL 7 was derived from the choice by Birkhoff & Frink [1948]. However, there in their proof (the proof of Lemma 4 , pp. 301-2,is

to a form axiom of is an error incorrect)

PART I , SET FORMS

104

which w a s l a t e r n o t e d and c o r r e c t e d by D i e n e r [19561.

McCoy

[1938] proved a s p e c i a l case of AL 7 and Fuchs [19491 gave a c o r r e c t p r o o f o f AL 7 .

( A c t u a l l y , Fuchs p r o v e d a s p e c i a l

c a s e , b u t h i s p r o o f e x t e n d s t o t h e g e n e r a l case v i r t u a l l y unchanged.)

The p r o o f g i v e n h e r e i s s i m i l a r t o D i e n e r ' s . Frattini

AL 9 i s due t o F r a t t i n i [1885] and Neumann [ 1 9 3 7 ] .

i n 1885 o n l y proved t h e theorem f o r f i n i t e g r o u p s and d i d n o t r e q u i r e t h e axiom of c h o i c e . The p r o o f g i v e n h e r e i s e s s e n t i a l l y t h e same a s t h e p r o o f g i v e n by Neumann i n 1937. K r u l l [1929] d e r i v e d a p r o p o s i t i o n which i s s i m i l a r t o AL 11 from t h e w e l l - o r d e r i n g theorem.

Schmidt [1953] p r o v e d t h a t

AL 7 , AL 9 , AL 1 0 , and AL 11 imply a form o f t h e axiom o f

choice.

AL 1 2 i s r e l a t e d t o a

(See a l s o Schmidt [ 1 9 6 2 ] . )

form g i v e n i n Schmidt [1962]. W e s h a l l prove M 3

-+

AL 6

AL'3

-+

-+

AL 7

AL 9

AL'3

-+

-+

-+

AL 8

AL 1 0

AL 11

-+

-+

-+

AC 3 ,

M 7,

M 7,

M 7 - + A L 1 2 + A C 2 . I t i s c l e a r t h a t AL 7

-+

AL 8

and s i n c e a n e l e m e n t i n

t h e b a s i s i s n o t a b s o l u t e l y d i s p e n s i b l e , AL 9

-+

AL 1 0 .

Being

a l g e b r a i c a l l y i n d e p e n d e n t i s a p r o p e r t y of f i n i t e c h a r a c t e r ,

so

M 7

-+

AL 1 2 .

THEOREM 6 . 1 4 :

M 3

PROOF:

= (L,{V,A})

Let

a , b E L, a i b , l e t

Suppose

c E L

3

_c [ a , b ]

= {D:D

{a) E in 9

8. L e t N . We shall

-+

AL 6.

be a c o m p l e t e l a t t i c e and f o r

[ a r b ] = Cd:d E L

and

a

d

L

b}.

i s compact, c i b and c & a . L e t a n d D # 0 and c & VD}. 3 # @ since be a non-empty n e s t c o n t a i n e d a s a s u b s e t show t h a t UN E 9 Clearly, i f N 0,

.

+

v

t h e n Ux Z 0 and U $ _ C [ a r b ] . Suppose c i U / s , t h e n s i n c e c i s compact t h e r e e x i s t dl,d2,...,dk,N1,N2,...,Nk, di E Ni

EN

such t h a t

c

i

v {dl,d2,...,dk}.

Since

f

is

96.

ALGEBRAIC FORMS

10 5

a nest, for some j E {1,2,...,k}l Ni 5 Nj for all i = 1,2,...,k, therefore c _c Nj. This is a contradiction has a since N j €9. Hence U l E a so that by M 3 , maximal element M.

v

a

v

M; then a h m s b and c L m. Suppose there Let m = is a d € [a,b] so that m < d 5; b. This implies M C M U {d} so that M U {d} Therefore, c (M U {d}) = d, q.e.d.

v

e$.

Before proceding with the proof that AL 6 a lemma due to Buchi [1952].

-+

AL 7 we prove

.f = (L,{V,AI) is said to be an algebraic lattice if L is the set of a l l subalgebras of an algebraic system. The meet of two elements of L is ordinary set intersection, but the join of two elements is the smallest subalgebra containing them both. (This is usually not ordinary set union.) It is easily verified that an algebraic lattice is a complete lattice. LEMMA 6.15: Every element in an algebraic lattice is the join of compact elements. Let .f = ( L , { V , A ) ) be an algebraic lattice and let p E q E L. We wish to show that the subalgebra generated by u , (the smallest subalgebra containing p ) , denoted by (p), is a compact element of L. Since ( p ) is a subalgebra and L is the set of all subalgebras of an algebra, ( p ) E L. Suppose M 5 L and ( p ) 5; a , then p E a This a@ a EM implies that there exists a finite subset N of M such that 1! E a Since a is a subalgebra it follows that aEN aEN (p) 5 a. Hence (11) is compact. a€N q.e.d. Let q E L; then clearly q = V ( p ) , P€ 9 THEOREM 6.16: AL 6 + AL 7. PROOF:

v

.

v

v .

v

v

PROOF: Let A be an algebraic system and (L,{V,A}) the algebraic lattice of the set of all subalgebras of A. Let B E L and let D be the intersection of all meet-irreducible

PART I, SET FORMS

106

subalgebras containing B. Suppose B D. Then there exists a compact element C E L such that C $! B and C 5 D. For suppose p E D B, then by 6.15, ( p ) is compact and clearly

-

(u) $ B

but

(p)

D.

5

Let

C = (p).

By AL 6, there exists an M E L such that B 2 M S. A, and M is maximal. Suppose M is not meet-irreducible. Then there exists a subset K 5 L such that M = A K and M K. Let N E K; then M N. Since M is maximal C S. N for all N E K. Therefore C s I\ K = M. This is a contradiction so that M is meet-irreducible. C

d M

B s M and D is the intersection of all meet-irreducible ideals which contain B, therefore D 5 M. Since C 5 D it follows that C 5 M and this again is a contradiction. Hence B = D , q.e.d. THEOREM 6.17:

AL 8

AC 3 .

-+

Let f be an arbitrary function whose domain is s and assume for all x E s , f(x) has at least two elements as follows: We define an algebra d = (A,{OU:u E A)) PROOF:

A

=

x E s

{(x,y):

OU(V,W) =

[

and

u

if

v

v

otherwise

=

y E f(x)}

(x,y),w = (x,z),y # z ,

.

A subset of A is a proper subalgebra if it is a function. It is clear that a subalgebra of d is meet-irreducible if and only if it is maximal. AL 8 implies, therefore, that there exists a maximal subalgebra B c A. If we let g = B, then g is a function, % ( g ) = 3(f) = s , and f o r all x E s , g(x) E

f

(x),

q.e.d.

THEOREM 6.18:

AL'3

+

AL 9.

PROOF: Let A be an algebraic system. Let D C_ A be the set of all absolutely dispensible elements and let be the set of all maximal proper subalgebras. We wish to prove D =

nn .

96.

is

.

n./C

x #

Suppose

< M.

x

M E1

Then t h e r e is a

x # D.

so t h a t

x # D.

NOW, suppose

closure operator. such t h a t

For any s e t

and

x g S.

x # M.

Clearly,

PROOF:

Let

of f i n i t e c h a r a c t e r .

x E A

For each

K = { ( y1 , y 2 ,

S c A

Assume

b E B.

maximal s u b a l g e b r a

-

of

A

P.

E A 1 a s follows:

P(S).

There e x i s t s a

PI

otherwise

c_

A

P.

is a

B

B , and

AL, 1 0 i m p l i e s t h a t t h e r e e x i s t s a

A

b g M.

such t h a t

M

and

P(N),

with the property

w e have

P(N

{b}).

Since

from t h e maximality o f

Thus e i t h e r

THEOREM 6 . 2 0 :

and

a property

P

i s a subalgebra.

B

{b], N * {b} = M

N = M U {b).

q.e.d.

s i n c e no p r o p e r s u b a l g e b r a c o n t a i n s

M c N

and

Therefore, t h e r e is a s m a l l e s t

P.

e v e r y p r o p e r s u b s e t of

If

S c_ A,

implies C(T)

which d o e s n o t have t h e p r o p e r t y

Suppose

A

d e f i n e n-ary o p e r a t o r s on

would have t h e p r o p e r t y A

C

does n o t have t h e p r o p e r t y

A

i s a s u b a l g e b r a i f and o n l y i f

basis for

T

M 7.

+

... , y n ) :yl , y 2 ,.. . ,Y,

B

be

finitary

f o r any

nk41c_ D,

f i n i t e s e t which does n o t have t h e p r o p e r t y f i n i t e set

a

AL'3

which c o n t a i n s

M

b e an a r b i t r a r y s e t and

A

C(S)

W e define a

Therefore

This implies

AL 1 0

THEOREM 6 . 1 9 :

C

a s follows:

P

P(C(T)).

.

n,/(

x #

Hence

let is

S _C A

S ; then

C ( T U { X I ) = A.

t h e r e i s a maximal s u b a l g e b r a

c_ N

M U {XI

.

x @ D, then there e x i s t s a set

If

C(T) c A

property of f i n i t e character

-

5 n,k

D

This implies

t h e smallest subalgebra containing

M

such t h a t

Therefore t h e smallest subalgebra containing

A,

P(S)

107

ALGEBRAIC FORMS

M

or

M

U {bl

M. Hence i s a maximal s u b s e t

P.

AL'3 -+ AL 11.

PROOF:

Suppose

Bo c A

i s a proper subalgebra.

A

has a f i n i t e b a s i s

B = {bl,b2,...,bn}

For any s e t

S

c_

A

we

PART I, SET FORMS

108

define C(S) as the smallest subalgebra containing S. Then c is a finitary closure operator. Define the property Pjf A, P.(S) bj g S. j = 1,2,...,n as follows: for any set S 1 P is a property of finite character for each j = 1,2,...,n. j A L ' 3 implies that we may construct the following finite sequence of subalgebras:

.

If b. E Bj,l

then

Bj = Bj-l, j = 1,2,...,n.

If

then

B

3

b. g Bj,l 7

containing

Bj-l

is a maximal subalgebra j as a subset and b . B j = 1,2,...,n. 1

By construction,each B is a proper subalgebra and j contains BjW1 5 Bj for each j = 1,2,...,n. Hence each B j Bo as a subset. We claim Bn is a maximal proper subalgebra containing Bo. For suppose Bn is not maximal. Then there exists a proper subalgebra D such that Bn c D. Since D # A, for some bk E B, bk g D. But then Bk c D which contradicts the definition of Bk. THEOREM 6.21: PROOF: to be Q.

AL 11

-+

M 7.

Similar to 6.19.

THEOREM 6.22:

AL 12

-+

Take the subalgebra of AL 11

AC 2.

PROOF: Let x be a set of non-empty pairwise disjoint sets. Let A = Ux. For each a E A we define a unary operation Oa on A such that for each b E A

Oa(b) =

(3u E x)a, b E u ,

a

if

b

otherwise

.

A subset B 5 A is algebraically independent iff all the elements of B belong to different elements in x. Thus, a basis for A is a choice set for x. It is also clear that the following variations of AL 12 are equivalent to the axiom of choice.

16.

ALGEBRAIC FORMS

109

AL’12: Any algebraically independent subset of an algebra can be extended to a basis. AL 13: Any subset of an algebra which generates the algebra contains a basis.

We shall next consider some statements about groups and related structures which are equivalent to the axiom of choice. For convenience we shall define the following algebras: DEFINITION 6.23 : (a) A groupoid is an algebra binary operation.

(A, 101) : where

(b) A cancellative groupoid, (A,( 0 ) ) satisfies the cancellation laws: (Va,b,c E A) [aob = aoc

-f

b = c)

A

o

is a

is a groupoid which

(aoc = boc

+

a = b)].

(c) An abelian (commutative) groupoid, (A,{o)) groupoid which satisfies the commutative law:

is a

(Va,b E A) (aob = boa) (d) A semigroup, (A,{o}) is a groupoid which satisfies the associative law: (Va,b,c E A) (ao(boc) = (aob)oc). (el A guasigroup, (A,{o}) is a cancellative groupoid such that (Va,b E A) (3x,y E A) (aox = b A yoa = b)

.

(f) A loop is an algebra (A,(o,e}) such that (A,{o)) is a quasigroup and e is the identity element: (Va E A ) (aoe = eoa = a). (The cancellation law implies that the identity element is unique.) (9) A group is an algebra (A,{o,e)) where (A,{o}) is a quasigroup, o is associative, and e is the identity element. (In a group, each element a has a unique inverse, a-l, such that aoa-’ = a-loa = e, and the cancellation laws hold. Frequently it is necessary to consider a group as .) the algebra (A,{o,e,-’})

110

PART I, SET FORMS

(h) A ring is an algebra (A,{+,=,O}) such that is associative and the (A,i+,O)) is an abelian group, distributive laws hold: (Va,b,c E A) [(a*(b+c) = a * b + a-c) A ((a+b) *c = a * c + b-c)]. (We call a ring commutative if

is commutative. )

(i) An integral domain, (A,{+,.,O}), is a commutative ring such that (Va,b E A) (a # 0 A b # 0 .+ a * b # 0). ( j ) A field is an algebra (A,{+,*,O,l}) such that {0},{*,1)) is an (A,{+,-,O}) is a commutative ring and (A abelian group.

-

-

(k) If

(A,{+,.,O,l}) is a field and there is an such that a + + a = na = 0, then the n times smallest positive integer n with this property is called the characteristic of the field. (Since a field is an integral domain, no divisors of zero, the characteristic is a prime.) If there is no such positive integer n such that na = 0, we say the field has characteristic zero.) n E w

-

...

{O}

The first result is due to Hajnal and Kertesz [ 1 9 7 2 ] who prove the following statement is equivalent to the axiom of choice. AL 14: On every non-empty set there exists a cancellative groupoid. THEOREM 6.24

:

WO 2

-+

AL 14.

PROOF: Suppose x is a non-empty set. If x is finite, say x has n elements, then we can take as our groupoid, the natural numbers modulo n. (In fact, this is a commutative ring with multiplicative identity). If x is infinite, it follows from the well ordering theorem that there is a set y such that x is equipollent to z = $',,(y) = the set of all finite subsets of y. (By WO 2 , there is an infinite ordinal a such that x M a . It is easy to show y,,(a) fi: a . ) Define the operation o on z to be symmetric ( u n v). difference. That is, if u , v E z , uov = ( u U v)

-

16.

ALGEBRAIC FORMS

111

It is easy to show that ( z , { o , p l ) ) is an abelian group. Thus, since x m z , we can define operations on x to make it an abelian group. THEOREM 6.25:

AL 14

PROOF: Let x be a that y m r(x) (Hartogs' z = x U y. AL 14 implies defined on z such that

+

WO 1.

non-empty set. Let y be a set such function) and x n y = pl. Let that there is a binary operation, +, (z,{+)) is a cancellative groupoid.

For each u E x there is a v E y such that u + v E y. If not, the function f such that for each v E y, f(v) = u + v is a 1-1 function mapping y into x. (The cancellation law implies f is 1-1.) Thus, y < x which contradicts the definition of y. Let w = y x y; then since y can be well-ordered, so can w. Define a function g:x -+ w so that for each u E x, g(u) is the smallest element in

It follows from the cancellation law that g is 1-1. Thus, x can be well-ordered. (Actually, only a one-sided cancellation law is required.) It is clear from the proofs of 6.24 and 6.25 that if in AL 14 we replace "cancellative groupoid" by any one of the following, the statement remains equivalent to the axiom of choice: (i) cancellative semigroup; (ii) cancellative abelian semigroup; (iii) quasigroup; (iv) loop; (v) group; (vi) abelian group; and (vii) ring. Also, for infinite sets, a tsanscendental extension of the field of rational numbers will, with the well-ordering theorem, give a field of any infinite cardinality. (The number of elements in a finite field is always the power of a prime.) So for infinite sets we can add the following to the above list: (viii) integral domain and (ix) field. Clearly, any other algebraic structure between a cancellative groupoid and a field will work. Krull [1929] noted that the axiom of choice implies that every commutative ring with a multiplicative identity element

112

PART I, SET FORMS

ha$ a 5-maximal proper ideal and Scott [1954] asked whether the converse holds. Hodges [1979] proved that the answer is yes. In fact he proved a stronger statement. DEFINITION 6.26: Suppose (A,{+,-,O,l}) is an integral domain (see 6.23(i)) with a multiplicative identity element 1, 0

# 1.

(a) a E A is called a & element if there is a b E A such that a - b = 1. (Since multiplication is commutative, b*a = 1 also. Thus, a unit element is an element which has a multiplicative inverse.)

(b) a E A is called irreducible if element and whenever a = b.c, either b element.

a is not a unit or c is a unit

(c) (A,{+,.,O,l}) is a unique factorization domain if it is an integral domain with multiplicative identity element 1, 0 # 1, satisfying the following properties:

-

a

i) If a E A {O} and a is not a unit element, then can be written as a product of irreducible elements.

ii) If a = b,.b 2...bj = cl*c2...~k where the b's and c's are irreducible elements, then j = k and the b's and c's can be reordered so that for each i there is a unit element ei such that bi = eici, i = 1,2, ...,j. (d) I _c A such that

is called an ideal if

I

is a subalgebra

(Va E A) (Vb E I) (a.b E I) I _c A is called a proper ideal if (Thus 1 f I if I is proper.)

I

is an ideal and

I # A.

(e) 1 5 A is called a prime ideal if I is an ideal and for all a, b E A if a * b E I then a E I or b E I. Hodges' [1979] form is the following: AL 15: Every unique factorization domain contains a i-maximal proper ideal.

ALGEBRAIC FORMS

16.

113

Using a proof similar to the proof of 4 . 3 , prove that M 3 + AL 15. (In 4 . 3 take y = {I c_ A: 1 g I

I

A

is an ideal}

Then, the same proof shows that THEOREM 6 . 2 7 :

AL 15

y

it is easy to

.

has ac_-maximal element.)

M 5(RA,L).

+

PROOF : Suppose (x ,5) is a ramified set. (That is, is a partially ordered set such that initial segments (XIS> are linearly ordered by 5 . ) We shall prove that x contains a 5-maximal linearly ordered subset. Let F be the field of rational numbers and let F[x] be the abelian polynomial ring with elements u E x as indeterminates. (That is, F 5 F[x] and each element of F[x] F is a sum of elements of the form aulu2...um where a E F, ui E x and the ui are not necessarily distinct.) F[x] is a unique factorization domain and for y C_ x if yF[x] is the set of all finite sums from {uv:u E y A v E F[xl] then yF[x] is a prime ideal in F[x]

-

.

x

Let L and let

be the set of all 5-linearly ordered subsets of

u

S = F[xl

yF[x].

YEL S

is a subalgebra of

F[x]

R = S-'F[x]

.

Let

= {u/s:

u E F[x]

A

s E S).

If addition and multiplication are defined in the natural way, R is a unique factorization domain. u/s E R is a unit (invertible) iff u E S. If, for each w E x, f w = {v E x:v s; w} then u/s E R is a unit iff (Vv E x)u g fF[x]

.

AL 15 implies that M. Let c = u/s E M I factors. Then

R

contains a 5-maximal proper ideal

so that u

and

s

have no common

114

PART I , SET FORMS

u = altl

+

a Z t Z +...+

a t

j j f

ai E F and ti i s a monomial i n n o t a u n i t , f o r i f it w e r e w e would have

where

if

u # 0

ti,

monomials,

x.

of

U

u E

u f! S , so

then

yF[x].

YE L

c is Thus,

x. Moreover, cc-l = 1 E M.

This implies t h e

have f a c t o r s i n some l i n e a r l y o r d e r e d s u b s e t

Therefore, i f

f i n i t e non-empty

c # 0,

t h e r e must b e a t l e a s t one

s - l i n e a r l y ordered set

A

c_ x

such t h a t

( i )e a c h monomial, t i , h a s a f a c t o r i n A, and ( i i ) e a c h A o c c u r s a s a f a c t o r o f some ti. The s e t A

element of

is

n o t u n i q u e , b u t t h e r e a r e a t most a f i n i t e number o f them, Let

A1,A 2 , . . . , A k .

E ( c ) = { v : ( ~ ~ E( w 1 )s i s k A v i s t h e < - l a r g e s t e l e m e n t of A i ) ] .

v E E(c)

If

then

c E fR.

c = 0

If

then

E(c) =

8.

We

define D = {v

E x : ( V c E M) ( c # 0

(3w E E(c)) ( v

5;

w

V

w < v)}.

i s a 5-maximal l i n e a r l y o r d e r e d s u b s e t The p r o o f depends on t h e f o l l o w i n g t h r e e lemmas:

W e s h a l l show t h a t

of

.+

x.

(1) D (2)

D

(3)

M

D

- M, i s a l i n e a r l y o r d e r e d i n i t i a l segment of

x,

DR.

By (1) and ( 3 1 , M = DR. I f D is n o t a c_maximal l i n e a r l y o r d e r e d s u b s e t o f x , t h e n , by ( 2 ) , t h e r e i s a which is < - l a r g e r t h a n e v e r y c o n t r a d i c t s t h e m a x i m a l i t y of

u E D.

Thus,

DR C GR,

v E x

which

M.

W e s h a l l s k e t c h t h e p r o o f s of

(l), ( 2 ) and ( 3 ) .

PROOF OF (1): D _C M. Suppose

is a

u E R

v E D and a

and

c E M

v

M.

Since

such t h a t

M

i s maximal t h e r e

u - v + c = 1.

Moreover,

ALGEBRAIC FORMS

96.

115

v S because {v} E L and v E {v)F[x]. Therefore, v is not a unit in R (u-v # l), so c # 0. By the definition of D, there is a w E E(c) which is comparable with u and c c E GR. If w s v, then c, v E vR, which implies the contradiction that 1 = U - v + c E GR. If v i w, then c, v E wk so again we have the contradiction 1 = u'v + c E GR. PROOF OF (2):

of

D

is a linearly ordered initial segment

x.

Suppose u , v E D _c M. Then u + v E M I so u + v is not a unit element. Therefore, there is a w E x such that u + v E GF[x]. This implies u , v E w' so u and v are comparable. Suppose u E D, v s u and v E x. If c is a element of M then there is a w E E(c) such that are comparable. If u 2 w then v s w so v E D. f then v, w E u , so v and w are comparable. This implies that v E D. PROOF OF ( 3 ) :

non-zero u and w If w s u, again

M _C DR.

-

Suppose c E M DR and suppose c # 0. No element u E D is in E(c); for otherwise c E f R _C DR by (2). Let v1,v2, ,vk be the distinct elements of E(c). Since the vi t D , there is a non-zero element wi E M such that for all u E E(wi) , u is incomparable with vi. Since F is an infinite field, we may choose scalars al,a2,...,ak E F such that in

...

d = c + a w

+ a2w2 +...+

akwk E

no monomial which occurs with a non-zero coefficient in c or some wi, vanishes in R. Suppose w E E(d) Then there is some vi such that vi 5 w. Also, there is a v E E(wi) f such that v i w. Therefore, v, vi E w, which implies that v and vi are comparable. This contradicts the choice of the vi, q.e.d.

.

It follows as a simple corollary that if in AL 15, "unique factorization domain" is replaced by either "commutative

116

PART I, SET FORMS

ring with multiplicative identity" or "integral domain with multiplicative identity" the resulting statement remains equivalent to the axiom of choice. We continue with a form which was given originally by Klimovsky [1962] and later extended and generalized by Felgner [1976]. AL

16:

Every group contains a 5-maximal abelian subgroup.

The proof that M 3 + AL 16 is similar to the proof of Alternatively, it is easy to show that M 7 + AL 16 because being abelian is a property of finite character and a c-maxima1 abelian subset of a group is a subgroup. 4.3.

DEFINITION 6.28: The group (G,{*,e,-1 3 ) is said to be free if there is a non-empty X 5 G such that:

(1) e t x I (2)

If x E X, then

x-l E X I

(3) If y E G, then either y = e or there is a unique f 0 ) and a unique sequence (yl,...,yn) such that n E w each yi E X U X - l , yi*yi+l-# e for all i, and = {x-I:x E X I . y = y1*y2*...BY, , where X

-

The set X is said to generate G and the representation specified in (3) is called the reduced form of y. Given any non-empty set X there is a free group which is generated by X I and any two such groups are isomorphic, with the isomorphism being the identity on X. Hence, we call such a group the free group generated by X. THEOREM 6.29:

AL 16

-+

AC 13(2).

PROOF: Let x be a set of non-empty sets. For each u E x, let FU be the free group generated by u. Let lU be the identity element of FU. Let G be the weak direct product of the FU1s. G =

If

E

x FU:{u

UEX

E x: f(u)

# lU}

is finite).

56.

ALGEBRAIC FORMS

117

If f, g E G, f*g(u) = f(u).g(u) and f - l ( u = (f(u))-l for all u E x. The identity element 1 is the function f such that for all u E x, f(u) = lU. Then ( G , { * 1,-1l ) is a Let A be a 5-maximal abelian subgroup of G. For group each u E x, let AU be the set of uth coordinates of elements Each AU is an abelian subgroup of Fu and no of A - l U l . For -if AU = {luJ and a E u we could extend AU AU to All - U [a], where [a] is the cyclic group generated by a, (all powers of a and a-1 ) thereby contradicting the maximality of A.

-

{lUl and w is in reduced form then w If w E AU must be a power, positive or negative, of a product of elements of u U u-’. Otherwise, AU would not be abelian. Let f(w) be the first factor of w, %(w) be the last factor of w and - ~ ~ . for all C(w) = { f ( w ) , ~ ( ~ ) , ( f ( w ) ) - ~ , ( ~ ( w ) ) Then w1,w2 E AU {lul, C(w,) = C(w2) and 1 2 (C(w,) n uI s 2. Therefore, for each u E x, define g(u) = C(w) n u where w is any element of AU {lull then g is the required choice function. DEFINITION 6.30: Suppose (G,{.,e, -1 1) is a group.

-

-

(a) (Va,b E G) (a,b) = a-lb-lab a and b.

of H

is called the commutator

(b) If H is the group generated by {(a,b):a,b E GI, is called the commutator subgroup of G.

*,...

( (alIa2,.. . fan-l),an) and ...,aklak+l) = e for all

(c) If (al,a ,an-l,an) = G has the property that (al,a2, ai E G I then G is called nil-k. all

(d) G is called metabelian if a,b,c,d E G.

(Nil-1 is abelian.) ((a,b),(c,d)) = e

for

(e) If k t 1 is a natural number and G I is the commutator subgroup of G I define = GI, and G (k+l) = (G(k)) G‘O) = G, G We say the solvability degree of G is k if G(k = {el and G(k-l) # { e l . (Solvability of degree 1 is abe ian and solvability of degree 5 2 is metabelian.)

PART I, SET FORMS

118

Felgner [1976] extended Klimovsky's results and showed that each of the following are equivalent to the axiom of choice. Let

k

2

1 be a natural number.

AL 17(k):

Every group contains a 5-maximal nil-k subgroup.

AL 18(k): Every group contains a 5-maximal subgroup with solvability degree s k. The proof that each implies the axiom of choice is similar to the proof of 6.29. Our next forms are statements about vector spaces.

We shall use the notation av shall use the same symbols + and

instead of Oa(v), and we 0 for V as for F.

The notions of independent, basis, etc. which were defined earlier (6.12) for algebras in general can be defined more simply for vector spaces. DEFINITION 6 . 3 2 : (V,{+,O] U {Oa:a E F))

Suppose F is a field and is a vector space over F.

(a) W _c V

is called a subspace if

(b) A _c V

is said to be linearly dependent if

(3n E

...,vn

w ) (3v1,

W

is .a subalgebra.

E A ) (3all...lan E F) O = alvl

and for some i, ai # 0. linearly independent.

Otherwise, A

+...+

is said to be

anvn'

ALGEBRAIC FORMS

56.

119

said

v = alul (V

+...+anun .

is said to be a linear combination of the

ui's.)

(d) B _c V is said to be a basis of V if B is linearly independent and generates V. ( A basis is a c-maximal linearly independent subset of V . ) (e)

T:V

+

W,

If W is also a vector space over T is called linear iff

(i) T(vl

+

(ii) T(av) Let

B

v2) = T ( v l )

=

aT(v)

+

T(v2)

for all

for a l l

v E V

F

and

vl, v2 E V.

and for all

a E F.

be the following statement.

B:

Every vector space has a basis.

AL 19: For every field over F, every subset of V basis. Let

F

F and every vector space V which generates V contains a

be a field:

AL 20(F): Every vector space V over F has the property that for all A _ c V , if A is linearly independent then A can be extended to a basis of V. AL 21(F): Every vector space V over F has the property that if S _c V is a subspace of V then there is a subspace S ' 5 V such that S n S ' = {O} and S U S ' generates V. It is clear that the axiom of choice implies B, for example AL 12 + B, but it was unknown up to the present time whether B implies the axiom of choice. However, Andreas Blass sent us a proof that B implies the multiple choice axiom. (Blass credits Dr. R. Harting for asking the question

120

PART I, SET FORMS

which Theorem 6 . 3 3 answers and for calling attention to the construction of the ground field, which was a key step in the It proof.) We shall include his proof here (Theorem 6 . 3 3 ) . is still unknown whether B implies the axiom of choice in NBGO. Bleicher [1964] proved that AL 20(F) and AL 21(F) each imply the multiple choice axiom and had some partial results involving AL 19. Halpern [1966] proved that AL 19 implies the axiom of choice. Armbrust 119721 proved that if F is a field of characteristic zero, then the multiple choice axiom implies AL 21(F). AL 19 B and AL'12 AL 20(F) -+ B. We Clearly AL 13 AC 2 , AL 20(F) -+ AL 21(F) MC 1, shall show that AL 19 B + MC 1, and if F is a field of characteristic zero then MC 1 AL 21(F). -f

-f

-+

-f

-f

-f

First, it is easy to see that if F is a field, AL 21(F). For suppose V is a vector space over AL 20(F) -f

F and S 5 V is a subspace. AL 20(F) implies that S has a basis B' and that B' can be extended to a basis B of F. Take S' to be the subspace generated by the elements in B B'. Then S n S ' = t o ] and S U S ' generates V.

-

THEOREM 6 . 3 3 :

B

-f

MC 1.

PROOF: Let x = {ui:i E I) be a non-empty set of pairwise disjoint non-empty sets. Let R be any field such that R (Ux) = pI. Let F be the field obtained from R by adding U x as a set of indeterminates. That is, each f E F is a rational function of the "variables" in Ux. For each i E I, we define the i-degree of a monomial to be the sum of the exponents of elements of ui in that monomial. A rational function f E F is called i-homogeneous of degree d if it is the quotient of two polynomials such that all monomials in the numerator have the same i-degree, n, while all monomials in the denominator have the same i-degree, n - d. The rational functions that are i-homogeneous of degree 0 for all i E I form a subfield K _c F. Thus, F is a vector space over the field K and we let V be the subspace which is generated by

Ux.

ALGEBRAIC FORMS

56.

121

By hypotheses V has a basis, B. For each i E I and t E ui we can express t as a finite linear combination of elements of B with coefficients in K:

where B(t) is a finite subset of B and for each b E B(t) , kb(t) is a non-zero element of K. If s is another element of the same ui as t then s =

c

b€B ( s )

kb(s)*b

If we multiply (1) by the element

.

s/t E K

we obtain

(3)

Since B is a basis each element in V is uniquely expressible as a linear combination of elements of B. Thus, it follows from ( 2 ) and ( 3 ) that B(s) = B(t) and kb(s) = (s/t)kb(t). This means that the finite subset B(t) _c B and the elements kb(t)/t depend only on i, not on the particular t 6 ui; we therefore call them Bi and kbi, respectively. Since kb(t) € K, kbi is i-homogeneous of degree -1 (and j-homogeneous of degree 0 if j # i). This means that when kbi is written as a quotient of polynomials in reduced form, some variables from ui must occur in the denominator. We define a multiple choice function g on x such that for each i E I, g(ui) = the set of those members of ui which occur in the denominator of kbi (in reduced form) for some b E Bi. Then g(ui) is a non-empty finite subset of ui, q.e.d. If a vector space V has a basis and all bases of V have the same cardinal number K, then K is called the dimension of V. Let B' be the following statement: B':

Every vector space has a dimension.

Then we have AC +. B' implies AC in NBGO

B MC 1. It is not known if B or if B' implies AC in NBGO.

+

+.

PART I, SET FORMS

122

(The statement B '

was suggested by Norbert Brunner.)

THEOREM 6.34: AL 19

AC 2 .

+

PROOF: Let x be a set of pairwise disjoint non-empty and let sets. Let R be any field such that R n (Ux) = F be the field obtained from R by adding U x as a set of indeterminates. (The same construction which was used in

Theorem 6 . 3 3 . ) =

Let V

{f: f

is a function, f:x

-+

{u E x: f(u) # 0)

is finite)

be the vector space of

For each

s E Ux

and

-s(u) =

F

and

.

F L X 1 over the field

u E x,

F.

define

s

if

s E u,

0

otherwise,

Is:

and let A = s E Ux}. Then it is easy to see that A is a linear combination generates V because each f E of s ' s . Thus, AL 19 implies that there is a B 5 A such - that B is a basis of V. Suppose s l , s 2 E B and there is a u E x such that s 1 (u) = s1 E u and s2(u) = s2 E u. Then (sil)sl ( s -1 2 ) s 2 = 0, contradicting the fact that B is linearly independent. Thus, for each u E x there is at most one E B such that s ( u ) # 0. On the other hand,since B generates V, for each u E x there is at least one f B such that s(u) # 0. Thus, { s : E Bl is a choice set for x.

-

-

s

s

THEOREM 6.35:

If

F

is a field, AL 21(F)

-t

MC 1.

PROOF: Let x be a set of pairwise disjoint non-empty sets. Let F be any field. Let L(u,F) = {alsl +...+

ansn: n E w,

ai E F

and

si E u3

L(u,F) is the set of all finite formal sums with indeterminates from u and coefficients in F. It can be considered to be a vector space over F, if we assume the usual conventions about addition and scalar multiplication. The set u is then a basis of L(u,F). Let LO(u,F) be the subspace of L(u,F)

56.

ALGEBRAIC FORMS

123

in which the sum of the coefficients of each finite sum is zero. That is, LO(u,F)

=

{alsl

+...+

ansn E L(u,F):al +

..+ an

= 01

.

Let V be the weak direct product of the L(u F ) for u E x and Vo be the weak direct product of the Lo u,F) for u E x.

v

=

X

{f E

is finite}.

L(u,F): {u E x: f(u) # 0 1

UEX

v0= {f E X L~(u,F):{u E x: f(u) #

01

is finite}.

UEX

Then Vo is a subspace of V, and is a subspace V ' c V such that

AL 21(F) implies that there

0 -

(i) V o

n

V;,

=

and (ii) V o U V;,

{O)

generates V.

In what follows we shall identify each function fs E V such that s if s E u , f (u) = 0 otherwise

s E Ux

with the

.

For each v E w' E V b such s1 = v1 + vi vi, v; E V b .

V there is a unique w E Vo and a unique that v = w + w'. Suppose sl, s2 E u E x, and s2 = v2 + v;, where vl, v2 E Vo and Then

s1

-

s2

=

(vl - v2)

+

'vi

Thus, by the definition of V o , we have so vi v; E V o n V b = {O}. Therefore, Consequently, for each u E x there is a that for each s E u there is a vo E Vo For each u E x, let g(u) be the finite u which generate the uth coordinate of multiple choice function on x.

-

If AL 21(F).

THEOREM 6 . 3 6 :

then

MC 1

-+

PROOF: Let F a vector space over

F

- "1,.

s1 - S 2 I v1 - v2 E V O l vi = vi. unique vu E V b such with s = vo + vu. set of elements in vu. Then 9 is a

is a field of characteristic zero

be a field of characteristic zero and V F. Let S be a subspace of V. We use

124

PART I, SET FORMS

MC 1 to prove that there is a linear mapping T from V to S such that TrS is the identity function. Then, if S ' = {v E V:T(v) = O}, the kernel of T, it is easy to prove that S ' is the required subspace. ( S ' is clearly a subspace. If u E S then T(u) = u and if u E S ' then T(u) = 0 so S n S ' = {O}. For any v E V, T(v) = s E S and T(s) = s. Consequently v = s + (v-s), where s = T(v) E S and v - s E S'.) It remains to be shown that such a linear mapping exists. Let f be a multiple choice function on P(V) { @ I . Let L be the set of all linear mappings from a subspace of V into S and let g be a multiple choice function on f?(L) {fl]. Define f(0) = g(pI) = 0. For each ordinal a we define a linear function ha and a set B a as follows:

-

-

ho = identity map on If

a

S,

Bo

= S.

is a limit ordinal, ha =

U h B and B
Ba =

U BB. B
If a = B + 1, let H B be the set of all linear extensions of h B to the subspace S a which is generated by Ba = B B U f(V BB). Suppose g(HB) = {kl,k2, kn}. Then

-

and ha:Sa element of

...,

is a linear mapping. Let S a = 81. Let T = h {a:V aO required linear mapping, q.e.d. -t

S

-

.

be the smallest Then T is the

a.

As a corollary to Armbrust's [1972] Theorem 6 . 3 6 , we have THEOREM 6 . 3 7 : If F then MC 4 (p) -t AL 21(F).

is a field of characteristic p

PROOF: The proof is similar to the proof of 6 . 3 6 . If is relatively prime to p then l/n exists in a field of characteristic p. If F is a field of characteristic p THEOREM 6 . 3 8 : and there is a choice function on the set of all non-empty finite subsets of F then AL 21 (F) -+ MC 4 (p)

.

n

56.

ALGEBRAIC FORMS

125

Follow the proof of 6.35 and for each u E x, construct vu E V;, where 0 # vu = alsl + a2s2 +...+ a s n n n' ai E F, si E u, C a i f 0 and no finite subsum of the i=1 a.s ' s is zero. Subdivide the si's into subsets, [sill such i i that s j E [si] iff a = ai. For some i = 1,2, n, the j number of elements in Isi] is relatively prime to p. Use the choice function to choose one of the ails from the finite set of coefficients of vu; then the corresponding set [si] is the required finite subset of u, q.e.d. PROOF:

...,

If AL 20 is ( 3 field F) AL 20(F) and similarly AL 21 is ( 3 field F) AL 21(F) then AL 20 AL 2 0 ( F ) and AL 21 AL 21(F). The next two forms, which are due to Bleicher [1964], will be stated with the prefix "There exists a field F" but remain equivalent to the axiom of choice if F is a parameter.

-

.+

AL 22: There is a field F such that for every space V over F if B is a basis for V then for subspace S _c V there is a subset B' _c B such that is the subspace generated by B' then S n S' = {Ol S U S' generates V. Let F be a field and let following statements: B(F) :

B(F)

Every vector space over

and F

BL(F)

vector each if S' and

be the

has a basis.

BL(F): If V is a vector space over F, B is a basis of V and X is any linearly independent subset of V then there is a subset B' c B such that X n B ' = 0 and X U B ' is a basis of V.

It is not known whether B(F) or BL(F) imply the axiom of choice but we can show AL 23 does. (The statement B above MC 1.) is: For each field F, B ( F ) , and Blass proved B -+

AL 23:

There is a field

F

such that

B(F)

and

BL(F)

We shall show that AL 12 A M 3 + AL 23 -t AL 22 + AC 2. We noted earlier that B(F) is a special case of AL 12 and the proof that M 3 + BL(F) is similar to the proof of 4.3.

PART I, SET FORMS

126

THEOREM 6.38:

AL 23

-+

AL 22.

PROOF: Let V be a vector space over a field F. Let be a basis of V and S a subspace of V. AL 23 implies S has a basis X. Then since X is linearly independent, AL 23 implies that there is a subset B ' 5 B such that X 0 B' = and X U B ' is a basis of V. The set B' satisfies the conclusion of AL 22.

B

THEOREM 6.39:

AL 22

-+

AC 2.

PROOF: Let x be a set of pairwise disjoint non-empty sets and let F be a field. Let V and Vo be the vector spaces constructed in the proof of 6.35. Then Vo is a subspace of V and B = Ifs:s E ux} is a basis for V, if fs is the function such that s E u,

s

if

0

otherwise

.

We shall identify B with ux. Now apply AL 22 taking Vo as the subspace of V and B as the basis, obtaining a subset B ' 5 B such that if S' is the subspace generated by B ' then Vo n S ' = {O} and Vo U S ' generates V. Let u E x and let w = u n B ' . We claim w has at most one element,for suppose s, t E w. Then s - t E Vo fl S ' = t o ) , which implies s = t. On the other hand w # pI because for any s E u E x there is a vo E Vo and v' E S' such that s = vo + v'. The sum of the coefficients of the terms in vo is 0, thus the sum of the coefficients of the terms in v' is 1. It follows that B' must contain terms from u for each u x. Therefore, w is a singleton, so B' is a choice set for x. I f F is a field of characteristic zero, then AL 21(F) can be stated in terms of groups.

DEFINITION 6.40: Suppose n

i

(G,{+,O})

(a) a E G has order n if na = 0 m. (na is an abbreviation of a + a

is a group. and if ma = 0, +...+ a, n times.)

16.

G

ALGEBRAIC FORMS

127

(b) G is said to be torsion-free if the only element in of finite order is 0.

(c) A group G is said to be divisible if (Va E G) (Vn E w {O}) (3b E G) a = nb.

-

(d) A group G is said to be uniquely divisible if (Va E G) (Vn € w {0)) (31b E G)a = nb. (For abelian groups, "torsion-free divisible" is the same as "uniquely divisible".)

-

(e) The group (G,{+,O)) group if there is a non-empty

L

(1) 0

(2) x E

is said to be a free abelian X c_ G such that

XI

x

x-1

-+

L

x - ~=

(Let

X.

{x-1 : x E X I )

( 3 ) y E G + y = 0 or there is a unique n E w {O} and a sequence ( y l , y 21...,yn) unique up to permutations such that

yi E X

each

y = yl

iy2

u

X-l,

no

+...+yn.

is the inverse of any

yi

The set

X

y

j'

and

is said to generate G.

(See 6.28). (f) If f:G1

-+

G2,

(Gll{+l,OII) f

f

(G2,{+2102})

are groupsl

is called a homomorphism if

(Va,b E G1) (f(a +1 b) in addition

and

is

= f (a)

1-1,

f

f2 f(b))

and

f(O1) = 02.

If

is called an isomorphism.

Since a vector space over a field of characteristic zero is an abelian, torsion-free, divisible group it follows from 6 . 3 6 that the following statement about groups is equivalent to MC 1. AL 24: If A is an abelian group and B is a torsionfree divisible subgroup of A then there exists a subgroup B' c_ A such that B fI B ' = {O} and B U B' generates A.

(It f o l l o w s from the proof of 6 . 3 6 that it is not necessary f o r A to be torsion-free o r divisible.) Armbrust [1972] asked if "torsion-free'' is omitted from AL 24, is the resulting statement AL 24T implied by MC l?

128

PART I, SET FORMS

Blass [1979] partially answers the question by proving that AL 25 is equivalent to the multiple choice axiom, and AL 26 and AL 27 are equivalent to the axiom of choice. AL 2 5 : I f G is a free abelian group, A is a uniquely divisible abelian group, B is an abelian group, f:A + B is a homomorphism mapping A onto B, and g:G -t B is a homomorphism, then there is a homomorphism h:G A such that foh = cj. (See Figure 6.1). -f

foh G

AL 26:

If G is abelian groups, €:A -+ B, and g:G -t B is a homomorphism h:G -+ A group is projective.)

=

g

FIGURE 6.1 a free abelian group, A and B are B is a homomorphism mapping A onto homomorphism, then there is a such that foh = g. (Every free abelian (See Figure 6.1).

AL 27: If G is a divisible abelian group, A and B are abelian groups, f:B + A is an isomorphism, and g:B -t G is a homomorphism, then there is a homomorphism h:A -+ G such that hof = g. (Every divisible abelian group is injective.) (See Figure 6.2).

hof

=

g

G

FIGURE 6.2

G f

Clearly, AL 27 + AL 24T, for if A is an abelian group, is a divisible subgroup of A , B = G, and the functions and g are identity functions then AL 27 implies there is

56.

ALGEBRAIC FORMS

129

a homomorphism h:A G such that hrG is the identity. Then the kernel of h = {a E A:h(a) = O} is the required subgroup of A. -f

THEOREM 6.41: (a) AL 25 * MC 1. (b) MC 1

-f

AL 25.

PROOF: (a) Let x be a set of non-empty pairwise disjoint sets and let w = Ux. Let G be the free abelian group generated by x. Let A be the vector space over the rationals generated by the elements of w as indeterminates and let B be the vector space over the rationals generated by the elements of x as indeterminates. Define functions f:A + B and g : G -t B as follows: If

a E u E x , f(a) = u

If

u E x, g ( u ) = u.

Extend f and g to all of A and G in the natural manner. Then G, A, B, f and g satisfy the hypothesis of AL 25. There exists a homomorphism h mapping G into A such that foh = g. Thus, if u E x, h(u)= rlal ir2a2 +...+ rnan‘ where the a ‘ s are distinct elements of w. However, i n .. f(h(u)) = g(u) = u, so C ri = 1 and at least one ai E u. i=1 (Some ails may not be elements of u , but then the sum of their coefficients must be zero.) Thus, we define a multiple choice function F on x such that for each u E x I F(U) = {al,a21...lanl n u. (b) Let G I A, B, f and g satisfy the hypothesis of AL 25. Let F be a multiple choice function on the non-empty subsets of A. For each generator x of G, let H(x) = F(f-l”{g(x)}). If H(x) has n elements define 1 a, q.e.d. h(x) = n a€H(x)

c

THEOREM 6.42: (a)

AC

1

-f

AL 26.

PART I , SET FORMS

130

AL 26

(b)

-f

S i m i l a r t o t h e proof of 6 . 4 1 (b), b u t t a k e

(a)

PROOF:

AC 2.

Then u n i q u e d i v i s i b i l i t y i s n o t

t o be a c h o i c e f u n c t i o n . required.

F

W e s h a l l p r o v e t h a t Az, 26 i m p l i e s b o t h MC 1 and t h e

(b)

axiom o f c h o i c e f o r f i n i t e s e t s .

The p r o o f t h a t AL 26 i m p l i e s

MC 1 i s t h e same as t h e proof of

x

Suppose u Ex, 2 and l e t

5

i s a s e t of f i n i t e s e t s such t h a t f o r e a c h

( u ] < w.

y =

For each

U 9 , .

Let

u Ex

y.

g e n e r a t e d by

6.41(a).

a E u E y

For each

q u = {v C_ u:2

u E x, let

IvI}

i

be t h e f r e e a b e l i a n g r o u p

G

a d d i t i v e group of i n t e g e r s modulo

let

IuI.

za = z U

w =

Let

be t h e

u

({u)

x

u)

.

UEY

(Then

(u,a) E w

p r o d u c t of t h e

iff za's

d i r e c t product of t h e

x

A = {3, E

a E u E y.)

Let

(u,a) E w

for

zUrs for

za:{(u,a)

A

b e t h e weak d i r e c t

and l e t

b e t h e weak

€3

u E y.

E w : $ ( u , a ) # 0)

is f i n i t e }

(u,a)Ew

X

B = {$ E

EY

A

and

B

zu:{u E y : $ ( u ) # 0)

is f i n i t e ) .

a r e a b e l i a n g r o u p s i f a d d i t i o n of f u n c t i o n s i s

d e f i n e d i n t h e n a t u r a l way.

For

@ l f @ 2 E B, (@l + @ 2 ) ( u ) =

$,(u) + $2 (u) I and t h e i d e n t i t y f u n c t i o n i s t h e z e r o f u n c t i o n : s i m i l a r l y f o r A. For each a E u E y , l e t defined a s follows (V(vrb) E

W)

3,

(VV

(vfb) = (uta)

E Y)$,(V)

3,

@fa)

and

@,

be t h e f u n c t i o n s

0

if

( u I d # (v,b),

1

if

(u,a) = (v,b)

0

if

u # v ,

1

if

u = v .

=

.

56.

ALGEBRAIC FORMS

131

a E u E y, ( J , ( u , a )) = $u and g ( u ) = $u. Extend f t o a l l o f A and g t o a l l of G i n t h e n a t u r a l manner. Then G I A , B , f and g s a t i s f y t h e h y p o t h e s i s of AL 2 6 . T h e r e f o r e , t h e r e is a homomorphism h:G -+ A s u c h t h a t f o h = g. Define f u n c t i o n s

f

and

u E y, then

Let

ni E z u

a s follows:

h ( u ) = J, =

'

Z

For

where

"iJ,(u,ai)

a1 .Eu

c ni

and

g

This follows f r o m t h e f a c t

l(modlu1).

i

that f ( h ( u ) ) = f ( $ I = g(u) = Since

IuI

2

m = min{ni:ai

is

2 , t h e r e must be a t l e a s t two d i s t i n c t n i ' s . E u ) . Then {a E u: c o e f f i c i e n t o f

Let

in

J,W?

i s a non-empty p r o p e r s u b s e t o f u . Thus, t h i s p r o c e s s r e d u c e s t h e o r d e r of t h e e l e m e n t s o f x s i m u l t a n e o u s l y . J,

m}

1.1

u E x, i f

For each

w e w i l l have t o a p p l y t h i s

= n,

-

1 t i m e s t o c h o o s e a n e l e m e n t from p r o c e s s a t most n C o n s e q u e n t l y , w e have a c h o i c e f u n c t i o n on x.

u.

THEOREM 6 . 4 3 :

(a)

M 3

(b)

AL 27

-+

-+

(a)

PROOF:

AL 2 7 . AC 2 .

Let

h y p o t h e s i s o f AL 2 7 .

a($)3 d ( f )

such t h a t

# fl

H

a ( h ) = A.

A* = {b

W e have i n

a E A

-

A*:

G

W e claim t h a t

$(h).

Suppose

y = x/no.

Let

+ ka: b E $ ( h )

A

such t h a t

to

s a t i s f i e s t h e h y p o t h e s i s of

H

n.

A

(Vb E B ) g ( b ) = @ I ( f ( b )1).

F o r some p o s i t i v e i n t e g e r

be t h e s m a l l e s t s u c h

y E G

and

h a s a 5-maximal e l e m e n t , h . F o r suppose

CASE 1. no

g0f-l E H .

because

so

M 3

satisfy the

$ i s a homomorphism from a subgroup o f

H = {@I:

H

B, f , a n d , g

GI A,

Let

n,

na E $ ( h ) .

h ( n o a ) = x E G.

0 i k < n,)

2

$(h).

Let

Choose

PART I, SET FORMS

132

bl

(bl + kla

+ k1a

= b2

+

k2a

I

+ (b2+k2a) =

bl

-+

=

b2

A

kl = k2

(bl + b2)+(kl + k2)a, if 0 s- kl + k 2 < n 0 ' (bl + b2 + noa)+ (kl + k2 - no)a, if n O s k l + k 2 < 2n0.

A* is a subgroup of A. Define h*:A* -+ G such that h*(b + ka) = h(b) + ky. Then h c h* and h*E H. This contradicts the maximality of h.

na f E(h) for all n E o is a homomorphism, -na f 3(h) for all CASE 2.

A* =

Define h*(b t na) contradiction.

b * na: b E B(h) =

h (b

.

Then

-

A

h* E H

{O}.

n E w

Then, since (01. Let

-

h

n E o and

h c h*,

a

(b) We shall f rst prove AL 2 7 + MC 1, and then prove that AL 27 implies the axiom of choice for finite sets.

Let x be a set of pairwise disjoint non-empty sets and let w = Ux. Let A = B = the free abelian group generated by x. Let Z be the set of integers and Q the set of rational numbers. For each u E x, let RU

For

Q E RU,

@ E QU:Ia E u:$(a)

-

let

[$I

=

I@ E

{lli

E

X sU: { u

G

is finite}.

X ( @ - @ I (a) = 0 ) . aEu be the weak direct product

R~:&($-$J)

Let sU= { [ $ I : $ E RU} and let of the Su's for u E x. G =

# 01 _c z

E x: ~ ( u #)

A

01

is finite].

UEX

Define

f:B

+

and

A

g:B

u ) = 2u =

+

u

G

as follows:

+ u.

u ) = lU, where

l U E G such that lU(v) = 1 0 1 ) and lU(u) = \, c 2 and $,(a) = 1. -L

-

U:a

where

S6.

A l l such f u n c t i o n s

$u

g

and

s a t i s f y t h e h y p o t h e s i s o f AL 27.

homomorphism

h:A

u E x, h(u) = lU/2.

2’ @ , ( a )

= 1/2.

such t h a t

G

-+

133

b e l o n g t o t h e same e q u i v a l e n c e c l a s s . B i n t h e n a t u r a l way. Then G , A , B , f

Extend g

t o a l l of

ALGEBRAIC FORMS

Then, f o r e a c h

@u E h ( u ) ( u ) ,

Therefore, i f

Define

There i s a

hof = g.

F ( u ) = { a E u: $,(a)

aEu all

then

E Q

-

-

Since

Z).

II, E h ( u ) ( u ) have t h e same v a l u e s on Q 2, f o r each u E x, F ( u ) i s a u n i q u e l y d e t e r m i n e d f i n i t e s u b s e t of u , so F i s a m u l t i p l e c h o i c e f u n c t i o n on x. To p r o v e t h e f i n i t e axiom o f c h o i c e l e t u E x, 2

f i n i t e s e t s such t h a t f o r a l l with

2

r;

(vl

then

t h a t of 6 . 4 2

(b).)

g e n e r a t e d by

x, Z

r a t i o n a l numbers.

r;

b e a s e t of

x

Iu/

v c u

and i f

( W e s h a l l g i v e a similar proof t o

v E x.

A = B = t h e f r e e a b e l i a n group

Let

t h e s e t of i n t e g e r s , and u E x,

For each

t h e set of

Q

let

Then, c o n t i n u e a s i n t h e f i r s t p a r t o f t h e p r o o f t o d e f i n e t h e group f:B

and t h e f u n c t i o n

G

+

g:B

-+

But h e r e , w e d e f i n e

G.

a s follows

A

.

(VU E x ) f ( u ) = l u l u

Extend

f

t o a l l of

t h e r e i s a homomorphism

h:A

@u E h ( u ) ( u ) :

Let

AL 27 i m p l i e s

i n t h e n a t u r a l way.

B

then

+

G

C

such t h a t

hof = g.

@,(u,a) = l / / u

.

Let

a ELI Q U ( u , a ) = @,(u,a) a E u,

let

For e a c h

-

(greatest integer

$,(u,a)).

m ( a ) = I U ~ $ ~ ( UE , Z~. ) Then

j < (u(

let

n

j

aFu = { a E u:m(a) = 1 ) .

c

Cm(a) = j1n.l 7 aEu lsj
r;

Inj(.

j , 1 5 j < IuI, Let

j,

5

For each

m ( a ) z 1 (modlul). Then

1 (modlul). such t h a t

be t h e s m a l l e s t s u c h

IuI j.

is not a Then

{a E u:m(a) = j , ) i s a non-empty p r o p e r s u b s e t of u. T h i s way w e c a n c o n t i n u e , a s i n t h e p r o o f o f 6 . 4 2 ( b ) , t o c o n s t r u c t

134

PART I, SET FORMS

a choice function on

x,

q.e.d.

Gratzer [19621, and Haft & Howard [1981] have studied a representation theorem for multi-algebras which is equivalent to the axiom of choice. DEFINITION 6.44: (A,{Ok:k E KI) is called a multialgebra if each Ok is a function from Ank to F(A) {B).

-

We shall assume that the multi-algebras we are considering have no constants. Suppose (B,{Ok:k E Kl) is an algebra. If E c -B x B is an equivalence relation then B/E is the set of equivalence classes of B with respect to E. That is: (Va E B) [a1 = {b E B:(a,b)

E El,

B/E = {Ia]:a E BI. An operation Ok on on B/E as follows: Ok/E([all

,. .., [an]) A

Then

B

can be extended to an operation

= { [ b l : (3a;

,... ,a;

E A) ((ai,ai))

Ok/E

E E

b = Ok(aj,...,a;)) 1.

(B/E, {Ok/E: k E K1)

is a multi-algebra.

AL 28: Let (A,{Ok:k E KI) be a multi-algebra. Then there is an algebra (B,{Oi:k E K}) of the same type as A and an equivalence relation E on B such that (A,{Ok:k E K 1 ) ;(B/E,{OA/E:k

E Kl).

AL 29 will be AL 28 with the added restriction that all the multi-operations Ok for k E K are unary operations. Clearly, AL 28 * AL 29. We shall show AC 1 -+ AL 28 and AL 29 * AC 1. Gratzer [1962] proved that the axiom of choice implies AL 28, but the proof we give here is due to Hoft & Howard [1981], as is the proof of 6.46.

ALGEBRAIC FORMS

16.

AC 1

THEOREM 6.45: PROOF:

Let

-+

(A,{Ok:k

AL 2 8 .

E

be a multi-algebra.

K})

X = {Ok(d):k E K

A

(alIbl)E(a2,b2) ((d,e))

O i

I

are defined on

= ((dlfe1),...,(dnfen)>

Oi(((d,e)))

=

(eofeo) if

Let

d E ff(Ok)).

Let F be a choice function on X. Let an equivalence relation E on B by

The operations

135

B = A

and define

x A

iff

al = a2'

B

so that if d, e E 3 ( O k ) ,

where

-...= en

eo = el -

(dofdo) otherwise, where

do

eo E Ok(d)

and

=

then

F(Ok(d)).

With these definitions, the multi-operations on

B/E

will

be Oi/E([((d,e))l)

= {[(d',ef)l:(3b

a ( o k ) (d',e') = Oi(d,b)}.

E

If we define g:B/E A such that g([(a b)l easy to prove that g is an isomorphism. +

THEOREM 6.46:

AL 29

+

)

= a

then it is

AC 1 -

Suppose x is a set of non-empty sets. Let A = { x } U (Ux). For each u E x, define a unary multioperation Ou on A as follows: PROOF:

OU

a) =

("

if

{x}

a

=

x,

otherwise

.

By AL 29 there is an algebra (B,{OA:u E XI), of the same type as A, an equivalence relation E 5 B x B, and an isomorphism f such that f:(B/E,{OA/E: Let

b E B

such that

U

E XI) 2 (A,{OU: u E XI).

f ([b]) = x

and define for each

u E x,

136

PART I, SET FORMS

This completes t h e s e c t i o n on algebra.

57.

CARDINAL NUMBER FORMS

All the statements we make about cardinals could just as well be statements about sets. However, the statements are simpler when we use the notion of a cardinal number, so we shall follow common practice and use cardinal numbers. For the sake of convenience, we shall repeat here some of the properties of cardinal numbers we will use. (Also, see 0.13 and 0.14 and the preceding discussion). 7.1: If x is a set, 1x1 is the cardinal number of We shall use the Greek letters, K , u , and v , and these letters with subscripts, for cardinal numbers. 7.2: [XI

=

K;

For every cardinal for all sets x and

1x1 s J y / iff

x

x.

there is a set x such that y, 1x1 = Iy( iff x = y ; and

K,

4 y.

7.3: For every set x, T ( x ) = {a E On: a < X I ( r is Hartogs' function). For each set x, T(x) is a set. If 1x1 = K, then I r ( x ) [ = K * . K* is the smallest cardinal of a well ordered set which is not 5 K . 7.4:

KO =

101

and for each ordinal

a, Na = ( w a I .

An

aleph is the cardinal number of an infinite well-ordered set. For each aleph, K , there is an ordinal a, such that K = Ha. If K = N then K * = K a+l' 7.5:

For every aleph, 8 , N2

7.6:

For every two alephs, N

= N.

and

N',

C; i h'

= N C;'

=

max K I N ' ) . 7.7: K

2

A cardinal number,

K,

KO' 137

is said to be transfinite if

138

PART I, SET FORMS

7.8: If K ~ K, ~ , p1 and u 2 are transfinite cardinals such that at least one is an aleph, then K ~ * sK P~l + P 2 implies K~ i ul or K~ s p2. 7.9: If K and 1-1 are transfinite cardinals and N an aleph such that K s K P then either K s K or K s p. 7.10: ( K ~ ) *= K * .

then

is

For each transfinite cardinal K, K * is an aleph and (In fact, if K , v # 0 and at least one is infinite

(K*W)*= K * * P * . )

7.11: For each transfinite cardinal K , there are no cardinals between K and K +

K

7.12:

For all cardinals

K,

K*

7.13:

For all cardinals

K,

K* s 22

K,

p and for all

7.14: For all cardinals nK = np -+ K = u .

K

+

K*

and

K*.

s 2ZK2.

2K

7.15: For all transfinite cardinals and is equal to 2 K .

. n E w

-

K,

K

-

{O).

exists

Georg Cantor may be considered by many to be the founder of set theory, but without any doubt, the work of Alfred Tarski is the foundation for this section. Almost all of the preliminary set theory stated above in 7.8-7.15 is due to Tarski and almost all the ideas in the proofs below are due to Tarski. Most of the forms of the axiom of choice given below were either due directly to Tarski or inspired by his work. Many of the forms listed below can be weakened, for example, by requiring one of the cardinals to be an aleph or by requiring two of the cardinals to be equal. In some instances these special cases also imply the axiom of choice. When this happens we shall list the special cases as primed forms. Then we shall prove the axiom of choice implies the standard form and the primed form implies the axiom of choice. Another variation can be obtained for forms which contain " h :So" in the hypotheses by replacing " K 2 KO" by " K is infinite". Since WO 1 A

K

is infinite

+

K

2 8

0'

57.

CARDINAL NUMBER FORMS

139

the resulting statement is equivalent to the axiom of choice. We shall not explicitly state these forms here. (The greek letters K , 1-1, v , and these letters with subscripts, represent cardinal numbers. The greek letters a, 8 , y,..., and these letters with subscripts represent ordinal numbers.) CN 1: CN'1: CN 2: CN'2: CN 3: CN'3: CN 4: CN'4:

(Tarski [1938a] noted that we can obtain an interesting equivalent formulation of CN 4 in the following way: It is that easy to prove in NBG' (*) I f x is finite then either x or there exist disjoint sets u and and u U v = x. and 1v[ <

[XI,

has at most one element v such that ( u I < 1x1

The converse of ( x ) is equivalent to CN 4, since as noted earlier, "1-1 t KO" in the hypothesis of CN 4 can be replaced by "1-1 is infinite". Thus, the following statement is equivalent to CN 4. CN 4F: A set x is finite iff x has at most one element or there exist disjoint sets u and v such that Iul < 1x1 < 1x1, and u U v = x. and

(vI

We can obtain forms equivalent to some of the other forms in this section in a similar way. For example,

140

CN 1F:

PART I, SET FORMS

A set

x is finite iff there exists a non-empty set u u u x < u x x.

x has at most one element or disjoint from x such that

CN 6F: A set x is finite iff x x < x x x. See Tarski [1938al . ) CN 5:

(VK)

(Vp

CN'5:

(VK

2

KO)

#

(K*K*

CN 6:

(VK

=

=

(K*K*

V K * l J = !J).

K

K*).

because

K

NO)

2

KO) ( K * p

Z

has at most one element or

6

K*

K.)

2

= K ).

(K

The statement: (VK z K O ) ( K = 21c) has been shown to be weaker than the axiom of choice by Halpern & Howard 119741 (using a Fraenkel-Mostowski model) and by Sageev [ 1 9 7 5 ] (using a Cohen model). CN 7:

(vK,p) ( K 2 =

CN 8:

(VK,p)

(K*

= p*

CN 9:

(VK

KO)

-I

2:

p2

-+

K

= 1-1).

-t

K

= p).

(K*

< 1-I <

+ K

< !J).

(311)

K

+

K*).

CN 1 0 :

( V K t p ) (K*

CN'10:

(VK)

(Va)

CN 11:

(VK,).I)

(K

CN'11:

( V a l ( V P ) (Ha <

CN12:

( V K Z K o ) ( V l - l r W ) [ K + ~ = K + V

CN'12:

(Val

< !J*

(K*

< K,+l

< p

(Vp,V)

u

-+

+

[R,

<

K*

-t

K

-+

< Ka).

).I*).

Ka+l

!J =

<

u*).

N, + v

*

(p=WV().ISKAWSKK.))].

(!J=w

-+

V (!J 5

Ka A W

(The following special case of CN 12 is true if and is therefore weaker than the axiom of choice: (VK

+

2 &o)(V!J)(K

CN 13:

(VK

> 0)

CN'13:

(Va)

(v!J,W)

CN 14:

(VK

2

(v!J,W)

[K*p

[K,*!J

K O ) (Vp)

=

= K 4-

K

K'W

= Na'W (

K

= ~ K*U

-+

-+

-+

IJ

+

P 5

(ll = V

(!J = W V !J 5 K ) .

V (!.I

5

K =

K,)

)]

2~

K).)

(!J S 5

K,

K

A V A W 5

<

K))].

K,))].

.

97.

(The statement: if K = 2~ for all axiom of choice.)

CARDINAL NUMBER FORMS

( V K , ~ )(K K

2

KO,

+

141

u

is true and is therefore weaker than the p i

K

+

K +

5;

K),

CN 19 CN 1, CN 6-7 and CN 15-18 are due to Tarski 119241. is a special case of CN 18. M. S. Lesniewski proved that CN 4 implies the trichotomy and the proof for CN 5 is similar. CN 2-3 follow from the work of Tarski. CN 8 and CN 10-11 are due to Sobocinski [1961b], and CN 9 to Sobocinski [1961cl. CN 1 2 and CN 2 0 were given by Tarski in Lindenbaum & Tarski [ 1 9 2 6 ] and CN 13-14 are similar to CN 12. CN 2 1 is due to Sierpinski [1946]. Each of CN 1 - 2 1 follows from the well-ordering theorem and properties of alephs and finite cardinals. The wellordering theorem implies that each infinite cardinal is an aleph. Then using 7.4-7.6 we easily obtain each of CN 1 - 2 1 . We shall now prove the converse, that is each of CN 1 - 2 1 implies some form of the axiom of choice.

PART I, SET FORMS

142 THEOREM 7.16:

CN'1

PROOF: Suppose implies K L K * or impossible. THEOREM 7.17:

s

K*

CN'2

Then by 7.8, K * K * = K + K * The second alternative is

tio.

P

K

WO 3.

-+

-f

K.

WO 3 .

PROOF: Suppose K 2 KO. Since K + K * s K + K * , CN'2 implies K + K * L K or K + K * i K * . The first alternative implies K * L K, which cannot happen, and the second implies K

L K*.

The proof that CN'3 WO 3 is similar to the proof of 7.17 with " + ' I replaced by ' I - ' ' . Each of CN'4 and CN'5 imply WO 3 because they each imply K i K * , for K P KO. -+

Even though CN 6 obviously implies CN 7, we shall include a direct proof that CN 6 -+ WO 3 . THEOREM 7.18: PROOF: K

+

CN 6

Suppose K*

= (K

K*)2

Thus, 7.8 implies K s is impossible, q.e.d. THEOREM 7.19:

WO 3 .

is transfinite.

K

+

-+

=

or

K*

CN 7

-+

+

4- 2 K K *

K2

K*

i K.

Then (K*)2

CN 6 P K*K*.

The second alternative

WO 3 .

PROOF: Let K 2 K O and let p = K"; then We shall show (1.1 + 1.1")~ = ( p - p * ) 2

.

p 2 = 1.1.

(1.I

+

u*)2 = p2 =

+

v +

= 1.I

+

= p.p*

21.I.p*

1.I-u* + pap*

implies

+

+

K

L

1.1

and

(1.I")Z

(v*)

by definition of p ,

p*

by 7.5,

because

u* is an aleph.

and p * = ( p * ) 2 , we obtain the desired result. Therefore, CN 7 implies 1.1 + 1.1* = p e p * . Since u* & p we obtain (by 7.8) p i p * . But K L 1.1 , SO K is an aleph. B u t since

1-1 = 1.'

CARDINAL NUMBER FORMS

17.

THEOREM 7.20:

CN 8

PROOF: Suppose implies that K * = K K

CN 9.

-f

Then

2 So.

K

+

143

(K*)*

=

(K

+

CN 8 and

K*)*.

so there is nothing between

K*

K*

K*.

THEOREM 7.21: PROOF:

C N . 9 -t CN 6.

Suppose =

K*

(K

2

K

P

)* 5

K

KO.

By 7.10,

K*

S

K2

+

K*.

CN 9 implies K * = K + K * or K + c * = K 2 + K * . The first 2 alternative implies K is an aleph, so by 7.5, K = K The 2 second alternative implies K i K2 s K + K * . BY 7.11, K = c or K~ = K + K * . The second equality implies K * 5 K~ which is impossible because ( K 2) * = K * (7.101, q.e.d.

.

THEOREM 7.22: CN'10 PROOF: Suppose for some a. Since THEOREM 7.23:

KO;

K z K*

WO 3 .

+

<

CN'11

K**,

then K * = K a and CN'10 implies K <

= %+1

K** K*.

WO 3.

-+

PROOF: Suppose K 2 KO. If K * < K + K*, then, by CN'11, K * * < (K + K * ) * = K * * , contradiction. Therefore, K * = K + K*, which implies K i K * . THEOREM 7.24: PROOF:

Let

CN'12 implies and K * + K 5

K

CN'12 K

P

which is impossible, or Each implies K i K*.

= K*

K*.

Replacing ' I + " by that CN'13 + WO 3 . THEOREM 7.25:

K i

p

PROOF: Let K and p 2 = p . (p

+

Then

No.

+

WO 3.

-c

K,

" - "

CN 14

s

K*

in the proof of 7.24 gives a proof

-+

WO 3.

be any cardinal and let

p*)2

K

= p.p*

= (ll

+ lJ*)

(v-Ll*).

p = rK0.

Then

PART I, SET FORMS

144

CN 14 implies that p * p * i p + p * . Thus, using 7.8, we obtain p * i p, which is impossible, or p i p * , q.e.d. THEOREM 7.26:

CN'15

-+

WO 3.

PROOF: Let K be any cardinal and let p = K O * ~ . Then and 2p = p . We have p < p + p * and p * i p + p*. If p* = p + p* then p i p * . Suppose p * p + p*. Then, by CN'15, K i p

u

+ Ll* < ( P + U * )

+ (!J+v*)

=

u

+

V*f

clearly a contradiction, q.e.d. The proof that CN'16 WO 3 is similar to the proof of 7.26 with p = K" and "+" replaced by " 'I . -+

THEOREM 7 . 2 7 : then K*

CN'17

-+

WO 3.

PROOF: Let K 2 N o ; then K* i K * + K . If K* = K * K i K * ; otherwise CN'17 implies the contradiction

+

K,

< K.

If "+" is replaced by a proof that CN'18 WO 3 .

in the proof of 7.27, we obtain Also, it is clear that CN 19 -+ CN 14.

' I - "

-+

THEOREM 7.28:

CN'20

-+

WO 3.

Suppose K 2 KO. We have K i K + K * and K * i K + K*. If strict inequality held for both then CN'20 implies a contradiction. Thus we must have either K = K + which cannot happen, or K * = K + K * which implies K s K * , PROOF:

q.e.d.

The proof that CN'21 implies WO 3 is obtained from the proof of 7.28 by replacing "+" by

"-".

The next set of statements we consider deal with subtraction and division of cardinal numbers. DEFINITION 7.29:

CN 22:

SUBTRACTION AND DIVISION

( V K , ~ )( K < LI

-+

p

-

K

exists).

K*,

57.

(That is, ( 3 1 ~()p = K < p

CN'22:

-t

(3v) (p = K

+

K

CARDINAL NUMBER FORMS

+ v)

(Va)( V p ) ( 8 , < p

(If in CN 2 4 we replace + p1 = K~ + p2", for a l l infinite K. choice. )

145

u). The statement, is provable in NBG'.) -+

p

-

K,

+

exists).

p 2 = K~

+

p1"

by the resulting statement holds if " K ~

K =

Thus, it is weaker than the axiom of

2 ~ ,

PART I, SET FORMS

146

CN 22-23 are due to Tarski, Lindenbaum & Tarski [1926]. Later, Sierpinski [1947b], published a proof of CN 22. CN 24 and CN 29-30 were given in Sobocinski 119651. CN 25-28 are related to the work of Tarksi and Sierpinski.

Just like the first twenty one statements, these statements follow easily from the well-ordering theorem. Below we give the proofs that each implies some form of the axiom of choice, THEOREM 7.30:

CN'22

WO 3.

-+

Suppose K 2 K O ; then K * s K * + K . If K * < K* + K then CN'22 implies that there is a unique cardinal p such that K * + p = K * + K . But K * + ( K * + K ) = K * + K , so we must have K = K * + K which is a contradiction. Therefore, K * = K * + K, which implies K i K * . PROOF:

Clearly, CN'23 THEOREM 7.31:

-+

CN'22.

CN'24

CN'23.

-+

PROOF: Suppose H, < p . Then there is a cardinal K such that K, + K = p . Since 0 < K we can apply CN'24 and obtain p = K. Thus, p - Ka = p . THEOREM 7.32:

CN'25 * WO 3.

PROOF: Let K 2 K O ; then K * i K + K * . If K * < K + K * then CN'25 implies that there is a v such that K + K * = K * * v . Then, using 7.8, K * i K , which is impossible, or v I; K*. The latter inequality implies K i

K

+

K*

=

K**V

=

K*.

Since CN'28 + CN'27 -+ CN'25 and CN'26 CN'25, it remains to be shown that each of CN'29 and CN'30 implies the axiom of choice. -+

THEOREM 7.33:

CN'29

-+

CN'26.

57. PROOF: J!N ;

Suppose

Since

Ha < p.

= l * ( N a * p ) , CN'29 implies

THEOREM 7.34:

K;p

CARDINAL NUMBER FORMS

CN'30

1

C

147 and

t(;p

p = K;p.

CN'26.

-+

PROOF: Suppose ka < p ; then Ka < K a - p , = l*(Ka*p). So, CN'30 implies p = R a m p .

1

p

and

The following group of statements are, for the most part, weakened forms of CN 6.

(VK) ( N a . ~= Ka.~L).

CN 32(Na) :

CN 33: ( 3 a ) ( V K t K O ) (There is no s-well-ordered set M cardinal numbers between K and K~ such that M g a . ) CN 34: ( ? a ) ( V K , ~ ) ( K * = p * of cardinal numbers between

There is no s-well-ordered set M K and p such that M z a ) . 2 2 U < Z K A ( 3 V , ) ( K = V 1 ) -t ( 3 V 2 ) (V = V,) 1 . -+

CN 35:

(VK,!J

t

KO) [ K

CN 36:

(VK,p

t

K O ) [ K < 1-1 < 2"

S

of

+

(3V)

(11 = K*V)I.

CN 31, while not explicitly stated in the literature, follows from the work of Tarski and Sierpinski; CN 32(Ka) is due to Alas [1968]; CN 33 first appeared in Rubin & Rubin 119631; CN 34 is a generalization of a form given in Sobocinski [1962a]; and CN 35-36 are due to Seki [1961]. Clearly, CN 6 implies each of CN 31-33; CN 8 -+ CN 34; CN 25 .+ CN 36; and CN 31 + CN 35. It remains to be shown that each of CN 32-36 imply the axiom of choice. THEOREM 7.35: PROOF:

Na * p

= p.

CN 32(Na)

-+

Suppose K t KO. CN 32(N ) implies a Na*K

(Ka'K)*

WO 3.

Let

p =

= [Na.K 2

(haOK)

-+

K a - K + ( K a * ~ ) * . Then (Ka.K)*]

(Ha*K)

*.

Using 7.8, we obtain K a . ~ s ( N a - ~ ) * or ( N a * ~ ) * s R a e " . second alternative is impossible and the first implies K an aleph.

The is

148

PART I , SET FOFNS

THEOREM 7.36:

-f

WO 3 .

but s u p p o s e

1-1 2

Suppose

PROOF:

aleph.

CN 33

1-1"

arbitrary ordinal, let numbers between

and

K

K

<-well-ordered

a

be a n

= p it ; 6 i a .

set

M

We

of cardinal

M g a.

such t h a t

K~

is not an

Let

and l e t

= K6

s h a l l show t h a t t h e r e i s a

P < a, l e t

For

-

%,a

- 1-I'%iB

If

1-1

W e s h a l l show t h a t CN 33 i s f a l s e .

B < y < a, v B < v

*

Y'

"%+$

-

-

Ngiy

I;

Then

N6ia

1-I'%iy

.

i Kgia

p-"KsiB

5

vB = vy.

f o r suppose

%+a

T h e r e f o r e , p.Ngiy t h a t 1-1 s Kgia or

'

I t f o l l o w s from 7.8

The f i r s t a l t e r n a t i v e i s

p-HgiB.

1-1

i m p o s s i b l e b e c a u s e w e assumed t h a t

w a s n o t an a l e p h .

The

N~.~.

s e c o n d a l t e r n a t i v e , by 7 . 9 , i m p l i e s Ngiy s p o r Kgiy 5; The f i r s t a l t e r n a t i v e c a n n o t happen b e c a u s e IJ*= K 6 and t h e

B < y.

s e c o n d c a n n o t happen e i t h e r b e c a u s e imp1i e s Let

VB

< v

Y'

M = {vg:B

t h a t €or e a c h

B < y

Thus,

B

< a};

a,

K

5

M ;a.

then

vB

I;

K

2

.

I t i s e a s y t o see

W e s h a l l show t h a t

€or some

vB

B < a.

e q u a l i t y cannot occur.

Suppose

K

Then

+

But t h i s e q u a l i t y i m p l i e s

p

iNgia

= p*NgiB l-I'NgiB

s

u

=

.

+ Kgia

Using 7.8, w e o b t a i n t h e c o n t r a d i c t i o n p N6+, or 2 f o r some 6, t h e n K6+B i u. On t h e o t h e r hand, i f K = v B 2 This implies 1-I -+ V"N6.+, = umKgiB i Ngia. ll'Ngia

s p*KgiB

f

.

Ngia

U s i n g 7 . 8 and 7.9 w e o b t a i n t h e c o n t r a d i c t i o n :

%+a

" %+a

Ll

T h u s , w e h a v e shown t h a t f o r e a c h

" %+a B

a,

%iB K

-

< vB <

K

2

,

q.e.d.

CARDINAL NUMBER FORMS

7.

149

The p r o o f o f 7.36 i s a l s o a p r o o f t h a t CN 34 for all

because, by 7.10, THEOREM 7.37:

PROOF:

’.

112 =

Let

CN 35

+

ko

K 2

H 0‘

;r

K

K*

+

WO 3,

= (K~)*.

WO 3.

and l e t

p =

K

80

.

Then

and

P

K i

By 7 . 1 2 ,

’*

2’

2

5

Therefore,

+ ’*

21-1 i 2’

If

2’

that

+ u,

2P

t h e n , by 7 . 1 5 ,

and a l s o

K,

2’

1-1

+

1.1* = 2”

2’

Since

i

22

1-1” = 22

i s an aleph.

+ u*

i 2’

2

If v

aleph.

2’

.

p*

and t h e r e f o r e

i s an

K,

then

2’,

i

which i m p l i e s

Suppose

= v By 7 . 8 , f i r s t i n e q u a l i t y i m p l i e s t h a t 2’, 2

P

.

it f o l l o w s from CN 35 t h a t there i s a v 2 v p* or v i 2’. The

’ + u*

= (2’)2,

such t h a t

P = 22

I;

+

21-1

I.r* = v 2

i (292

= 2P

.

This implies , p i P

If

p

+

aleph.

P* = 2’,

s 21-1.

t h e n , by 7 . 1 5 ,

Suppose p i

1-1‘

+ ’*

’+

P*

implying

P* = 2’,

K

i s an

< 2’.

w e c a n a p p l y CN 35 a g a i n . T h e r e f o r e , t h e r e i s 2 a vl s u c h t h a t p + p * = (v,) Now w e are f i n i s h e d b e c a u s e by 7 . 8 , v1 s p o r v1 i y * . I f t h e f i r s t inequality holds we Since

= p,

.

would h a v e

u*

i

’+

which i s impossible. and t h e r e f o r e

K,

P * = (v1)2

i

1-12 =

’

The second i n e q u a l i t y i m p l i e s t h a t

is an aleph.

p,

PART I, S E T FORMS

150

The proof that CN 36 + WO 3 is similar. Whenever CN 35 is used in the proof of 7.37, use CN 36 and the same results follow. Our next three statements involve the notion of "immediate successor". DEFINITION 7.38:

IMMEDIATE SUCCESSOR

( 3 ~ ( )K < v < p).

(a)

K

c1 p

iff

K

< 1-1 A

(b)

K

<2 p

iff

K

< 1-1 A (Vv)( K < v

(c)

K

<3 1-1

iff

K

<

K

<1

u

7

(VV)( v < 1-1

A

It follows from 7.11 that if K + K*. Thus, (*)

(VK

2

K O ) (31-1)( K

+

p s v).

+

v s

K).

is transfinite then

K

U)

is provable in NBGO. If we replace ''<1'' in ( * ) by ' 1 < 2 ' 1 , the resulting statement, CN 37, is equivalent to the axiom of choice. (Tarski I19541.1 If in ( * ) we replace 'r
(CN 38 * CN'38 because

K O ) (vU,V)( K < l.l

CN 39:

(VK

CN'39:

(VK 2 No) ( K *

2

(CN 39 * CN'39

Ka < 1

< K*

because

+ K*

K

A K

*

<3

<3

K**

S

V

*

V S

K* -b

p).

K ) -

K**.)

CN 37 is due to Tarski [1954]; CN 38 to Sudan [19381; and CN 39 is related to work given in Sobocinski [1962b].

7.

CARDINAL NUMBER FORMS

151

Using 7.4-7.6, it is easy to see that the well-ordering theorem implies each of CN 37-39. THEOREM 7.39:

CN'37

+

WO 3.

PROOF: Suppose K 2 KO. CN'37 implies that K * has a 2-successor, u . But K * < K * * . Therefore, by 7.38(b), )1 i K * * . Since there is nothing between K * and K * * , p = K * * . Thus, K * < 2 K**.

+

If K * < K + K * then by 7.38(b), K** s K + K*. This inequality, by 7.5 and 7.8, implies the contradiction K** s K or K * * i K * . Thus, we must have K * = K + K* which implies K i K * . Clearly,

s

K*

K

K*.

(This proof is credited to A. Levy by Truss [1973].) THEOREM 7.40: PROOF: that either

CN'38

Let

K

K-K,

=

+

WO 3.

and suppose K * = Ha. K-K,+~ or K - 8 , C1 K*N,+~.

CN'38 implies However,

2 No

+ Ka+l

K * H a i K'Ka

S

KC'K,+l

Therefore, either (1)

K*&,

= K-K,

+ K,+l

(2)

KwK,

+

- K'Na+l

or Na+l

.

If (1) holds then Ka+l i K*H,. But Ka+l b' K and If ( 2 ) holds, then by 7.8, which contradicts 7.9. Ka+l $ K a ,

either

5

K - K ~ , which is impossible, or

THEOREM 7.41:

CN'39

-+

K

s

q.e.d.

WO 3.

PROOF: It follows from 7.5 and 7.8 that the conclusion of CN'39 is false. Thus, the hypotheses is false which implies K * = K * + K. Consequently, K s K * . We continue with statements involving cardinal exponentiation. Let

v

1 be a cardinal,

CN ~O(V): ( V K , p )

(K"

<

uv

+

K

< p)

.

PART I, SET FORMS

152 CN 41:

( 3 >~ 1) ( V K ~ , K[ (~V)v ) (1 < v

CN 42:

( V K , ~ , V )( K # 0

CN’42:

( v K , ~ )(va) (K

<

A K’

# o

V

K

+

<

A K’

s

+

K;

v

< K ~ )* K~ <

~ ~ 1 .

p < v).

&a

K

LJ

+

p < NJ.

CN 40(v) and CN 42 were given by Tarski in Lindenbaum Tarski L19261, and CN 41 is a generalization of CN 40.

&

For v > 1, CN 4 0 ( v ) * CN 41. The trichotomy and monotonicity laws for cardinal exponentiation imply CN 4 0 ( v ) and CN 4 2 . We shall show that CN 41 and CN 42 each imply the well ordering theorem. CN 41

THEOREM 7.42:

-+

WO 3. M’H0

Suppose K 2 Ho. Let p1 = 2 ; then Let h = K p 1 , K1 = A s k * and K 2 = A -t ( A . ( . k * ) ’ ) ) * . p’P1 = p l . Suppose 1 < v s p, then PROOF:

K;

=

[A

2

A. ((A* (h*)p)*)V-l

2

A - ( ( (A*)”*)v-l

4-

(A*(A*)’)*]V

(by 7.5)

(because A’

V K~

We shall show next that V

(K1)*

=

((x.h*)’)*

=

( A * (A*)”)*

<

V

K~

=

A)

.

s (A.(x*)p)* 5;

Therefore, K~

<

K

~

.

K:

<

K~V

K2

<

V

K2

.

because

(K,)* V $

K V~ .

That is,

A*A*

<

A

+ (A.(A*)V)*.

CN 4 1 implies that

CARDINAL NUMBER FORMS

57.

A*

By 7.8, either

X

i

which is impossible, or K i X, we obtain the desired result.

Since

X*(X*)')*.

(

A,

i

THEOREM 7.43:

153

CN'42

+

WO 3.

NO Let )J P No and let K = '2 Then K' = K i KK*. If K = K ~ * , then since K * 5 K ~ *, we would have K * i K which is impossible. Therefore K' K ~ and * K # 0, so CN'42 implies 1.1 < K * , q.e.d.

.

PROOF:

n

be the greatest lower bound of K and 1.1, and K U 1.1 their least upper bound. Tarski [1949] has shown that the following statement is provable in NBGO: Let

K

(*) K

+ 1.1

=

1.1

If

K

n

11)

(K

n +

1.1

exists then

(K

u

exists and

U 1.1

K

1.1).

(If w = K n 1.1 then K = w + K~ and 1.1 = w + 1 . 1 ~ for some K~ and 1 . 1 ~ . It can be shown that K U 1.1 = v + K~ + ul. It was pointed out by Arthur L. Rubin that the statement: (VK,)J) ( K n p exists) , is weaker than the axiom of choice.)

"+" is replaced by

We have shown that if

"."in

the

(*),

resulting statement is equivalent to the axiom of choice. CN 43:

(VK,u)

n

~ m 1 . 1 = (K

(K

n

n

K

exists then

1.1

K

U 1.1

exists and

u 1.111.

( V K ) ( V a ) [If

CN'43: K * N ~ =

[If

~.I).(K

N~).(K

u

K

n

Na

exists then

K

U

Ha

exists and

N,)I.

The trichotomy, T1, implies that for all cardinals K and l . I r K I; p or 1.I < K . Thus, K fl 1-I = min(K,P), K U 1.I = rnax(~~1.I)and CN 43 follows. THEOREM 7.44:

v* K

I;

PROOF: Let Ka+l. Let

U Ka+l

= V

(1)

K'Kcr+l

(2)

(K

n

CN'43

w zr K O K

= v

+

+

WO 3.

and let K a : then

01 K

be chosen so that n = Ka and

+ Na+l= (V

sa)'Ka+l -

V'na+l

N~+~)*( u KN ~ + ~= )K;(W

-k

Ka+l-

+ K ~ + ~= )V

* H ~+

K

~

.+

~

PART I, SET FORMS

154

CN'43 i m p l i e s K - K , + ~ = from (1) and ( 2 ) t h a t

(K

v'K,+l

cC+l

u

fi,+l)*(~

i V'K,

Using 7.8, w e o b t a i n t<

n

.

+ Ka+l

v s Ka+l.

or

i v-K,

So it f o l l o w s

But

K,+l 6 K, so t h e f i r s t i n e q u a l i t y c o n t r a d i c t s T h e r e f o r e , w e a r e l e f t w i t h v s K,+l.

6 v

7.9.

and

DEFINITION 7.45: (a)

y,,(X)

(b)

If

i s t h e c l a s s of a l l f i n i t e s u b s e t s of

x

1x1 =

i s a s e t and

K,

then

e ( K )

=

X.

lp
I t i s e a s y t o prove t h e f o l l o w i n g t h r e e f a c t s a b o u t

in

e

NBGO:

.

7.46:

(Va)e(K,) = K,

7.47:

(VK,p)e(K+p)

7.48:

( V K , ~ () K = K * N + ~ ~ ( K - K , ) = e(K)-K,.

= e(K)*e(p).

The f o l l o w i n g t h r e e forms a r e due t o A l a s [1968,19701.

(CN'46 i s a s p e c i a l c a s e of CN 46 b e c a u s e f o r any a l e p h , e(K) = K

and

e ( K

+ [ e ( ~ ) l * #)

HI

e(K).)

It f o l l o w s from t h e w e l l o r d e r i n g theorem and p r o p e r t i e s of a l e p h s t h a t e a c h of CN 44-46 h o l d s . THEOREM 7.49: PROOF: e ( K

+

K,)

Let

= e(K)

CN'44 K zt

+

N .,

KO

-+

WO 3.

and l e t But, e ( K ) * N ,

K,

= [ e ( ~ ) ] * . By CN'44, = e ( K

+

tia)

(by 7.46

97. and 7.47). Na 5 e(K), K

5

CARDINAL NUMBER FORYS

155

By 7.8, Therefore, e(K)-sa = e(K) -t Ka. which is impossible, or e(K) 5 Ka, which implies

Ka.

THEOREM 7.50:

CN 45

WO 3 .

+

PROOF: Let K = p * K o . Then K + K* s K*K* and by 7.46, 7.47 and 7.48, e ( + ~ K * ) = e(K*K*). CN 45 implies K + K* = K-K*. Using 7.8 we obtain K i K * , which implies P ~ ~ ' +4 wo 6 3.

THEOREM 7.51: PROOF:

K

i

SK*.

e(K + [e(~)]*)= [e(~)]*.

The following three forms, CN 47-49, are due to Tarski [1938,1939]

.

We shall prove AC 1 T 1

A

A

T 1 * CN 47

AC 1

A

WO 1

+

+

CN 48

+

CN 49 * CN 48.

It is clear that CN 47 case of CN 47. THEOREM 7.52:

WO 3 .

AC 1

A

+

CN 48 because CN 48 is a special

T 1

-+

CN 47.

Let a be a PROOF: Let s # fl and suppose Is1 = K . successor ordinal such that ( 2 " ) * 5 Na. Define a function f on w a as follows: For B E w a , f(B) =

{X:X

5

Uf"B A

1x1

K).

Then define t = U f"wa ' We shall show that t is the required set. Suppose x E t. Then for some B < w a , x E f(B). So 1x1 < K which implies K L 1x1. Also, x c U f " B _c u f"wa= t. Conversely, suppose x 5 t 1x1 < K . For each u E x, let

and K L 1x1. T 1 implies BU = {B:u E f(B)) and let

156

PART I, SET FORMS

be the smallest element of Then 2IR' 6 2Ix'6 2 K . Thus,

BU.

yu

(i)

IRI

R = {yu:u E x).

Let

< 2K.

Since R is a set of ordinal numbers, R is either finite or IR[ is an aleph. Thus, it follows from (i) and the definition of K, that

8 = UR. We shall show 6 wa. If $ E R then This implies / $ I < K,. Since a is a successor ordinal ) @ I 5 Let AB be the set of all 1-1 functions w mapping into C is a set a-1' Let C = { A g : $ E R). of non-empty sets. Thus, AC 1 implies that there is a choice function g on C such that for each B E R, gB E AB.

Let

6

wa.

It is clear that

8

< U ((8)

BER such that

define a function h is a 1-1 function mapping Therefore, (iii) 8 < R

u

BER

x

8). For each

h(B,y)

=

(B,g,(y)). r

({Bl

x

B)

into

u

B E K

y E

({B)

Then x

h

w ~ - ~ ) .

BER

x w ~ - ~ .

It follows from (ii) and (iii) that

Consequently, we obtain

Now we shall show x E t. For every u E x. There is a B i 8 such that u E f(B). Therefore, u E U f " ( 8 + 1 ) , which implies x 5 U f"(8+1). By (iv), 8 < w a . Since o is a limit ordinal, 8 + 1 < w a . We have shown above that 1x1 < K . So it follows from the definition of t that x E t, q.e.d. Let P be the statement obtained from CN 47 by replacing " I s I L 1x1" by "Ixl < Is)". Dana Scott noted that the only place the axiom of choice is used in proving P is in proving (iii), and therefore, (iv). That is, for any successor ordinal a , w a is not the union of < w, ordinals, each of

57. CARDINAL NUMBER FORMS

157

which is < ma. (ma is regular.) It follows from the work of Paul Cohen that this statement does not imply the axiom of choice in NBG. However, we do not know if P is provable in NBGO or NBG. Before proving CN 48 lemma (Tarski [1939]) :

+

WO 3, we shall prove the following

LEMMA 7.53: Let t and v be sets such that I v I then there is a set w with the following properties:

5

(ti;

(i) w _c t, (ii) w (iii) w

v, can be well ordered.

PROOF: Since IvI i Itl, there is a 1-1 function f mapping v into t. Define a function g such that for each ordinal number a , g(a) =

f (g"a)

if

g"a E v,

otherwise

.

As long as g(c1) E t, g is 1-1. Therefore, there is an ordinal c1 such that g(a) = u & t. Let y be the smallest such ordinal. Then w = g"y is the required set. THEOREM 7.54:

CN 48

+

WO 3.

Let s be a non-empty set. CN 48 implies there is a set t such that t w {x 5 t:ls( L IxI} = v. Since IvI = It[, t and v satisfy the hypothesis of 7.53. Let w be a set satisfying (i), (ii) and (iii) of 7.53. Since w _c t and w & v we must have I s / s Iw/. By 7.53 (iii), w can be well-ordered, q.e.d. PROOF:

THEOREM 7.55: PROOF: and let K = that s = {x Since K = K So s k {x _c

T 1

A AC

1

A

WO 1

+

CN 49.

Let t and p satisfy the hypothesis of CN 49, (ti and s = {x _c t:p { 1x1). Then T 1 implies _ct:lx/ < u ) . Let u be a set so that [ul = p. + 1, t = a U {to}, where to a and la1 = K. a:IxI 5 p } . For each x _C a such that 1x1 5 p,

158

PART I , SET FORMS

t h e r e i s an i n j e c t i o n , gx

Define

on

u

fx,

mapping a s u b s e t of

u

onto

x.

so t h a t

fx(v)

if

gxW =

a(fx)

v E

I

otherwise.

x E s w e can f i n d a f u n c t i o n x 5 t , and x, y E s , x # y i m p l i e s s 4 t U , which i m p l i e s I s ( s K’.

T h e r e f o r e , u s i n g AC 1, f o r e a c h gx

mapping

gx # g y .

u

onto

Thus

C o n v e r s e l y , WO 1 i m p l i e s b o t h ordered.

Since

p i

K,

and

K

c a n be w e l l

p

it f o l l o w s from 7.6 t h a t

This implies t h a t t h e r e i s a function

h:t

x

u

W

P-K

t.

= K.

Let

f E tU. Define

$ ( f ) = Ih(f(v),v):v E ul. Then

$:t

U

Therefore,

4 s.

THEOREM 7 . 5 6 :

PROOF: p 2 2.

that

t = jl,

Let

It1 =

Therefore,

s

Let

take

p = 0

CN 4 9

2’

K = K.

+

K

s Is/,

q.e.d.

CN 48.

be a non-empty set and l e t and i f

KO ; t h e n

P = 1 K’

=

K.

I s [ = p.

t = {PI. Suppose L e t t be a s e t s u c h

take

CN 49 i m p l i e s

t = {x 5 t:IsI & 1x1).

CN 50-52 are t h e l a s t t h r e e forms i n t h i s s e c t i o n .

CN 51:

If

J

domain

i s a set and J

then

A

and

B

a r e functions with

If

57.

CN 52:

CARDINAL NUMBER FORMS

If J is a set, A for each j E J, 2

159

is a function with domain i IA.1 then

J

and

3

I,u Ajl

I

JEJ

x

Ajl

j EJ

-

CN 5 0 was stated without proof in Lindenbaum & Tarski [19261 by Tarski,and later Tarski 119481 published a proof of its equivalence to the axiom of choice. CN 51 is frequently referred to as Konig's Theorem or Lemma. Konig [1905] proved it from the axiom of choice for countable J and later Zermelo [1908] proved it for the general case. CN 52 is a variation of CN 51 which is due to Arthur L. Rubin. We shall prove first that CN 50 is equivalent to the well ordering theorem. THEOREM 7.57: PROOF:

WO 1

CN 50.

u

Ai. It is clear that CN 50 iEw finite, and we may assume without loss of each Ai is infinite. WO 1 implies that the of an infinite set is an aleph. Let 1x1 = H a l Y = Bi. It follows from properties of

Suppose x ,<

holds if x is generality that cardinal number [Ai]= K and Bi alephs that for each

I pi

+

,u

1EW n € w

u ~l

isn

=

maxCK

' i

:i

i

nI.

I iEw UAil

K *K = O Y K s K c l . This contradicts the hypothesis that Ka 1 U Ai I . Y igw Thus we must have 8, 21 c1 for some n, which implies If

1x1

< c1

for all

i f then

y

i

a, so

s

lirn , u Ail'

Before proving that CN 50 implies the well ordering theorem we shall prove a preliminary lemma. It follows from 7.13 that for all transfinite y f r(y) < p(?'(p(y))). However, we need a little stronger result. LEMMA 7.58: There is a function F such that for each transfinite set y, F is a 1-1 function mapping r(y) into Y B(Ip(YJ(Y)) ) .

PART I, SET FORMS

160

PROOF: Let cx E T(y) and let mapping a into y. Let Af = { z : ( 3 B ) ( B Then

Af E y(?(y)). Q

=

a

i

A z =

f

be a 1-1 function

f"0)1.

Let

{Af: ( 3 a E r(y) (f:a 6 y) 1 5 ?(f'(y)).

There is a natural mapping, I), from Q onto T(y), namely, for each Af E Q, let $(Af) = a(f). Thus, we can construct a 1-1 function h mapping ? ( r (y)) into $'(a). (For each Define a function g such that for u 5 r(y) , h(u) = $-"'u.) each cx E r(y), g(a) = {a}. Then g:r(y) 6 P(r(y)). NOW, for each a E T ( y ) define F (a) = h(g(cl)), then F is Y Y the required function. CN 5 0

THEOREM 7.59: PROOF:

Let

x

-+

WO 3.

be transfinite.

Construct a sequence of

sets such that yo = x and for each i E W , yi+l = P(Yi). Let u = U T(yi). It follows from 7 . 5 8 that iEw u < u (yi x {i}). If u < U (yi x {i)), then it follows iEw iEw from CN 50 that there is an n E w such that u < U (yi x {ill. Let ui _c yi such that u fi: u (ui x {i}). isn iin Then J u =~ C Juil = max J u1.I = Iuj I. isn i s n This implies

T(yj) 6 u 6 y j

which is a contradiction.

Thus

u (yi x {i}), which implies x d u, q.e.d. iEw We shall prove that AC 1 + CN 5 1 A CN 5 2 and CN 5 1 V CN 52 -+ AC 6. u

fi:

THEOREM 7.60:

AC 1

-+

CN 52.

PROOF: Suppose J and A satisfy the hypothesis of CN 52. We may assume without loss of generality that the and A.'s are pairwise disjoint. For each j E J, let u 7 j v be distinct elements of A (Use AC 1 to choose these j. j elements.) For each x E U Aj, there is a unique i E J j EJ

CARDINAL NUMBER FORMS

7.

such that

x E Ai.

if if if

X

Then the function is an injection, THEOREM

x

Define Fx E

G

7.61:

Aj j€J

as follows:

j = i, j # i ~ x # u 1' j# ir \x=u 1 .

such that for each

AC 1

+

161

x E

U A., G(x) jEJ

=

Fx,

CN 51.

PROOF: Suppose J , A and B satisfy the hypotheses of CN 51. We may assume, without loss of generality that the A . ' s are pairwise disjoint and I B . 1 2 2 for all j E J. 7 3 Using a proof similar to the proof of 7.60, it is easy to show

I

U A 3 1 2 I,XBjl. JEJ jEJ Suppose F: *j = j EJ projection of F onto B

jgj.

u

For , j E J,

let

That is, F . (x) =

1'

3

Fjr A.: A . 3

7

+

B

F

be the j F(x) ( 1 ) and

1'

Since A. < B. F . ('A cannot be surjective. Thus, using 3 I' 3 j AC 1, we can choose t E B F rA."A Let j j j i j ' g = i(j,tj):j E J}. Then g E Bj. Consequently, there is 7EJ a unique j E J and x E Aj such that F(x) = g. In particular,

-

,x

t = g ( j ) = F(x) ( 1 ) = Fj(x) E FjrAj " A 1' j which is absurd. It is easy to see that each of CN 51 and CN 52 implies AC 6. Let {x : j E J } be a set of non-empty sets. Without j loss of generality, we may assume 1x.I z 2 for each j E J. 3 Then CN 51 implies xj # 8. (take B = x and j j jEJ A = 8 for each j E J) and CN 52 implies xj # 8 i€J j because U xj # 8 .

x

j €J

x

This Page Intentionally Left Blank

58.

FORMS FROM

TOPOLOGY, ANALYSISl AND LOGIC

In preceding sections we tried to make the text complete in itself so the reader would not have to refer to the literature for definitions and proofs. Since this section contains material from several different branches of mathematics, to continue this procedure would make the section unnecessarily long. Here we shall refer the reader to the literature for many of the definitions and subject matter proofs. We shall first consider some statements from logic. The first two are Lowenheim-Skolem theorems and the third is a compactness theorem. The proofs that these three theorems are implied by the axiom of choice are due to Tarski (Tarski & Vaught [ 1 9 5 7 ] ) , Malcev [1936] and Henkin [1949]. The proofs given here that they imply the axiom of choice are due to Vaught [1956]. In the following three propositions, "formula" means a "first order formula". LOG 1:

A formula having a model in a set of cardinality

also has a model in a set of cardinality 1-1 K O 5 1J. 5 K .

K

if

LOG 2 :

A formula having a model in a set of cardinality K O also has a model in a set of cardinality K , for all K 2 KO.

LOG 3:

If Q is a set of formulas in which the set of individual constants has cardinality K and every finite subset of Q has a model, then Q has a model in a set whose cardinality is not greater than Y + h o . 163

164

PART I, SET FORMS We shall show that THEOREM 8.1:

LOG 3

LOG 3 -+

-f

LOG 2

-+

LOG 1

-+

CN 6.

LOG 2 .

PROOF: Let $ be a formula which has a model in a set x where 1x1 = KO. Let K be a cardinal such that KO i K . Let C be a set such that I C I = K and C n x = For each a, b E C, a # b, we define the formula qab as l ( a = b). Let A = {Jlab:a, b E C A a # bl and let Q = U A. Since each finite subset of A has a finite subset of C as a model, + has x as a model, and x II C = p, it follows that each finite subset of Q has a model. Thus, Q satisfies the hypotheses of LOG 3. Therefore, Q has a model in a set K + N o = K. whose cardinality is not greater than By construction of Q, the cardinal number of the model cannot be less than K . Hence, )I has a model in a set of cardinal number K , q.e.d.

a.

The proof that LOG 2 -f LOG 1 uses the Lowenheim-Skolem Theorem (see Skolem [1929]) which is provable in NBGO: Every consistent set of formulas (with no individual constants) has a model in a set of cardinality KO. THEOREM 8.2: LOG 2 LOG 1. -f

PROOF: Suppose the formula I+ has a model in a set of cardinality K . Then I+ is consistent. Therefore, by the Lowenheim-Skolem Theorem, I+ has a model in a set of cardinality KO. S o by LOG 2 , @ has a model in a set of any cardinality greater than No, namely p where N o i p s K. THEOREM 8.3:

LOG 1

-f

CN 6.

PROOF: Let K be an arbitrary transfinite cardinal K'No K'K0)2 - 2K'&0'2 number. Let = 2K'hO; then u 2 = ( 2 = 2 and So i K i u. Consider the following formula: = )I S:

(Vx,y)(3z)R(x,y,z) A

98.

FORMS FROM TOPOLOGY, ANALYSIS AND LOGIC

An algebra ( A I R ) , relation,is a model for

165

where A is a set and R a ternary S iff R is a 1-1 function mapping

A x A into A (so that ) A ) *= [A)). Clearly, model of cardinality u . Therefore, by LOG 1, S 2 of cardinality K . Therefore, K = K , q.e.d.

S

has a has a model

Our next statement is due to Klimovsky [19561. LOG 4:

If A is a set of sentences of first-order logic and B _C A is a consistent subset, then there exists a 5-maximal consistent set C such that B 5 C _c A. Since the property of being consistent is a property of finite character, M ' 7 .+ LOG 4 . THEOREM 8.4:

LOG 4

.+

AC 2 .

PROOF: Let t be a set of pairwise disjoint non-empty sets. Let A be the following set of sentences:

Let

1.

(]lx)ix;

2.

a # b;

3.

ua;

B

€or each for each

for each

u E t.

a, b E U t , a # b.

a E u E t.

consist of all sentences of the first two types.

Any

c-maximal consistent extension of B determines a choice function on

t.

Dzik [198l] also has a logical equivalent of the axiom of choice, but since it is a special case of AL'3 we will not include it here. theorem is frequently used in logic. We shall show that $og' theorem together with the Boolean prime ideal theorem, BPI, is equivalent to the axiom of choice. As was mentioned earlier in Section 6, in 1967 Halpern & Levy [19711 proved that BPI does not imply the axiom of choice in NBG, and it was shown by Blass [1977] that $06' theorem does not imply the axiom of choice in NBG.

$06'

DEFINITION 8.5: a b i n a r y relation on

Let A and X be non-empty sets, R A, 7 a filter on p(X), and M = (AIR).

PART I, SET FORMS

166

(i) For

fl,f2 E Ax

define:

Since X E 7 and F is closed with respect to finite intersections, it is easy to see that S is an equivalence Let relation on .'A [f] = {g E Ax: f S g } and let Ax/.$ = { [ f ] : f E AX 1 . (ii) For

[f], [g] E Ax/F,

[flR' Igl

iff

define

R'

as follows:

{ u E X: f(u) R g ( u ) } E .$.

Then we define the reduced power Y/'M

= (A~/.F,RI).

Using the axiom of choice, Eos [ 1 9 5 5 ] proved the following: LT: If A, X, R and M are defined as in 8.5 and .$ is an ultrafilter on X then MX/Y and M are elementarily equivalent. (Actually, $os proved the theorem for any relational system M which has arbitrarily many finitary relations. But LT is sufficient for our purposes.) The proof of the following theorem is due to P. E. Howard [19751.

THEOREM 8.6:

LT

-k

BPI

+

AC 2 .

PROOF: Let X be a s e t of pairwise disjoint non-empty sets such that for each y E XI y n X = 8. Let A = X U (UX) and define R on A so that

t R y t R t TU

R v

if t E Y E X , if t E U X , otherwise.

58.

167

FORMS FROM TOPOLOGY, ANALYSIS AND LOGIC

Suppose X

has no choice function.

Define

15

p(X)

so

that I

=

{u

5

X:

u

has a choice function).

Then I is an ideal, and since X has no choice function, I is a proper ideal. Using BPI' we extend I to a maximal BPI'.) (proper) ideal J. (It was shown in section 6 that BPI = {X u: u E J). Then 9 is an ultrafilter. Let

-

By LT, MX/F and M are elementarily equivalent, and by the definition of R, M I- (Vy) (jt)t R y. Thus Mx/F 1 ( V y ) (3t)t R y. In particular, let ix be the identity Then, it follows that for function on X and let y = [i,]. some function f E AxI {u E X: f (u)R ix(u)1 E 7 . That is, Z = { u E X: f(u) E u} E 7 . This implies that X Z E J. But clearly Z has a choice function, so Z E J. This contracts the fact that J is a proper ideal. Hence, the assumption that X has no choice function leads to a contradiction, q.e.d.

-

Blass [1977] proves that the statement UP:

All ultrafilters are principal.

LT and that is consistent with NBG. It is clear that UP i UP. Thus in Blass' model of NBG, LT is true and AC AC is false. -+

-+

Probably the most well-known and important topological equivalent of the axiom of choice is the Tychonoff Compactness Theorem. (See, for example, Kelley [1955] for definitions of topological terms.) TOP 1: The product of compact spaces is compact in the product topology. Tychonoff [1935] derived TOP 1 from a maximal principle and later J. Xelley [1950] proved the converse. L. E. Ward 119621 showed that a weaker version of TOP 1 is equivalent to AC . A.

TOP 2 : The product of mutually homeomorphic compact spaces is compact in the product topology.

PART I, SET FORMS

168

Then 0. T. Alas [1969] proved that a weaker version of TOP 2 is still equivalent to AC. TOP 3 : The product of a set of mutually homeomorphic topological spaces {(Xi,Ti): i E I}, where for each i E 1 has three elements, is compact in the product topology. Ti Clearly TOP 1 + TOP 2 proof that TOP 3 + AC. THEOREM 8.7:

TOP 3

-+

+

TOP 3.

We shall g i v e the

AC 7.

PROOF: Let {Xi: i E I) be a set of equipollent sets. a be an element such that a f U Xi. For each i E I, i61 let Yi = Xi U {a] and define Ti = {fl,{a},Yil. Then {(Yi,Ti): i E I) satisfies the hypothesis of TOP 3 . Let

Let

x

Yi be given the product topology and define iC1 Zi = {f: f E W A f (i) E Xi}. Then {Zi: i E I} is a family of closed sets with the finite intersection property, Consequently, by TOP 3 , Zi = Xi is non-empty. iEI i€I If in TOP 1 and 2 the topological spaces are T 1, the resulting statements are equivalent to the axiom of choice. However, if the topological spaces are T 2, it was shown by $05 & Ryll-Nardzewski 119541 that the resulting statements are equivalent to BPI. (See also Rubin & Scott [1954]. ) We next give a result of Franklin & Thomas 119711.

W =

r)

x

DEFINITION 8.8: (a) A function f from one topological space to another is called perfect if the image of a closed set is closed and the inverse image of each point is compact. (b) An onto function is irreducible if the image of a proper closed subset of the domain is a proper subset of the range. (c) If

1EI

is a union of topological spaces

X is called a free union if 0 5 X is open iff n Xi is open in Xi for each i E I.

Xi, 0

X = , U Xi

58.

FORMS FROM TOPOLOGY, ANALYSIS, AND LOGIC

169

TOP 4 : Any perfect onto map from one topological space to another has an irreducible restriction to a closed subspace of its domain. TOP 4 is provable from a maximal principle. THEOREM 8.9:

TOP 4

+

AC

2.

PROOF: Let {Xi: i E I] be a family of non-empty pairwise disjoint sets and fox each i € I, let Ti be the (a set is open iff its complement cofinite topology on Xi is finite). Let X = U Xi be the free union. I f the index i61 set I is given the discrete topology then the function f:X I such that for each u E Xi, f ( u ) = i, is perfect and maps X onto I. Thus, TOP 4 implies that there is a subspace Y 5 X on which f is irreducible. The intersection of Y with each Xi is a singleton, q.e.d. ~

-+

N. Brunner (19831 studies statements of the following form: TOP 5 ( P ) :

Each

T 2 (Hausdorff) space has the property

TOP 6(P,Q): Each T 2 space which has the property has the property Q. X

P

P. also

For P and Q, Brunner considers the following, where is a space and T 5 p(X) is its topology. W 1:

T

can be well ordered.

W 2:

X

can be well ordered.

D 1: X is covered by a well ordered family of closed, discrete sets. D 2:

X

is covered by a well ordered family of discrete

T

has a well orderable base.

sets. B 1:

T such that W B 2 : There is a function f:X x W can be well ordered and for each x E X, f"({x} x W ) is a neighborhood base at x. -+

170

PART I , SET FORMS

There i s a f u n c t i o n

B 3:

b e w e l l o r d e r e d and f o r e a c h L 1:

f:X x W x E X,

-+

T

such t h a t

n f " ( 1x1

x

W) =

can

W

{XI.

Each open c o v e r of

X

has a w e l l orderable

Each open c o v e r of

X

has a w e l l orderable refine-

subcover. L 2:

ment. has a dense w e l l o r d e r a b l e s u b s e t .

S:

X

C:

Each f a m i l y

0

_c

of p a i r w i s e d i s j o i n t s e t s c a n b e

T

w e l l ordered. A 1:

0 c T

If

A 2:

If

covers

_c T

0

t h e r e i s a mapping

X

f:X

-+

0

f:X

+

T

x E X, x E f ( x ) .

such t h a t f o r a l l

such t h a t f o r a l l

covers

t h e r e i s a mapping

X

x E X, x E f ( x )

and

f"X

refines

0.

I t i s e a s y t o show t h a t t h e f o l l o w i n g i m p l i c a t i o n s h o l d

( p r o o f s a r e g i v e n i n Brunner [19831):

Figure 8.1 Also,

if

P'

+

and

P

Q

+

Q'

then

and TOP 5 ( Q ) * TOP 5 ( Q ' ) ; and i f NBGO

TOP 6 ( P , Q )

P

+

Q

-+

TOP 6 ( P ' , Q ' )

then

It i s a l s o clear t h a t TOP 6 ( P , Q ) f o r a l l P and

TOP 6 ( P , Q ) .

AC -* TOP 5 ( Q )

+

THEOREM 8 . 1 0 :

t h e axiom of c h o i c e .

Q.

Each of t h e f o l l o w i n g i s e q u i v a l e n t t o

§8.

FORMS FROM TOPOLOGY, ANALYSIS, AND LOGIC

1.

TOP 6(P,Q), where P = D 1, D 2, B 2, B 3, and Q = W 1, W 2, B 1, L 1, L 2, S, C, or A 1. Q

=

2. TOP 6 ( P , Q ) , where P W 1, W 2, B 1, S, or C. 3.

or

TOP 5 ( Q ) , where

Q

=

I

=

L 1, L 2, or

A 1

171

or

A 2,

and

W 1, W 2, B 1, L 1, L 2, S, C,

A I.

PROOF: It follows from Figure 8.1 and the remarks preceding the theorem that to prove 1. it is sufficient to prove that each of TOP 6(P,Q), where P = D 1 or B 2 , and Q = L 2, C, or A 1, implies AC. Let X1 be a set and T1 = ?(X,); then (X1,T1) is a topological space satisfying D 1 and B 2. If (X1,T1) satisfies L 2 or C then X1 can be well ordered. Let Y be a set of non-empty pairwise disjoint sets such that no one is an element of another, X2 = Y U (UY) and T2 = ?'(X2) then (X2,T2) satisfies D 1 and B 2. Suppose (X2,T2) satisfies A 1. The set 0 = {{a,u}: a E u E Y) is an open cover of X2. Suppose f:X2 -+ 0 is an A 1 mapping. For each u E Y, let g(u) = f(u) i u } . Then g is a choice function on y.

-

To prove 2. it is sufficient to show that TOP 6(L 1,C) implies AC. Let X be a set with the discrete topology and X* = X U {XI its one point compactification. The topology T* of X * is all open subsets of X and all subsets U 5 X* such that X* U is a closed compact subset of X. In our U is finite. Thus, (X*,T*) satisfies L 1. If case X* (X*,T*) satisfies C then {{a}: a E XI can be well ordered. Therefore, X can be well ordered.

-

-

Part 3 . follows from

1.

THEOREM 8.11: Each of the following is equivalent to the multiple choice axiom, MC. (See section 5. We shall use MC 2.) the fact that MC 1

.+

172

PART I, SET FORMS 1.

TOP 6(P,Q), where P = D 2, L 1, L 2 , A 1, Q = D 1, B 2, or B 3 .

and

2.

TOP 6(D 2, A 2).

3.

TOP 5(Q), where

Q = D 1, B 2, B 3,

or

or

A 2,

A 2.

PROOF: To show that MC implies each one of the above TOP 5(Q) for statements it is sufficient to show that MC 2 Q = D 1 and B 2. +.

Let (X,T) be a T 2 space. MC 2 implies TOP 5(D 1).

Since points are closed

To prove TOP 5 ( B 2), we use the fact that M C 2 implies that there is a well-ordered family of finite sets W such that T = UW. Define f:X X W + T so that if (a,E) E X x W,

f(a,E) =

ntu E E: a E

UI

if

a E UE,

otherwise.

If a E 0 E T, then there is an E E W such that 0 E E. Thus, a E f(a,E) 5 0, thereby proving TOP 5 ( B 2). To prove the converse, it is sufficient to prove that each of TOP 6(D 2, B 31, TOP 6(L 1, B 3 implies MC. Let X be an infinite set with the discrete topology and X* = X U 1x1 with the topology T*, its one point compactification. (See 8.10). Then (X*,T*) sat sfies L 1 and D 2. Suppose W is a well ordered family of sets such that f:X* x W + T* is a B 3-function. Then IIIf(x,u): u E

wl

= {XI.

Therefore, X = U{X* c. f(X,u): u E W} and for each u E W, X* f(X,u) is finite. This, completes the proof of part 1. of the theorem.

-

Let

To prove 2. we shall show that TOP 6(D 2 , A 2) +. M C 1. Y be a family of pairwise disjoint infinite sets, such

58.

FORMS FROM TOPOLOGY, ANALYSIS, AND LOGIC

t h a t no one i s an e l e m e n t o f a n o t h e r .

u*

173

u E Y, let

For each

b e t h e one p o i n t c o m p a c t i f i c a t i o n o f t h e d i s c r e t e

u. 8 . 8 ( ~ ) )a,n d l e t

t o p o l o g y on

T

-

0 = {u*

Let c o v e r of

{ a } : a E u E Y).

Suppose

X.

f

+

0.

For each

of

f

T

That is,

0.

and

-

(X,T)

refines

f"X

u E Y, l e t g(u) = u f ( u ) . By t h e d e f i n i t i o n T I g ( u ) i s a f i n i t e s u b s e t of u. To c o m p l e t e t h e

and

8.

g(u) #

p r o o f w e must show t h a t

Since

v E Y

and an

a E v

u E f ( u ) , so

v = u.

This implies

there is a But

for

x E X, x E f(x)

so

is a n open

0

i s a n A 2-mapping

such t h a t f o r a l l

f:X

Then

UY

and

Y

X = Y U ( U Y),

and

X

b e t h e f r e e u n i o n (see The sets

b e i t s topology.

a r e both d i s c r e t e subsetsof satisfies D 2.

E Y)

X = u{u*: u

Let

f"X

refines

c_

f (u)

such t h a t

a E g(u).

v*

-

0,

{a}.

Thus

g

is the required multiple choice function. Since

TOP 5 ( Q )

TOP 6 ( P , Q )

+

for a l l

P

and

Q, t h i s

c o m p l e t e s t h e p r o o f of t h e t h e o r e m . THEOREM 8 . 1 2 :

Each o f t h e f o l l o w i n g i s e q u i v a l e n t t o PW.

1.

TOP 6(P,W l ) , where

P = W 2 , B 1,

or

S.

2.

TOP 6 ( B 1 , Q ) , where

Q = W 2 , D 1,

or

D 2.

3.

TOP 6(S,W 2 ) .

PROOF: P = B 1, S, property

P

F i r s t we note t h a t or

W 2

PW

because i f

PW

also have t h e p r o p e r t y

To c o m p l e t e t h e p r o o f of l . , l e t

let (X,T)

T = p(X).

Then

Since

TOP 6 ( P , Q )

p(X)

W 2 + D 1 + D 2,

holds, spaces with t h e

X

b e w e l l o r d e r e d and

TOP 6 ( B 1, D 2 )

+

it f o l l o w s t h a t TOP 6 ( B 1, W 2 )

base element

B

if

a , b E D, a # b, a R b

such t h a t

B

n

D = {a},

However,

because i f

h a s a w e l l o r d e r e d b a s e l t h e n any d i s c r e t e s e t ordered a s follows:

If

S.

c a n be w e l l o r d e r e d .

TOP 6 ( B 1, W 2 ) + TOP 6 ( B 1, D 1) + TOP 6 ( B 1, D 2 ) .

we c l a i m

for

W 1.

s a t i s f i e s W 2 , B 1 and

(X,T)

s a t i s f i e s W 1, t h e n

implies

D

T

c a n be w e l l i f f the f i r s t

precedes t h e f i r s t

174

PART I , SET FORMS

base element

such t h a t

B'

n

B'

D = {b).

Thus, i f

c o v e r e d by a w e l l o r d e r e d f a m i l y of d i s c r e t e sets, TOP 6 ( B 1, W 2 )

Suppose

c a n be

T h e r e f o r e , t o p r o v e 2 . it i s s u f f i c i e n t t o

w e l l ordered. prove t h a t

is

X X

X

-+

PW.

is a w e l l ordered set.

Let

be t h e

T

a E X and f E 2', let -111 paof = f ( a ) . Then 2 = {pa { i } : i E C 0 , l ) A a E X I i s a s u b b a s e f o r t h e t o p o l o g y and Z c a n b e w e l l o r d e r e d . Therep r o d u c t t o p o l o g y on

fore,

(2',T)

2'.

For each

B 1.

satisfies

Thus, W 2 i m p l i e s

2'

can be

w e l l ordered. W e also n o t e t h a t t h e above s p a c e a l s o s a t i s f i e s

S

because i f

E 2x:f-1t1{1~ i s f i n i t e }

D =

then

D

i s d e n s e and c a n be w e l l o r d e r e d . + PW, q.e.d.

Consequently,

TOP 6(S,W 2 )

THEOREM 8 . 1 3 :

Each o f t h e f o l l o w i n g imply

1.

TOP 6 ( P , D l),

2.

TOP 6 ( C , Q ) ,

3.

TOP 5 ( D 2 ) .

PROOF:

P = B 2, B 3, S

where

where

PW.

Q = W 1

or

or

C.

W 2.

To p r o v e 1., it i s s u f f i c i e n t t o p r o v e

TOP 6 ( S , D 1) PW and TOP 6 ( B 2 , D 1) -+ PW. Suppose X i s a w e l l o r d e r e d s e t and T i s t h e l e x i c o g r a p h i c a l o r d e r The s p a c e (ZX,T) i s compact and it t o p o l o g y on 2'. 1 s a t i s f i e s S b e c a u s e t h e s e t D = {g E 2': g 11) i s f i n i t e ) +

-

i s d e n s e and w e l l o r d e r e d .

s

Let

11

be t h e l e x i c o g r a p h i c

o r d e r i n g r e l a t i o n on 2'. F o r e a c h g E 2', let X c X = {h E 2 : g < h l and l e t g = { h E 2 :h < g } . function and

f

from

2'

x

to

D

T

a s follows:

h E D

$

if

h

h'

if

g < h,

2'

if

g = h.

g,

If

Define a g

E 2'

58.

g E 2',

Then f o r e a c h

neighborhood system a t

Since 2'

g.

Since W 1 TOP 6(C,W 2 ) +

Now, it i s e a s y t o see how t o 2'.

i s compact e a c h c l o s e d , d i s c r e t e s u b s e t

(2',T)

is finite,

w e l l o r d e r e d by be w e l l o r d e r e d .

Moreover, e a c h f i n i t e s u b s e t of

g.

Thus, i f

is

2'

s a t i s f i e s D 1, 2'

(2',T)

can

T h i s c o m p l e t e s t h e p r o o f o f 1.

W 2

i t f o l l o w s t h a t TOP 6(C,W 1) TOP 6(C,D 1 ) . I n p a r t 1, w e p r o v e d -f

TOP 6(C,D 1) + PW,

+

D 1,

t h e r e f o r e , p a r t 2. h o l d s .

from Theorem 8 . 1 2 p a r t 2 . b e c a u s e MC 2

COROLLARY 8.14: PROOF:

is a subbase f o r t h e

f " ( { g } x D)

construct a B 2 function f o r

of

175

FORMS FROM TOPOLOGY, ANALYSIS, AND LOGIC

-f

TOP 5(D 2 )

TOP 5(D 2 )

+.

TOP 5(D 2 )

-+

I t f o l l o w s from 8.10-8.14 s t a t e m e n t s are e q u i v a l e n t t o TOP 5 ( Q ) , f o r a l l

-f

follows

TOP 6 ( B l , D 2 ) .

PW.

I t f o l l o w s from 8 . 1 1 t h a t MC 2

from 8 . 1 3 t h a t

P a r t 3.

+.

TOP 5(D 2 ) and

PW.

t h a t each of t h e following AC

in

NBG:

( i n our list).

Q

TOP 6(P,W l),

for

P

#

TOP 6(P,W 2 1 ,

for

P

# w 1, w 2.

TOP 6 ( P , D l),

for

P

# W 1, W 2 ,

for

P

# W 1, W 2 , B 1,

W 1.

D 1.

TOP 6 ( B 1 , D 2 ) . TOP 6 ( P , B l),

S I C.

TOP 6 ( P , Q ) f o r P = D 2 , L 1, L 2 , A 1, A 2 , and Q = B 2 , B3. TOP 6 ( P , Q ) ,

for

P = D 1, D 2 , B 2 , B 3 , A 2,

and

Q = L 1 , L2. TOP 6 ( P , Q ) , f o r P = D 1, D 2 , B 2 , B 3 , A 2 ,

and

Q = S r C. TOP 6 ( P , A l), TOP 6(D 2 , A 2 ) .

for

+

P = D 1, D 2 , B 2 , B 3 , A 2 .

PART I , SET FORMS

176

T h e r e are s e v e r a l unanswered q u e s t i o n s .

For example, whether

f o r Q = B 1, B 2 , L 1, L 2 , A 1, o r TOP 6 ( W 2 , Q ) , i m p l i e s AC. S e e B r u n n e r [1983] f o r d e t a i l s .

2,

A

H o w e s [1980] c o n s i d e r s a n o t h e r p r o p e r t y w h i c h i s u s e f u l i n topology: Each p o i n t o f

B 2':

h a s a n e i g h b o r h o o d b a s e which c a n b e

X

w e l l o r d e r e d and t h e w e l l o r d e r i n g i s c o m p a t i b l e w i t h s e t (By t h i s w e mean t h a t i f

inclusion.

u 5 v

o f t h e n e i g h b o r h o o d base and

u

and

then

u

are e l e m e n t s

v

follows

v

in

the w e l l ordering.) H o w e s [1980] and Vaughan [1975] p r o v e t h a t ( V X ) B 2 '

-

MC.

( T h e r e i s an e r r o r i n H o w e s p r o o f b u t i t i s e a s i l y c o r r e c t e d . )

at

-

MC 2

THEOREM 8 . 1 5 :

p E X

PROOF: L e t Then MC

(VX)B 2'.

and l e t

b e a n y n e i g h b o r h o o d base

T

i m p l i e s t h a t f o r some o r d i n a l

p.

U S, where e a c h a<X l e t u, = each a < A,

T =

X I

i s a f i n i t e non-empty s e t .

S, nSa.

U = {ucl:a < XI

Then

w e l l o r d e r e d n e i g h b o r h o o d base a t

For

is a

p.

N e x t , w e have t o f i n d a w e l l o r d e r i n g which i s c o m p a t i b l e with

5. Towards t h i s e n d w e s h a l l f i n d ucl, uB € U'

such t h a t i f a function

f:X

+

X

and

5U

U'

a subset

u, _C uB t h e n

c1 2

B.

Define

such t h a t : f ( 0 ) = 0;

if

0 <

ci

< X , f ( a ) = m i n I y : ( V B < a) ( u

y

2 Uf ( B ) ) }

L e t U' = { u ~ (:a ~ <) XI. Then it i s c l e a r t h a t r e q u i r e d neighborhood base.

Conversely, l e t

X

b e a s e t and

T

U'

*

is t h e

t h e f i n i t e complement

( T h a t i s , a set i s o p e n i f i t s complement i s (P,s) b e a w e l l o r d e r e d n e i g h b o r h o o d base f o r Some p o i n t p E X. D e f i n e a f u n c t i o n f:X { p ) + P so t h a t € o r e a c h y € X , y # p , f ( y ) = A , where A i s t h e 5 - f i r s t t o p o l o g y on finite.)

Let

X.

-

98.

FORMS FROM TOPOLOGY, ANALYSIS, AND LOGIC

177

-

element of P such that A 5 X {y). For each A E %(f), -1 f-1"{A3 is finite and X {p} = lJ f CAI. Thus, X is

-

It

AEP

a well ordered union of finite sets, thereby proving MC 2. The last part of this section deals with equivalents of the axiom of choice in analysis. There is no sharp dividing line between analysis and topology. There may be some difference of opinion whether, for example, the Krein-Milman theorem is analysis, but these results seemed to fit together so we shall include them here. Most non-constructive principles used in real analysis are weaker than the axiom of choice since they deal with sets of cardinality 5 2". Here we will consider more general statements. We shall mention several well known non-constructive theorems in analysis. HB is the Hahn-Banach theorem, KM the Krein-Milman theorem and AL is Alaoglu's theorem.

is

HB: Let p be a real-valued function defined on the vector space X over the reals, W, such that p(x+y) 2 p(x) t p(y) for all x, y E X and p(ax) = ap(x) for each a B 0, a E W , (p is called a sublinear functional). Suppose that f is a linear functional defined on a subspace S _c X and that f ( s ) 5 p(s) for all s E S. Then there is a linear functional F defined on X such that F(x) 5 p(x) for all x E X and F(s) = f(s) for all s E S.

KM: Let K be a compact convex set in a locally convex topological vector space X. Then K has an extreme point. (An extreme point is a point which is not an interior point of any line segment which lies in K). If X is a vector space, a nonnegative real-valued if for all function 11 11 defined on X is called a x , y E X and all a E W:

PART I, SET FORMS

178

If X is a normed vector space, X*, the dual of X, is the space of bounded linear functionals on X. A base at the point O* of X* for the weak* topology is given by sets of the form Cf E X*: f(xi) < 6 , i = 1,2,...,n} where { X ~ , X ~ ~ . . . ,is X ~a~ finite subset of X. For each f E X*, [If[]= supif (XI:llxil s 11. AL: The unit sphere S * = {f E X*:jjf]l I; 1) in the dual of a normed vector space X over the reals is compact in the weak* topology. The proofs of HB, KM, and AL in NBGO -k AC are given, for example, in Royden [1963]. We obtain variations of KM and AL by replacing "compact" by "convex-compact" (Some authors call convex-compactl quasicompact.) If X is a topological vector space, Y _c X is called convex-compact if whenever Ci, for i E I, are closed convex sets in X , and { Y n Ci:i E I] has the finite intersection property, then ( Y n Ci) # a. Let KMC and ALC be the statements iEI obtained from KM and AL, respectively, by replacing "compact" by "convex-compact". Then

.

n

KMC

+

KM

and

(KMC is a l s o provable in NBGO

+

AL

ALC.

-t

AC.)

The following results are known: I

BPI f AC

in

I1

HB f BPI

in NBG.

+ KM

I11

HB

IV

BPI

V

ALC

VI

AL

+

-

f AC

HB.

NBG. (Halpern

in

NBG'.

I

&

Levy 119711)

(Pincus [19721 [1974])

($og pl Ryll-Nardzewski 119511 and

Luxemburg [19621) HB.

BPI.

(Luxemburg [19691) (Rubin

&

Scott [19541)

We shall include results due to Bell Edwards [1975] and Lembcke [1979].

&

Fremlin 119721,

58.

FORMS FROM TOPOLOGY, ANALYSIS, AND LOGIC

179

AN 1: The unit sphere in the dual of a normed vector space over the reals has an extreme point.

AN 1 is a well-known theorem about normed vector spaces. Bell & Fremlin [1972] prove that it implies the axiom of choice. AN 2: Let X be a real vector space, Y a subspace and S a subset of X. Suppose p:X +IR is a sublinear functional on X and f:Y IR a linear functional such that f (y) 2 p(y) for all y E Y. Then the set Z of all p-dominated linear extensions of f to X has an element g such that for all h E Z with g ( s ) 2 h(s) for all s E S , we have g ( s ) = h(s) for all s E S. That is, g is S-maximal in Z. +

AN 3 : Same hypotheses a s AN 2 (no subset S is required) but the conclusion is that Z has an extreme point. Andenaes [1970] proved AN 2 and Bonsall [1954,19561 proved AN 3, while Lembcke [1979] proved thatAN 2 + MC and AN 3 + AC. We shall prove first that AN 3 THEOREM 8.16:

AN 3

+

+

AN 1.

AN 1.

PROOF: Let X be a real normed vector space with norm p. Let f be the 0-functional on Y = { O } . Then the unit ball of X* is the set Z of AN 3 . Thus, AN 3 + AN 1. The proof of the following theorem is immediate. THEOREM 8.17: 1. AL + KM -+ AN 1. 2.

ALC

+

KMC

+

AN 1.

COROLLARY 8.18: 1. BPI + KM + AN 1. 2.

HB

+

KMC

+

AN 1.

PART I, SET FORMS

180

PROOF: above.

The proof follows from 8.16 and

The following proof is due to Bell THEOREM 8.19:

AN 1

-t

&

V

and

VI

Fremlin 119721.

AC 1.

PROOF: Let {Ai:i E I) be a family of pairwise disjoint sets and let A = U A,. Define iEI s = If E I RA :sup 2’ If t) I 5 1) iEI tEAi

w

= Cf EIRA:sup

x

=

iE1

If E I R ~ :( v E > 0 ) (It E A: lf(t) I > € 1 and

Then by :

W

x

If tEAi

and

X

/jgjjx =

is finite)

c

sup (f(t)1 < m1. iEI tEAi

are normed vector spaces with norms defined

,.X

SUP

LEI tEAi

Ig(t) I,

for

E X.

g

Also, S is the unit sphere of W. But W is isometrically iaomorpnic to the dual X* of X, and therefore, S may be regarded as the unit sphere of X*. AN 1 implies that S has an extreme point e. We shall show that for each i E I there is a unique t E Ai such that e(t) # 0. First suppose there is an i E I such that e(t) = 0 t E Ai. Let u E Ai and choose f, g E S such that

f o r all

f(u) = 1, g ( u ) = -1 f(t) = g(t) = e(t)

for a l l

t E A

{u}.

Then e = (f + g ) / 2 and e # f, e # g. This contradicts the fact that e is an extreme point of S .

181

98. FORMS FROM TOPOLOGY, ANALYSIS, AND LOGIC Next, suppose there is an elements u, v E Ai such that Define f, g E S as follows:

i E I and two distinct e(u) # 0 and e(v) # 0.

Then e # f, e # g and e = (f + g)/2, again contradicting the definition of e as an extreme point. by

Now we define a choice function, F, on {Ai: i E I}, F(Ai) is the unique t E Ai such that e(t) # 0. Each of the following is equivalent to

COROLLARY 8.20: AC : AN 1, AN 3 ,

AL

+

KM, ALC

+

KMC, BPI

+

KM,

and

HB

f

KMC.

Lembcke [1979] gives a proof that AN 2 MC 1 which is similar to the Bell & Fremlin [1972] proof, Theorem 8.18, AC. AC 1. Thus it follows that NBG I- AN 2 that AN 1 -+

..

-+

In the same paper Lembcke gives the following form: AN 4 : Let X and Y be T 2 spaces, f:X + Y a continuous map, and v an (inner) regular Borel measure on Y. If l~ is inner regular with respect to the system of sets { f " K : K is a compact subset of XI, then there is a regular Borel -In (See, measure v on X such that p = vf, where vf = v o f for example, Halmos [ 1 9 5 0 1 for terminology.)

.

Lembcke [1970,1977] proves that AC -+ AN 4 and the following proof is given in Lembcke [19791. THEOREM 8.21:

AN 4

-+

MC 1.

PROOF: Let {Ai:i E 1) be a family of non-empty, pairwise disjoint sets and let X = U Ai. Give X and Y = I iEI

PART I, SET FORMS

182

t h e d i s c r e t e topologies.

Define

f:X

such t h a t f o r each

Y

+

a E Ai 5 X , f ( a ) = i , and d e f i n e 1-1 on Y t o be t h e c o u n t i n g measure, t h a t i s , f o r e a c h S _C Y , p ( S ) = I S 1

is f i n i t e , otherwise

p(S) =

if

S

+m.

i s a r e g u l a r B o r e l measure on Y . A set i s A l l compact s u b s e t s o f Y are images o f compact s u b s e t s o f X. Thus, AN 4 i m p l i e s t h a t t h e r e i s a r e g u l a r Borel measure v on X s u c h t h a t v f = U. T h e r e f o r e , f o r e a c h i E I , v ( A i ) = p ( { i } ) = 1. S i n c e compact Then

p

compact i f f i t i s f i n i t e .

v

s u b s e t s a r e f i n i t e and t h e r e is an

a E Ai

is i n n e r r e g u l a r , f o r each v({a}) # 0. Let

si = sup{v ( { a } ) : a E A i l .

i E I , there is a

Then f o r e a c h

f i n i t e number o f a E Ai s u c h t h a t v ( { a } ) = si. on {Ai:i E I} s u c h t h a t f o r e a c h i E I, g(Ai) = { a E A i : v ( i a ] ) Then

i E I

such t h a t

Define

g

= s . 1. 1

is a multiple choice function.

g

Thus, w e have NBG

1 AN

-

4

AC.

The n e x t form, AN 5 , i s a theorem o f c l a s s i c a l a n a l y s i s . Edwards 119751 proved i t i m p l i e s t h e axiom o f c h o i c e .

Before

s t a t i n g AN 5 w e g i v e some d e f i n i t i o n s . (XI<) is a r e a l vector l a t t i c e i f DEFINITION 8.22: i s a v e c t o r s p a c e o v e r IR, ( X , r ) i s a l a t t i c e , and t h e vector operations preserve

2.

(That i s , f o r a l l

and f o r a l l p o s i t i v e r e a l numbers x

+ z

i;

+ z and

y

ax s ay.)

n

such t h a t

-ne

if

An e l e m e n t

x E X

be an o r d e r - u n i t if f o r e a c h integer

a

i;

y

e E X

x, y, z E X then

i s s a i d to

t h e r e is a p o s i t i v e

x s ne.

i;

x

X

(Alternatively, i f

x

i s a normed l a t t i c e ,

e E X i s c a l l e d a n order-un& if for IIxI] s 1, x i; e. For o u r p u r p o s e s it d o e s n ' t m a t t e r which d e f i n i t i o n i s u s e d . ) all

x E X

such t h a t

Our i n t e r e s t w i l l b e i n r e a l v e c t o r l a t t i c e s o f bounded real v a l u e d f u n c t i o n s . If f u n c t i o n s w i t h domain A ,

f

and

g

are real valued

FORMS FROM TOPOLOGY, ANALYSIS, AND LOGIC

58.

f

i

4

f ( a ) s g ( a ) , Va E A ,

if

E

( f V g ) ( a ) = m a x ( f ( a ) , g ( a ) ) , Va ( f A g ) ( a ) = min(f ( a ), g ( a ) ) (f + 9)

(a) = f ( a )

+

,

A,

Va E A ,

g ( a ) , Va E A ,

.

( b f ) ( a ) = b ( f ( a ) ) , Vb E R , Va E A The c h a r a c t e r i s t i c f u n c t i o n on an order-unit

183

A,

xA,

i s an example o f

i n both s e n s e s .

AN 5: L e t X be a r e a l v e c t o r l a t t i c e w i t h a d i s t i n g u i s h e d o r d e r - u n i t ( d e n o t e d by l), Y a s u b l a t t i c e o f X , and Z F

t h e s e t of a l l p o s i t i v e l i n e a r f u n c t i o n a l s that

F ( 1 ) = 1.

on

t h e convex s e t 2 which i s Y - minimal i n Z . M' i s any o t h e r e x t r e m e p o i n t o f Z s u c h t h a t y E Y, t h e n

for a l l

M ' ( y ) = M(y)

such

X

Then t h e r e e x i s t s a n e x t r e m e p o i n t

M

of

(That i s , if M ' ( y ) i M(y)

y E Y.)

for a l l

The n e x t theorem i s due t o Edwards [19751. AN 5

THEOREM 8 . 2 3 :

PROOF:

+

(L,s)

Let

AL'1.

b e a l a t t i c e w i t h i d e n t i t y and a t

l e a s t one o t h e r e l e m e n t .

Then t h e r e i s a s e t

c o l l e c t i o n of s u b s e t s o f

A,

xs

For e a c h S 5 A l e t S. Then f o r a E A ,

W e s h a l l denote Let

in

e.

that

For

J

A

-

S

2, such t h a t

A

be t h e c h a r a c t e r i s t i c function f o r

by

S'

.

be a proper i d e a l i n

j

and l e t

L

b e i t s image

Then t h e r e i s a Boolean a l g e b r a i d e a l

$

=

x

n 2.

(Take

= {S

5 A:

D e f i n e an e q u i v a l e n c e r e l a t i o n f , g E IRA f R g

and a

(ZI_C>

(L,s)

iff

(3T E R

on

f)

S

lRA

5

x 5 fp(A) TI.)

a s follows:

(3K E %) (Va E K') ( f ( a ) = g ( a ) ) .

such

184

PART I, SET FORMS

Let [f] = {g E Z?:f R q} and let X be the vector space generated by {[xs]:S 5 A } . Then X is a real vector lattice Let Y = {[x,] :S E de} and let Z with order-unit 1 = [x,]. be the set of all positive linear functionals F on X with F(1) = 1. AN 5 implies that 2 has a Y-minimal extreme point, M. First we note that for each characteristic S 5 A , M([xSI) is 0 or 1. Since [x,] 0 i M([xS]) i 1. Suppose 0 < M([xS]) < 1 for For [f] E X, let qs = f[S + 01s' and gsl = Then f = g + qsl. Therefore, for

This contradicts the fact that M

K E

X

Let

then

98

=

2

[xKl

=

[xa1 ,

- %. Then

( x , ~ ) such that

1 5 %.

some frS'

S

+

5 A. OTS.

is an extreme point of

Let % = IS 5 A:M([xS]) = 1). Then 21 algebra ultrafilter in F ( A ) . Moreover, % if

function xs + [xsll = 1,

Z.

is a Boolean = because

n 21

so M([xKI) = 0.

% is an ideal in the lattice We claim that d is a 5-maximal Suppose not. Suppose there is

proper ideal containing f. an ideal 3 3 % . Then if we repeat the above argument with 3 instead of f we would obtain an extreme point M' of Z such that M'([xS]) = 0 for all S € 3, contradicting the fact that M is a Y-minimal extreme point of 2 , q.e.d. In the above proof, the function M is a two-valued measure. Thus, AN 5 is equivalent to the following:

AN 6: If (L,s) is a Boolean algebra with identity 1, and K is a sublattice of L, then there exists a two-valued ) 1. K-minimal measure p on L such that ~ ( 1 = If "K-minimal" is omitted f r o m AN 6 then the resulting statement is equivalent to BPI.

PART 11, CLASS FORMS INTRODUCTION

We now turn our attention to the class or global forms of the axiom of choice. J. von Neumann [1925] listed CWO 10 as an axiom of set theory (Axiom IV 2) and stated that the well ordering theorem can be derived from it. Then in 1928 (von Neumann [19281),he proved that CWO 10 implies that the universe can be well ordered. Gijdel [19401 used axiom E, which is equivalent to our CAC 1, as an axiom of set theory and Bernays [1941] used CAC 4 as an axiom. Bernays also proved that CAC 4 was equivalent to CAC 5 . Since Rubin & Rubin 119631, other people who have worked with the class form of the axiom choice include Easton 119641, Felgner 11971, 1976a1, Felgner & Flannagan [1978], Flannagan [1975], Fleisher [1973], Gaifman [1968], Grigen [1972], Harper & Rubin 119771, Howard, Rubin & Rubin [1978], and Isbell & Wright 119661. For each set form of the axiom of choice in Part I which has the form: For every

x

there exists a

y

such that

$~(x,y),

we can obtain a class form, There is a function F

such that for every x, $(x,F(x)).

For example, see CWO 1, GEdel's axiom E, and the following form of the trichotomy: There is a function F such that f o r all sets x and y, F(x,y) is a 1-1 function of a subset of x into a subset of y such that either a(F(x,y) ) = x or %(F(x,y) ) = y.

185

186

PART 11, CLASS FORMS

Other ways of obtaining class forms are CMS 1-4, CMR 8 (A), CHI and CWO 12. However, common method of obtaining a class form from changing the word "set" to "class". Most of we consider are obtained in this way.

illustrated by by far the most a set form is by the class forms

The only class forms we are considering are class forms of the well ordering theorem, the axiom of choice, and the maximal principles. The letter "C" will precede the name of of each proposition (AC becomes CAC, WO becomes CWO and M becomes CM) and the letter "C" will also follow the number of each theorem and definition in Part 11. The axiom of foundation plays a moke important role in proving equivalences of class forms than it did for set forms. To prove equivalences f o r set forms, we frequently had to construct collections of sets which had certain properties. In the corresponding class forms these sets may become classes and cannot be elements. Another type of proof which is used for the set forms but only works for the class forms with the axiom of foundation is to deduce from the fact that there is a choice function on the class of all non-empty subsets of a class, that the class can be well ordered. However, we give here two forms of the well ordering theorem, CWO 5 (Each proper class is equipollent to On) and CWO 12(P(On) can be well ordered) such that for every other class form S that we give here we can prove in NBGO that CWO 5 S CWO 12. Even though the axiom of foundation is frequently needed to prove equivalences, the only proof that we have to give in NBG is that CWO 12 CWO 5. -+

-+

-+

§l. THE WELL ORDERING THEOREM

CWO 1-4 are class forms of WO 1, CWO 5 is a class form

of WO 2, CWO 6 ( m ) - 8 are class forms of WO 4 ( m ) - 6 and CWO 9 is a class form of MC 2.

There is a function F(x) well orders x.

such that for each set

F

CWO 1:

There is a relation R ordered by R.

x,

such that every set is well

CWO 2:

Every class can be well ordered. universe, V, can be well ordered.)

CWO 3:

(Equivalently, the

There is a relation R which well orders V in such a way that each proper initial R-segment of V is a set. CWO 4 :

CWO 5: V

Each proper class is equipollent to

On.

(Equivalently,

On.) Let

m E

w

such that

m

2

1.

For each class X there is a function F defined U F(a) = X. such that for each a E On, F ( a ) < m and aeon There is an m E w {O} such that CWO 6 ( m ) .

CWO 6 ( m ) :

on

On

CWO 7:

-

-

For each class X there exists an m E w {O} and a function F defined on On such that for each a E On, F(a) 4 m and U F(a) = X . aEOn CWO 9: Every class X is the union of a well ordered class of finite sets. CWO 8:

187

PART 11, CLASS FORMS

188

I n e a c h of CWO 6 ( m ) - 9 , w e o b t a i n an e q u i v a l e n t s t a t e m e n t by r e p l a c i n g " X " by " V " . Thus, e a c h i m p l i e s t h a t t h e u n i v e r s e i s a w e l l o r d e r e d union of f i n i t e sets. The n e x t form, CWO 1 0 , i s a c l a s s form of P R O J g i v e n i n s e c t i o n 5.

I t was g i v e n a s a n axiom of s e t t h e o r y i n von

Neumann [1925

,

Axiom I V 21 a s a r e p l a c e m e n t f o r t h e w e l l

o r d e r i n g theorem and i n von Neumann [1928] it was proved t h a t CWO 1 0

+

CWO 1 0 :

CWO 3.

Every p r o p e r c l a s s c a n be mapped o n t o

every proper c l a s s

X, V

<*

The f o l l o w i n g s t a t e m e n t

V.

(For

X.)

i s a c l a s s form o f t h e

CT

t r i c h o t o m y , b u t it i s e a s i l y s e e n t o b e e q u i v a l e n t t o a form of t h e w e l l o r d e r i n g theorem.

CT:

Each p a i r of p r o p e r c l a s s e s i s

equipollent.

I t i s n o t known whether t h e s t a t e m e n t :

i s comparable.

Each p a i r o f p r o p e r c l a s s e s

i m p l i e s a c l a s s form of t h e axiom of c h o i c e .

Felgner

&

Flannagan [1978] have shown t h a t t h e s t a t e m e n t : On

c a n be i n j e c t e d i n t o e v e r y p r o p e r c l a s s .

d o e s n o t imply t h e s e t form o f t h e axiom o f c h o i c e i n NBG. CWO 11, CWO 1 2 , and CPW a r e c l a s s forms o f PW, CLW i s a

c l a s s form of LW, and CCW i s a c l a s s form o f WO 8. CWO 11:

For e a c h p r o p e r c l a s s

CWO 1 2 :

F(0n)

CPW:

X,

p(X) <

X.

c a n be w e l l o r d e r e d .

The power c l a s s o f a w e l l o r d e r e d c l a s s c a n be w e l l

ordered. CLW:

Every l i n e a r l y o r d e r e d c l a s s can b e w e l l o r d e r e d .

CCW: Every c l a s s on which t h e r e is a c h o i c e f u n c t i o n w e l l ordered.

The f o l l o w i n g i m p l i c a t i o n s a r e immediate.

can be

51.

CT

-

CWO 5

-

CWO 5

CWO 3 + CWO 2

-+

4

CWO

THE WELL O R D E R I N G THEOREM

-

CWO 10;

-+

CWO 6 ( n )

CWO 6 ( m )

-+

CWO 7 + CWO 8 -+

CWO 1;

-

if

The p r o o f t h a t

CWO 8

-+

CWO 5

i s s i m i l a r t o t h e proof

I n f a c t , t h e proof h e r e i s a l i t t l e e a s i e r

v

and w e a u t o m a t i c a l l y have +

CPW,

CCW;

CWO 9 ;

+

b e c a u s e it i s s u f f i c i e n t t o t a k e t h e CLW

-

CPW -+ CWO 12.

o f Theoxem 1.1. "V"

CWO 3 CWO 5;

m s n;

CWO 6 ( m )

CWO 5 + CWO 11

+

CWO 6 ( 1 )

189

CWO 12

and

+

-+

PW.

"X"

o f CWO 8 t o be

v 5 V.

The p r o o f s t h a t

CWO 4 a r e s i m i l a r t o t h e p r o o f s f o r

( S e e Theorems 5 . 5 and 5 . 7 and

t h e c o r r e s p o n d i n g set forms. n o t i c e t h a t CWO 1 2

x

The p r o o f t h a t CWO 1 2

+

CWO 4 i s

T o complete t h e proof o f e q u i v a l e n c e s w e g i v e t h e

i n NBG.)

f o l l o w i n g two t h e o r e m s . CWO 1

THEOREM 1 . 1 C :

PROOF:

For each

o r d i n a l number

a

u

+

CWO 1 2 .

5 On

let

such t h a t f o r a l l

be t h e smallest B E u, B < a. F o r e a c h

p(u)

xa = {u _c On: p ( u ) = a}. Then s i n c e x 5 ?(a), xa i s a s e t . L e t F b e t h e f u n c t i o n g u a r a n t e e d by CWO 1 s u c h

a, l e t

t h a t f o r each

w e l l order

a E On, F(x,) w e l l o r d e r s xa. Then w e c a n by a r e l a t i o n R a s f o l l o w s : f o r u , v 5 On,

?(On)

THEOREM 1 . 2 C :

PROOF: by

R.

CWO 9

Suppose

+

i s a c l a s s which i s l i n e a r l y o r d e r e d

X

i s a w e l l o r d e r e d u n i o n of A l i n e a r o r d e r i n g on a f i n i t e s e t i s a w e l l

CWO 9 i m p l i e s t h a t

f i n i t e sets. ordering.

CLW.

Thus, u s i n g

In Figure 1.1C,

R

X

we can w e l l o r d e r

X.

w e summarize t h e r e s u l t s o f t h i s s e c t i o n .

A l l i m p l i c a t i o n s were p r o v e d i n NBG'.

All t h e f o r m s i n t h e s e t @ are e q u i v a l e n t i n NBGO, where

@

= {CT, CWO 4, CWO 5, CWO 6 ( m ) , CWO 7, CWO 8 , CWO 101.

PART 11, CLASS FORMS

190

y

qcw y CWO 2 CLW

(3)

@

+ CWO 1

ccw

3\

(3)

f

Y

CPW 4 CWO 12

11

FIGURE 1.1C

@ in NBG; thus all the We have shown that CWO 12 statements in Figure 1.1C are equivalent in NBG. (It was shown by Howard, Rubin & Rubin I19781 that the implications (1) and ( 2 ) are not reversible in NBGO, but a strongly inaccessible cardinal was used to prove ( 2 ) . Felgner & Jech [1973] proved that the implications ( 3 ) are not reversible in NBGO. The proofs in Felgner & Jech were for the corresponding set forms, but it is easy to see that they hold for the class forms also. It follows from model (e) in Felgner & Jech I19731 that ( 4 ) is not reversible in NBGO). -+

To conclude this section we shall show that CWO 2 along with DC, the Principle of Dependent Choices (see Part I, Section 2 ) implies CWO 3 . Clearly, CWO 3 -P DC. THEOREM 1.3C:

CWO 2

A

DC

-+

CWO 3 .

PROOF: Let W be an irreflexive relation which well orders each set. Suppose W does not well order V. Then there is a non-empty class A such that A has no W-first element. Define a relation R as follows: R Since

=

A

{(x,y)

E A x A: (y,x) E W}.

has no W-first element, $(R) = A. Therefore, DC implies that there is a function f such that for each n E w, (f(n),f(n+l)) E R. But, by the This definition of R, (f (n+l),f (n)) E W for each n E w. implies the contradiction that the set {f(n): n E w } has no W-first element, q.e.d.

g(R) 5 a(R) so

52.

THE AXIOM OF CHOICE

We can obtain class forms of the axiom of choice, CAC k, from the corresponding set forms, AC k, for 1 i k I 15, but k # 6,7,9 by substituting the word "class" for "set" in the appropriate places. AC 20 and 21 are class forms of AC 4 and 18, respectively, which were proved to be equivalent to a set form in NBG (see Part I, 55). AC 6 and 7 have no class analogues because the direct product of a proper class of nonempty sets is empty. AC 17 has no class analogue for similar reasons - there does not exist a class which has a proper class as an element. AC 9, 18, and 19 do have class analogues, but they are written in a different form. We do not know if AC 16(n,k) has a class form. We shall list the CAC k forms below. CAC 1: If S is a class of non-empty sets, there is a function F such that for each x E S, F(x) E x. G6del's [1940] Axiom E was the statement: E: There exists a function F set x, F ( x ) E x.

in

such that for each non-empty

Clearly, E -+ CAC 1, and CAC 1 + E CAC 1 may be taken to be V {g}.

-

because the class

S

CAC 2 : If T is a pairwise disjoint class of non-empty sets, there is a class C which consists of one and only one element from each set in T. CAC 3 : For every function F there is a function G such that for every x, if x E $(F) and F ( x ) # then G(x) E F(x). 191

192

PART 11, CLASS FORMS

For every relation R there is a function F such that $ ( F ) = a ( R ) and F 5 R. (See AC 20 in Section 5 ) .

CAC 4 :

For every function F there is a function G such that g ( G ) = ?(F) and for every x E B ( G , F(G(x)) = x.

CAC 5:

If S is a class of pairs of equ pollent sets then there is a function which associates with each pair a bijection mapping one onto the other.

CAC 8:

Let R be a relation such that for each x, y E a ( R ) , R"{x} = R"{y}; then there exists a relation S such that for is a bijection mapping R"{x} each x, y E B(R), S"i(x,y)) onto R"{y}.

CAC 9 :

We do not know if the statement obtained from AC 9 by changing the word "set" to "class" is equivalent to a class form of the axiom of choice.

n

Let CAC 10(n):

2

2

be a natural number.

is a class of infinite sets, there is a function F such that for each x E S, F(x) is a decomposition of x into a set of disjoint sets of size between 2 and n. CAC 11: CAC

If

S

There exists a natural number n 10 (n)

.

2 2

such that

CAC 12: I f S is a class of infinite sets there is a natural number n z 2 and a function F such that for each x E s , F(x) is a decomposition of x into a set of disjoint sets of size between 2 and n.

Each of the set forms AC 10(n), 11 and 12 remain equivalent to the set form of the axiom of choice if the size of each of the sets in the partition is exactly n. However, for the class forms, we were not able to prove equivalence so we shall state the latter forms explicitly. in NBG', Let

n

2 2

be a natural number.

12.

THE AXIOM OF CHOICE

193

If S is a class of infinite sets, there is a function F such that for each x E S, F(x) is a decomposition of x into a set of disjoint sets of size exactly n. CAC 10E(n):

Similarly, for CAC 11E and CAC 12E. is the same as CAC 10 (2). Let

m

2

We have CAC lOE(2)

1 be a natural number.

CAC 13(m):

If S is a class of non-empty sets there is a function F such that for each x E S, F(x) is a non-empty subset of x with F(x) < m. There is a natural number

CAC 14:

m

5

1

such that

CAC 13(1n).

If S is a class of non-empty sets then there is a natural number m 2 1 and a function F such that for each x E S, F(x) is a non-empty subset of x with F(x) < m. CAC 15:

CAC 18 and 18C are class forms of AC 18 (See AC 21 in Part I, Section 5 . ) , CAC 19 is a class form of AC 19. CAC 1 8 : For any non-empty class A , and for every function B whose domain is A and Ba is a non-empty set for each a E A

x E

0 U Xarb aEA b€Ba

such that

-

There is a function F

(Va E A) [F(a) E Ba

A

For any non-empty class function B whose domain is A

CAC 1 8 C :

x E

U

Xarb

a E A bEB [a1

such that

-

with domain A

x E XarFta)1 A,

for every class-valued

There is a function F

(Va t A ) [F(a) E BIal

A

-

with domain

x E Xa,F(a)I .

(See Definition 0 . 4 ( k ) for the definition of a class-valued function).

A

PART 11, CLASS FORMS

1 94

CAC 1 9 :

For any non-empty c l a s s A , e a c h o f whose e l e m e n t s i s

non-empty

,

n

U

a€A

bEa

x E

b

There e x i s t s a f u n c t i o n

(Va E A ) [ F ( a ) E a

such t h a t

A

-

F

w i t h domain

x E F ( a )1.

A

CAC 2 2 and 23 a r e c l a s s forms of t h e m u l t i p l e c h o i c e

axiom, MC 1. If

CAC 2 2 :

function

i s a c l a s s of non-empty s e t s t h e n t h e r e i s a

S

F

such t h a t f o r each

There e x i s t s a f u n c t i o n

empty set

i s a non-empty

x.

f i n i t e s u b s e t of CAC 23:

x E S, F(x)

F

s u c h t h a t f o r e a c h non-

i s a non-empty f i n i t e s u b s e t o f

x, F(x)

x.

CAC 2 4 i s a class form o f MC 3. CAC 24:

If

i s a c l a s s of i n f i n i t e s e t s , t h e n t h e r e i s a

S

x E S , F(x) i s a p a r t i t i o n x i n t o p a i r w i s e d i s j o i n t f i n i t e s e t s , e a c h o f which h a s a t l e a s t 2 elements. function

F

such t h a t f o r each

of

The p r o o f s t h a t CAC 1 - 5 a r e e q u i v a l e n t a r e t h e same a s t h e c o r r e s p o n d i n g p r o o f s f o r t h e s e t forms w i t h one e x c e p t i o n . The proof t h a t AC 3 * AC 4 (see 2 . 7 ) d o e s n o t h o l d f o r t h e

x E $(R)

c l a s s forms b e c a u s e f o r some r e l a t i o n d e s c r i b e d i n CAC 4 ) class. that

(where

{y: ( x , y ) E R }

Therefore, i f we could find a r e l a t i o n $(R)

and f o r a l l

= a(S)

t h a t CAC 3 * CAC 4 would f o l l o w .

S

5

R

such

is a s e t , t h e proof

x , S"{x} Let

is the

R

may b e a p r o p e r

PR

be t h e f o l l o w i n g

statement: PR:

that

For every r e l a t i o n a(S)

LEMMA 2 . 1 ~ :

PROOF: that

x S y

-

Let

R

t h e r e is a r e l a t i o n

and f o r a l l

= a(R)

NBGIR

x R y

such

S _CR

x , S"{x} i s a set.

PR.

be a n a r b i t r a r y r e l a t i o n ; d e f i n e A

(Vz) [ x R z * p ( y )

5 p(z)l.

S

so

It follows

from 0 . 8 and 0 . 9 t h a t t h e c l a s s o f a l l e l e m e n t s w i t h a

§

given rank i s a set;

so

is t h e required relation.

S

I t i s also c l e a r t h a t CAC 4

-+

THEOREM 2 . 2 C :

CAC 4 .

PROOF:

CAC 3 A PR

Let

195

THE AXIOM O F CHOICE

2.

-+

PR.

b e a n a r b i t r a r y r e l a t i o n and l e t

R

S

be

t h e a s s o c i a t e d r e l a t i o n whose e x i s t e n c e i s g u a r a n t e e d by PR. x E a(S)

For each

define a function

CAC 3 i m p l i e s t h e r e i s a f u n c t i o n

x E $(F),

if

# 9

F(x)

so t h a t

F

such t h a t f o r each

G

G(x) E F ( x ) .

then

F ( x ) = S"{x).

is t h e

G

required function, q.e.d. I t was shown i n Howard, Rubin & Rubin [19781 t h a t CAC 3

d o e s n o t i m p l y CAC 4 i n NBGO. W e s h a l l show n e x t t h a t e a c h o f CAC 1-5 i s e q u i v a l e n t t o t h e class form o f t h e w e l l o r d e r i n g theorem. C l e a r l y CWO 1

-+

CAC 1

THEOREM 2.3C:

PROOF:

set

x

CAC 1. +

CWO 1.

F i r s t , w e h a v e CAC 1

+

WO 1.

F o r e a c h non-empty

define Sx = {R

X

x: R

Sx # 9

WO 1 i m p l i e s t h a t S = {Sx:x E Vl.

5x

w e l l orders

f o r each set

x.

x}. Define

CAC 1 i m p l i e s t h e r e i s a f u n c t i o n

t h a t f o r each

Sx E S , G(Sx) E Sx.

F(x) = G(Sx).

F

For each

G

x E V

such

define

is t h e required function.

We c a n a l s o p r o v e t h e f o l l o w i n g theorem i n NBGO. CWO 3 + CAC 4 .

THEOREM 2 . 4 C :

PROOF:

Let

R

be an a r b i t r a r y r e l a t i o n .

t h a t there is a r e l a t i o n function

F

S

so t h a t f o r e a c h

which w e l l o r d e r s x E $(R)

F ( x ) = S - f i r s t element of F

CWO 3 i m p l i e s V.

Define a

, R"{x}.

is t h e required function. The p r o o f s o f t h e f o l l o w i n g are t h e s a m e a s f o r t h e

c o r r e s p o n d i n g set f o r m s .

( S e e P a r t 1 S e c t i o n 2 and Theorem 5.23.)

PART 11, C L A S S FORMS

196

CAC 1 CAC 1

-+

CAC 1

+

CAC 4

.+

CAC 8 ;

CAC 10E (n)

-

+

CAC 10(n)

+

CAC 13(n-1) , n 2 2;

CAC 10 (n)

+

CAC 11

14

CAC 13(n)

+

CAC 1 8 C

CAC 1 8

NBG

CLW

+

-+

CAC

-

+

CAC 1 5

CAC 19

...

-+

+

CAC 12

CAC 22

CAC

-+

-+

CAC 15;

CLW;

1;

CWO 4 .

The following implications are easy to prove: CAC 11

-

CAC 12

-+

CAC 22

.+

CAC 23;

CAC 24

-+

CAC 22

CAC 10E(n)

+

CAC 14

-

.+

CAC 15;

CAC 1 CAC lO(2) ( = CAC lOE(2)) ; (similar to the proof of 2.12) i CAC 1 1 E .+ CAC 12E CAC 11. -+

We can also prove the following implications in NBG': THEOREM 2.5C :

(a) If n 2 2, CAC 13(n) (b) CAC 14 -+ CAC 11. (c) CAC 22 + CAC 24.

-+

CAC 10(n+l).

(a) Let S be the class of all infinite sets. CAC 13(n) implies that there is a function F such that for each x E S , F(x) is a non-empty finite subset of x such that F(x) 6 n. For each ordinal a and x E S , let PROOF:

Then x = {G(x,a): G ( x , c i ) f fl A ci E On} is a well ordered P partition of x into non-empty sets each of which has s n elements, but some elements of x may only have one element. P If there are an infinite number of G ( x , a ) ' s with one element, combine them in pairs. If there are a finite number, k of them and k is even, again combine them in pairs. If k is odd, k 2 3 and n t 3, combine the first three G (x,a)' s which are singletons and then combine the remaining ones in pairs. Finally, if k = 1 or n = 2, we have to combine the

12.

197

THE AXIOM OF CHOICE

G ( x ~ c ~ ) ' s which a r e s i n g l e t o n s w i t h o t h e r s e t s i n t h e

p a r t i t i o n and i n t h i s c a s e w e may end up w i t h a s e t i n t h e p a r t i t i o n which h a s

+

n

1 elements.

(We can carry out t h i s

procedure s i m u l t a n e o u s l y f o r each set i n

because each

S

p a r t i t i o n has a defined w e l l ordering). T h i s same p r o c e d u r e may b e u s e d t o p r o v e b o t h (b) and ( c ), q . e . d .

-

Thus i t f o l l o w s from 2.5C and p r e c e d i n g r e s u l t s t h a t f o r n

2

2 , CAC 1 3 ( n )

CAC 22

.+

CAC 1 0 ( n + l ) ;

CAC 11;

1.t

and

CAC 24.

W e s h a l l show n e x t t h a t THEOREM 2.6C:

PROOF:

of CAC 9.

U

CAC 1 4

Let

CWO 5

+

CWO 5

-+

CAC 9

+

CAC 4.

CAC 9 .

be a r e l a t i o n s a t i s f y i n g t h e h y p o t h e s e s

R

CWO 5 i m p l i e s t h a t t h e r e i s a b i j e c t i o n

On.

(R"{x) x { X I ) o n t o

F

x E a(R)

For each

mapping

let

xEa(R) Fx = F r ( R " ( x 1

X

By h y p o t h e s i s , f o r a l l

{XI).

Y,(Fx)

=

2 ( F y ) , and by d e f i n i t i o n of

%(Fx)

n

% ( F Y I = pI.

g(FX) onto F-'~G

y

Let

g(Fy)

G

R"{Y) x { y l ,

R"{x} x

{XI

Then onto

q.e.d. CAC 9

THEOREM 2 . 7 C :

PROOF:

Let

R

+

CAC 4.

be a r e l a t i o n .

Define

2 =

[

U

x E.%R) Then, j u s t a s i n t h e p r o o f o f 2 . 8 , w e show t h a t f o r e a c h

x E $(R),

R"{x)

relation

S

Choose function

X

2

2.

Applying CAC 9 , w e c o n s t r u c t a

such t h a t f o r a l l

x E $(R) F

and

b e t h e u n i q u e b i j e c t i o n mapping

XY

which p r e s e r v e s o r d e r .

i s a b i j e c t i o n from

yx°Fx

x, y E $(R),

F , z(Fx) , % ( F Y I c - On

,

a E R"{x)

and

x, y E a ( R ) ,

f E 2;

then define a

so t h a t

F = { ( y , b ) : b E y A (39 E Z ) S " { ( x l y ) ) ( a , f ) = ( b , g ) } .

R"{x}lw.

198

F

PART 11, CLASS FORMS

is the required function, q.e.d.

The last class form of the axiom of choice we are considering in this section is due to Hilbert [1923]. It is a metamathematical statement and has no set counterpart. Hilbert stated it in the following form: There is a function F such that for each property if F(P) has the property P then every x has the property P .

P

However, such a statement is meaningless in the system of set theory in which we are working. F is some sort of metamathematical function which has not been defined. Through discussions with A. Levy and R. Vaught we were able to come up with a more meaningful formulation. To do this we introduce the metamathematical notion of the iota operator. DEFINITION 2.8C: free variable then:

If

is a formula with at least one

@

The statement CH below is the contrapositive of Hilbert's statement. CH: There is a formula such that

We shall prove

CWO 3

THEOREM 2.9C:

-f

with at least three free variable

I$

CH

CWO 3

-t

+

CAC 4 .

CH.

PROOF: Let $(Z,X,Y) mean Z is the X-first element of Y. CWO 3 implies that there is a relation X which well orders the universe. Therefore, for each non-empty class Y, (iZ)g(Z,X,Y) E Y, q.e.d. THEOREM 2.10C:

CH

-f

CAC 4 .

12.

THE AXIOM OF CHOICE

199

Let R be an arbitrary relation, and let @ be a formula and X a class which satisfies CH. We define a function F such that for all x E a ( R ) , PROOF:

Then it follows from CH that CAC 4.

is a function which satisfies

F

We shall summarize the results from Section 1 and 2 , Part 2, in Figure 2.1C. All the statements in each of the following sets are equivalent.

@

= {CT,CWO 4, 5, 6(n), 7, 8, 1 0 1 ,

@

= {CAC 4 ,

@

= {CWO

5,

= {CAC 11E, 12E1. = {CAC

1.

1, E, CAC 1, 2, 3 , 8, 10(2), 18, 19).

= {CAC 10(m), 13(m-l)},

@ @

2

18C1.

G;

@

n

11, 12, 14, 151.

= {CAC 22, 23, 24).

m

2

3.

PART 11, CLASS FORMS

200

CPW

I I

CCW 6 - 3 CWO 3

CWO 3

<

DC

CAC 9

CWO 2

CWO 12

CH

.i

PR

CAC 10E ( m ) , m s 3

CLW

1 1

CPW

CWO 12 FIGURE 2.1C

52.

THE AXIOM OF CHOICE

201

The implications (1) and (2) in Figure 2.1C are not reversible in NBGO, but (2) used the existence of a strongly inaccessible cardinal.

The implications (3) are not

reversible in NBG. We have shown in Part 11, section 1, that NBG 1- CWO 12 + CWO 4. Thus, all the class forms of the axiom

of choice given in Figure 2.1C are equivalent in NBG.

This Page Intentionally Left Blank

53.

MAXIMAL PRINCIPLES

The first four maximal principles we give here are obtained from M1-4 by changing the word "set" to "class". CM 1: If R is a partial ordering relation on a non-empty class X and if every subclass of X which is linearly ordered by R has an R-upper bound, then X has an R-maximal element. CM 2: If R is a partial ordering relation on a non-empty class X and if every subclass of X which is well ordered by R has an R-upper bound, then X has an R-maximal element. CM 3 : If every non-empty nest which is a subclass of a nonempty class has its union an element of X I then X has a c-maximal element. CM 4: If every non-empty well ordered nest which is a subclass of a non-empty class X has its union an element of X, then X has a 5-maximal element. The following four.statements are also class forms of Ml-4. CMS 1: If R is a partial ordering relation on a non-empty class X and if every subset of X which is linearly ordered by R has an R-upper bound then either X has an R-maximal element or X has a subclass which is a proper class and which is linearly ordered by R. CMS 2 is obtained from CMS 1 by replacing "linearly ordered" by "well ordered".

203

204

PART 11, CLASS FORMS

CMS 3: If every non-empty nest which is a subset of a nonempty class X has its union an element of X, then either X has a 5-maximal element or there is a subclass of X which is a nest and a proper class. CMS 4 is obtained from CMS 3 by replacing "nest" by "well ordered nest". Clearly CM n + CMS n for n = 1,2,3,4. It is also easy to see that CMS n + CM n for n = 3 , 4 because if a proper class is linearly ordered by 5 , (that is, a nest) then its union is a proper class and cannot be an element. (Howard, Rubin & Rubin [1978] have shown that if n = 1 or 2, CMS n does not imply CM n in NBGO if there exists a strongly inaccessible cardinal.) Later we shall show that they are all equivalent in NBG. There are also the two corresponding types of class forms for M l(Q,U). (See pp.37-8 for the possibilities for Q and U. 1 CM l(Q,U): Every non-empty class which is Q-ordered by R in which each U-ordered subclass has an R-upper bound, has an R-maximal element. CMS l(Q,U): Every non-empty class which is Q-ordered by R in which each U-ordered subset has an R-upper bound, either has an R-maximal element or contains a U-ordered subclass which is a proper class. CM l(P,L) is CM 1 and CM 1(P,W) is CM 2. Similarly for CMS l(P,L) and CMS l(P,W). It is clear that CM l(Q,U) -+ CMS l(Q,U) for all Q and U. We proved in Part I, Section 3 that the M l(Q,U) are equivalent if Q = TR, P, RA, or F and U = C, A S K , T R K , L, D, or W; or if Q = D and U = W. (See pp. 3 7 - 4 1 . ) These same proofs may be used to prove that the corresponding forms of CMS l(Q,U) are all equivalent. However, these proofs do not all hold for the stronger forms, CM l(Q,U). For these forms we were only able to prove the following in NBGO: (In figure 3.1C, "(Q,U)" means "CM 1 (Q,u)I' )

.

23.

(See Harper

&

MAXIMAL PRINCIPLES

205

Rubin [1977]. )

We shall also consider class forms of M5, M 5(Q,U), M 6-7, M 8(A) , M 9(A), M 10-11, M 15 (A:B) , M 16(A:B) , M 17, and A, the anti-chain axiom. (See pp 276-278.) The following implications are easy to prove: CM 5 CWO 5 * CM l(TR,W) + CM 2 -+ CM 1, and CM 1 -+ CMS 1.

-+

CM 1,

THEOREM 3.1C: The following statements are equivalent: CMS 1, CMS 2, CMS 3, CMS 4, CM 3, CM 4, CM 6 and CM 17. (CM 8 ( K ) is the same as CM 6 . ) PROOF:

The following implications are easily seen to

hold : CMS 2 CMS 2

-+

-+

CMS 1 CMS 4

+ -+

CMS 3 CMS 3

,

First, we will prove that CMS 3 -+ CMS 2. Let L be a non-empty class and R a partial ordering relation on X. Define Y as follows: Y =

IS C

p ( X ) :(Vx E

s ) (R

well orders

x)

(Vx,y E s ) (x is an R-initial segment of y

is an R-initial segment of

x)}.

A

y

V

PART 11, CLASS FORMS

206

Let z 5 Y be a nest, then U z EY. Therefore, Y satisfies the hypothesis of CMS 3. It follows that either Y has an R-maximal element of Y has a subclass which is a nest and a proper class. Suppose the first alternative holds. Then Y has an R-maximal element t. It follows that U t can be well ordered by R. The hypothesis of CMS 2 implies that U t has an R-upper bound, b. Thus, b is an R-maximal element of X. If the second alternative holds then there exists a nest Z 5 Y such that Z is a proper class. Then, UUZ is a subclass of X which is a proper class and well ordered by R.

-

-

As was announced earlier CM 3 CMS 3 and CM 4 Thus, the first six statements given in the theorem are equivalent.

We shall show next that

CM 6

-f

CMS 3

and

CMS 1

+

CMS 4.

CM 6.

Assume CM 6 holds and let X be a non-empty class. Then CM 6 implies that there exists a 5-maximal subclass Z 5 X such that Z is a nest. If Z is a proper class then X has a subclass which is a nest and a proper class. If Z is a set, the hypotheses of CMS 3 implies that U Z EX. In this case, U Z is a 5-maximal element of X. Next, assume class. Define Y Y = Cx

CMS 1 holds and let as follows:

X

be a non-empty

5 X: x is a 5-maximal nest in x n ~ ( u x ) )

Y is partially ordered by inclusion. Clearly, CMS 1 -t M 1 -+ M’6 and M ’ 6 a 5-maximal nest y such that

It is clear that y E Y. Since f’ that

Let s 5 Y be a nest. implies that there is

y is a 5-upper bound of s . We claim that and U are monotone, it follows from (i)

MAXIMAL PRINCIPLES

53.

(ii) ' ~ ' ( u u s ) _c For any set

Thus,

p(Uy)

z,

207

B Therefore, from (ii) we obtain

= z.

u'f'(z)

so it follows from

= 'f'(UUs),

(i) that

y E Y.

CMS 1 implies that either Y has a 5-maximal element which is clearly a 5-maximal nest in X, or Y contains a subclass S which is a nest and a proper class. U S is clearly a nest, and we claim it is a 5-maximal nest contained in X. For suppose not. Suppose there exists a u E X such that U S U {u} is a nest. Then, either v _C u for all v E U S or there exists a v E U S such that u c v. In U U S _c u, which is a contradiction because the first case u is a set and S is a proper class. In the second case, v E U S implies there is an x E S such that v E x. Since x U {ul 5 U S U {u), x U {u] is a nest. Moreover, since u c v E x, x U {u} _c X n ~ ( U X ) . But this is a contradiction because x E Y, and therefore x is supposed to be a c-maximal nest in x n B ( u x ) .

Finally, we shall show that CM 17

-

CMS 3.

Suppose CM 17 holds and X is a non-empty class. We define a relation R on X x On as follows: If s , t E X (s,a)R(t,B)

-

(S C

t

A c1 < B ) V

( s = t, s is a 5-maximal element of

Define a function F

on

A

X

x

On

x

A

as follows:

(s',a')R(s,a)~.

It is easily shown that F(S,a)

=

F(ttB)

F(s,a) E F(t,B)

-

( S i a )

=

(t,B)t

(s,a)R(t,B)

(see 4 . 4 6 ) .

a. < 6).

208

PART 11, CLASS FORMS

CM 17 implies that {F(s,o):(s,a) E X X On} has a 5-maximal subclass Q such that for all distinct s, t E Q, s E t or t E s. Therefore, X x On has a 5-maximal subclass Z which is linearly ordered by R U I, where I is the identity relation on X x On. Suppose X has no 5-maximal element. $(Z) is a nest. If D ( Z ) is a set,then by the hypothesis of CMS 3 @(Z) E XI and therefore, UZ(Z) would be a 5-maximal element of X. Consequently, if X has no 5-maximal element, a ( Z ) is a subclass of X which is a nest and a proper class. Conversely, suppose CMS 3 holds. Let X be a class. For each set x, let TC(x) be the transitive closure of x. (That is, TC(x) = x U U x U UUx U...) Define Y = {y 5 X: y is a 5-maximal subset of X n TC(y) such that E is connected on y } . CMS 3 + M 3 + M 17 which implies Y # PI. If Y has a 5-maximal element then CM 17 holds. If Y has no 5-maximal element then CMS 3 implies that there is a nest Z 5 Y such that 2 is a proper class. We claim that U Z is a 5-maximal subclass of X such that E is connected on UZ. It is clear that U Z 5 X and that E is connected on U Z. Suppose U Z is not 5-maximal. Then there is an

-

x E X U Z such that E is connected on U Z U {XI. Therefore, for all y E U Z either y E x or x E y. Since U Z is a proper class U Z $ x; therefore, there is a y E UZ such that x E y. Consequently, there is a t E Z such that y E t. We have Z _c Y; thus t E Z implies t E Y. Since x E y E t, x E TC(t). Since E is connected on U Z U {XI and t 5 U Z , it follows that t U {XI is E-connected. However, t is a maximal subset of X n TC(t) which is E-connected; therefore, x E t. This implies the contradiction that x E UZ. Now we have completed the proof of Theorem 3.1C. Our next concern are the forms CM 5 ( Q , U ) where the possibilities for Q and U are A , TR, AS, C, TR&C, AS&C, D (directed), RA, and F (see pp 37-8).

93.

209

MAXIMAL PRINCIPLES

We note first: that facts ( l ) - ( l O ) on pp 42-43 all hold for CM 5 (Q,U) Moreover, the proof that M 5 (TR,AS) + AC 2 (pp 43-4 ) also gives a proof that CM 5(TR,AS) + CAC 2. The equivalences given on p 78 for M 5(TR&C,U) also hold for CM 5(TR&C,U) whexe U = AS, AS&C, P, L, D and RA. Since each of the propexties AS, TR, C, AS&C, TRLC, P , L, D and RA are properties of finite character it follows that CM 7 implies CM 5(Q,U) for each of the 50 pairs of values for Q and U given in Theorem 4.11 and for the 6 pairs of values given on p 7 8 Thus, it remains to show that each of CM 5(D,L), CM 5(F,L) and CM 5(TR&C,L) implies some form of the axiom of choice. It is easy to see that a modification of the proofs on p 43 of M 5(DIL) + M 5 and M 5(F,L) -+ M 2 give us CM 5(D,L) + CMS 1 and CM 5(F,L) + CMS 2, respectively.

.

.

THEOREM 3.2C:

CM 5(TR&C,L)

+

CWO 12.

PROOF: First, we can prove using the first part of the proof of Theorem 5.8, that CM 5(TR&C,L) implies that a linearly ordered class of non-empty sets has a choice function, CACLo

.

For each set u _c On, we define the rank of u, p(u) to be the smallest ordinal number a such that for all B E u, B < a. Every subset of On has a rank and the class of all subsets with a given rank is a set. For each a € On, let xa = (u _c 0n:p (u) = a). Each xa can be linearly ordered by a relation R such that if u # v, u R v iff the €-first element of (u U v) (u n v) is an element of u. As mentioned above, CM 5(TR&C,L) + CACLo, and clearly, CACLo ACLo * LW. LW implies each x can be well ordered. a Let Ra be the set of all well orderings of x and let X = {RcL:a E On). CACLo implies that X has aachoice function, F. Then P(0n) can be well ordered by a relation S defined as follows:

-

+

u

S

v

iff

p ( u ) < p(v) V ( p ( u ) = p(v) = a A u F ( R )v). 01

In Figure 3.2C below we summarize the results f o r the 56 statements of the form CM 5(Q,U). All implications are

210

PART 11, CLASS FORMS ( I n F i g u r e 3.2CI " ( Q , U ) "

p r o v a b l e i n NBGO.

means

"CM 5 ( Q , U ) " . )

where

U = TR,AS,C,TR&C,AS&C,P,L,D,

+.

(TR,AS) (D,RA) * (D,L) ;

(D,U)

(Flu)

(TR&C,U)

-+

(TR,P)

CWO 12, (C,U)

CMS 1, f o r

-+

(AS,L),

(C,AS)

.

CM 5

(P,u),

-

U = C, AS&C, TR&C, o r L.

U = C, AS&C, TR&C, L, D, o r W.

U = A S , A S K , P , L, D,

for

-

RA.

CAC 2.

-+

CMS 2, f o r

-+

or

U = P , D, L,

for

(ASIC);

(C,AS&C)

for

u

=

-

RA.

or

or

RA.

(AS,AS&C).

c, AS&C, TR&C,

or

L.

FIGURE 3.2C

We have shown i n S e c t i o n 1 t h a t CWO 12 CWO 5 i n NBG CAC 2 -+ CWO 5 i n NBG. We h a v e shown CMS 2 i n NBGO. Next w e shall i n Theorem 3.1C t h a t CMS 1 show t h a t CAC 4 CMS 3, CMS 1 CWO 12 and CWO 5 -+ CM 7. -+

and i n S e c t i o n 2 t h a t

-

-+

THEOREM 3.3C: PROOF:

-+

CAC 4

-+

CMS 3.

Let

X

be a c l a s s which s a t i s f i e s t h e h y p o t h e s i s

o f CMS 3 and l e t

R

be a r e l a t i o n defin ed a s follows:

R = { ( x , ~ ) :x 5 X I s E X ,

and

( v t ) (t E

x

+

t

C s)}.

§3.

F

CAC 4 i m p l i e s t h e r e i s a f u n c t i o n

If there is a nest

a(F) = a(R).

211

MAXIMAL PRINCIPLES

n

5

such t h a t

R

cX

such t h a t

n

h a s no

s t r i c t 5-upper bound, t h e n Un, which by t h e h y p o t h e s i s o f CMS 3 i s a n e l e m e n t of X , i s a g m a x i m a l e l e m e n t o f X . So suppose e v e r y n e s t n h a s a s t r i c t S-upper bound; t h e n F ( n ) i s a s t r i c t 5-upper bound o f n. D e f i n e a f u n c t i o n G on On such t h a t G ( m ) = F ( G " a ) f o r a l l a E On. Then it i s e a s y t o see t h a t $ ( G ) 5 X , Z ( G ) i s a n e s t and a p r o p e r c l a s s , q.e.d. THEOREM 3.4C:

PROOF:

CMS 1

+

CWO 12.

u _C On, p ( u )

For e a c h

is the smallest ordinal

a s u c h t h a t f o r e a c h 6 E u, 6 < a . A s w e mentioned i n t h e proof o f 3.2C, e a c h s u b s e t of On h a s a r a n k , and t h e c l a s s of a l l s u b s e t s o f On of a g i v e n r a n k i s a s e t . For e a c h

a E On, so e a c h

l e t y, = {u _C On:p(u) S a}. CMS 1 y, c a n be w e l l - o r d e r e d . Let = {R

R,

and l e t

U

X =

5 y,

R,.

X

y,:

Let

S

R

+ M

w e l l orders

1

-P

WO 1,

y,),

be a r e l a t i o n d e f i n e d on

X

a €On

so t h a t i f Then

R, R ' E X I R S R'

and

X

t o see t h a t

S X

iff

l i n e a r l y o r d e r e d by

u

y,

i s an e x t e n s i o n of

h a s no S-maximal e l e m e n t .

implies t h a t there is a proper c l a s s orders

R'

s a t i s f y t h e h y p o t h e s i s o f CMS 1.

S.

Then

= P(On),

UY

T h e r e f o r e , CMS 1

such t h a t Y is i s a r e l a t i o n which w e l l Y 5 X

q.e.d.

,€On CWO 5

THEOREM 3.5C:

+

CM

I.

PROOF : L e t X be a c l a s s and P a p r o p e r t y o f f i n i t e CWO 5 i m p l i e s t h a t t h e r e i s a b i j e c t i o n F s u c h character. that

F:On

X.

Let

a.

be t h e smallest o r d i n a l number

{a,) h a s t h e p r o p e r t y P. ( I f no s u c h o r d i n a l number e x i s t s t h e n fl i s t h e 5-maximal s u b c l a s s of X w i t h such t h a t

t h e property P.)

Define a f u n c t i o n

R.

It i s easy

G

on

On

such t h a t :

PART 11, CLASS FORMS

212

F ( a O ) otherwise.

is a 5-maximal subclass of

Then $ ( G ) q.e.d.

X

with the property P ,

To complete the discussion of the CM 5(Q,U) forms,we shall look at CM 5(A,UN) and CM 5(A,UN(n)). Since these statements come up again later and this notation is rather cumbersome, we shall give these statements new numbers, CM 20(n). Let

be such that

n E w

n

2

2.

CM 20(n) : If X is a class and R an n-ary relation there exists a 5-maximal subclass Y 5 X such that Yn 5 R. CM 5(A,UN) is CM 20(2) which we shall denote by CM 20. The following implications are proved in a similar manner to the corresponding implications for the set forms: CM 7

+

CM 20 (n)

CM 20(n)

-+

+

CM 20

CM 20 (n-1), for

+

n

2

3,

CM 5(A,C) * CM 5.

(CM 20 (n) + CM 20 (n-1) because if R is an (n-1)-ary relation then R x V is an n-ary relation. A 5-maximal subclass Y such that Yn 5 R x V induces a 5-maximal subclass Z

(=

Y) I such that

Znml 5 R - )

The class form of the anti-chain axiom, CAI is as follows : CA: For each class X and partial ordering relation R there is a 5-maximal class Y 5 X which is an R-anti-chain. The property of being an anti-chain is a property of finite is a special case of CA character so CM 7 -+ CA. CM a(:)

53.

(R

is the

CM 8

(E)

-t

MAXIMAL PRINCIPLES

so

relation),

_C

CA

CM 8 ( z )

+

CWO 1 2 .

J u s t a s i n t h e p r o o f o f Theorem 3.2C,

PROOF:

a r e l a t i o n which l i n e a r l y o r d e r s x u = Iu _C On:p(u) = u]. Define a r e l a t i o n

S

x = y

A

u R v

t-

(x,u) = {(y,v) E Y:(y,v)S(x,u)]

( x , u ) E Y , and

-K .

A

5 P(on)

p(u)

u E x).

A

p(v).

i

t-

5 (y,v) tz = C(x,u):(x,u) E Y I .

Let

be

R

is a s e t f o r each

t-

(x,u)S(y,v)

( x , u ) , ( y , v ) E Y. implies t h a t Z

let

and l e t

so t h a t

Y

iff

Y(0n)

Y = { ( X , U ) : #~ x

Let

on

(X,u)S(y,v)

property

W e s h a l l show

CM 8 ( K ) .

-+

CWO 1 2 .

THEOREM 3.6C:

Then

213

i f f (x,u)

f o r each CM 8 ( K )

c o n t a i n s a 5-maximal s u b c l a s s

with the

G

Then w e c a n d e f i n e a c h o i c e f u n c t i o n

F

on

p(Y(0n)) cfl} so t h a t i f )3 # x 5 P ( 0 n ) , F ( x ) = u , where u ti s t h e u n i q u e e l e m e n t o f x s u c h t h a t ( x , u ) E G. The f u n c t i o n F c a n b e u s e d t o w e l l o r d e r e a c h x a l f o r u E On. Since P ( 0 n ) = U x a l t h i s i m p l i e s P ( 0 n ) c a n be w e l l aEOn o r d e r e d , q.e.d. Our n e x t c o n s i d e r a t i o n w i l l be t h e c l a s s forms 9 ( A ) , 1 0 , 11, 1 2 , 1 3 , and 1 4 .

The c h o i c e s f o r

J, Dml

5, K, 5 ,

and t h e i r complements

(1)

x

(2)

X K Y

iff X 5 Y v Y 5 X ,

(3)

X J Y

iff

(4)

A class

iff

D Y

X

x

rl Y =

Dm

and

are

D, K,

where

8,

x ~ Y A Y ~ X A X n Y # O A X U Y # v I

has t h e property

m

E X I xi's

xl, x2,...,xm

A

CM 8 ( A ) ,

distinct,

n

Dm

i f f for a l l

xi = g.

i=l

W e n o t e d above t h a t CM 8 ( K ) is t h e same as CM 6 and t h a t CA

+

CM 8

CM 8 ( J ) J

for

+

5

-f

CWO 12.

CM 8 ( K )

and

K

(Theorem 3 . 6 C ) - The p r o o f t h a t i s s i m i l a r t o t h e proof of 4.24. (Substitute for

K

i n 4.24.)

CM 8 ( 6 )

i s M 1 9 and i s

PART 11, CLASS FORMS

214

e q u i v a l e n t t o a s e t form of t h e axiom o f c h o i c e i n NBG. Similarly for

CM 8(5,),

CM 1 2 ,

and

(See 5 . 2 1 ) .

CM 1 3 .

CM 1 4 i s e q u i v a l e n t t o a s e t form of t h e axiom o f c h o i c e i n

NBGO.

(See 4.31) The proof t h a t CM 8(Dm) CAC 13(m-1) f o r i s s i m i l a r t o t h e proof of 4.27. Clearly, CM 8(Dm) + CM 1 0 CM 11. The CM 8(A) s t a t e m e n t s which remain a r e t h o s e i n which A = D o r J . W e s h a l l show t h a t

m

2

-+

2

-

-f

CM 11

CAC 4

+

CM 8(5)

and

The proof t h a t CM 11 THEOREM 3.7:

PROOF:

For any set

Let

+

+

CAC 4 .

i s d u e t o A. L. Rubin.

CAC 4

CAC 4 .

be any r e l a t i o n and l e t

R

= {{(O,a),(l,a,b)l:(a,b)

u

and f o r any

):(

m E w,

m

2

E R}. 2 , let

wu = w

b e t h e s e t of a l l m-element s u b s e t s o f u

and

= (t E ( ? )

m,u

Then

-+

CM 8(D)

X1

and l e t For each such

z

CM 11

+

.

m

u

x

w

U'

d e f i n e t h e f o l l o w i n g t h r e e sets

t h e element of

t

with the largest

f i r s t c o o r d i n a t e is unique

we define the classes

X2

and

X

a s follows:

x = x l u x 2' CM 11 i m p l i e s t h a t t h e r e i s a n a t u r a l number

c_-maximal s u b c l a s s For e a c h

a E a(R),

Y 5 X

let

such t h a t

Y

m

2 and a h a s t h e p r o p e r t y Dm. 2

53.

MAXIMAL P R I N C I P L E S

215

E Yl.

u = {b:{(O,a),(l,a,b)l

By definition, u 5 R"{a}. It follows from the maximality of Y that u # 0, and since Y satisfies D ~ , < m. If IuI = 1 define F(a) to be the unique element of u. Otherwise, define

IuI

We claim that each m-element subset of v is an element For suppose not. Suppose Of Zm,u* is an m-element subset of t = {(nl,b1),(n2,b2 j,...,(nm,bm)} v such that n1 = n2 and for all i = lr2,...,m, ni i n1' Then for each i = 1,2,,..,m, yi = C(m,u)l

-

(fmIu(nirbi) zm,u 1 E Y , m and t E fmlu(niIbi) zmIu. This implies f l yi # which i=l contradicts the fact that Y satisfies Dm. This proves the claim. x

-

NOW, we claim that there is an

n E w such that there is a unique b E u with (n,b) E v. To prove this claim we note z m. (Otherwise we that because of the maximality of Y, could add an element to v with a larger first coordinate.) If /v1 2 m, then, for example, {b E u:(m,b) E v} has one element. (Otherwise there is an m-element subset of v which ) This proves the second claim. is not an element of z m,u' Let no be the smallest n which satisfies the second claim. Then we define F(a) to be the unique b such that (no,b) E v. This proves CAC 4 , because we have given a constructive procedure which gives us F(a) for each a E . 9 ( ~ ) , q.e.d.

(vI

THEOREM 3 . 8 C : PROOF: s

U

Let

= s U {(s,u)l

(sIu) $ t.

CM 8 ( 5 )

+

.

CM 8 (D)

X be a class. where for each

For each s E XI define t E X I if t # s then

(We can always find such an element

u

since we

PART 11, CLASS FORMS

216

could replace

X x {a}

a.)

f o r any

If

s # t,

s, t E X , t h e n su $ tU. Thus, su D tU i f f su j tU. s # t i m p l i e s su D t U i f f s D t . CM 8(5) i m p l i e s

where Also,

{su:s E X }

that

5.

by

X

Hence,

h a s a 5-maximal s u b c l a s s w i t h t h e p r o p e r t y

h a s a 5-maximal s u b c l a s s w i t h t h e p r o p e r t y

X

D,

q.e.d. THEOREM 3.9C: PROOF:

Let

CM 8 ( D )

-+

CAC 4 .

be a r e l a t i o n .

R

Define

Two e l e m e n t s i n X , { ( O , u ) , ( u , v ) } i f f u # u'. Thus, a c-maximal s u b c l a s s o f X w i t h t h e p r o p e r t y D p r o d u c e s a f u n c t i o n s a t i s f y i n g CAC 4 . X = ~ ~ ( O , u ) , ( u , v ) ~R: uv ) .

and {(O,u'),(u',v')~

are disjoint

I t i s a l s o t r u e t h a t CM 9 ( J ) CM 9 (J)

proof t h a t 4.24,

* CM 9 (z)

and t h e p r o o f t h a t

p r o o f of 4.25.

CM 9

K,

z.

+ CAC

The

2.

CM 9 ( A )

-+

+

CM 8 ( A ) ,

,

W e s h a l l show t h a t CM 7 -+ CM 9 ( A )

5,

(K)

i s s i m i l a r t o t h e p r o o f of CAC 2 i s s i m i l a r t o t h e

CM 9 ( z )

Clearly,

+

for a l l

for a l l

A.

A = D,

b, J,

Since each of these properties a r e p r o p e r t i e s of

f i n i t e c h a r a c t e r , t o show t h a t

CM 7

s u f f i c i e n t t o prove t h a t be l i n e a r l y o r d e r e d .

implies t h a t t h e universe can

LV:

CM 9 ( A ) ,

-+

it i s

c a n be l i n e a r l y o r d e r e d .

V

LEMMA 3.10C:

PROOF: X

CM 7

If

CM 7 X

+

LV.

is a c l a s s of ordered p a i r s we s h a l l say

i s c i r c u l a r i f t h e r e e x i s t s a f i n i t e sequence o f l e n g t h

n > 1 of ordered p a i r s i n ( X ~ - ~ , X ~such ) , that

X , ( xo,xl)

x o = xn

no s u c h sequence e x i s t s

and

X

has t h e property

P

iff

X

-

,...,

( x1,x2) ,( x2,x3) {xo,xl, x 1 n. n

...,

is c a l l e d non-circular.

X

be a property defined f o r each class that

,

Let

If P

o f o r d e r e d p a i r s so is non-circular. P is

X

a p r o p e r t y of f i n i t e c h a r a c t e r . CM 7 i m p l i e s t h a t t h e r e is a c-maximal

t h e property

P.

W e claim t h a t

R

c l a s s , R, w i t h is a reflexive, anti-

§ 3.

MAXIMAL PRINCIPLES

211

symmetric, transitive, and connected relation on V. Clearly R is reflexive and anti-symmetric. To show R is transitive, suppose (x,y),(y,z) E R and (x,z) !f R. This implies, by is circular. Since the maximality of R, that R U t(x,z)l (x,y) and (y,z) are elements of R, it follows that R must have been circular. This is a contradiction. Thus (x,z) E R and R is transitive. Suppose (x,y) JZ R, x # y; then R U {(x,Y)} is circular. There exists a finite sequence of ordered pairs (zorzljr(zlIz2)f . . . , ( z n - 2 f ~ n - l ) f ( ~ n - l f ~ O )in R U {(x,Y)j such that n > 1, the zi's are distinct and one of the ordered pairs in the sequence is (x,y). There is no loss of generality in assuming ( z o , z , ) is (x,y). Then the following ordered pairs are in R: ( ~ , z ~ ) , ( z ~ , z ~ ) , . . . , ( Z ~ - ~ , ~ ~ - ~ ) , ( Z ~ - ~Since , X ) . R is transitive, we must have (y,x) E R. Thus, R is connected, so R linearly orders V, q.e.d. Figure 3 . 3 C below illustrates the relationships between the statements CM 8 ( A ) , 9 ( A ) , 1 0 , and 11 in NBGO. CM 7

-+

CM 9 (A)

CM 9 ( J )

CM 8 ( 2 ) CM 8 ( J ) CM

8(Dm)

-+

-+

-+

-

CM 9 ( k ) CM 8 ( D )

CM 8 ( k ) CM 10

+

LV -t

-+

+

+

A

CM 8 ( A )

,

for all A;

CAC 2; CAC 4 CWO

CM 11

-t

CM 8 ( K )

.+CM 6

* CWO 12:

12; -+

CAC 4.

FIGURE 3.3C

Since CWO 12 CWO 4 in NBG, and CWO 4 CM 7 in NBGO, all the statements in Figure 3 . 3 C are equivalent to CWO 4 in NBG. Moreover we have CM 14 M 14 in NBGO and each of CM 8 (E), CM 8 (Em), CM 12 and CM 13 is equivalent to a set form +

+

..

of the axiom of choice in NBG. If of D, K

x

and or J

Y are distinct, non-empty sets exactly one holds between them. However, if X and Y

PART 11, CLASS FORMS

218

are classes there is another possibility, namely Consequently, we define another relation, U . DEFINITION 3.11C:

Let

X

and

(i) X U Y

iff

X u Y = V.

x 6Y

iff

x u

(ii)

Y

X U Y = V.

be classes.

Y # V.

is dual to D . That is, X D Y iff If x and Y are distinct, non-empty, non-universal and non-complementary classes, exactly one of the properties D, K , J, or U holds between them. El 5 and are the complements of D , K , J and U , so E = K V J v U , i ? = D v J V U , ~ = D V K V U , and 6 = D V K V J. For the sake of completeness we shall also consider the six properties, D v K (DK), D V J (DJ), D V U (DU), K v J ( K J ) , K v U ( K U ) , and J v U (JU). The property

(v

- x) u - Y).

U

(V

It would be impossible to state the form CM 8 ( A ) where or 6 is substituted for A because it would be necessary to construct a class whose elements are classes. To avoid this difficulty we restate CM 8(A) as follows: U

CMR 8 ( A ) : For every relation R and class X a 5-maximal subclass Y 5 X such that for all x # y then R"{x) A R"{y).

there exists x, y E Y, if

CMR 8(A) is equivalent to a class form of the axiom of - choice if A = D , 6, J , 5 , K , E l Dm, Dm, D!m, U , 6, D K , D J , DU, K J , KU, or J U . (The property as

D!

m

is excluded because it is the same

D.)

The following propositions are also equivalent to a class form of the axiom of choice. CMR 9 ( A ) : The universe can be linearly ordered and for every relation R and class X there exists a 5-maximal subclass Y 5 X such that for all x, y E Y, if x # y then - -R"{x] A R"{y), where A = D , E l J , 3 , K , K , Dm, Dml Dim, U , U , D K , DJ, D U , K J , K U , or J U .

93.

MAXIMAL PRINCIPLES

219

We shall not spend much time on the CMR 9 ( A ) statements because CMR 9 ( A ) is the same as LV A CMR 8 (A). However, CMR 9 ( E ) is of some interest because of historical reasons. CMR 10: There exists a natural number

m

2

2

such that

CMR 8(Dm). CMR 11: For every relation R and class X there is a natural number m 2 2 and a 5-maximal subclass Y - 5 X such that for all y1,y2,...,ym E Y, yi's distinct, n R"{Yi} = i=1 111

a.

The statement: There exists a natural number CMR 8 (6,)

m

.

2

2

such that

is also equivalent to a class form of the axiom of choice. (We shall show later CMR 8 (D,) CMR 8 (Em).) However, as we mentioned earlier for the set forms, if in CMR 11 we replace m m " n R"{yi} = 8'' by " n R"{yJ # the resulting statement i=1 i=1 is provable in NBGO. (See the discussion preceding definition 4.28.) This was the reason for introducing D!,.

..

a",

CMR 12: There exists a natural number CMR 8(s!m).

m

1

2

such that

CMR 13: For every relation R and class X there exists a natural number m zr 2 and a 5-maximal subclass Y 5 X such that for all

y1,y2,...,ym E Y,

m

n

i=1

R"{Yi) # 8 .

CMR 14: If X is a class and R a relation there exists a 5-maximal subclass Y _C X such that fo; all natural numbers ... m 2 2 and for all y1,y2,...,y, E Y, n R"{Yi} # i=l

a-

Even though CM 14 is equivalent to M 14 in NBGO proved by A. L. Rubin that CMR 14 CM 7 in NBGO.

-

it was

There is no advantage to introducing "MR" forms for the set forms of the axiom of choice, because they are c ear Y

PART 11, CLASS FORMS

220

equivalent to the corresponding "M" forms.

However, for the

class forms it gives us a method for handling collections of proper classes. Each CMR 8 form implies the corresponding CM 8 form. (Define R so that u R v iff v E u. Then, for each x, R"{x) = x.) We shall show that each of CMR 8(A), for A = D, 5, J, 5, u, 6 , DK, DJ, DU, KJ, KU or JU is CM 5, and equivalent to CM 20, and that CMR 8(K CMR 8(i?) CA. The proof that CMR 8(J -+ CMR 8(?) is similar (Substitute J for 5 and i? for K to the proof of 4 . 2 4 . in 4.24.)

+.

-

THEOREM 3.12C : (a) CM 5 * CMR 8(K) (b) CA CMR 8(i?).

-

PROOF: (a) Suppose CM 5 holds and let R be an arbitrary relation. Define a relation S for all x, y E a(R) such that x

S

y

. R"{x)

5

R"{y]

is a partial ordering so by CM 5 there exists a 5-maximal subclass Y Zfi(R) such that Y is linearly ordered by R. Y is the required class. S

Conversely, let R be a partial ordering relation on a class X. Then we claim that for all x, y E X, (1)

x R y

*.

R"{y] _C R"{x).

For suppose x R y and z E R"{y). Then z R y holds. Since R is transitive, z R x holds, which implies R"{y) 5 R"{x]. Now suppose R"{y} 5 R"{x]. Since R is Thus, reflexive, y E R"{y) which implies that y E R"{x}. x R y holds. Therefore, it follows from (1) that CMR 8(K) implies CM 5. The proof of (b) is similar.

MAXIMAL PRINCIPLES

83.

THEOREM 3.13C:

u , ij,

DJ,

DK,

PROOF: relation

CM 2 0

*

KU,

or

D U , KJ,

Let

CMR 8 ( A ) ,

b e a r e l a t i o n and

R

x S y

-

x = y

(x # y

V

where

A = D,

5,

J,

5,

JU.

such t h a t f o r e a c h

S

221

a class.

X

Define a

x, y E $ ( R ) , A

R"{x} A R"{y})

CM 2 0 i m p l i e s

Y

X

Y

5

X h a s a 5-maximal s u b c l a s s Y s u c h t h a t T h i s i s a n e q u i v a l e n t way o f s a y i n g t h a t Y i s a

S.

5-maximal s u b c l a s s o f

x # y

then

such t h a t f o r a l l

X

x, y E Y , i f

R"{x} A R"{y}.

To p r o v e t h e c o n v e r s e w e u s e t h e f a c t t h a t e a c h o f t h e s e

r e l a t i o n s are u n i v e r s a l .

That i s , g i v e n any symmetric

relation

R

and any o f t h e above r e l a t i o n s

relation

S

such t h a t i f

x # y, x R y

A,

iff

t h e r e is a

S"{x} A S"{y}.

( A s i m i l a r remark h o l d s f o r t h e s e t forms and was u s e d by

Kurepa [1952]. For D and D t h i s was p r e v i o u s l y o b s e r v e d by S z p i l r a j n - M a r c z e w s k i [1945].) THEOFlEM 3.14C: CMR 8 ( D )

(a)

-+

r e p l a c e d by D K , DU

U

hold f o r

and

-

*

(The proof i s s i m i l a r i f

(a)

Let

.

be a r e l a t i o n .

R

(32) [y =

( T h a t i s , ( x , z ) f? R

or

Define a r e l a t i o n

{x,z}

(x,z)

L

A

u

(x,z) E

R-l

or

x =

2.)

because

E S''{xI,

If

X

equivalent:

5

$(R)

S

U I].

x, y E a ( S ) and x # y t h e n S " ( x 1 $ S " { y l but ix} S"{y}. I f x # y then S ~ ~ E XnI S ~ ~ I Y _c I{ { X , y i i . If

{XI

5

x E a(R),

such t h a t f o r each x S y

is

is By d u a l i t y , an a n a l o g o u s p r o o f

CM 2 0 .

r e p l a c e d by J , J U o r J K . w i l l h o l d f o r 6 and D J . ) PROOF:

D

By d u a l i t y , a n a n a l o g o u s p r o o f w i l l

J.

KU.)

CMR 8(;)

(b)

(The p r o o f i s s i m i l a r i f

CM 2 0 .

or

= $(S)

t h e n t h e f o l l o w i n g s t a t e m e n t s are

PART 11, CLASS FORMS

222

(i) s"{x} D ~ " { y ) , f o r a l l

n

( i i ) s"{x}

ii u E-',

(iii) ( x l y >

x

(v) If

n

(x,y) E R

(iv)

x 5

x

8,

~ " { y =)

R-

n

(R

1

,

x, y E

x # Y.

x , y E X I x # Y.

for a l l for a l l

x , y E X, x # y.

for a l l

x , y E X , x # y.

u

R - ~ )

I.

i s a 5-maximal s u b c l a s s o f t h e

X

x,

such t h a t

B(R)

( i )h o l d s and

then

i s a 5-maximal s u b c l a s s o f

Y = X - 2

such t h a t

$(R)

Y X Y S R .

(b) L e t each

x E $(R)

b e a r e l a t i o n and d e f i n e

R

J u s t a s i n p a r t ( a ) w e h a v e t h a t if S"{x) $ S " { y )

then

X E$(R)

If

S

such t h a t f o r

,

x , y E $(S)

x # y,

and

S " { x } fl S " { y ) _C { { x , y j } .

and

t h e following statements a r e

= $(S),

equivalent: ( i ) s ~ ~ I xE} s " { y I ,

(ii) S"{x)

n

s"{y} # 8,

(iii) (x,y> E R

(iv)

x

x

x

for a l l

c (R

n

R - ~ )

x , y E X , x # y.

for a l l for a l l

R-',

n

x , y E X , x # y.

u

x, y E

x,

x # y.

I.

Then, j u s t a s i n p a r t ( a ) , w e c o n s t r u c t a s - m a x i m a l s u b c l a s s Y 5 a ( R ) s u c h t h a t Y x Y 5 R, q.e.d. The f o l l o w i n g c o r o l l a r y f o l l o w s from Theorems 3.13C and

3.14C. COROLLARY 3.15C:

U,

E l DK,

CM 20

D J , DU, KJ, KU,

THEOREM 3.16C:

-

and

LV A CM 2 0

CMR 0 ( A )

,

for

JU. -f

CMR 9 ( E ) .

A = D,

E l J, 5 ,

5 3.

PROOF:

MAXIMAL PRINCIPLES

Substitute iT

for

The proof that CMR 9 ( z )

+

D

223

in 3.13C.

CAC 4 is similar to the proof

of 4.25. THEOREM

3.17C:

CMR 9 ( i i ) -+ CAC 4 .

PROOF: Let L be a relation which linearly orders Define a relation S as follows:

If

(x,y),(z,w)

E a(S)

V.

then

CMR 9 ( K ) F 5 a(S)

implies that there exists a c-maximal subclass such that if s, t E F and s # t then S " { s l i?S"{t}. Clearly, a(F) = $(R) and F _C R. If (x,y),(x,z) E F then S"{(x,y)} K S"{(x,z)). But, it follows from the definition of F that this cannot happen unless y = z. Thus, F is a function which satisfies CAC 4 .

-

Clearly, CMR 9 (A) LV A CMR 8 ( A ) , for all A. Since CM 7 -+ LV, it follows that CM 7 CMR 9(A) for each A which is a property of finite character. Even though each of CM 8 (Ern) , CM 12 and CM 13 is equivalent to a set form of the axiom of choice in NBG, we shall prove that each of CMR 8 (Ern) , CMR 12, and CMR 13 is equivalent to a class form. We shall show that each of CMR 8 (D,) , CMR 8(6,), CMR 8 (E!m) , is equivalent to CM 20(m); and each of CMR 10, 11, 12, 13 is equivalent to CM 20. -+

The following implications hold for logical reasons. CMR 8(D,) CMR 8(6!,) The proof of gives us Dm and

CMR 10 * CMR 11,

-+

-+

CMR 12 * CMR 13.

CMR 8(Dm) is similar to the proof that CMR 8(5,) Theorem 4.21. A proof similar to the proof of 3.13C that CM 20 (m) implies each of CMR 8 (A), for A = D m' D!,. -+

PART 11, CLASS FORMS

224

We shall show next that CMR 8(Dm) a preliminary definition and lemma.

-+

DEFINITION 3 . 1 8 C : If P is a property of is a relation and X is a class then

R

P,[X] for all

iff

P[R"{xl},R"{x2},

x1,x2,...,xm E X

such that the

but first

CMR 8(Dm),

... xi's

rn

classes,

R"{xm}l are distinct.

Using this notation we could have stated CMR 8(A) as follows: For every relation R and class X there exists a 5-maximal subclass Y such that AR[Y].

m

5

LEMMA 3.19C: For every relation R and natural number 2 there exists a relation S such that

-

DmRIXl for every class

S

PROOF: Let as follows.

iff

Dms[Xl

I

X.

R

and

m

satisfy the hypothesis.

It follows immediately from 3.19C that CMR 8(Dm) COROLLARY 3.20C:

For each natural number m

-+

Define

CMR 8(Dm). 2

2,

§

3.

Theorems 3.21C-3.25C THEOREM 3.21C:

-

CM 2 0 ( m )

225

MAXIMAL PRINCIPLES

a r e a l l due t o A .

Rubin.

L.

m

F o r e a c h n a t u r a l number

2,

2

CMR 8(6!,).

-

The p r o o f t h a t CM 20 ( m )

PROOF:

+

CMR 8 D!m)

is s i m i l a r

t o t h e p r o o f o f 3.13C. Conversely , suppose

class and

R

CMR 8 (D! m )

any r e l a t i o n .

s

x

y

X

S

< m,

ym

x E y, y

iff

be any so t h a t

Let

holds.

Define a r e l a t i n and

CMR 8 ( 5 ! m ) i m p l i e s t h a t t h e r e i s a 5-maximal

x1 , ~ 2 , . . . , ~ m E Y ,

S " { x i } = {y

5

xi

X:

E

n

PI.

s"{xi} #

i=l

< m,

y, y

R.

subclass

m such t h a t f o r a l l

5

and

ym

5

Y _C X

Since

R},

m

n

it f o l l o w s t h a t

S " { x i ] = { { x l , x 2 , . . . ,xm}}.

for a l l

i=l x1,x2,...,x

that

is a 5-maximal s u b c l a s s o f

Therefore,

m

m E Y , { x ~ , x ~ , . . . , x ~5} R.

Y

This implies

such t h a t

X

Ym _C R ,

q.e.d. !, i n

The n e x t r e s u l t shows t h a t t h e f a c t o r i a l ,

D! m

behaves l i k e a f a c t o r i a l . THEOREM 3.22C:

-

CMR 8(6!,)

For a l l n a t u r a l numbers

I\

2

2,

I\

CMR 8 ( E i ) .

i=2

m (

m

m

CMR 8 ( D i ) = CMR 8 ( D )

A CMR

8(D3) A . .

i=2

.* CMR

S(6,)).

m

CMR 8 (Di) , i t f o l l o w s i=2 from 3.21C t h a t it i s s u f f i c i e n t t o p r o v e t h a t CM 20(m) +

To p r o v e

PROOF:

CMR 8 ( D k )

f o r each

CMR 8 ( 6 ! m )

+

k , 2 5 k 5 m.

I\

Since

CM 2 0 ( m ) + CM 2 0 ( k ) ,

u s e a p r o o f s i m i l a r t o t h e p r o o f o f Theorem 3.13C t o p r o v e t h a t CM 2 0 ( k )

*

CMR 8

(Ek).

Conversely, l e t assume

m

I\

i=2

CMR 8 ( D i ) .

X

be a c l a s s , Define

Y1 = X

R

a r e l a t i o n , and

n

a(R).

For e a c h

k,

226 2

I

PART 11, CLASS FORMS k < m,

that

define

Yk+l

Ek+l,RIYk+l]. Ym

as a 5-maximal subclass of

Yk

such

is the required class, q.e.d.

The next theorem is the basis for proving the remaining equivalences and it is the most difficult of these theorems to prove. THEOREM 3.23C: For all natural numbers m CMR 8(Dm) + CMR 8(EmV1).

2 3,

PROOF: Let X be a class and R a relation. class Y and natural number n ;1 1, let

We define a class

Y

For any

as follows:

(Note that Y is definable in NBGO since it can be expressed as a union of sets). For each u E Y, the rank of u is the smallest n E w such that u E Yn. Next, we define a class Z such that

A U _C W U

({W}

X

X ) A Em-l,R [It:(w,t)

Also, define a relation S (y,u) E s Let

W

iff

E

ul1)l.

such that $ ( S )

(y E u E Z ) v u = {y}.

be a 5-maximal subclass of

Y

Y

=

-

and

such khat Dm,s[Wl.

53.

MAXIMAL PRINCIPLES

227

(I)

We see first that E [W] + 5 Z . This is because m,S m u E n S"{Yi} u = ~Y1lY2~".IYm~. i=l The next claim is that W has at least m elements. Suppose not. Let w be any m - 1 element subset of Y containing W. Let u = w U {(w,a)l such that a E X. Then D [u] and W c u. This contradicts the maximality of W. m,S E W, where (wi,xi> Suppose ( W ~ , X ~ ) , ( W ~ ~,...,(wmlxm) X~) are distinct. Let t be a wi of maximal rank and let Since v E Z , it follows that for i i i m}. v = {(wi,xi):l

-

all i = 1,2,...,m, either wi = t or (wi,xi) E t. Since 1 t I i m - 1, we must have that l$(W) I L m and W = t U ({tj x T). Since 5 Z, it follows that T is a c-maximal subclass of X such that ~m-l,RITl, q.e.d.

():

COROLLARY 3.24C: The following statements are equivalent for each m 2 3: CM 20(m) , CMR 8(Dm) , CMR 8(Em), and CMR 8(6!,).

-

PROOF: By 3.2OCl CMR 8(Dm) CMR 8(Em). By 3.22C, It follows from 3.23C that CMR 8(6!,) + CMR 8(Em). CMR 8(Em) + CMR 8 (Di), and from 3.22C that i=2 m E\ CMR 8(Ei) + CMR 8(E!m). Thus, CMR 8(Dm) CMR 8(Em) i=2 To complete the proof, in 3.21C we proved CMR 8(6!,). CM 20(m) +,CMR 8 (E!m), q.e.d.

;\m

-

-

THEOREM 3.25C: The following statements are equivalent: CM 20, CMR 10, 12, 13.

-

CMR 12. To PROOF: It follows from 3.24C that CMR 10 complete the proof we shall show that CM 20 CMR 12 CMR 13 CM 20. -f

-+

-f

CM 20 CMR 12. It follows from Theorem 3.13C that CM 20 -+ CMR 8(E) , and clearly CMR 8 (E) CMR 12 (take m -+

-f

Clearly CMR 12

+.

=

CMR 13.

CMR 13 + CM 20. Let X be a class and R a relation. There is no loss of generality in assuming R is symmetric

2).

PART 11, CLASS FORMS

228

and r e f l e x i v e .

Define a r e l a t i o n

E

(x,y)

x E y , y < w,

iff

S

and

m

n

# 8, w h i c h i n t u r n

S"{y}

Therefore,

-Y

x

and a

2

2

E

YI

E Y,

5

{x,y} x {x,y}

(Y U { X I )

x

R.

5-maximal.

is

Y

u {XI)

(Y

and

X, y

implies

We c l a i m

Y x Y _C R .

x E

suppose

such t h a t f o r a l l y1,y2,...,ym

X

# 8. T h i s i m p l i e s t h a t f o r a l l

f l S"{yi} i=l

S"{x}

5

Y

m

5

y x y

CMR 1 3 i m p l i e s t h a t t h e r e i s a n a t u r a l number

5-maximal

x 6 X,

so t h a t i f

S

5

R.

For

R.

Then,

S, i f y1,y2,...,ym-1 E Y, using t h e d e f i n i t i o n of m- 1 ~ " { x ln ( I s " { y i l 2 I y 1 , y 2 , . . . , y m - 1 , x 3 # B. Since Y w a s i=1 Thus, 5-maximal w i t h t h i s p r o p e r t y , x E Y , c o n t r a d i c t i o n .

CMR 1 3

+

CM 2 0 ,

q.e.d. CMR 11

THEOREM 3.26C:

Let

PROOF:

-+

CMR 8 (6).

be a c l a s s a n d

X

a sequence of c l a s s e s

Xn,

a relation.

S

u s i n g t h e c o n s t r u c t i o n i n Theorem 3.23C. S2 = S.

Let

construct a relation 2.

R = {(x,y):(3n

E

Dn,Sn[Zn].

such that Dn,sn Now d e f i n e a r e l a t i o n

w) ( 3 z , w ) [ x = ( n , z ) , y

(The d e f i n i t i o n of

R

then

is

5

Zn

U s i n g 3.19C,

-

Tn

Sn and

5 Xn+l

[Zn+ll D n + l ,s ~ + ~ -

is 5-maximal f o r t h e p r o p e r t y

for a l l classes

X2 = X

Let

b e a c l a s s so t h a t i f

Xn+l

5-maximal f o r t h e p r o p e r t y

W e define

and a s e q u e n c e o f r e l a t i o n s

[Z]

iff

D

Xn

we

n,Tn

[Z]

R:

= ( n , w j , and ( z , w )

is c o n s t r u c t i v e a n d , s i n c e

R

E T,]}.

c a n be

e x p r e s s e d a s a u n i o n o f s e t s , i t i s d e f i n a b l e i n NBGO.) CMR 11 i m p l i e s t h e r e i s a n

- (R)

Y

such t h a t

- (Tm)

Zlll

i s 5-maximal s u c h t h a t

d ~ f. i n i t i o n of

-

P,ll,S l Z , l l l .

m Let

D,,,[Y].

Tmp

zm 5

L'(Sm)

2

2

and a 5-maximal

Zm = { z : ( m , z )

Dm,Tm[Zml.

i s 5-maximal

Usinq t h e d e f i n i t i o n of

Sm,

E Y}.

Then

BY t h e

such t h a t

w e now c o n s t r u c t a

111

-

9 ,

S

which i s '-maximal

with the property

bs[Z2],

q.e.d.

§ 3.

COROLLARY 3.27C:

The f o l l o w i n g s t a t e m e n t s a r e

CM 2 0 , CMR 1 0 , 11, 1 2 , 13.

equivalent:

CMR 10

PROOF: CM 20

229

MAXIMAL PRINCIPLES

.+

CMR 11

-+

f o r l o g i c a l reasons.

By 3.15C,

The c o r o l l a r y i s a c o n s e q u e n c e of 3.25C

CMR 8 ( 6 ) .

and 3.26C. To c o m p l e t e o u r d i s c u s s i o n o f t h e s e C M R . s t a t e m e n t s , w e s h a l l p r o v e t h a t CMR 1 4 A.

L.

-

Rubin.) CM 7

THEOREM 3.28C:

( T h i s proof i s a l s o due t o

CM 7 .

-

CMR 1 4 .

Since t h e f i n i t e i n t e r s e c t i o n property i s a

PROOF:

p r o p e r t y o f f i n i t e c h a r a c t e r it f o l l o w s t h a t CM 7 Conversely, l e t f i n i t e character.

X

Define a r e l a t i o n

(x,y) E R

iff

R

x E y , y .c: w,

CMR 1 4 g i v e s u s a 5-maximal s u b c l a s s

m

x ~ , x ~ , . . . , Ex ~ Y,

R"{xi}

i=1

#

Y

has t h e property

p r o p e r t y of f i n i t e c h a r a c t e r , f o l l o w s from t h e d e f i n i t i o n of subclass of

X

Y

a p r o p e r t y of

on

so t h a t

X

and

5

CMR 1 4 .

P(Y).

X

such t h a t f o r a l l

This implies t h a t

pI.

{ x ~ , x ~ , . . . , x ~ }h a s t h e p r o p e r t y s u b s e t of

P

b e a c l a s s and

-+

P, so t h a t every f i n i t e P.

Y

Thus, s i n c e

P

has t h e property Y

with t h e property

that P,

Y

is a It

P.

is a c-maximal

q.e.d.

W e summarize o u r r e s u l t s f o r t h e CMR s t a t e m e n t s i n ( A l s o see Theorems 3.1C,

F i g u r e 3.4C.

3.4C and F i g u r e 3 . 3 C . )

The s t a t e m e n t s i n e a c h o f t h e f o l l o w i n g sets a r e equivalent:

@ @

@

= {CM

= {CM 2 0 ( m ) , CMR 8 ( A ) } , where A =

0

Dm,6m,6!mand

= {CM 2 0 , CMR 1 0 , 11, 1 2 , 1 3 , CMR 8 ( A ) ) ,

J,

@

7 , CMR 1 4 } .

5,

U,

El

DK,

= {CM 5 , CMR 8(K) 3. = {CAI CMR 8 ( i ? ) } .

D J , D U , K J , KU, J U .

m

2

3.

where A = D I D ,

PART 11, CLASS FORMS

230

@j =

{CMR 1, 2 , 3 1 4, CM 3 1 4, CM 6 (CM 8(K))r CM 1 7 1 cMS l(D,W), CMS l(Q,U)},

and

where

Q = TR, P I M I F,

U = C , A S K , L, D, W.

CMR 9 ( A ) CM 9 ( A )

..

LV A CMR 8 ( A ) ,

LV A CM 8 ( A ) ,

f o r all

for all

A

A

FIGURE 3 . 4 C To conclude this section we shall consider the forms CM 15(A:B)

and

CMR 1 5 ( A : B ) .

CM 15(A:B): Any class which contains a 5-maximal subclass with the property A contains a 5-maximal subclass with the property B.

We shall use the notation introduced in Definition 3 . 1 8 C to state CMR 15(A:B). CMR 15(A:B) : For any relation R and class X I if there csists a 5-maximal subclass Y 5 X such that A R [ Y ] then there esists a 5-maximal subclass 2 5 X such that B R [ Z ] .

is defined to be LV A CM 1 5 ( A : B ) , and c’PII; 1 6 (A:B) is defined to be LV A CMR 1 5 ( A : B ) . It i s easy to sce that the following implications hold for all A and B. A l s o , CM 1 6 ( A : B )

5 3.

231

MAXIMAL PRINCIPLES

F i r s t , w e s h a l l c o n s i d e r t h e forms CM 1 5 ( A : B ) ,

where t h e

A and B , A f B , a r e D , D, J , 5, K , o r K. i s M 1 9 and i s e q u i v a l e n t t o a set form of t h e axiom

choices f o r CM

8(5)

o f c h o i c e i n NBG.

Thus, by ( * ) above, CM 8 ( A : D )

c a n n o t be

s t r o n g e r t h a n a s e t form of t h e axiom o f c h o i c e , and s i n c e CM 8 ( A : D )

implies

M 8(A:D),

is equivalent to a

CM 8 ( A : E )

s e t form i n NBG. By u s i n g s i m p l e m o d i f i c a t i o n s i n Theorems 4 . 3 5 , 4 . 3 8 , 4 . 4 0 , and 4 . 4 1 (with A = K o r c o r r e s p o n d i n g r e s u l t s f o r CM 1 5 ( A : B ) .

t h a t any c l a s s

X

K),

we obtain the

( W e may always assume

i n any one o f t h e s e theorems i s d i f f e r e n t

from V , f o r X c o u l d be r e p l a c e d , for example, by Using t h e s e r e s u l t s w e o b t a i n : J, o r

D;

5,

+

CM 8 ( B ) , f o r A = K ,

-+

CM 8 ( i ? ) .

CM 1 5 ( A : B )

+

CM 8 ( B )

CM 1 5 ( D : 5 )

* CM 8 ( K ) .

CM 1 5 ( A : J )

+

CM 8 ( K ) , f o r A = D

CM 1 5 ( A : K )

+

CM 8 ( K ) ,

for

A =

K

or

J.

(4.40)

CM 1 5 ( R : B )

+

CM 8 ( D ) ,

for

B = D

or

5.

(4.41)

CM 1 5 ( A : B )

4.36,

B =

X

{O).)

o r D.

(4.35) CM

15(5:K)

(4.36)

, for

A = D or

D;

B = K or

E.

(4.38)

(4.38)

D.

or

(4.38)

The f o l l o w i n g theorems g i v e us t h e r e s u l t s f o r t h e remaining s t a t e m e n t s : and

CM 1 5 ( A : J ) ,

where

CM 1 5 ( 3 : B ) , A = K

or

where

-

K.

B = D, J , o r

K;

PART 11, CLASS FORMS

232

THEOREM 3 . 2 9C :

(a)

CM 15(3:K)

-f

CM 8 ( K ) ,

(b)

CM 1 5 ( A : J )

+

CM 8 ( K ) ,

PROOF:

Let

be a c l a s s .

X

d i s t i n c t elements.

for

u1 = (11)

x

u)

u

Y = Iuo:u E

xl u

or

Iul:u

Y

d

be

Let

u

E XI

{YOfY1L

u, v E X:

J V1t

(2)

yo K uo

and

y1 K ul,

(3)

yo J u1

and

y1 J u o f

(4)

YO D Y 1 r

(5)

ui K vi

iff

u K v, i = 0

or

(6)

ui J vi

iff

ui

vi

u

I t f o l l o w s from ( 1 ) - ( 4 ) t h a t

of

and

define

Each of t h e f o l l o w i n g i s e a s y t o v e r i f y f o r a l l

(1) uo

E.

Cafcrd~,

yo = { a r b ) , y1 = Cc,d).

and d e f i n e

5

a , b, c

Let

u E X,

For each

A =

with the property

iff

{yo,yl}

5.

Using

1,

ii

v, i = 0

or

1.

i s a 5-maximal s u b c l a s s we obtain

CM 1 5 ( 3 : K )

t h a t Y h a s a 5-maximal s u b c l a s s , 2, w i t h t h e p r o p e r t y K . I t f o l l o w s from (1) t h a t i f u i , v E 2 t h e n i = j . j T h e r e f o r e , u s i n g ( 5 ) w e see t h a t {u E X : ui E 2 1 i s a 5-maximal s u b c l a s s o f X w i t h t h e p r o p e r t y K . This proves (a). TO prove ( b ) w i t h

A =

5,

we let

2'

b e a 5-maximal

s u b c l a s s o f Y w i t h t h e p r o p e r t y J. By ( 4 ) t h e r e i s a t I f yo E 2 ' t h e n by ( 2 ) uo 6 Z' f o r most one yi E 2 ' . any

u E X.

Then u s i n g ( 6 1 ,

c-maximal s u b c l a s s of y1 E 2 ' .

I f no

yi

X

w

= {U E X: u1 E 2 ' )

with the property

-

K.

E Z', t h e n by (l), u o E 2 '

is a

Similarly, i f iff

u1 E Z'

53.

so

MAXIMAL PRINCIPLES

A =

If

z,

then w e note t h a t A =

2,

K;

t h u s t h e proof

CM 8 ( K ) .

CM 1 5 ( K : J )

PROOF:

be a c l a s s and suppose

X

-+

a, b, c

= {ua:u E subclass of

XI Y

u

{yo,yll. W e c l a i m Iyo,yll i s a 5-maximal with t h e property K. C l e a r l y yo K y,. It

i s a l s o e a s y t o see t h a t

y . J ua

Thus, CM 1 5 ( K : J )

subclass

Z

iff

u K v.

THEOREM 3.31C:

{ u E X: ua E Z}

-

CM 1 5 ( 3 : D )

Z

has t h e property

with the property

choice function

F

D

is a

K.

CWO 1 2 .

-+

The p r o o f i s s i m i l a r t o t h a t o f 3.6C,

PROOF: Z

u, v E X,

with t h e property

X

and

h a s a 5-maximal

Y

For a l l

J.

Therefore,

5-maximal s u b c l a s s of

the class

implies t h a t

with t h e property

u E X

for a l l

1

i = 0,l. ua J va

UX.

and

ua = u U { a ) , y o = { a , b ) , y1 = { a , b , c }

Y

is t h e

q.e.d.

THEOREM 3.30C: Let

K.

i s a 5-maximal

{yO,ylf

-

with t h e property

Y

same a s t h a t f o r

Let

-

i s a g a i n a 5-maximal s u b c l a s s w i t h t h e p r o p e r t y

W

subclass of

of

233

-

J.

because

Thus, a 5-maximal s u b c l a s s

can be used t o c o n s t r u c t t h e

o f 3.6C.

W e i l l u s t r a t e o u r r e s u l t s w i t h t h e t a b l e i n F i g u r e 3.5C.

-

K

D

-

K

D

CM 15(A:B)

FIGURE 3.5C

K

PART 11, CLASS FORMS

234

T h e l e t t e r s i n t h e body of t h e t a b l e i n d i c a t e t h e CM 8 f o r m w h i c h i s i m p l i e d by t h e CM 1 5 ( A : B )

there is a that

For e x a m p l e ,

form.

i n t h e f i r s t r o w , t h i r d column.

K

CM 1 5 ( D , 5 )

+

X's

The

CM 8 ( K ) .

T h i s means

a r e f o r t h e CM 1 5 ( A : A )

s t a t e m e n t s w h i c h a r e p r o v a b l e i n NBGO and t h e CWO 1 2 m e a n s CM 1 5 ( 5 : D )

CWO 1 2 .

-+

C o m b i n i n g Figure 3.5C w i t h the fact t h a t CM 8 ( B )

* CM 1 5 ( A : B )

for a l l

.+ -

B, w e obtain

and

A

THEOREM 3 . 3 2 C : CM 8 ( K ) CM 8 (X) CM 8 ( D ) CM 8 ( D )

-+

CM 8 ( J ) CM

-+

8(5)

,

for a l l

A

# i?.

CM 1 5 ( A : K ) ,

for a l l

A

# K.

CM 1 5 ( A : D ) ,

for a l l

A

# D , 3.

CM 1 5 ( A : K )

CM l 5 ( J : D ) CM 1 5 ( A : J )

+

+

,

+

CM 15(A::)

CM 8 ( 3 )

-+

CM 1 5 ( D : j )

CM 8 ( 5 )

-+

CM l 5 ( ? : ? )

( T h e possible values f o r

CWO 12.

for -+

for a l l

CM 8 ( K ) , A =

D,

J , K.

D,

J,

6

B #

# J.

CM 8 ( K ) .

* CM 8 ( D ) . are

A

D,

T h u s i t f o l l o w s t h a t a l l CM 1 5 ( A : B )

and

A

5,

K

or

forms with

ii). A

# B

a r e e q u i v a l e n t t o a c l a s s f o r m of t h e a x i o m of

c h o i c e i n NBG.

(See F i g u r e 3 . 3 C . )

F i n a l l y , w e t u r n o u r a t t e n t i o n t o t h e CMR 1 5 ( A : B ) statements. B.

Clearly

We have s h o w n

CMR 8 ( K )

-

D,

U,

J,

3,

CMR 8 ( K )

CA ( 3 . 1 2 C ( b ) )

c,

B =

CMR 1 5 ( A : B )

6 , fi,

-+

-+

CMR 8 ( A )

and

+.

CM 2 0

for

J U (3.15C).

A = D, We shall

B' e x c e p t f o r CMR 1 5 ( J K : B )

B , A # B , there i s a

CMR 8 ( B ' )

,

,

w h i c h a r e p r o v a b l e i n NBGO.

T o accomplish t h i s m o n u m e n t a l t a s k ,

relation

and

A

CM 5 ( 3 . 1 2 C ( a ) ) ,

and A

for a l l

CMR 1 5 ( A : B )

DK, D J , DU, K J , KU, and

show b e l o w t h a t for a l l such that

-

CMR 8 ( B )

R

w e shall construct

i = 1 , 2 , . ..,9, s u c h t h a t w h e n

9

f o r any c l a s s

relations,

CMR 1 5 ( A : B )

X

Si,

is a p p l i e d w i t h

and

§

MAXIMAL PRINCIPLES

3.

Si we obtain a 5-maximal subclass Yi for some B'.

_C

X

235 such that Bi[Yi],

In the 9 constructions which follow we assume X is a given class and R is a given relation. We shall also assume without loss of generality that for all x E X, R"(x} is neither empty nor the universe. Under these assumptions, if R"{x) and R"{y) are not complementary, exactly one of D, K, J, or U holds between them. If R"{x} and R"{yI are complementary then both D and U hold. In what follows we shall replace the statement "Y is a 5-maximal subclass such that AS[Y]" by "Y is A(S)-maximal".

1.

For each

x E X, = ({O}

S;{x)

x

R"{x}) U {(l,x)I.

If x, y E X, x # y, then S;{x} A S ; { y ) , for A = li, ii, and (= DJ). Thus, for each x E X, 1x1 is A(S 1 )-maximal, for A = K, U, or KU; and X itself is A(S 1)-maximal for A = K, 5 or DJ. We also have that for x # y, R"{x} D R"{y}, for B = D, DK, DU, Or

(a) S;{x]

B S;{y)

iff

(b) S;{x}

B SYfyI iff

R"{x} 6 R"{y}, for B =

5,

JU, JK

5.

Or J.

If A = K, U , KU, I?, or DJ, the hypothesis of CMR 15(A:B) is satisfied. Thus, there is a subclass Y which is B(S1)-maximal. It follows from (a) that Y is D(R)-maximal, if B = D, DK, DU or 2; and it follows from (b) that Y is D(R)-maximal if B = 5, JU, JK, or J. Thus, we have proved the following: If

A

=

K,

u , KU, li, ii or

DJ,

-

CMR 15(A:B)

-+

CMR 8(D),

for

B = D, DK, DU,

or

J,

CMR 15(A:B)

+

CMR 8(6),

for B = D , JU, J K ,

or

J.

Using the fact that D and U are duals of each other (i.e., X D Y iff (V X) U (V Y)) and K and J are self-dual, we obtain, in addition:

-

-

PART 11, CLASS FORMS

236

If

A = K, D, DK,

CMR 15(A:B) CMR 15(A:B) If

Let

2.

+

CMR 8(U), CMR 8(5),

JU,

for for

B = U, KU, B = ij, DJ.

A = D, DK, 6, JU,

CMR ~s(A:B) CMR 15(A:B)

2 ( F J f V.

+

2, 5 ,

F

+ +

CMR 8(U), CMR 8(6),

for for

B = 5 , DU, B = J, JK.

be any 1-1 function mapping x E X,

X

into V

such that

For each

Sll{x} = F" (R"{x}). 2 Then

{XI

S:{x) fi S ! j { y ) for all x, y E X. Thus, for each x E XI is U(S2)-maximal, and X is c(S2)-maximal. Moreover, S;tx) A S;ty)

iff

R"tx} A R"{y),

for

A = K

or

i?.

It follows that

CMR 15 (A:B)

+

c,

and B = K

or

iT;

CMR 8(B), for A = D or 6, and B = K

or

E.

CMR 8 (B), for A = U or

and by duality, CMR 15(A:B)

-+

3. L e t x E X and let z o , z1 be distinct sets such that for all x E X, zi f (x,j), for i, j = 0,l. Then, for i = 0,1, we define S"{

3 (

x,i))

= Rl'ix) x ti}, S " { z . } = V

3

1

(R"X

x

{i}).

Then the following seven statements are easy to verify, if x, y E X, i # j , i, j = 0,l. (a)

s;{zo)

u s;hll.

(b) S;{zi) D S;{(x,i)]. (c) S;Czi1

K S;t(x,j)l.

(d) S;t(x,O)l

D S;{(y,l)).

(e) S;{(x,i)I

D S;{(y,i)1

iff

R"{X}

(f)

K S;{(y,i))

iff

R"{x) K R"{y).

S;{(x,i)}

D R"{y}.

5 3.

MAXIMAL PRINCIPLES

237

Therefore, {zo,zl} is A(S3)-maximal for A = U, E l E, DU, {zi} for i = 0,1, are 6(S3)-maximal; KU and JU; a ( S , ) and each {zi), i = 0,1, is J(S )-maximal. 3

-

For maximal.

B = D, DJ, K, DK, JK,

and

5,

let

YB

be

B(S3)-

S;{zl) # 0, there is at (i) B = D. Since S;{zo} most one zi E Y,,. If zi E YD, then it follows from (c) that no (XI]) E YDl with j # i. It follows from (b) and (e) that {x E X:(x,i) E YD} is D(R)-maximal.

If no zi E YD, then by (d), classes, Yo and Yl, such that Thus, each of D (R)-maximal.

{x E X:(x,i)

E YD}

YD DS3 for

is a union of disjoint [Yo] and

i

=

0,1,

D

s3 is

[Y1l.

(ii) B = DJ. The proof is similar to (i), but instead of finishing with a DJ(R)-maximal class, we finish with a z(R)-maximal class. (This follows from (g). ) (iii) B = K. By (d), if there is a (x,i) E YK, then no (ylj) E YK if i # j . By (a), (b) and (d), if (x,i) E YK then zj E YK if i # j, but zi @ YK. Therefore, YK contains exactly one zi and a class of (x,j)'s, with j # i. It follows from (f) that {x E X:(x,j) E YKI is a K(R)-maximal class. (iv) B = DK. In this case YDK contains exactly one zi and two classes Yo and Y1 as in part (i). Thus, fx E X:(x,i) E YDK) for i = 0,1, are each DK(R)-maximal. (v) B = JK. This is similar to (i). There is at most one zi f YJKl but, because of (g), we end with a D(R)maximal subclass instead of a JK(R)-maximal subclass.

where

(vi) B = 3 . In this case Ys = ~ z o , z lU~Yo U Y1, Yo and Y1 are defined in (i). Again, because of

238

PART 11, CLASS FORMS

(g), {x E X:(x,i) subclasses.

of

E Yg),

i = 0,l

are each

DK(R)-maximal

Using the results obtained in (i)-(vi) and the duality D and U, we list below the new results obtained from S3. CMR 8(D) , A =

5,

J, JU, DU.

CMR 15(A:U) * CMR 8(U), A =

c,

J , DJ, DU.

CMR 15(A:D)

+

CMR 15(A:DJ)

+

CMR 8(K) , A = J , U,

fi,

CMR 15(A:JU)

+

CMR 8 ( K ) , A = J , D,

5 , DK, DU.

CMR 15(A:K) CMR 15(A:DK)

+

CMR 8(K), A = J ,

+

CMR 8(DK), A =

z

5 , J, JU, DU.

CMR 8(5) , A = J , DU.

* CMR 8 (DK),

CMR 15(A:;) 4. Let define

-+

DJ, KU, DK, JU, DU.

5, J, DJ, DU.

CMR 15(A:KU) +CMR 8(KU), A = CMR 15(A:JK)

E,

KU, DU.

A = J, DU.

be any set such that = R"{x) S"{x} 4

X

{O},

z f X.

For each

x E X,

Si{z) = {(z,l)).

Then { z ) is A(S4)-maximal for A = D, J, K, KU, JU Suppose YB is a B(S4)-maximal subclass.

or

JK.

Then the following are true. (i)

{x E X: x E YD)

is

D(R)-maximal.

(ii)

{x E X: x E YDJ)

is

E(R)-maximal.

(iii)

{x E X: x E YDK}

is

DK(R)-maximal.

(iv)

{x E X: x E YDu)

is

D(R)-maximal.

(v)

{x E X: x E Y5)

is

DK(R)-maximal.

(vi)

{x E X: x E Y E )

is

E(R)-maximal.

Using (i)-(vi) and duality, the additional results we obtain from this example are as follows:

239

MAXIMAL PRINCIPLES

53.

CMR 15(JK:B) * CMR 8 ( B ) , B = D , U, DK, KU. CMR 1 5 ( A : D U ) * CMR 8(D), A = J, JK. CMR 15 (JK:j) CMR 1 5 (A:i?) 5.

y

L

-+

CMR 8 (DK). CMR 8 ( E ) , A

+

=

J, K, KU, JU, JK, DJ, DK.

Let Z be a non-empty, non-universal class and suppose X. Then we define,

It is easy to see that {y} is A ( S 5 )-maximal for A = J , DJ, and DU. (We are assuming R"{x} # V for all x E X.) Suppose Y is U(S5)-maximal. Since S;{X}

Y

- {y}

is

6 s;{y}

iff

R"{x}

V

R"{Y},

;(R)-maximal.

Using example 5 and duality we obtain the following additional results: CMR 1 5 ( A : c )

+

CMR 8 ( c ) , A = J, DJ, DU.

CMR 1 5 ( A : E )

+

CMR 8 (E) , A = J, JU, DU.

6.

Suppose

fl #

Z

cV

and

S"{x} = R"{x}

& X.

z x

Define

V, S " { z }

= V

Z.

X

The range of S is V x V. We are going to modify S so that its range is V. Let F be a function mapping V onto $(S). Then for each u E .fJ(S),

s;{ul Since

F

= F-l"s"{ul.

is a function,

if

yl, y2

F - ~ I IAYF~- ~ I ' Y iff ~

F

-1

,I

y1

5 S(S)

y1 A y2,

u F - ~ ~ Iiff Y ~ y1 u y2

=

I

for A

v

x

v.

=

D

or

K.

PART 11, CLASS FORMS

24 0

For all x E X, S;{x} J S : { z } . Thus { z ) is A(S 6 )-maximal for A = D, K, DK, DU, KU and 5. Suppose YB is B ( S 6 ) maximal. We have for all x, y E X, x # y, s;{xl

B s ; { y } iff

Therefore, for

B

= fit

R"{x] B R"{y},

i?, G I J , DJ, JU

{x E X: x E YB} The additional results using

7.

Let

In

g X. Then for each f01l o w s :

B = D, K , J , U.

and

JK,

is

B(R)-maximal.

S6

and duality are listed below.

be the set of all integers, T x E X and

z

(See 4 . 4 3 ) i, j E In.

for

U R"{X}, and X€X i E In, we define S 7 as =

S"{(x,i)} 7

= R"{x} x {i} x {l,Z},

S;{(z,i)}

=

T

x

{j E In: j

i

i}

x

{l}.

The following are easy to verify for

x, y E X,

53.

MAXIMAL PRINCIPLES

241

If Z = {(z,i): i E In} then 2 is A(S,)-maximal for A = K, DK, JK, KU, 6, and 5 . Suppose YB is B ( S 7 )-maximal, B = D, DU, J or JU. i) B = D. It follows from (d) that there is at most one (z,i) E YD. If no (z,i) is an element of YD then {x E X:(x,i) E YD} is D(R)-maximal for each i E In. If (z,i) E YD then for each j > i, {x E X:(x,j) E YD} is D (R)-maximal. The proof is analogous if "D" is replaced by "DU". (ii) B = J. Using (c), (d) and (e), we see that there is exactly one (z,i) E YJ. It also follows from (e) that if (x,i), ( y , ] ) E YJf then i = j . Thus, there is a j E In, j 2 i, such that if (x,k) E YJ then k = j . Then, by (a) and (b), {x E X:(x,j) E YJ} is JU(R)-maximal. Similarly, if "J" is replaced by "JU" we obtain a JU(R)maximal class. Using (i), (ii) and duality, we obtain the following results : CMR 15(f:B)

+

CMR 8(B), B

=

D, U, DU.

CMR 15(KU:U) * CMR 8 (U). CMR 15(DK:D) * CMR 8(D). CMR 15 (JK:B) * CMR 8(JU) , B = J, JU. CMR 15fJK:DJ) 8.

For each

-+

CMR 8(DJ).

x E X, let Si{x} = (R"{x}

X

t o } ) U {(O,l)}.

Then, X is JK(S8)-maximal. Let Y be Then Y is K(R)-maximal. Therefore, CMR 15(JK:K)

-+

CMR 8(K).

K(S8)-maximal.

PART 11, CLASS FORMS

242

9.

Let

B #

2 c

V

and

z @ X.

For each

x E X,

To obtain S 9 from S, we modify S as we did in example 6. We have a ( S ) = V x V x V and X(Sg) = V.

If hold :

x, y E X, i # j, i, j = 0,1,

the following relations

We see that ~(z,O),(z,l)~ is a z(Sg)-maximal subclass and are each D(Sg)-maximal. Suppose YB for i = O,l,{(z,i)} is B(Sg)-maximal. if

If B = K, then there is exactly one (z,i) E YK, i # j, {x E X:(x,y) E YK) is K(R)-maximal.

so

If B = DK, there is at most one (z,i) E YDK. If (z,i) E YDK, then {x E X:(x,j) E YDKj is DK(R)-maximal if j # i. If no (z,i) E YDK, then {x E X:(x,i) E YDK} is DK(R)-maximal for i = 0,l. Using this example and duality we obtain the following results: CMR 15(5:B) CMR 15(D:DK)

+

CMR 8(B), B = K, DK, KU.

+

CMR 8(DK).

CMR 15 (U:KU) * CMR 8 (KU). We summarize our results in Theorem 3.33C.

53.

MAXIMAL PRINCIPLES

243

THEOREM 3.33C: CMR 15(A:B) CMR 8(B), for all A and B, A # B, except A = JK and B = f j or fi and those listed below.

-

(i) CMR 8(DJ) + CMR 15(A:DJ) + CMR 8 ( i ? ) , for A = J, U , fi, KU, and DU. (ii) CMR 8(JU) * CMR 15(A:JU) + CMR 8(K), A = J, D, D l DK, and DU. (Possible values f o r A and DK, DJ, DU, KJ, KU, JU.) PROOF:

B

are

D,

5,

for

K,

J, j , U,

fi,

Corollary 3.15C and examples 1-9 above.

Thus, it follows from 3.33C and preceding results that CMR 15(A:B) is equivalent to a class form of the axiom of choice f o r all A and B, A # B, except for the two statements CMR 15(JK:E) and its dual, CMR 15(JK:c). A. L. Rubin has shown that these two statements are provable in NBGO. LEMMA 3.34C: PROOF:

Let

CMR 15(JK:6) X

A

CMR 15 (JK:c).

be any class and suppose Y

5X

is a

c-maximal subclass such that JKR[Y]. Let Z = {z E

Then, we claim that such that fjR[W].

X:(Vy E Y)R"{zI

W = Y U Z

6 R"{y}}.

is a 5-maximal subclass of

X

To show ERIWl, it is sufficient to show that for all zl, z 2 E z Y, R"{z~} 5 ~ ~ ~ { z ~ } . zl, z 2 E z Y. Suppose Since zl, z 2 E Z and zl, z2 Y, there must be yl, y2 E Y such that

-

-

(1) R"{yl} U R"{zl}

Moreover, since Y1' Y2 E (2)

R"{Y1).

and

R " ( y 2 } U R"{z2}.

YI

5 R"{Y2}.

It follows from (1) and ( 2 ) that there is a such that v E R"{zl} r l R"{z21. (R"{yl] U R"{y2}) v E V

-

244

PART 11, CLASS FORMS

Therefore,

R"{zlI 13 R"{z21

and

ER[W].

It is clear from the definition of W that it is a $-maximal subclass of X such that ER[W]. This proves CMR 15(JK:E). We obtain CMR 15(JK:;) by duality. We summarize the results of Part I1 below. The statements in each of the following s e t s are equivalent:

@

9

=

{CT, CWO 4 , 5, 6(n), 7, 8, lo}, n

2

1.

= {CAC 4 , 5, 18Cl.

,B = d

{CWO 1, E, CAC 1, 2, 3, 8, 10(2), 18, 191.

@ = {CAC 10 (m), 13 (m-1)1,

m z 3.

9

= {CAC 11E, 12E1.

@

= {CAC 11, 12, 14, 151. = ICAC 22, 23, 24).

@

= (CM 7 , CMR 141.

(3 {CM 20(m), CMR 8 =

@

(A)

I , where A

-

and

m

2

3.

= {CM 20, CMR 10, 11, 12, 13, CMR 8(B1), CMR 15(A2:B2),

A2 # B2, CMR 15(A3:DJ)

A2

=

5,

U,

D, E, J ,

5,

K, E l

B2 = D, and

c,

J,

B1 = D,

5,

6,

J, j , U,

B2 = E l

KJ, JU;

A4 =

t,

,

CMR 15(A4:JU) 1, where

DJ, DK, DU, JK, JU, KU; U,

c,

5 , K,

= CMR 15 (A:DJ).

@(A)

= CMR 15(A:JU).

z,

DJ, DK, DU, JK, JU,

DK, DU, JK, KU, but

are omitted;

G(A) @)

-

= Dm, Dm, D !,,

U,

A3 = D, E,

5,

E l DJ, KJ, KU.

= {CM 5, CMR 8(K), CMR 15(A:K)

1, where

A

# K.

KU;

A2 = J K

K,

E,

DK,

53.

@

= { C A I CMR 8 ( ? ) ,

( P o s s i b l e values of

GI

U,

in

(3and

@

are

A #

x.

D,

D,

J,

3,

K,

?,

D J , DK, DU, J K , JU, K U . ) =

@

where

CMR 1 5 ( A : z ) I , A

245

MAXIMAL P R I N C I P L E S

{CMS 1, 2 ,

3 , 4 , CM 3 , 4 , 6 ( C M 8 ( K ) ) , CM 1 7 ,

, CMS l ( Q , U ) 1, where U = C , A S & C , T R & C , L , D, W . CMS l ( D , W )

CPW

3

ccw-CWO

CWO 2

Q = TR, P I RA, F

&

\

CH

a w

\1

m

J.

NBG

CWO 1 2

CWO l2-B

(1) (2) (3)

m z 3. See Figure 3 . 1 C for the remaining CM l ( Q , U ) forms. CM 5 is CM 5 ( P , L ) . See Figure 3 . 2 C for the remaining CM 5 ( Q , U ) forms.

(4) (5)

See Theorem 3 . 3 2 C for the CM 1 5 ( A : B ) forms.

(A,A')

=

(J,J),(V,D>,(G,~),(KU,DK),(DU,DU).

and

This Page Intentionally Left Blank

L I S T OF THE S E T FORMS

I n p a r e n t h e s e s f o l l o w i n g e a c h of t h e s t a t e m e n t s w e s h a l l g i v e t h e name of t h e p e r s o n who i s g i v e n c r e d i t f o r p r o v i n g i t e q u i v a l e n t t o t h e axiom o f c h o i c e (when a p p l i c a b l e ) and t h e d a t e f o l l o w e d by t h e page number where i t o c c u r s i n t h e t e x t . Rubin 1963 r e f e r s t o t h e 1963 e d i t i o n o f " E q u i v a l e n t s " and Rubin 1985 r e f e r s t o t h e p r e s e n t work. THE WELL ORDERING THEOREM

(1)

WO 1:

Every set can b e w e l l - o r d e r e d .

WO 2 :

Every set i s e q u i p o l l e n t t o an o r d i n a l number.

WO 3:

Every s e t i s e q u i p o l l e n t t o a s u b s e t of a n o r d i n a l number.

Let

m

(1)

m

be a n a t u r a l number,

WO 4 ( m ) :

(1)

For every set

x

and a f u n c t i o n f o r every

1.

a

t h e r e e x i s t s an o r d i n a l number f

B < a

2

d e f i n e d on and

u

a

f(B)

such t h a t

= X.

f (B)

S

m

(Levy 1962; 1)

B
WO 5:

T h e r e e x i s t s a n a t u r a l number WO 4 ( m ) .

WO 6:

m

2

1

such t h a t

(Levy 1 9 6 2 ; 1)

x

For every set

t h e r e e x i s t s a n a t u r a l number

m

2

1,

a, and a f u n c t i o n f d e f i n e d on a f o r e v e r y B < c1 and f ( B ) = x. B
an o r d i n a l number such t h a t WO 7:

f(B)

R

of

u

m

x, x

For e v e r y s e t ordering

<

i s f i n i t e i f f f o r each w e l l a l s o w e l l o r d e r s x. (Tarski

x , R-l

1 9 2 4 a , 1938a; 5 )

247

248

. L I S T O F SET FORMS

wo 8 :

Every s e t on which t h e r e i s a c h o i c e f u n c t i o n can be ( B r u n n e r , Rubin 1985; 6)

w e l l ordered.

THE AXIOM OF CHOICE

s a s e t o f non-empty s e t s , t h e r e i s a f u n c t i o n (Zermelo such t h a t f o r e v e r y x E s , f ( x ) E x.

If

AC 1:

s

f

1904; 7

t

If

AC 2 :

i s a p a i r w i s e d i s j o i n t s e t ofnon-empty s e t s , c which c o n s i s t s of one and o n l y one

t h e r e i s a set

element from e a c h s e t i n For e v e r y f u n c t i o n

AC 3 :

f

t.

(Russell 1906; 7)

there is a function

t h a t f o r e v e r y s, i f x E B ( f ) g(x) E f ( x ) . (Zermelo 1 9 0 4 ; 7 )

r

For e v e r y r e l a t i o n

AC 4 :

that

a(f) =

a(r)

and

For every f u n c t i o n

AC 5 :

a(q) = x(f)

that

there is a function

5 r.

f

0, f

then

such

(Bernays 1 9 4 1 ; 7 )

t h e r e is a function

f

such

g

f(x) #

and

and f o r e v e r y

g

such

x E a(g) , f ( g f x ) ) = x.

(Bernays 1941: 7 ) The C a r t e s i a n p r o d u c t of a set o f non-empty s e t s i s

AC 6 :

non-empty.

(Zermelo 1 9 0 4 ; 8 )

The C a r t e s i a n p r o d u c t of a s e t of non-empty sets of

AC 7 :

t h e same c a r d i n a l i t y i s non-empty.

(Ward 1962; 8 )

AC 8 :

If

s

i s a s e t o f p a i r s o f e q u i p o l l e n t sets t h e n t h e r e i s a f u n c t i o n which a s s o c i a t e s w i t h e a c h p a i r a b i j e c t i o n mapping one o n t o t h e o t h e r . (Pincus 1974; 8)

AC 9 :

If

s

i s a s e t of s e t s o f t h e same c a r d i n a l i t y t h e n

t h e r e i s a f u n c t i o n which a s s o c i a t e s w i t h e a c h p a i r a b i j e c t i o n mapping one o n t o t h e o t h e r . Let

n

2

2

AC 1 0 ( n ) :

( P i n c u s 1974; 8 )

b e a n a t u r a l number. If

s

i s a s e t of i n f i n i t e s e t s , t h e r e i s a

function

f

such t h a t f o r e a c h

d e c o m p o s i t i o n of

s i z e between

x

x E s , f (x)

is a

i n t o a set o f d i s j o i n t sets o f

2 and n.

( S i e r p i n s k i 1965, Levy 1962;8)

THE AXIOM OF CHOICE

249

AC 11:

There exists a natural number n I 2 such that AC 10(n). (Sierpinski 1965, Levy 1962; 8)

AC 12:

If s is a set of infinite sets there is a natural number n 2 2 and a function f such that for each x E s , f(x) is a decomposition of x into a set of disjoint sets of size between 2 and n. (Sierpinski 1965, Levy 1962, Rubin 1963; 8)

Let

m

1 be a natural number.

2

AC 13(m):

If s is a set of non-empty sets there is a function f such that for each x E s , f (x) is a non-empty subset of x with f(x) < m. (Levy 1962; 8)

AC 14:

There is a natural number (Levy 1962; 8)

AC 15:

If s is a set of non-empty sets then there is a natural number m 2 1 and a function f such that for each x E s, f(x) is a non-empty subset of x with f(x) < m. (Levy 1962, Rubin 1963; 8)

Let

and

n

AC 16(n,k) :

k

m

I

be natural numbers with

1

such that

n > 2

and

AC 13(m).

1 < k

n.

If s is an infinite set then there is a set tn of n-element subsets of s such that each k-element subset of s is a subset of exactly one element of tn. (Frascella 1965,1966, Rubin & Rubin 1967; 9)

AC 17:

If s is a set, t = p ( s ) * {(a}, F is the set of all functions from t to s and g is a function from F to t then there is an f E F such that f(g(f)) E g(f). (Tarski 1951; 9)

AC 18:

For any non-empty set A, and for every function B whose domain is A and Ba is a non-empty set for each a E A,

(Collins 1954, Farrah 1955; 9)

L I S T OF SET FORMS

250

AC 19:

For any non-empty set non-empty,

A, each of whose elements is

n

n f w t

u b = c1 a€A .bEa fEC(A) a€A

where C(A) is the set of all choice functions on (Collins 1954, Farrah 1955, Rubin 1963; 9)

A.

(Va)DC(a): For each ordinal a , if R is a relation between subsets and elements of a set x such that (Vy 5 x) (y < a +. (3u € x)(y,u) E R) then there is a function f mapping c1 into x such that ( V B < p)(f"B,f(B)) E R. (Levy 1964; 19)

THE TRICHOTOMY y, x < y, x = y

y < x.

T 1:

For all sets x and (Hartogs 1915; 21)

T 2:

For every two non-empty sets, there is a mapping of one onto the other. (Lindenbaum 1926, Sierpinski 1948; 21)

If

5

n

T 3(n):

2

or

is a natural number,

If x is a set of cardinal numbers such that 1x1 = n then there exists K , p E x such that K < p or < K. (Tarski 1964; 22)

'

T 4:

For all cardinals K~ i K~ or K~ 5

', K ~

if 1-1 i K ~ K~ , s '2 . (Kruse 1962; 24)

T 5:

For all cardinals K and p , if p < 2 K or u s K. (Sobocinski 1961a; 27)

T 6:

For all cardinals K

T 7:

i

K

K

or

s

K.

For all cardinals and if for all

vl

implies

v1

= v2,

then either

then

and alephs N, if (Sobocinski 1961a; 27)

< ZK

K,

K ~ , K

and then

(Sobocinski 1961b; 28)

~ and ,

v2, K~

K~

s. K~

p,

if

+ vl or

K

0

= K~ K~

K 5

u

then

s K ~ K~ , < 1~

+ v2

i

K

1'

= p

MAXIMAL P R I N C I P L E S

T 8:

For a l l s e t s

<

fp,,(y) T 9:

x

and

y , p , w ( x ) 4 ?<,(y)

or

( A l a s 1968; 2 9 )

y<,(x).

F o r a l l sets

251

x

and

y,

x d p
?,,(y)

x.

4

( A l a s 1968; 29) For a l l sets

T 10:

x

and

< Pc3,,(y) o r

y, x

y

< ?<,(x).

( A l a s 1968; 2 9 )

MAXIMAL PRINCIPLES M 1:

If x

i s a p a r t i a l o r d e r i n g r e l a t i o n on a non-empty s e t and i f e v e r y s u b s e t o f x which i s l i n e a r l y o r d e r e d R

by

h a s a n R-upper bound, t h e n

R

element. M 2:

If

h a s an R-maximal

i s a p a r t i a l o r d e r i n g r e l a t i o n on a non-empty s e t

R

x

and i f e v e r y s u b s e t of

R

h a s an R-upper bound, t h e n

element. M 3:

x

(Bourbaki 1939, Tukey 1940; 32)

x

which i s w e l l - o r d e r e d by x

h a s a n R-maximal

(Kneser 1950, S z e l e 1950; 33)

I f e v e r y non-empty n e s t which i s a s u b s e t of a non-empty

set x h a s i t s u n i o n a n e l e m e n t of x , t h e n x h a s a c-maximal e l e m e n t . (Kuratowski 1922; Zorn 1935; 33) M 4:

I f e v e r y non-empty w e l l - o r d e r e d n e s t which i s a s u b s e t o f a non-empty s e t then

x

x

h a s i t s union a n e l e m e n t of

h a s a 5-maximal e l e m e n t .

x,

(Kuratowski 1922,

Hausdorff 1 9 2 7 ; 3 3 ) M 5:

If

R

i s a p a r t i a l o r d e r i n g r e l a t i o n on

e x i s t s a 5-maximal s u b s e t of o r d e r e d by M 6:

M 7:

x, then there

which i s l i n e a r l y

(Hausdorff 1 9 1 4 ; 33)

R.

For e v e r y s e t

x

x , t h e r e e x i s t s a 5-maximal s u b s e t of

which i s a n e s t .

(Hausdorff 1 9 0 7 , 1909; 33)

For e v e r y set

and e v e r y p r o p e r t y

x

P

of f i n i t e

c h a r a c t e r , t h e r e e x i s t s a 5-maximal s u b s e t o f has t h e property Bourbaki 1940; 3 4 )

P.

x

x

which

( T e i c h m i i l l e r 1939, Tukey 1 9 4 0 ,

L I S T OF SET FORMS

252

M'1:

If

i s a p a r t i a l o r d e r i n g on

R

x

and if e v e r y s u b s e t

of x which i s l i n e a r l y o r d e r e d by R h a s a n R-upper bound and i f y E x t h e n t h e r e i s a n R-maximal e l e m e n t z E x

such t h a t

y R z.

(Bourbaki 1939, Tukey 1940,

Kuratowski 1922; 36) M'2:

If of

R

i s a p a r t i a l o r d e r i n g on

x

and i f e v e r y s u b s e t

x

which i s w e l l o r d e r e d by

R

h a s a n R-upper

bound and i f z E x

t h e n t h e r e i s a n R-maximal e l e m e n t

y E x

such t h a t

y R z.

(Kneser 1950, S z e l e 1950,

Kuratowski 1 9 2 2 ; 36) M'3:

I f e v e r y non-empty n e s t which i s a s u b s e t o f a non-empty

set

x

h a s i t s union a n e l e m e n t o f

x

and i f

t h e n t h e r e i s a 5-maximal e l e m e n t z E x y 5 z. (Kuratowski 1922, Zorn 1935; 36) M'4:

y E x

such t h a t

If e v e r y non-empty w e l l o r d e r e d n e s t which i s a s u b s e t of a non-empty s e t x h a s i t s union i n x and i f t h e n t h e r e i s a 5-maximal e l e m e n t z € x s u c h that y c - z. (Kuratowski 1922, Hausdorff 1927; 36)

y E x

MIS:

I f R i s a p a r t i a l o r d e r i n g o n x and y _c x s u c h t h a t y i s l i n e a r l y o r d e r e d by R , t h e n t h e r e e x i s t s a 2-maximal

set

z

5x

such t h a t

y

5

(Hausdorff 1 9 1 4 ,

z.

Kuratowski 1922; 36) M'6:

For e v e r y s e t

x

5-maximal n e s t

z

5:

there is a

and e v e r y n e s t

y

5

y _c z.

x

such t h a t

x

(Hausdorff

1907,1909, Kuratowski 1922; 36) M'7:

For e v e r y s e t

x , f o r every property

c h a r a c t e r , and € o r e v e r y s e t t h e property that

y c z

y _C x

of f i n i t e

P

such t h a t

P, t h e r e i s a 5-maximal s e t and

z

has t h e property

P.

y

z _C x

has such

(Teichmiiller

1939, Tukey 1940, Bourbaki 1940; 37) M l(Q,U) :

Every non-empty s e t which i s Q-ordered by which e a c h U-ordered

R

in

s u b s e t h a s a n R-upper bound,

h a s an R-maximal e l e m e n t , where Q = TR, P, RA o r F and U = C , A S X , T R & C , L, D , o r W; U = W.

(Harper

&

and Q = D and

Rubin 1 9 7 6 , F e l g n e r 1 9 6 7 ; 37)

MAXIMAL PRINCIPLES

M 5(Q,U):

253

Every s e t which i s Q - o r d e r e d by

contains a

R

5-maximal s u b s e t which is U-ordered by Q = A

or

or

RA; and

Q = C

RA;

and

U = R A ;

U = D

or

UN(n) , n 2 2 .

P, L , D , or U = C , A S K , T R & C , o r L;

U = AS, A S K ,

and

and

Q = R A Q = A

and

W;

U = C , AS&C, T R & C , L ,

and

P

and

D , RA o r F

Q =

Q = D

or

Q = AS

RA;

D, o r

R, where

U = AS, C , A S & C , T R & C , P , L , D ,

and

TR

and

U = D ; Q = F U = UN o r

(Wallace 1 9 4 4 , G o t t s c h a l k 1 9 5 2 ,

Hljft & Howard 1 9 7 3 , H a r p e r & Rubin 1 9 7 6 ; 4 1 , 4 5 )

M'5(Q,U):

I n e v e r y Q-ordered s e t , e v e r y U-ordered s u b s e t c a n

be e x t e n d e d t o a 5-maximal U-ordered s e t , where and

u

a r e t h e same a s i n M 5 ( Q , U ) ,

Q = A

M 8 (A) :

U = TR.

and

( H a r p e r & Rubin 1 9 7 6 ; 4 4 )

E v e r y s e t h a s a 5-maximal s u b s e t w i t h t h e p r o p e r t y A; where

A = D (Vaught 1 9 5 2 ;

(Kurepa 1 9 5 2 ; 68);

A = D

m

Levy 1 9 6 2 , Rubin 1 9 6 3 ; 7 1 ) ; (A. M 9(K):

Q

and a l s o

Rubin 1985;

Rubin & H .

68); or

A =

Dm

A = D V

5,

-

or

J

J

(Chang 1 9 6 0 ,

5

V

J

V

5

V K

V

K*

54)

Every s e t c a n be l i n e a r l y o r d e r e d , and e v e r y s e t h a s a 5-maximal

subset with t h e property

-

K.

(Kurepa

1952; 54) M 10:

T h e r e e x i s t s a n a t u r a l number

m

2

2

such t h a t

M 8(Dm).

(Chang 1 9 6 0 , Levy 1 9 6 2 ; 5 7 )

M 11:

For e v e r y s e t such t h a t property

M 12:

h a s a 5-maximal

property

m

2

2

s u b s e t which h a s t h e

(Chang 1 9 6 0 , Levy 1 9 6 2 , Rubin 1 9 6 3 ; 5 7 )

m

t

2

such t h a t

(Rubin 1 9 6 3 ; 5 8 )

For e v e r y s e t such t h a t

M 14:

t h e r e e x i s t s a n a t u r a l number

T h e r e e x i s t s a n a t u r a l number M 8(;!,).

M 13:

x Dm.

x

x

E!m.

x , t h e r e e x i s t s a n a t u r a l number

m 1 2

has a c-maximal s u b s e t w h i c h h a s t h e (Rubin 1 9 6 3 ; 5 8 )

Every s e t c o n t a i n s a 5-maximal s u b s e t w i t h t h e f i n i t e i n t e r s e c t i o n property.

(Klimovski 1 9 5 6 ; 6 0 )

254

M 145:

L I S T OF SET FORMS

Every c l a s s c o n t a i n s a 5-maximal s u b c l a s s w i t h t h e f i n i t e i n t e r s e c t i o n property.

(Klimovski 1 9 5 6 , Rubin

1985; 61) M'14:

If

y

5 x

and

y

has the f i n i t e i n t e r s e c t i o n property

t h e n t h e r e i s a c-maximal set z with t h e f i n i t e i n t e r s e c t i o n p r o p e r t y s u c h t h a t y 5 z 5 x. (Klimovski 1956; 6 1 ) m y s e t which c o n t a i n s a 5-maximal s u b s e t w i t h t h e property

c o n t a i n s a 5-maximal s u b s e t w i t h

A

t h e property or

B , where B = D,

and

A J,

# B, A 5, o r

= D,

6, J, 3,

K

(Rubin 1963,

K.

1 9 8 5 ; 65)

Every set c a n be l i n e a r l y o r d e r e d and any s e t which c o n t a i n s a 5-maximal s u b s e t w i t h t h e p r o p e r t y A

c o n t a i n s a 5-maximal s u b s e t w i t h t h e p r o p e r t y

B , where

A

# B,

A , B = D,

5,

J,

5,

K

?.

or

(Rubin 1963; 6 5 ) .

M 17:

Every s e t

x

c o n t a i n s a 5-maximal s u b s e t y s u c h t h a t s , t E y , s # t , e i t h e r s E t o r t E s.

for e v e r y

(Rubin 1963; 6 9 ) M 18:

Suppose

x

i s p a r t i a l l y o r d e r e d by

t E x, let -+ t = { s E x: t R s

A

R.

For each

t # s}.

Suppose e v e r y R - l i n e a r l y o r d e r e d s u b s e t of R-upper bound. h a s no R - f i r s t

x

has an

t E x such t h a t ( B l a i r & Tomber 1 9 6 0 , Rubin

Then t h e r e i s a element.

1963; 7 1 ) M 19(Q,U):

Every s e t

x

such t h a t

and e v e r y U-ordered

x

i s Q - o r d e r e d by

R

s u b s e t h a s a n R-upper bound

h a s t h e p r o p e r t y t h a t e v e r y f u n c t i o n f from x t o x such t h a t f o r a l l u E x , u R f (u), h a s a f i x e d p o i n t , where values a s i n

72)

Q

M l(Q,U).

and

U

have t h e same

(Abian 1983, Rubin 1985;

EQUIVALENCE UNDER EXTENSIONALITY AND FOUNDATION

255

FORMS EQUIVALENT TO THE AXIOM OF CHOICE UNDER THE AXIOMS OF EXTENSIONALITY AND FOUNDATION PW:

The power set of a well-ordered set can be well-ordered. (H. Rubin 1960; 73)

LW:

Every linearly ordered set can be well-ordered. 1960; 73)

INJ:

If x is a non-empty set and A x < A. (von Neumann 1925; 73)

(H. Rubin

is a proper class then

PROJ:

If x is a non-empty set and A is a proper class then A can be mapped onto x. (von Neumann 1925; 74)

MC 1:

If s is a set of non-empty sets then there is a function f such that for each x E s, f(x) is a non-empty finite subset of x. (Levy 1962, Felgner Jech 1973; 74)

&

MC 2:

Every set is the union of a well-ordered set of finite sets. (Levy 1962, Rubin 1985; 74)

MC 3:

If s is a set of infinite sets, then there is a function f such that for each x E s , f(x) is a partition of x into pairwise disjoint finite sets, each of which has at least 2 elements. (Sierpinski 1965, Levy 1962, Rubin 1985; 74)

A:

Every partially ordered set contains a 5-maximal antichain. (Kurepa 1952; 74)

P ACQ:

P MCQ:

For each set x, if x is a P-ordered set of non-empty sets such that each u E x can be Q-ordered, then there is a function f such that for each u E x, f(u) E u, where P = LO and Q = pI, LO or DLO; and P = pI and Q = LO or DLO. (Howard ti Rubin 1977; 7 8 ) . For each set x, if x is a P-ordered set of non-empty sets such that each u E x can be Q-ordered, then there is a function f such that for each u E x, f(u) is a non-empty, finite subset of u, where P = LO and Q = 8 , LO or DLO; and P = 0 and Q = LO or DLO. (Howard & Rubin 1977; 78)

256

L I S T O F SET FORMS

i s a s e t and f i s a f u n c t i o n s u c h t h a t f o r e a c h f ( u ) i s a non-empty s e t of l i n e a r o r d e r i n g s o f

If x u E x

ORD:

{f ( u ): u E x }

u , then

has a choice function.

(Truss

1 9 7 2 ; 78) If

W-ORD:

i s a s e t and

x

f

i s a function such t h a t f o r

e a c h u E x, f ( u ) i s a non-empty s e t o f w e l l - o r d e r i n g s of u , t h e n { f ( u ) : u E x } has a choice function. (Truss 1972; 78) Every t r a n s i t i v e and c o n n e c t e d set c o n t a i n s a 5-maximal

TCL:

l i n e a r ordered set. MC 4 ( n ) :

s

If

(Howard

&

Rubin 1 9 7 7 ; 7 8 )

i s a s e t of non-empty sets t h e n t h e r e i s a f such t h a t f o r a l l x E s , f ( x ) i s a f i n i t e s u b s e t o f x and ( I f ( x ) I , n ) = 1.

function non-empty

( B l e i c h e r 1963, 1965; 83) For every set

INF 2 :

x , i f every i n f i n i t e s u b s e t o f

structured, then

can be w e l l ordered.

x

x

is

(Brunner

1982; 84) P

UQ:

The u n i o n o f a P - o r d e r e d f a m i l y of s e t s , e a c h of w h i c h c a n be Q - o r d e r e d , c a n b e w e l l - o r d e r e d , LO

and

LO, o r

Q = LO, DLO, WO DLO;

and

P = LO

or and

DWO;

Q =

where

P = WO

0.

0

or

Q =

0,

P =

and

(J. E.

Rubin

1982; 85) M 5 (Q,U) :

Every s e t which i s Q-ordered by 5-maximal s u b s e t w h i c h i s U Q = T R , A S , AS&TR, D o r RA

R

contains a

o r d e r e d by R , where and U = VC. (Rubin

1985; 8 6 ) M 19:

AC 2 0 :

Every c l a s s h a s a 5-maximal s u b c l a s s w i t h t h e p r o p e r t y

-D .

(Kurepa 1 9 5 2 , Rubin 1 9 6 3 ; 8 7 )

For e v e r y r e l a t i o n

R

whose domain i s a s e t ( i t s

r a n g e may b e a p r o p e r c l a s s ) t h e r e i s a f u n c t i o n f s u c h t h a t D ( f ) = $(R) and f 5 R. (Bernays 1941, Rubin 1 9 6 3 ; 8 8 )

257

ALGEBRAIC FORMS

AC 2 1 :

F o r a n y non-empty s e t whose domain i s

n

u

bEBa

aEA

A,

and f o r e a c h r e l a t i o n

B

A,

x

= f E xu Ba xa , f ( a ) aEA

where Ba- i s d e f i n e d t o be B"{a} f o r e a c h ( C o l l i n s 1 9 5 4 , F a r r a h 1 9 5 5 , Rubin 1 9 6 3 ; 8 8 ) (Va)DCC(a) :

a , if

For each o r d i n a l

a E A.

i s a r e l a t i o n between X such t h a t

R

s u b s e t s and e l e m e n t s o f a c l a s s

(Vy 5 X ) ( y < a -+ (3u E X ) ( y , u ) E R) t h e n t h e r e i s a f u n c t i o n f mapping a i n t o x s uch t h a t

(VB < a ) ( f " B r f ( 6 ) ) E R. T 11:

{ a E On: T ( N a )

(Felgner 1976a; 89)

< ZNa}

A

i s unbounded.

(Kruse

1963; 90) T 12:

{a E On: T ( K a )

A

T(2

i s unbounded.

)}

(Kruse 1963;

90) T 13:

For a l l c a r d i n a l s

K

and

s u c h t h a t K , p I; 2 I a 1 (Kruse 1963; 90) T 14:

{ a E On: T ( Ka)

6 E On.

A

1 ~ , i f there is an ordinal a

then

K

2

c o f ( o ~ + 2~ 6) )

1-1

or

p 2 K.

i s unbounded f o r e a c h

(Kruse 1963; 91) ALGEBRAIC FORMS

AL 1:

E v e r y l a t t i c e w i t h a n i d e n t i t y e l e m e n t and a t l e a s t o n e o t h e r e l e m e n t c o n t a i n s a 5-maximal p r o p e r i d e a l . ( S c o t t 1953; 95)

AL'1:

E v e r y p r o p e r i d e a l of a l a t t i c e w i t h a n i d e n t i t y e l e m e n t c a n be e x t e n d e d t o a 5-maximal p r o p e r i d e a l . 1953; 95)

AL'2:

If

A

a E A,

i s an a l g e b r a i c system, but

B

a s u b a l g e b r a , and

a f B , t h e n t h e r e e x i s t s a 5-maximal B as a s u b s e t , b u t d o e s n o t

s u b a l g e b r a which c o n t a i n s contain

a

(Scott

as a n e l e m e n t .

( B l a i r 1 9 5 7 ; 95)

LIST OF SET FORMS

258

AL'3:

I f x i s an a r b i t r a r y s e t , C a f i n i t a r y c l o s u r e o p e r a t o r , P a p r o p e r t y of f i n i t e c h a r a c t e r , and y a C-closed s u b s e t o f x which h a s t h e p r o p e r t y P, t h e n t h e r e i s a 5-maximal C-closed s u b s e t o f contains

y

x

a s a s u b s e t and h a s t h e p r o p e r t y

which P.

( S c o t t 1 9 5 3 , Rubin 1963; 95)

AL'4:

i s an a r b i t r a r y s e t ,

y

5 x,

t h e r e i s a 5-maximal s u b s e t

z

of

If

x

and

AL 5:

and

x

R : y

then

such t h a t

y

5

z

( B e t h 1959; 9 7 )

R : z.

is a Boolean a l g e b r a and S _C B such t h a t 0 f S , t h e n t h e r e e x i s t s a 5-maximal p r o p e r i d e a l d i s j o i n t If

B

from AL 5(V) :

(Mrowka 1 9 5 5 ; 98)

S.

S _C B s u c h t h a t i s closed with respect t o V ( j o i n ) , t h e n t h e r e e x i s t s a ?-maximal i d e a l d i s j o i n t from S. (Howard 1 9 7 7 ; 9 9 )

If

i s a Boolean a l g e b r a and

B

0 f S

and

S

i s a Boolean a l g e b r a and S 5 B s u c h t h a t 0 f S and S i s c l o s e d w i t h r e s p e c t t o ( c o m p l e m e n t a t i o n ) , t h e n t h e r e e x i s t s a 5-maximal

AL 5 ( - ) :

If

B

-

i d e a l d i s j o i n t from AL 1D:

(Howard 1977; 1 0 0 )

S.

Every d i s t r i b u t i v e l a t t i c e w i t h an i d e n t i t y e l e m e n t and a t l e a s t one o t h e r e l e m e n t c o n t a i n s a 5-maximal proper i d e a l .

DAL 1D:

(Klimovski 1958; 1 0 1 )

Every d i s t r i b u t i v e l a t t i c e w i t h a z e r o e l e m e n t and a t l e a s t one o t h e r e l e m e n t c o n t a i n s an u l t r a f i l t e r . (Klimovski 1958; 1 0 1 )

AL 6:

=

If

a

i

b,

(L,{V,Aj)

c

i s compact, c s b

e x i s t s an e l e m e n t and i f t h e r e i s a c

AL 7 :

S

d.

i s a c o m p l e t e l a t t i c e , a , b , c E L,

m E L d E L

and

c

a

then there

such t h a t

a

m

i

b , c P! m

such t h a t

m < d

5

b

5

then

( D i e n e r 1956; 1 0 2 )

Every s u b a l g e b r a i s t h e i n t e r s e c t i o n of a l l m e e t i r r e d u c i b l e s u b a l g e b r a s c o n t a i n i n g it as a s u b s e t . (McCoy 1 9 3 8 , B i r k h o f f 8, F r i n k 1 9 4 8 , Fuchs 1 9 4 9 , D i e n e r 1 9 5 6 , Schmidt 1953; 1 0 3 )

259

ALGEBRAIC FORMS

Every p r o p e r s u b a l g e b r a h a s a p r o p e r m e e t - i r r e d u c i b l e

AL 8:

s u b a l g e b r a c o n t a i n i n g it a s a s u b s e t .

( D i e n e r 1956;

103) The s e t of a b s o l u t e l y d i s p e n s i b l e e l e m e n t s is e q u a l t o

AL 9 :

t h e i n t e r s e c t i o n of t h e s e t o f a l l Ejmaximal p r o p e r subalgebras.

( F r a t t i n i 1885, Neumann 1937, Schmidt

1953; 103) AL 10:

F o r e v e r y e l e m e n t i n a b a s i s t h e r e e x i s t s a 5-maximal s u b a l g e b r a which d o e s n o t c o n t a i n i t a s an e l e m e n t . (Schmidt 1953; 103)

AL 11:

If

A

h a s a f i n i t e b a s i s and

B c A

is a subalgebra

t h e n t h e r e e x i s t s a 5-maximal p r o p e r s u b a l g e b r a which contains

a s a subset.

B

( K r u l l 1929; Schmidt 1953;

103) AL 1 2 :

Every a l g e b r a h a s a b a s i s .

AL'12:

Any a l g e b r a i c a l l y i n d e p e n d e n t s u b s e t of a n a l g e b r a c a n be e x t e n d e d t o a b a s i s .

(Schmidt 1962; 1 0 3 )

(Schmidt 1962; Rubin 1985;

109)

AL 13:

Any s u b s e t of a n a l g e b r a which g e n e r a t e s t h e a l g e b r a contains a basis.

AL 1 4 :

On e v e r y non-empty s e t t h e r e e x i s t s a c a n c e l l a t i v e groupoid.

AL 15:

(Hajnal

&

Kertesz 1972; 110)

Every unique f a c t o r i z a t i o n domain c o n t a i n s a 5-maximal proper ideal.

AL 1 6 :

(Schmidt 1 9 6 2 ; Rubin 1985; 1 0 9 )

(Hodges 1979; 1 1 2 )

Every group c o n t a i n s a 5-maximal a b e l i a n subgroup. (Klimovsky 1962, F e l g n e r 1976; 1 1 6 )

Let

k

2

1 be a n a t u r a l number,

AL 1 7 ( k ) :

Every group c o n t a i n s a 5-maximal n i l - k subgroup. ( F e l g n e r 1976; 1 1 8 )

AL 1 8 ( k ) :

Every group c o n t a i n s a 5-maximal s u b g r o u p w i t h s o l v a b i l i t y degree

B:

5

k.

Every v e c t o r s p a c e h a s a b a s i s .

( F e l g n e r 1976; 1 1 8 ) ( B l a s s 1984; 1 1 9 )

LIST OF SET FORMS

260

AL 1 9 :

For every f i e l d every s u b s e t of basis.

Let

F

and e v e r y v e c t o r s p a c e V o v e r which g e n e r a t e s V c o n t a i n s a

F V

F,

( B l e i c h e r 1 9 6 4 , H a l p e r n 1966; 1 1 9 )

be a f i e l d ,

AL 2 0 ( F ) :

Every v e c t o r s p a c e that for all then

A

c_

over

V

V, i f

A

F

has the property

is l i n e a r l y independent

c a n be e x t e n d e d t o a b a s i s o f

A

V.

(Bleicher 1 9 6 4 ; 119) B':

Every v e c t o r s p a c e h a s a d i m e n s i o n .

AL 2 0 :

There e x i s t s a f i e l d

F

(Brunner; 1 2 1 )

such t h a t

AL 20 ( F )

.

( B l e i c h e r 1964; 125)

Let

F

be a f i e l d ,

AL 2 1 ( F ) :

Every v e c t o r s p a c e V o v e r F h a s t h e p r o p e r t y t h a t i f S 5 V i s a s u b s p a c e of V t h e n t h e r e i s a subspace

S'

-V c

generates

S U S'

such t h a t V.

S

0

S'

= I01

and

( B l e i c h e r 1964; Armbrust

1972; 1 1 9 ) AL 21:

There e x i s t s a f i e l d

s u c h t h a t AL 2 1 ( F ) .

F

(Bleicher

1 9 6 4 ; 125) AL 2 2 :

There is a f i e l d V

over

subspace if S

5

S

such t h a t f o r every v e c t o r space

F B

V

is a b a s i s f o r V t h e n f o r e a c h t h e r e is a s u b s e t B ' 5 B s u c h t h a t

i s t h e s u b s p a c e g e n e r a t e d by

S'

n

if

F

S ' = {O)

S U S'

and

generates

B'

then

V.

(Bleicher

1964; 1 2 5 ) AL 23:

There i s a f i e l d

F

F

has a b a s i s l a n d i f

B

i s a b a s i s of

s u b s e t of that

X

n

V,

B' =

V

such t h a t e v e r y v e c t o r space o v e r V

and

i s a v e c t o r space over X

then t h e r e i s a subset and

( B l e i c h e r 1964 ; 1 2 5 )

X U B'

F,

i s any l i n e a r l y i n d e p e n d e n t €3'

_C B

is a basis of

such V.

CARDINAL NUMBER FORMS

261

AL 2 4 :

If A is an abelian group and B is a torsion-free divisible subgroup of A then there exists a subgroup B' A such that B n B' = { O } and B u B' generates A. (Bleicher 1 9 6 4 , Armbrust 1 9 7 2 ; 1 2 7 )

AL 2 5 :

If G is a free abelian group, A is a uniquely divisible abelian group, B is an abelian group, f:A B is a homomorphism mapping A onto B, and g:G -+ B is a homomorphism then there is a homomorphism h:G + A such that foh = q. (Blass 1 9 7 9 ; 1 2 8 ) -+

AL 26:

If G is a free abelian group, A and B are abelian groups, f:A -+ B is a homomorphism mapping A onto B, and g:G -+ B is a homomorphism then there is a homomorphism h:G -+ A such that foh = g. (Blass 1 9 7 9 ; 1 2 8 )

AL 2 7 :

If G is a divisible abelian group, A and B are A is an isomorphism, and abelian groups, f:B g:B -+ G is a homomorphism then there is a homomorphism h:A -+ G such that hof = g. (Blass 1 9 7 9 ; 1 2 8 ) -+

AL 28:

Let (A,{Ok:k E K ) ) be a multi-algebra. Then there is an algebra (B,{OL:k E K } ) of the same type as A and an equivalence relation E on B such that (B/E,{Oi/E:k E K I ) . (Gratzer 1 9 6 2 , (A,[Ok:k E K } ) H6ft & Howard 1980; 1 3 4 )

AL 29:

Let (A,{Ok:k E K ) ) be a multi-algebra such that all the multi-operations Ok for k E K are unary operations. Then there is an algebra (B,{Oi:k E K}) of the same type as A and an equivalence relation E on B such that (A,{Ok:k E K } ) G (B/E,tOi/E:k E K}). (Gratzer 1 9 6 2 , H6ft & Howard 1 9 8 0 ; 1 3 4 ) CARDINAL NUMBER FORMS

CN 1:

(VK

CN'1:

( V K 2 H0) ( K * K * = K

> 0) (Vp

2

KO)

( ~ * = p K

+

K*).

+

p).

(Tarski 1 9 2 4 ; 1 3 9 )

(Tarski 1 9 2 4 ; 1 3 9 )

LIST OF SET FORMS

262

CN 1F:

CN 2:

CN'4:

x is finite iff x has at most one element or there exists a non-empty set u disjoint from x such that u u x < u x x. (Tarski 1938a; 140)

A set

( V ~ , p ~ zv N 0 )

(VK

No)

2

(K

s p

[K

+

=

K*

+

u

-+

[K

s p

V K i

v)].

(Tarski;

.

(Tarski; 139)

&),

]

.

(Tarski;

(Lesniewski, Sierpinski 1947;

K*).

139) CN 4F:

CN 5:

set x is finite iff x has at most one element or there exist disjoint sets u and v such that ]u( < 1x1 and I v / < 1x1, and u u v = x. (Tarski 1938a; 139)

A

(VK) (vp

2 No) ( K * V = K

V K-U = p).

(Lesniewski,

Sierpinski 1947; 140) CN'5:

%) (K*K*

(VK z

(Lesniewski, Sierpinski 1947;

= K*).

140) CN 6: CN 6F:

(VK

2

L

t Z O ) ( ~= K ) .

(Tarski 1924; 140)

A set

x is finite iff x has at most one element or x < x x x. (Tarski 1938a;140)

CN 7:

( Y K t p ) (K2 = p2

-+

K

= p).

(Tarski 1924; 140)

CN 8:

( V K , ~ )( K * = U* *

K

= p).

(Sobocinski 1961b; 140)

CN 9:

(VK

2

KO)

( 3 ~ ( )K * <

i

p <

K

+

K*).

(Sobocinski 1 9 6 1 ~ ;

140) CN 10:

( V K ~ (~K *) < p *

CN'10:

(VK) (Val

(K*

<

+ K

Ka+l

< p). +

K

<

(Sobocinski 1961b; 140)

Ka).

(Sobocinski 1961b; 140)

C A R D I N A L NUMBER FORMS

CN 11:

( V K , ~ )( K < p

CN'11:

( V a ) ( V v ) ( Ka

CN 1 2 :

(VK 2

CN'12:

CN 1 3 :

CN'13:

< p

KO) (vp,U)

(p = v V

CN 1 5 :

CN'15:

CN 1 6 :

CN'16:

CN 1 7 : CN'17: CN 1 8 : CN'18: CN 1 9 : CN 2 0 :

CN'20:

CN 2 1 :

+

[K

p =

A

( V a l (Vp,w) [Ha

+

p = K

(u

Ka

a

+

A

v

i

(VK > 0 ) ( v p , V )

[K*p

=

K'V

( T a r s k i 1926;

140)

V (p i

( V a l (VP,V) [ H a * u =

+

K

V

v s K))].

K

= v

(Sobocinski 1961b;

~;v

140).

v

+

140)

-f

(Tarski 1926; 140) -+

Ka) ) I , -f

140)

(Sobocinski 1961b;

< p*).

fia+l

-+

(p i

( T a r s k i 1926;

CN 1 4 :

< p*).

K*

-+

263

(Tarski 1926; 140)

(p = V

(p =

v v

V

(u

(11 i

5

tia

A V

K

A

i K))].

v s Ha) ) I .

264

CN'21:

CN 22: CN'22: CN 2 3 : CN'23: CN 24:

CN'24:

CN 25: CN'25: CN 26: CN'26: CN 2 1 : CN ' 27 : CN 2 8 : CN'28: CN 29:

c"'29:

LIST OF SET FORMS

CARDINAL NUMBER FORMS

CN 30:

~ ~ - u pl ~ = )b 2 ] . (Val ( V K ) ( V u l f u 2 +

CN 3 1 :

vl

(VK

2

CN 3 2 ( N a ) :

CN 33:

u21.

=

< lJl

K1

< V2

A K2

-

A K1'p2

(Sobocinski 1965; 146)

-+

CN'30:

<

2 HO)[(O

(VKlrK2)(V!J1rlJ2

265

[(Na <

2 No)

u1

A Ka*p2

p2

A K <

=

K . V ~ )

(Sobocinski 1965; 146)

K O ) (311)

2 = p ).

(K

=

(VK) (Ha'K

( 3 a ) ( V K 2 Ho)

2

Ka'K

(Tarski, S i e r p i n s k i ; 147) ( A l a s 1968; 147)

).

( T h e r e i s no s - w e l l o r d e r e d s e t

c a r d i n a l numbers b e t w e e n

and

K

M & a.)

such t h a t

K~

of

M

( T a r s k i 1 9 2 4 ; Rubin 1 9 6 3 ; 1 4 7 ) CN 34:

( 3 a ) ( V K , ~ )( K * = p*

of c a r d i n a l numbers between M 2 a). CN 35:

CN 36:

(VK,p

2

2

1-1

and

K

M

such t h a t

( S o b o c i n s k i 1962a, Rubin 1985; 147)

KO) [K 5 2 (3v2) (p = v,)].

(VK,p

ordered set

T h e r e i s n o <-well

-+

KO)

p <

(K

(3Vl)

2

vl)

=

+.

( S e k i 1961; 147)

zK

<

5 p

[K

A

(3v) (p =

-f

( S e k i 1961;

K*V)].

147)

K O ) (3111(k. <2 v ) .

CN 3 7 :

(VK

CN'37:

(Val ( 3 p ) ( H a <2

CN 3 8 :

( V K r p 2 K O ) [ K <1 lJ (Sudan 1938; 1 5 0 )

CN'38:

( V a ) ( V v ) ( ~ * k= v * K ~ +V ~ " . H a a 150)

CN 39:

(VK

2

2

Ho) ( V p , v )

( T a r s k i 1954; 1 5 0 )

p).

(K

(Tarski 1954; 150)

-+

< p

A K

<1

= V * p V V'K

(VV) (V*K

V'u)].

<1 V * K & + ~ ) . ( S u d a n 1 9 3 8 ;

<3 v

+

v

2 p).

+

K).

(Sobocinski

196213; 1 5 0 ) CN'39:

Let

( V K 2 No) ( K * 1962b; 150)

<

K*

+

K**

K

-+

K

< p).

5 K*

(Sobocinski

v > 1 be a c a r d i n a l ,

CN 4 0 ( v ) : CN 41:

(VK,lJ)

(Kv

<

l.Iv

-+

(311 > 1) ( V K ~ , K [~( V) v ) (1 < v ( T a r s k i 1926; 151)

( T a r s k i 1926; 151) p

-+

K:

<

KV ~

)

-+

K~

<

K ~ ] .

266

LIST OF SET FORMS

CN 42:

(VK,~,")

CN'4Z:

( V K , ~ )(Va) ( K # 0 151)

CN 43:

( V K , ~ ) [If ~

CN'43:

# 0

(K

* =p

=

(K

n

K

K,

-+

n

Kcr)

U

* (K

e x i s t s and

U p

K

exists then

Nu

( T a r s k i 1926;

Ma).

p <

( T a r s k i 1949, Rubin 1963; 153)

U p)].

K

( T a r s k i 1926; 151)

p < v).

e x i s t s then

K

( V K ) (Va) [ I f K.8,

+

<

A K'

fl ' ) - ( K

(K

< K"

A K'

Q )1

u

e x i s t s and ( T a r s k i 1949, Rubin 1963;

.

K

&,

153) CN 44:

+

(VK,p 2 K O ) [e(K+p) = e ( K )

( A l a s 1968, 1970;

e(p)].

154)

CN'44: CN 45:

( V K 2 K O ) (VCC) [ e ( K 1970; 154)

KO)

2

(VK,'

(e(K)

+ H,)

= e(K)

+

= e(w) A K 5 p

-f

N ],

K

.

( A l a s 1968,

( A l a s 1968,

= p).

1970; 154) CN 46:

(VK,p z k o ) [e(K+p) =

v e(K+p) = e ( p ) ] .

e ( K )

(Alas

1968, 1970; 154) CN'46:

( V K 2 NO) ( e ( K 154)

CN 47:

( V s # 8 ) ( i t )( t 1939; 155)

CN 48:

( V s # 8 ) ( 3 t ) ( t k {x 1939; 155)

CN 49:

+

[ e ( K ) ] * ) = [ e ( ~ ) ] * ) . ( A l a s 1968,1970;

= {x C -

t: I s /

b

1x1)).

( T a r s k i 1938,

5

t: I s (

b

1x1)).

( T a r s k i 1938,

2 K O A P 5 It1 * I{X ( T a r s k i 1938,1939; 155)

(Vt)(vD)[ltl

.

I t 1'1

5

t : p f' I X I ) I =

CN 50: ( T a r s k i 1926,1948; 158)

CN 51:

If

is a s e t and

J

domain

J

and

A

B

are f u n c t i o n s w i t h

then

(Konig 1905, Zermelo 1908; 158) L" 5 2 :

If

is a s e t ,

J

1

I

,

jc,T

A ' '

3 '

1

x

i s a f u n c t i o n w i t h domain

A

j E J, 2

f o r each

i

Ajl

j€J

lAi/ J

J

and

then

( K o n i g 1905, A . L .

Rubin;

159)

FORMS FROM L O G I C

267

FORMS FROM L O G I C LOG

1: A formula having a model in a set of cardinality K also has a model in a set of cardinality p if N 5 2 K. (Vaught 1956, Tarski 1957, Malcev 1936, 0 Henkin 1949; 163)

LOG 2 :

A formula having a model in a set of cardinality No also has a model in a set of cardinality K , for all K 2 No. (Vaught 1956, Tarski 1957, Malcev 1936, Henkin 1949; 163)

LOG 3:

If Q is a set of formulas in which the set of individual constants has cardinality K and every finite subset of Q has a model, then Q has a model in a set whose cardinality is not greater than K + No. (Vaught 1956, Tarski 1957, Malcev 1936, Henken 1949; 163)

LOG 4:

If

A

is a set of sentences of first-order logic and

B 5 A is a consistent subset, then there exists a 5-maximal consistent set C such that €3 5 C c - A. (Klimovski 1956; 165) FORMS FROM TOPOLOGY

TOP 1: The product of compact spaces is compact in the product topology. (Tychonoff 1935, Kelley 1950; 167) TOP 2:

The product of mutually homeomorphic compact spaces is compact in the product topology. (Tychonoff 1935, Ward 1962; 167)

TOP 3 :

The product of a set of mutually homeomorphic topological spaces {(Xi,Ti): i E 11, where for each i E I, Ti has three elements, is compact in the product topology. (Tychonoff 1935, Alas 1969; 168)

T O P 4:

~ n y perfect onto map from one topological space to another has an irreducible restriction to a closed subspace of its domain. (Franklin & Thomas 1971; 169)

LIST OF SET FORMS

268

TOP 5 ( P ) :

TOP 6 ( P , Q ) :

B 2':

Each T 2 (Hausdorff) space has the property P, where P is any one of the properties given in the list on pp 1 6 9 - 1 7 0 . (Brunner 1 9 8 3 ; 1 6 9 ) Each T 2 space which has the property P also has the property Q, where P and Q are chosen from the list on pp 1 6 9 - 7 0 and (i) Q = W I, P f W 1; (ii) Q = W 2 , P # W 1, W 2 ; (iii) Q = D 1, P # W 1 W 2 , D 1; (iv) Q = D 2 , P = B 1, B 2 , B 3 , L 2 , A 2 ; (v) Q = B 1, P # W 1, W 2 , B 1, S , C (vi) Q = B 2, B 3 , P = D 2 , L 1, L 2 , A 1, A 2 ; (vii) Q = L 1, L 2 , P = D 1, D 2 , B 2 , B 3 , A 2 ; (viii) Q = S, C , P = D 1, D 2 , B 2 , B 3, A 2 ; (ix) A = A 1, P = D 1, D 2 , B 2 , B 3 , A 2 ; and (x) Q = A 2 , P = D 2. (Brunner 1 9 8 3 ; 1 6 9 )

Each topological space X has the property that each point of X has a neighborhood base which can be well ordered and the well ordering is compatible with set inclusion. (Howes 1 9 8 0 , Vaughan 1 9 7 5 ; 1 7 6 ) FORMS FROM ANALYSIS

AN 1:

The unit sphere in the dual of a normed vector space over the reals has an extreme point. (Bell & Fremlin 1972; 179)

AN 2 :

Let X be a real vector space, Y a subspace and S a subset of X. Suppose p:X +IR is a sublinear functional on X and f:Y + l R a linear functional such that f ( y ) I p(y) for all y E Y. Then the set Z of all p-dominated linear extensions of f to X has an element g such that for a l l h E Z and g ( s ) 2; h(s) f o r all s E S, we have g(s) = h(s) for a11 s E S. That is, g is S-maximal in Z. (Andenaes 1 9 7 0 , Lembc1:e 1 9 7 9 ; 1 7 9 )

F9RMS FROM ANALYSIS

269

AN 3 :

Let X be a real vector space and Y a subspace. Suppose p:X +IR is a sublinear functional on X and f : Y +IR a linear functional such that f(y) 2 p(y) for all y E Y. Then the set Z of all p-dominated linear extensions of f to X has an extreme point. (Bonsall 1954, 1956, Lembcke 1979; 179)

AN 4:

Let X and Y be T 2 spaces, f:X + Y a continuous map, and U an (inner) regular Borel measure on Y. If 1-1 is inner regular with respect to the system of sets {f"K: K is a compact subset of X I , then there is a regular Borel measure V on X such that ?J = vf. (Lembcke 1970, 1977, 1979; 181)

AN 5:

Let X be a real vector lattice with a distinguished order-unit (denoted by l), Y a sublattice of X , and Z the set of all positive linear functionals F on X such that F (1) = 1. Then there exists an extreme point M of the convex set Z which is Y - minimal in Z. (Edwards 1975; 183)

AN 6:

If (L,s) is a Boolean algebra with identity 1, and K is a sublattice of L then there exists a two-valued K-minimal measure 1-1 on L such that U(1) = 1. (Lembcke 1970,1977,1979; 184)

Each of the following is equivalent to the axiom of choice:

AL + KM, ALC + KMC, BPI 1972; 181)

+

KM,

and

HB

+

KMC.

(Bell

&

Fremlin

This Page Intentionally Left Blank

LIST O F THE CLASS FORMS

In this list, in parentheses following each statement we shall give the page number where it occurs in the text. THE WELL ORDERING THEOREM CWO 1: There is a function F such that for each set F(x) well orders x. (187)

x,

CWO 2 :

There is a relation R ordered by R. (187)

CWO 3 :

Every class can be well ordered. (Equivalently, the universe, V, can be well ordered.) (187)

CWO 4:

There is a relation R which well orders V in such a way that each proper initial R-segment of V is a set. (187)

CWO

Let

5:

such that every set is well

Each proper class is equipollent to V = On.) (187)

m E w

such that

On.

(Equivalently,

m z 1.

CWO 6(m):

For each class X there is a function F defined on O n such that for each ci E On, F(c1) < m and F(c1) = X. (187) c1 €On There is an m E w {O) such that CWO 6(m). (187)

u

CWO 7: CWO 8:

-

-

For each class X there exists an m E w {O} function F defined on On such that for each ci E On, F ( a ) < m and IJ F ( a ) = X. (187) aEOn

271

and a

LIST O F THE CLASS FORMS

272

Every c l a s s

CWO 9 :

i s t h e u n i o n of a w e l l o r d e r e d c l a s s

X

of f i n i t e sets. CWO 1 0 :

Every p r o p e r c l a s s c a n be mapped o n t o proper c l a s s

CT:

(187)

<*

XI V

V.

(188)

X.)

Each p a i r of p r o p e r c l a s s e s are e q u i p o l l e n t .

CWO 11:

For each proper c l a s s

CWO 1 2 :

P(0n)

(For every

X , ?(X)

can be w e l l prdered.

<

X.

(188)

(188)

(188)

The power c l a s s of a w e l l o r d e r e d c l a s s c a n be w e l l

CPW:

ordered.

(188)

CLW:

Every l i n e a r l y o r d e r e d class c a n b e w e l l o r d e r e d .

CCW:

Every c l a s s on which t h e r e i s a c h o i c e f u n c t i o n c a n b e w e l l ordered. (188)

(188)

THE AXIOM OF CHOICE CAC 1:

If

S

i s a c l a s s of non-empty sets, t h e r e i s a

function

F

such t h a t f o r each

x E S, F(x) E x .

(191) E:

There e x i s t s a f u n c t i o n

set CAC 2:

F(x) E x .

x, If

T

F

s u c h t h a t f o r e a c h non-empty

(191)

i s a p a i r w i s e d i s j o i n t c l a s s of non-empty sets,

there is a c l a s s

which c o n s i s t s of one and o n l y

C

one e l e m e n t from e a c h set i n CAC 3 :

For w e r y f u n c t i o n

F

t h a t f o r every

if

G ( x ) E F(x)* CAC 4 :

(191)

there i s a function x E $(F)

and

G

such

F ( x ) # JJ

then

(191)

For e v e r y r e l a t i o n that

CAC 5:

x,

T.

a ( F ) = $(R)

R

and

t h e r e is a f u n c t i o n F

5

R.

F

such

(192)

For every f u n c t i o n F t h e r e i s a f u n c t i o n G such t h a t $(GI = %(F) and f o r e v e r y x E a ( G ) , F ( G ( x ) ) = x. (192)

THE AXIOM OF CHOICE

273

CAC 8 :

If S is a class of pairs of equipollent sets then there is a function which associates with each pair a bijection mapping one onto the other. (192)

CAC 9 :

Let H be a relation such that for each x, y E a(R), R"{x) k R " { y ) ; then there exists a relation S such that for each x, y E a ( R ) , S"{(x,y)} is a bijection (192) mapping R"(x] onto R"{y).

Let

n

2

be a natural number.

2

CAC lO(n):

If S is a class of infinite sets, there is a function F such that for each x E S , F(x) is a decomposition of x into a set of disjoict sets of size between 2 and n. (192)

CAC 11: There exists a natural number CAC 10(n). CAC 12:

Let

n

n

such that

2

2

(192)

If S is a class of infinite sets there is a natural number n 2 2 and a function F such that for each x E S, F(x) is a decomposition of x into a set of disjoint sets of size between 2 and n. (192) 2

be a natural number.

2

CAC 10E(n):

If S is a class of infinite sets, there is a function F such that for each x E S, F(x) is a decomposition of x into a set of disjoint sets of size exactly n. (193)

CAC 11E:

There exists a natural number CAC 10E(n). (193)

CAC 12E:

If S is a class of infinite sets there is a natural number n B 2 and a function F such that for each x E S, F(x) is a decomposition of x into a set of disjoint sets of size exactly n. (1931

Let

m

2

n

2

2

such that

1 be a natural number.

CAC 13(m):

If S is a class of non-empty sets there is a function F such that for each x E S, F(x) is non-empty subset of x with F(x) < m. (193)

il

274

CAC 14:

LIST OF THE CLASS FORMS

There is a natural number

m

2

1

such that

CAC 13(m).

(193) CAC 15:

If S is a class of non-empty sets then there is a natural number m 2 1 and a function F such that for each x E S , F(x) is a non-empty subset of x (193) with F(x) 6 m.

CAC 18:

For any non-empty class whose domain is A and each a E A,

x E

u

n

a€A bEBa

domain

A

-

Xarb

A, and for every function B Ba is a non-empty set for

There is a function

F

with

such that (Va E A) [F(a) E Ba A x E Xa,F (a)

(193) CAC 18C:

For any non-empty class A, for every class-valued function B whose domain is A, x E

n

aEA

domain

IJ XaIb * There is a function bEB [a1 A

such that (Va E A) [F(a) E B[a]

F

with

A

CAC 19:

For any non-empty class A, each of whose elements is non-empty, x E fl (J b *There exists a function F with domain aEA bEa A such that (Va E A) [F(a) E a A x E F(a) I. (194)

CAC 2 2 :

If S is a class of non-empty sets then there is a function F such that f o r each x E S r F(x) is a (194) non-empty finite subset of x.

CAC 23:

There exists a function F such that for each nonempty set x, F(x) is a non-empty finite subset of x. (194)

*

MAXIMAL PRINCIPLES CAC 24:

CH:

275

If S is a class of infinite sets, then there is a function F such that for each x E S, F(x) is a partition of x into pairwise disjoint finite sets, (194) each of which has at least 2 elements.

There is a formula such that

ax) ( V Y ) ( Y

# B

$

+

with at least three free variables (IZ)+(Z,X,Y) f Y ) .

(198)

MAXIMAL PRINCIPLES CM 1:

If R is a partial ordering relation on a non-empty class X and if every subclass of X which is linearly ordered by R has an R-upper bound, then X has an (203) R-maximal element.

CM 2 :

If R is a partial ordering relation on a non-empty class X and if every subclass of X which is well ordered by R has an R-upper bound, then X has an R-maximal element. (203)

CM 3:

If every non-empty nest which is a subclass of a nonempty class X has its union an element of X I then (203) X has a 5-maximal element.

CM 4 :

If every non-empty well ordered nest which is a subclass of a non-empty class X has its union an element of X, then X has a c-maximal element. (203)

CMS l(2):

If R is a partial ordering relation on a nonempty class X and if every subset of X which is linearly ordered (well ordered) by R has an R-upper bound then either X has an R-maximal element or X has a subclass which is a proper class and which is linearly ordered (well ordered) by R. (203)

CMS 3(4):

If every non-empty nest (well ordered nest) which is a subset of a non-empty class X has its union an element of X I then either X has a 5-maximal element or there is a subclass of X which is a nest (well ordered nest) and a proper class. (204)

276

L I S T O F THE CLASS FORMS

CM l ( Q I U ) :

c l a s s w h i c h i s Q - o r d e r e d by R i n s u b c l a s s has an R - u p p e r bound,

E v e r y non-empty

which e a c h U - o r d e r e d

Q = F, R, P , or

h a s an R - m a x i m a l e l e m e n t , w h e r e

T R , and U = W , L , A S & C , T R & C , C , o r D ; Q = D

(204)

U = W.

c l a s s w h i c h i s Q - o r d e r e d by R i n w h i c h each U-ordered s u b s e t h a s a n R - u p p e r bound, e i t h e r h a s an R - m a x i m a l e l e m e n t or c o n t a i n s a U-ordered s u b c l a s s w h i c h i s a proper c l a s s , w h e r e Q = T R , P , RA, o r F and U = C , A S & C , (204) T R & C , L , D , o r W ; Q = D and u = W.

CMS l ( Q , U ) :

CM 5:

E v e r y non-empty

I f R i s a p a r t i a l o r d e r i n g on X , then there e x i s t s a 5 - m a x i m a l s u b c l a s s of X w h i c h i s l i n e a r l y ordered by

R.

(205)

E v e r y c l a s s w h i c h i s Q - o r d e r e d by

CM 5 ( Q l U ) :

R

c - m a x i m a l s u b c l a s s which i s U - o r d e r e d Q = A

or

D, or RA;

TR

o r RA;

Q = D

or

Q = C

P

and

U = A S , A S K , P I L , D,

and

U = C , A S & C , T R & C , L , o r RA;

and

of CM 7 :

U = C , A S & C , T R & C , L , D, o r

For e v e r y c l a s s X

D;

X

or Q = F

(205)

W.

X I there e x i s t s a 5 - m a x i m a l

w h i c h is a nest.

For e v e r y c l a s s

R, w h e r e

U = C , A S K , TR&C,

Q = RA and U = C , AS&C, TR&C, L , o r

and

contains a by

U = A S , C , ASLC, T R K , P , L,

and

Q = AS

L , D , o r RA;

CM 6 :

and

subclass

(205)

of f i n i t e s u b c l a s s of X

and every p r o p e r t y

character, these e x i s t s a c-maximal P. (205)

P

w h i c h has t h e p r o p e r t y CM 8 ( A ) :

E v e r y c l a s s has a c-maximal

property

A, w h e r e

A = D,

subclass w i t h t h e K, J,

K, 3 or

D ~ m,

(205) CM 9 ( A ) :

E v e r y c l a s s can be l i n e a r l y ordered and CM 8 ( A ) , where

CM 1 0 :

A = D,

K, J,

z,

3 , or

T h e r e exists a natural number CM 8 ( D m ) .

(205)

m

Dm, m

;r

2

2

2.

(205)

such t h a t

2

2.

MAXIMAL PRINCIPLES

CM 11:

For e v e r y c l a s s m z 2

t h e r e e x i s t s a n a t u r a l number

X

such t h a t

h a s a _c-maximal s u b c l a s s which h a s

X

Dm.

t h e property

(205)

For e v e r y r e l a t i o n

R

c-maximal s u b c l a s s

Y

CMR 8 ( A ) :

A = D,

KJ,

El

KU,

and c l a s s

5 X

x # y

if

x, y E Y ,

277

J, 5, K , or JU.

then

2, Dm,

there e x i s t s a

X

such t h a t f o r a l l

-

R"{x) A R"{y),

-

D m f D!mf

U,

El

where

DK,

D J , DU,

(218)

The u n i v e r s e can be l i n e a r l y o r d e r e d and f o r e v e r y

CMR 9 ( A ) :

relation

R

subclass

Y

and c l a s s

5

X

t h e r e e x i s t s a 5-maximal

X

such t h a t f o r a l l

x, y E Y,

if

x # y t h e n R"{x} A R " t y } , where A = D , 5, J , j , K, K, D ~ , D ~ D!,, , u , 6, DK, DJ, DU, KJ, K U , o r JU

-

-

(218) CMR 1 0 :

There e x i s t s a n a t u r a l number CMR 8 (D,)

CMR 11:

.

m

2

2

such t h a t

(219)

For e v e r y r e l a t i o n R and c l a s s X t h e r e i s a n a t u r a l number m 2 2 and a 5-maximal s u b c l a s s Y 5 X such t h a t f o r a l l y 1 , y 2 , . . . , y m E Y , y i ' s d i s t i n c t , m

CMR 1 2 :

There e x i s t s a n a t u r a l number CMR 8

CMR 13:

(E!m).

2: 2

such t h a t

(219)

For every r e l a t i o n R n a t u r a l number m 2: 2 Y _C X

m

and c l a s s X t h e r e e x i s t s a and a 5-maximal s u b c l a s s

such t h a t f o r a l l

y1,y2,...,ym

E

YI

m

CMR 14:

i s a c l a s s and 5-maximal s u b c l a s s Y If

X

numbers

m

m

2

2

R

a relation there e x i s t s a

5X

and f o r all

such t h a t f o r a l l n a t u r a l y1fy2f...fym E Y ,

LIST OF THE CLASS FORMS

278

CM 1 5 ( A : B ) :

Any c l a s s which c o n t a i n s a 5-maximal s u b c l a s s w i t h t h e property

c o n t a i n s a 5-maximal s u b c l a s s

A

with the property

B , where

(The p o s s i b l e v a l u e s f o r

5,

K

K).

and

A

A

#

and

B

and

B

B

# E. D,

are

5,

J,

(230)

For any r e l a t i o n R and c l a s s X I i f t h e r e e x i s t s a 5-maximal s u b c l a s s Y 5 X such t h a t AR[Y]

CMR 1 5 ( A : € 3 ) :

t h e n t h e r e e x i s t s a 5-maximal s u b c l a s s

5X

2

s u c h t h a t BR[Z], where A # B and i t i s n o t t h e c a s e t h a t A = J K and B = E l o r A = J K and B = U. (The p o s s i b l e v a l u e s f o r A and B a r e D, D, J, 3, K, K, u, ij, DJ, DK, DU, JK, JU,

-

KU.)

and CM 1 6 ( A : B ) :

(230)

Every c l a s s can be l i n e a r l y o r d e r e d and CM 1 5 ( A : B ) , where

and

A

CM 1 5 ( A : B ) . CMR 1 6 ( A : B ) :

have t h e same v a l u e s a s i n

B

(230)

Every c l a s s can be l i n e a r l y o r d e r e d and CMR 1 5 ( A : B ) ,

where

A

and

v a l u e s a s i n CMR 1 5 ( A : B ) . CM 1 7 :

Every c l a s s

t E s. CA:

have t h e same

(230)

c o n t a i n s a 5-maximal s u b c l a s s

X

that for a l l

B

s, t

E

Y,

s # t,

either

s E t

Y

such

or

(205)

Every p a r t i a l l y o r d e r e d class c o n t a i n s a _c-maximal a n t i chain.

(205,212)

LIST OF FORMS RELATED TO THE AXIOM OF CHOICE

In the text we mention a number of statements related to the axiom choice, some weaker, some stronger and some whose relative strength is unknown. We shall list them in the six categories described below.

I.

This list of statements has two parts, (a) those statements such that AC + S in NBG and S -+ AC in NBGO; and (b) those statements S such that AC -t S in NBGO and S AC in NBG. Each of these statements have been discussed in the text,so here we shall give its name followed in parentheses by the page number where it occurs in the text. S

+

(a)

1.

(73)

2.

PROJ

3.

M 1 9 (87) AC 2 0 ( 8 8 ) AC 2 1 ( 8 8 ) (Va)DCC(a) ( 8 9 )

4.

5. 6.

(b)

INJ

(74)

1. PW

(73)

2.

LW

(73)

3.

7.

MC 1 ( 7 4 ) MC 2 ( 7 4 ) MC 3 ( 7 4 ) M 8(i?) (74) M 15(A:E), for

8.

ACQ,

for

DLO.

(78)

4.

5. 6.

P

A = D,

Dl

J,

5,

K.

(77)

P = j3, Q = LO, DLO; P = LO, Q =

279

0 , LO,

LIST O F FORMS RELATED TO THE AXIOM OF CHOICE

280

9.

P MCQ, for

P

=

j3, Q = LO, DLO; P = LO, Q =

8 , LO, DLO.

(78)

10.

ORD

11.

W-ORD

12.

TCL ( 7 8 )

13.

INF 2

14.

Up

(78)

(78)

(84)

for

P = LO, Q = LO, WO; P = WO,

Q = LO.

(85)

15.

9' M 5(UIVC), for

16.

M 5(C,UN)

17.

M 5(U,VC)

18.

T 11

19.

T 12

(90)

20.

T 13

(90)

21.

B

22.

AL 2 0 ( F )

(119)

23.

AL 2 1 ( F )

(119)

24.

B'

25.

AL 2 5

26.

TOP 6 ( P , Q ) , where P = D 2 , L 1, L 2 , A 1 or A 2 and Q = D 1, B 2 or B 3 ; and P = D 2 and Q = A 2.

U = AS&TR, AS, TR.

(86)

(86)

for

U = RA, D.

(86)

(90)

(119)

(121)

(128)

(169I 1 7 2 ) Q = D 1, B 2, B 3, or

27.

TOP 5 ( Q ) , where

28.

TOP 6 ( P , Q ) , where P = W 2 , B 1, or S and Q = W 1; and P = B 1 and Q = W 2 , D 1 or D 2. (169,173)

29.

TOP 6(S,W 2 )

30.

TOP 6 ( P , Q ) , where P = B 2 , B 3 , S , or C

and P = C

and

A 2.

(169,172)

(169,173) Q = W 1 or

W 2.

and Q = D 1;

(169,174)

LIST OF FORMS RELATED TO THE AXIOM OF CHOICE

31.

TOP 5 ( D 2)

32.

AN

4

281

(169,174)

(181)

11. In this set we shall list statements S such that AC implies S in NBG', but it is not known if S implies AC in NBG. In parentheses following each statement we shall give the page number where it is mentioned in the text. 1.

(The Nielson-Schreier Theorem). group is free. (xiii)

2.

(Basis Theorems for Vector Spaces). (a) There exists a field F such that every vector space over F has a basis. (125)

Every subgroup of a free

(b) There exists a field F such that for every vector space V over F, if B is a basis of V and X any linearly independent subset of V, then there is a subset €3' _c B such that X n B' = 0 and X U B ' is a basis of V. (125) either p(x) < p(y)

3.

For all sets x and y RY)< B(x). (xiii)

4.

For all infinite cardinals

5.

(The Partition Principle). For each set P of x, P < x. (xiii,7)

6.

(T 3 ( ~ ) ,for K an infinite cardinal). If K is an infinite cardinal, then for every set of cardinals x such that 1x1 = K there exist vl, p 2 E x such that 1-11< 1-12 or

7.

1-12 < I l l .

K2

=

K

3

.

(xiii) x

and partition

(22)

+ Reflexive)) . For any set x and relation R there exists a 5-maximal subset y 5 x such that R is (M 5 (A,TR

transitive and reflexive on 8.

K,

or

y.

(41,44)

If B is a Boolean algebra and S _C B such that S is closed with respect to A (meet) then there is a _c-maximal proper ideal I such that I n S 5 {O}. (101)

282

9.

LIST OF FORMS RELATED TO THE AXIOM OF CHOICE For each cardinal K 2 No there is a cardinal 1-1 such that K < p and (Vv) ( v < p * v s K ) (Ie. K < 3 11). (150)

10.

For each pair of cardinals bound of K and p , K n p ,

11.

(TOP 6 ( w 2,Q)). Each T 2 space which has the property W 2 also has the property Q, where Q = B 1, B 2 , L 1, L 2 , A 1, or A 2. (169,176)

K

and p , the greatest lower exists. (153)

111. The statements in this list are either equivalent or implied by the Boolean Prime Ideal Theorem (BPI). None are provable in NBG. In parentheses following each statement we shall write E, if it is equivalent, and I, if it is implied by BPI, followed by the page number where the statement appears in t h e text. The statement 1. below is BPI.

1.

(BPI) Every Boolean algebra contains a proper prime (5-maximal) ideal. (E; xii,99)

2.

(BPI') If I is a proper ideal in a Boolean algebra then I can be extended to a proper prime (5-maximal) ideal. (E; 101)

3.

If B is a Boolean algebra and S 5 B such that 0 S and S is closed with respect to A (meet) then there is a proper prime (5-maximal) ideal disjoint from S . (E;101)

4.

Every distributive lattice with an identity element and at least one other element contains a prime ideal. (E; 101, 102)

5.

(The Tychonoff Compactness Theorem for Hausdorff (T 2 ) spaces.) The product of compact T 2 spaces is compact in the product topology. (E;167,168)

6.

(The Hahn-Banach Theorem, HB) Let p be a real-valued function defined on the vector space X over the reals IR such that p is a sublinear functional. Suppose f is a linear functional defined on a subspace S 5 X such that

LIST OF FORMS RELATED TO THE AXIOM OF CHOICE

f(s) 5 p(s) for all s E S. functional F defined on X all x E X and F(S) = f(s)

7.

283

Then there is a linear such that F(x) 5 p(x) for for all S E S. (I;177)

(Alaoglu's Theorem) (a) (AL) The unit sphere in the dual of a normed vector space over the reals is compact in the weak* topology. (E; 178) (b) (ALC) The unit sphere in the dual of a normed vector space over the reals is convex-compact (quasi-compact) in the weak* topology. (I (ALC * HB); 178)

8.

If (L,s) is a Boolean algebra with identity 1, then there exists a two-valued measure p on L such that p(1) = 1. ( E ; 184)

IV. Each of the statements in the list below is implied by the 2~ = K Principle which is the first statement in the list. As far as we know none are provable in NBG. In parentheses following each statement we give the page number where it occurs in the text. 1.

2.

3.

4.

Each of the statements listed below is implied by the axiom of choice in NBGO, but does not imply the axiom of choice in NBG. As far as we know, none are provable in NBG. In parentheses following each statement we give the page number where it appears in the text. V.

LIST OF FOKMS RELATED TO THE AXIOM OF CHOICE

284

1.

(The Principle of Dependent Choices, DC) If R # @ is a relation such that %(R) _c a ( R ) then there is a function f with domain such that for all n E w, (f (n),f (n+l)) E R. (19)

2.

(The Kinna-Wagner Principle, KW 1

-

KW 2)

(a) (KW 1) If x is a set of sets such that for each u E x, lul B 2 then there is a function f such that for each u E x , @ # f(u) C u. (83) (b) (KW 2) For each set x there is an ordinal c( such that x < ~ ( c c ) . ( 8 3 . Also see statement (1) on p 25). 3.

4.

5.

6.

Every set can be linearly ordered.

(54,651

P (The Axiom of Choice for Well Ordered Sets, ACQ. ) For each set x, if x is a P-ordered set of non-empty sets such that each u E x can be Q-ordered, then there is a function f such that for each u E x, f (u) E u, where P = WO and Q = /a, LO, DLO, or WO; and P = /a or LO and Q = WO. (78,851 P (The Multiple Choice Axiom for Well Ordered Sets, MCQ). For each set x, if x is a P-ordered set of non-empty sets such that each u E x can be Q-ordered then there is a function f such that for each u E x, f (u) is a non-empty finite subset of u, where P = WO and Q = @, LO, DLO or WO; and P = 0 or LO and Q = WO. (78)

wo (Uwo)

The union of a well ordered set of well ordered

sets can be well ordered.

(85)

The following implications hold in NBGO between the statements in 4 . , 5., and 6. above.

LIST OF FORMS RELATED TO THE AXIOM OF CHOICE

285

ACWo

wo

A'DLO ~

7.

If x is a set such that for each u E x, u can be linearly ordered, then there is a function f such that for each u E x, f ( u ) linearly orders u. (82)

8.

( $ 0 ~ Theorem, ' LT) Let A and X be non-empty sets, R a binary relation on A, M = (A,R), and 7 an ultrafilter on X, then MX / y and M are elementarily equivalent.

(166) 9.

(The Krein-Milman Theorem) (a) (KM) Let K be a compact convex set in a locally convex topological vector space. Then K has an extreme point. (177) (b) (KMC) Let K be a convex compact (quasicompact) convex set in a locally convex topological vector space. Then K has an extreme point. (178)

286

LIST OF F O N G RELATED TO THE AXIOM OF CHOICE

K,+l

10.

For each ordinal

11.

(INF 1) For every set is Pinfinite, then x

a,

i

2

.

(26)

x, if every infinite subset of can be well ordered. (84)

VI. This list contains the remaining statements related to the axiom of choice and are mentioned In parentheses following each statement we shall number where it appears in the text, followed by statement describing its relative strength. 1.

x

which are in the text. give the page a short

(The Generalized Continuum Hypotheses, GCH). For each cardinal K z K O , there is no cardinal p such that K p 2K. (xii; GCH AC in NBGO, but AC does not imply GCH in NBG.) -f

2.

(The Continuum Hypotheses, CH). The cardinality of the set of real numbers is t$. (xii; GCH * CH in NBG', but AC and CH are independent in NBG.)

3.

(UP) All ultrafilters are principal. and AC * 1 U P ) .

4.

(PR) For every relation R there is a relation S 5 R such that a ( S ) = d ( R ) and for each x E $(S), S " { x } is a set. (194; NBG t- PR but PR is not provable in NBGO.)

5.

(LV) The universe can be linearly ordered. ( 2 1 6 ; A class form of the axiom choice implies LV in NBGO, but LV does not imply any set form of the axiom of choice in NBG and LV is not provable in NBG.)

6.

For each pair of proper classes X and Y, either X < Y or Y d X. (188; A class form of AC implies 6. and 6 . implies a set form of AC in NBGO, but it is not known if 6. implies a class form of AC in NBG.)

7.

If S is a class of sets of the same cardinality then there is a function which associates with each pair of elements of S a bijection mapping one onto the other. (192; A class form of AC implies 7. and 7. implies a set form of AC in NBGO, but it is not known if 7 . implies a class form of AC in NBG.)

(167;

UP

-+

LT

L I S T O F FORMS RELATED TO THE AXIOM OF CHOICE

8.

287

On can be injected into each proper class. (188; A class form of AC implies 8. in NBGO, but 8. does not imply any set form of in NBG.)

AC

in

NBG

and 8. is not provable

This Page Intentionally Left Blank

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INDEX OF AUTHORS

Author

Pages A

Abian, A.

72

Alaoglu, L Alas, 0. T.

177 29, 147, 154, 168 179

Andenaes, P. R. Arrnbrust, M. K. Artin, E.

120, 124, 127 31, 32 B

Bachmann, H.

3

Banach, S.

177

Bell, J. L. Bernays, P. A. Beth, E. W. Birkhoff, G. Blair, R. L. Blass, A. Bleicher, M. N. Bonsall, F. F.

179, 180, 181 xi, xii, xv, 10, 19, 185

Bourbaki, N. Brouwer, L.

32, 47 31

Brunner, N. Buchi, J. R.

vii, 6, 84, 122, 169, 170, 176

97 93, 103 70, 96 vii, 101, 119, 125, 128, 165, 167 83, 120, 125 179

105

305

306

Author

INDEX OF AUTHORS Pages C

Campbell, P. Cantor, G. Chang, C. C. Chevalley, C. Cohen, P. J. Collins. G. E.

31, 32 1, 7, 138 56 31 xii, 140, 157 10

D Diener, K. H. Dzik, W.

103

96, 165

E

Easton, W. B. Edwards, D. A.

xii, 185 178, 182, 183 F

Farah, E. Felgner, U. Flannagan, T . B. Fleisher, I. Fraenkel, A. Franklin, S. P. Frascella, W. Frattini, G. Frenlin, D. H. Frink, 0. Fuchs, L.

10

xii, 19, 39, 54, 73, 74, 85, 88, 89, 116, 118, 185, 188, 190 88, 185, 188 185 xii, xxvii, 54, 73, 83, 140 168 10

104

179, 180, 181 48, 103 104

G Gaifman, H. Gauntt, R. J . G8de1, K . Gottschalk, W. H . Gratzer, F. Grighin, V. N.

xii, 185 xxv, 79 xi, xii, xv, xvi, 2, 185, 191 41, 45 134 xii, 185

INDEX OF AUTHORS

Author Hahn, H. Hajnal, A. Halmos, P. R. Halpern, J. D. Harper, J. M. Harting, R. Hartogs, F. Hausdorff, F. Henkin, L. Hessenberg, M. Hilbert, D. Hodges , W. Hoft, H. Howard, P. E.

Howes, N. R.

Pages H 177 110 181 xii, 54, 65, 73, 83, 99, 120, 140, 165, 178 37, 41, 185, 205 119 xxv,16, 19, 21, 76, 111, 137 31, 46, 1 6 9 163

32 198 vii, 112 41, 134 vii, xii, 41, 44, 73, 77, 78, 79, 99, 134, 140, 166, 185, 190, 195, 204 176 I

185

Isbell, J. R.

J Janiszewski, Jech, T. J. Jensen, R.

S.

31 54, 73, 74, 190 xii K

Kelley, J. L. Kertesz, A. Kinna, W. Klimovsky, G. Kneser, H. Konig, J. Krein, M.

307

167 110 83 60, 101, 116, 118, 165 32, 47 159 177

308

INDEX OF AUTHORS

Author

Pages

Kripke, S. Krull , W. Kruse, A. H. Kuratowski, C. Kurepa, G.

Lauchli, H. Lembcke, J. Lesniewski, M. S. Levi , B , Levy, A.

Lindenbaum, A. Go;, J. LGwenheim, L. Luxemburg, W. A. J.

K xii 104, 111 22, 24, 90, 91 31, 32, 36, 47, 83 54, 62, 65, 74, 221

xiii 178, 181 14 1 7 xii, xxv, 1, 2, 10, 19, 56, 57, 74, 83, 84, 99, 151, 165, 178, 198 xii, 21, 146, 152, 159 166, 168, 178 163, 164 178 M

Malcev, A. Mazurkiewicz, E. McCoy, N. H. Millman, D. Moore, R. L. Mostowski, A. Mrowka, A.

163

31 104 177 32 xii, xxvii, 54, 73, 83, 84, 140 98 N

Neumann, B. H. Nielson, J.

104 xiii

P Peano, G. Pincus , D

.

7 10, 178

I N D E X OF AUTHORS

Author Royden, H. L. Rubin, A. L.

Pages R 178

viii, 62, 1 5 3 , 1 5 9 , 1 8 5 , 1 9 0 , 195,

Rubin, H.

204,

10, 62, 178,

Rubin, J.

309

214,

71,

75,

225,

229,

243

102, 147, 168,

185

10, 37, 41, 147,

219,

74,

185,

Russell, B.

xi, 9, 1 0

Ryll-Nardzewski, C.

168,

7 3 1 7 5 1 7 7 1 19, 8 5 1

190, 195,

204,

205

178

S

Sageev, G. Schmidt, E.

xii, 1 4 0 , 1 5 0 9

Schmidt, J.

93,

Schreier, 0 .

xiii

Scott, D. S .

xxvi, 93,

104 96,

102, 112,

156, 168,

178

Seki, S.

1 47

Shoenfield, J. R. Sierpinski, W.

xii

Skolem, T. Sobocinski, B.

163, 164

Solovay, R. Specker, E. Sudan, G.

xii xii

Szele, T. Szpilrajn-Marczewski, E.

xii, 1 0 , 21, 10,

21,

29,

141, 146, 141,

146,

147 147,

150

150 3 2 1 47 221

T

Tarski , A.

xii, 5, 10, 1 9 , 2 1 , 2 2 , 84, 139, 141,

146,

147,

150,

153,

151,

159,

163

Teichmuller , 0.

32

Thomas, B. V. S.

168

155,

138,

152,

310

INDEX OF AUTHORS

Author

Pages T

Tomber, M. L.

70 78, 82, 84, 150, 151

Truss, J. Tukey, J. W. Tychonoff, A.

Vaughan, J. E. Vaught, R. von Neumann, J.

32 167

176 54, 163, 198 xi, xv, xxvi, 74, 185, 188 W

Wagner, K. Wallace, A. D. Ward, L. E. Weston, J. D. Whitehead, A. N. Wright, F. B.

83 41 10, 167 50 10 185 Z

Zermelo, E. Zoretti, L. Zorn, M.

xi, xxvii, 1, 9, 10, 12, 48, 159

31 31, 32

INDEX OF SYMBOLS

Page -

xv xv xvi xvi xvi xvi xvii xvi i xvi i xvii xviii xviii xix xix xix xix xix F:X

-f

Y,

F:X

onto

> Y

xx xx xx xx xx xx xxii xxii xxiii

F:X 6 Y X <* Y, F:X < * Y X k Y, F:X .= Y X 6 Y,

F [XI Ro S (X,R) (X,R)

(Y,S)

w, O n 311

312

I N D E X O F SYMBOLS

Page -

xxiii xxiii xxiii xxiv xxv xxv xxvi xxvi xxvi xxvi xxvi 24 29 31 38 43 45 50 53 56 51 60

f .i.p. K* -+ t

62

I-

13

NBG

71

LO, DLO

D-infinite DWO

WO,

18 84 85

vc

86

T

90 91

(K)

cof ( A ) (A,(Ok:k

E K))

a V b , a ~ b 0,

1

93 94

94

313

INDEX OF SYMBOLS

Symbols

Page

ha, V a a€A a€A

94

C-closed

95 97

R: X

97

-*a

98

(u)

105 115 150

c W

<2’ < 3

C1I

P<,(X),

154

e(K)

W 1, W 2, D 1, D 2, B 1, B 2

169

B 3, L 1, L 2, S,

170 177

C,

A 1, A 2

llx II Y-minimal K-minima 1

u,

v

DK, DJ, DUI KJ, K U I JU PR [ X I m

A CMR 8(Cm) i= 2 A ( s ) -maximal

183 184 218 218 224 225 235

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INDEX OF TECHNICAL TERMS

Page -

Technical Terms A Abelian (commutative) groupoid Absolutely dispensible Addition (cardinals) Alaoglu ' s theorem (AL) Aleph Algebra Algebraic lattice Algebraic system Ant ichain Anti-symmetric Atoms

109 103 xxvi 178 xxvi 93 105

93 74 xx i xv

B Basis (algebraic system) Basis (vector space) Bijection Boolean Algebra Boolean Prime Ideal Theorem (BPI) Bore1 Measure

103 119 xx 98 99, 101 181

C 95 109 xxv xix

C-closed Cancellative groupoid Cardinal number Cartesian product

315

316

INDEX OF TECHNICAL TERMS

Technical Terms

Page

C 110 183

Characteristic (field) Characteristic function Choice function Circular Class Class difference Class valued function Cofinality index Commutative ring Commutator Commutator subgroup Compact (lattice) Compact space Compactification, one point Complement (class) Complement (lattice) Complement (Relation) Complete Complete (lattice) Composite product Connected Consistent Convex-compact

6

216 xv xvii xx 91 110 117 117 95

167 171 xvii 98 43 48

95

xx xxi 165 178 D

Dedekind-infinite Dependent (algebraically) Dependent choice, principle (DC) Direct product Directed relation Discrete topology Disjoint, pairwise disjoint Divisible (group) Division (cardinals)

84 103 19, 89 xix 34, 38 171 xvii 127 xxvi, 144

INDEX OF TECHNICAL TERMS

Technical Terms

317

Page

D

xvii 178

Domain Dual (vector space) E

xvi xvi xx xx xxvi xvi 177

Element Empty set axion Equinumerous Equipollent Exponentiation (cardinal) Extensionality, axiom of Extreme point F f-chain Field Filter, proper (lattice) Finitary closure operator Finite Finite character Finite complement topology Finite intersection property (f.i.p.) First (R-first) Fixed-point theorems Forest Foundation, axiom of Free abelian group Free group Free union (topology) Function, 1-1 function

50

110 94

95 xxiii, 5 33

176 60 xxi 72 38 xviii 127 116 168 xx

G Generators (algebraic system) Greatest lower bound (lattice) Greatest lower bound (K-greatest lower bound)

102 94

xxi

318

INDEX OF TECHNICAL TERMS

Technical Terms

Page G

109 109

Group Groupoid H Hahn-Banach Theorem (HB) Hartogs' function Homeomorphic Homomorphism (group)

177 xxv 167 127

Ideal, proper (integral domain) Ideal, proper (lattice) Identity element Image Immediate successor (cardinals) Independent (algebraically) Indeterminates Indexed classes Infimum Infinite Infinity, axiom of Initial ordinal Initial R-segment Injection Integral domain Intersection Inverse Irreducible (integral domain) Irreducible (topology) Irreflexive Isomorphism (group) Isomorphism (partially ordered sets)

112 94 94 xix 150 103 120 xix xxi xxiii, 5 xviii xxiii xxii xx 110 xvi, xvii xix 112 168 xxi 127 xxii

I N D E X OF TECHNICAL TERMS

319

Page -

Technical Terms J

Join

94 K

Kinna-Wagner principle (KW) Krein-Milman theorem (KM) Kruse's function

83 177 24

L Largest (R-largest) Last (R-last) Lattice Least upper bound (lattice) Least upper bound (R-least upper bound) Limit ordinal Linear functional Linear ordering Linearly dependent (vector space) Linearly independent (vector space) Loop $05 Theorem (LT) Lzwenheim-Skolem Theorem Lower bound (R-lower bound)

xx i xxi 93 94 xxi xxiii 119 xxi 118 118 109 166 163, 164

xx i

M Maximal (R-maximal) Meet Meet-irreducible Metabelian Minimal (R-minimal) Minimal (Y-minimal) Mode 1 Multi-algebra Multiple choice axiom Multiplication (cardinals)

xxi 94 102 117

xxi 183 163 134 74 xxvi

320

INDEX OF TECHNICAL TERMS

Page

Technical Terms M

10

Multiplicative principle N

xxiii

Natural number Nest Nielson-Schreier Theorem Nil-k Norm (vector space) Normal

33

xiii 117 177 49 0

126 182

Order (group) Order-unit Ordered n-tuple Ordered pair Ordinal Ordinal number

xix xvii xxii xxii P xvii xxi xxii xiii

Pairing axiom Partial ordering Partially ordered class Partition principle Perfect (topology) Polynomial ring Power class Power set axiom Prime ideal (integral domain) Prime ideal (lattice) Proper class Proper subclass

168 113

xvi xvii 112 95

xvi xvi

Q Quasi-group

109

I N D E X OF TECHNICAL TERMS

Technical Terms

321

Page R

38 xvii

Ramified Range Rank Real vector lattice Recursion theorem Reduced power Reflexive Regular (ordinals) Relation (binary) Relation (n-try) Replacement, axiom of Restriction Ring

xxiv, 209, 226 182 xxiii 166 xx i 156, 157 xvii xx xviii xix 110 S

109 xv xx i 117 84 93 xvi 177 118 xxvi, 144 xxii xxiii xxi xx xxi

Semigroup Set Smallest (R-smallest) Solvability degree Structured Subalgebra (proper) Subclass Sublinear functional Subspace (vector space) Subtraction (cardina1s) Successor class Successor ordinal Supremum Surjection Symmetric T Torsion-free (group) Transfinite (cardinal)

127 xxvi

322

INDEX OF TECHNICAL TERMS

Technica1 Ternis

T

Page

xxiv xxiii xxii xviii xxi

Transfinite induction Transfinite (set) Transitive (class) Transitive closure Transitive (relation) Tree Trichotomy Tychonoff Compactness Theorem

38 21

167 U

94 xvii 112 127

Ultrafilter Union axiom Unique factorization domain Uniquely divisible (group) Unit element (integral domain) Unit set Universal Un iversa1 ( relation) Universe Unordered n-tuple Unordered pair Upper bound (R-upper bound)

112

xvii 221 45

xix xix xvii xx i V 86 118

Vacuous (relation) Vector space W

123, 132

Weak direct product Weak* topology Well ordering

178 xxii

Zero element (lattice)

94

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