Hausdorff Gaps and Limits: 132

HAUSDORFF GAPS AND LIMITS

STUDIES IN LOGIC AND

THE FOUNDATIONS OF MATHEMATICS VOLUME 132

Honorary Editor:

P. SUPPES, Stanford

Editors: S. ABRAMSKY, London J. BARWISE, Stanford K . FINE, LosAngeles H. J . KEISLER, Madison A . S. TROELSTRA, Amsterdam

NORTH-HOLLAND AMSTERDAM LONDON NEW YORK *TOKYO

HAUSDORFF GAPSAND LIMITS

Ryszard FRANKIEWICZ Institute of Mathematics Polish Academy of Sciences Warsaw, Poland Pawel ZBIERSKI Department of Mathematics Warsaw University Warsaw, Poland

1994

NORTH-HOLLAND AMSTERDAM LONDON *NEWYORK *TOKYO

ELSEVIER SCIENCE B.V. Sara Burgerhartstraat 25 P.O. Box 211, lOOOAE Amsterdam,The Netherlands

ISBN: 0 444 89490 X @

1994 Elsevier Science B.V. All rights reserved

No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the publisher, Elsevier Science B.V., Copyright & Permissions Department, P.O. Box 521, 1000 AM Amsterdam, The Netherlands. Special regulations for readers in the U.S.A. - This publication has been registered with the Copyright Clearance Center Inc. (CCC), Salem, Massachusetts. Information can be obtained from the CCC about conditions under which photocopies of parts of this publication may be made in the U.S.A. All other copyright questions, including photocopying outside of the U.S.A., should be referred to the publisher. No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. This book is printed on acid-free paper. Printed in The Netherlands

To Professor C, ell-Nardzewski

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PREFACE Gaps and limits - the notions exposed in the title - are two phenomena occuring in the Boolean algebra P(w)/fin. Both were discovered by F. Hausdorff in the mid 1930’s. In this book we want to show how they can be used in solving several kinds of mathematical problems. Also, we try to convince the reader that they are of interest of their own. The elements of P(w)/fin can be viewed as the open-closed sets in the remainder space w* = P [ w ] \ w . Thus, the Stone-Cech compactification of w is, as a matter of fact, the main subject of the book. We barely indicate some of its numerous connections with (as well as aplication to) various fields of mathematics: set theory, topology, Boolean algebras, analysis, number theory. Since many problems concerning the space w* are undecidable, i.e. they can be proved to be consistent with set theory, we widely use the forcing technique in the text. Since this technique is not commonly known, we give a short exposition of the forcing method in Chapter 11. We generally omit the proofs there except those concerning the iterated forcing. We refer the reader to Kunen’s text book “Set theory: An Introduction to Independence Proofs”, North Holland, Amsterdam 1979, for complete proofs. Each chapter (except the last one) ends with a number of problems. Some of them are only easy exercises, the more difficult are supplied with hints, a few are used in some proofs. R. Frankiewicz, P. Zbierski

vii

TABLE OF CONTENTS Notation and terminology . Chapter 1 Boolean Algebras $1Introduction . $2 Formulas . $3 Atoms $4 Complete algebras . $5 Homomorphism and filters $6 Ultrafilters $7 Extending a homomorphism $8 Chains and antichains Problems . Chapter 2 Gaps and Limits $1 Introduction . $2 Dominance $3 HausdorfF Gaps $4 The ParoviEenko theorem . $5 Types of gaps and limits . Problems . Chapter 3 Stone Spaces $1The Stone representation . $2 Subalgebras and homomorphisms $3 Zero-sets $4 The Stone-cech compactification $5 Spaces of uniform ultrafilters $6 Strongly zero-dimensional spaces $7 Extremally disconnected spaces . Problems

...

Vlll

.

xi

.

1

. . . .

2 5 6 9 15 18 21 24

.

. . . . 1

31 35

.

40

. . .

43

. . .

63 68 69 72 76 80 83 87

. . .

. .

48 58

ix

TABLE OF CONTENTS

Chapter 4 F-Spaces $1Extending a function $2 Characterization of countable gaps $3 Construction of ParoviEenko spaces $4 Closed sets in the space w* $5 On the ParoviEenko theorem $6 On P-sets in the space w* . $7 Character of points . Problems . Chapter 5 x-Base Matrix $1Base tree $2 Stationary sets $3 c-Points Problems . Chapter 6 Inhomogeneity $1Kunen's points $2 A matrix of independent sets $3 Countable sets in F-spaces . 54 Inhomogeneity of products of compact spaces Problems . Chapter 7 Extending of Continuous Functions $1Weak Lindelof property . $2 A long convergent sequence $3 Strongly discrete sets Problems . Chapter 8 The Martin Axiom $1Continuous images . $2 The space /3[wl] $3 On the ParoviEenko theorem . $4 Gaps $5 Homomorphisms of C ( X ) . Problems . Chapter 9 Partitions of Antichains $1Partition algebras . $2 Complete algebras . $3 Partition algebras under MA

. . *

93 96 98 104 106 108 117 121 123 132 136 142

143 148 153 157 159 161

167 170 170

179 186 191 197 202 209 209 218 22 1

X

TABLE OF CONTENTS

$4 More on partition algebras Problems . Chapter 10 Small P-Sets in w* $1 Proper forcing 52 On P-filters with the ccc Problems . Chapter 11 Forcing $1 Set theory and its models $2 Forcing . $3 Complete ernbeddings $4 Cardinal numbers . $5 Selected models $6 Iterated forcing $7 The Martin Axiom . Bibliography . Index .

234 244 243 256 264 259 275 2 79 281 283 289 299 294 297

NOTATION AND TERMINOLOGY LOGICALSYMBOLS. The sign + denotes the implication, while = is the logical equivalence. The symbols V, A denote the negation, disjunction and conjunction, respectively. The universal quantifier (for every ...) is denoted by V and the existential one (there exists ...) by 3. 1,

OPERATIONS ON SETS. If R is a family of sets, then U R denotes the union and R the intersection of the family R, respectively. Thus, for an arbitrary Z, the following equivalences hold zEUR and

ZEn R

3AER(zEA)

=

VAE R(z € A ) .

I f R = { A , B } consistsofsets A a n d B , t h e n U R = A U B a n d n R = A n B . The union of an indexed family {A, : i E I}is denoted by U{A, : i E I } or by UiErAi (similarly for the intersection). The difference of the sets A, B is denoted by A \ B. The symbol 8 denotes the empty set. We say that the family R is disjoint, if R # 8 and the following implications hold if A E R, then A # $3, if A , B E R and A # B, then A n B = 8. The symbol P ( X ) denotes the power set of X , i.e. P ( X ) = { A : A

X}.

FUNCTIONS. The symbol dm(f) denotes the domain and rg (f) the range of a function f, respectively. The image of a set A dm (f)under f is denoted by f [ A ] . Similarly, the counter image of a set B under f is denoted by f-' [B]. As usual, f l A is the restriction of f to a subset A C dm (f).

xi

xii

NOTATION AND TERMINOLOGY

Let ( A , < A ) and ( B , ~ Bbe) partial orderings. A one-to-one B is called an order embedding if the following condition is

ORDERINGS.

function f : A satisfied x


y

E

f(x) < B (y)

In addition, if f is onto B(f : A

for arbitrary x,y E A .

"3 B ) , then we say that f is an order isomorphism.

<

For a subset S E A and an element a E A we write a 5 S, if a x holds for each x E S. Similarly, S 5 T means that I 5 y, for each x E S and y E T .

ORDINAL NUMBERS.Following von Neumann a set a is called an ordinal if it satisfies the following conditions (for arbitrary x, y): if x E a , then x a , if x,y € a , thenx E y or x = y or y E x .

<

The class On of all the ordinals is well ordered by inclusion and we usually write a p for a C p. The strict ordering a < p is equivalent then to the membership (Y E p. Hence, each ordinal a is the set of all smaller ordinals a = { p E On : p < a } . Each (linear) well ordering is order isomorphic to a unique ordinal (ordered by inclusion). For any a E On the set a 1 = a U { a } is an ordinal and it is an immediate successor of a i.e. there is no ( such that a < ( < a 1. Ordinals X which are not of the form a 1 are called limit. Thus, X is limit if the condition a < X implies a 1 < A, for each a E On. The smallest infinite ordinal is denoted by w and is referred to as the set of natural numbers or nonnegative integers, since the consecutive elements of w are Q, = 0, { Q,} = 1, { Q, , { Q,}} = 2 , . . . etc. Each nonempty set A On has the supremum sup A = U A and also the infimum infA = A.

+

+

+

+

n

CARDINAL NUMBERS. An ordinal K E On is called a cardinal if no a < K has the same power as IF. i.e. if there is no one-to-one function from a onto IF.. For each set A there is a unique cardinal IF. having the same power as A and we write K = cardA in this case and say that K is the cardinality (or the power) of A . Numbers i E w are cardinalities of finite sets and w itself is the cardinality of all countable sets. Consecutive infinite cardinals are denoted by symbols

Thus, w1 denotes the smallest uncountable cardinal.

...

NOTATION AND TERMINOLOGY

Xlll

If a function f has an ordinal X as its domain (i.e. X = dm (f) E On), then we call f a sequence of length X and often write f = ( f ( a ): a < A). For limit ordinals X E On we define the notion of cofindity cf(X) = inf{cardA :

A

X and supA = A}.

Thus, cf(X) is a cardinal nongreater then X and simultaneously it is the length of a shortest sequence f = ( f ( a ): a < p) such that supa..@ f ( a ) = A. Cardinals K satisfying cf(n) = n are called regular. For example, w and all nonlimit (i.e. of the form K = wa+l) cardinals are regular. For arbitrary sets A , B the set B A consists of all functions f : A + B. The cardinality of B A is denoted by n’, where n = cardB and X = cardA. The cardinality of the continuum (that is, the cardinality of the set W of real numbers) is denoted by c. Thus, c = cardl3 = card{O,l}” = cardP(w). The continuum hypothesis c = w1 is abbreviated as CH and the generalized continuum hypothesis Va(2”’a = w a + l ) as GCH. For X 1 w let K<’ = sup{n’ : y < A}. In the case of A 5 n we have = card{A

K

: cardA

< A}.

If GCH holds below a given regular n, then n fulfills the equalities

n
n = Fn.

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Chapter 1

Boolean algebras $1. INTRODUCTION. At the beginning we recall the notion of a field of sets. A family F , consisting of (some) subsets of a given space X , is called a field of sets, if F is closed under set-theoretical operations of union, intersection and complementation i.e. for arbitrary A , B E F the sets A U B , A n B and X \A belong to F . In addition, we assume always that the empty set (Band the whole space X are in F . The family P ( X ) = { A : A 5 X } of all subsets of a (nonempty) set X and the two element family {(B, X } are examples of fields. For an arbitrary field F of subsets of X , we have always

{(B,X)c F

c P(X).

Boolean algebras can be introduced as generalizations of fields of sets. Let us consider a system of the form

A = (A; + ,

.,

-

, 0,1),

”+”

in which A is a nonempty set, and ”.” are binary operations on A , ”-” is a unary operation on A , 0 and 1 are some fixed (distinct) elements of A . The system A is called a Boolean algebra, if for arbitrary elements a , b, c E A , the following equations are satisfied (axioms of Boolean algebras): (1) (2) (3) (4) (5)

a+b=b+a a+(b+c)=(a+b)+c a . ( b C) = a . b a . c

+

a.b=b.a u . ( b . C) = ( U . b ) . c u ( b . C) = ( U b ) . ( U

+

+

+

+ C)

a.n=a a .(-a) = 0

a+0=a a+(-a) =n

”+”

Operations and ”.” are called Boolean sum and product, respectively and is called a Boolean complementation. Elements 0 and 1 are called (Boolean) zero and unit, respectively. Usually, we identify A with the underlying set A (thus, we write e.g. a E A instead of a E A etc.). ”-”

1

CH.1 BOOLEAN ALGEBRAS

2

Any field F , of subsets of a space X , is a Boolean algebra if we define A + B = A U B , A . B = A n B , - A = X \ A , 0 = (3 and I = X . EXAMPLES: 1. The field P ( X ) ,for any X

# (3.

It is often called the power-set algebra of X .

2. Let X be an infinite set. The family F C ( X ) consisting of all finite subset of X and their complements is a field of sets (called the field of finite-cofinite subsets of X ) . 3. Let X be a topological space. The family B ( X ) consisting of all open-closed subsets of X is a field of sets. If X is connected, then B ( X ) = { ( 3 , X } . If X is zero-dimensional, then B ( X ) is a base (for open as well as for closed sets).

4. The family L of all Lebesgue measurable sets on the real line W,or - more generally - in an n-dimensional Euclidean space W”, is a field and even a ufield i.e. it is closed under countable unions Un<w A,, and hence, under countable intersections.

52. FORMULAS. We prove here some algebraic laws valid in all Boolean algebras. The proofs are obtained directly from axioms (1) - (5) by simple calculations. Let a , b, c , . . . denote arbitrary elements of a fixed Boolean algebra A. 2.1.

a+a=a

and a . a = a .

Indeed, wehavea = a + 0 = a + ( u . ( - a ) ] = ( a + a ) . ( u + ( - u ) ) = ( u + a ) . l = a + a . Similarly, a = a . 1 = a . ( u ( - a ) ) = a . a . . ( - a ) = a . a 0 = a a .

+

2.2. We have a a

u+I=I

+

+

and u . @ = @ .

+ 1= a + ( a + ( - u ) )

=(a

+u) + (-a) = a +(-u)

0 = a(u * ( - a ) ) = ( a . a)(-.) = a * ( - a ) = 0. 2.3.

a+b=b

u . ~ = u .

Multiplying the first equality by a we get

Analogously, adding b to the second equality, we obtain

=

I and similarly,

52. FORMULAS

3

Every field of sets is (as every family of sets) partially ordered by inclusion, which can be characterized by the following equivalences

ACB

G

=

AUB=B

A ~ B = A .

Thus, the formula 2.3 is a Boolean counterpart of the second equivalence. Hence,

5 as follows

we introduce in the algebra A a relation a

5b

a

5

2.4. The relation

+b =b

(equivalently, a . b = a ) .

(Boolean ordering) partially orders A.

5 a , since a + a = a . If a 5 b and b 5 a , i.e. if a + b = b and 5 b and b 5 c, i.e. a + b = band b+ c = c, then = ( a + b) + c = b + c = c, i.e. a 5 c .

Indeed, we have a

b + a = a , then obviously a = b. If a a + c = a + ( b +c)

2.5. If a

5 b, then for each x in A we have a + x 5 b + x and a . x 5 b . x.

In fact

(a

+ x) + ( b + x) = ( a + b) + (x + x) = b +

2

and (U

. Z) . ( b . Z) = ( U

. b)

. (Z. Z) = u . x

2.6. 0 5 a 5 I, i.e. 0 is the smallest and I is the greatest element, respectively.

+ 1 = I.

Because CD + a = a and a 2.7.

a

We have a,b

+ b = sup{a, b}

and a . b = inf{a, b}.

5 a + b, since a+(a+b)=a+b

and

b+(a+b)=a+b.

If x is an upper bound of { a , b}. i.e. a , b 5 x, then (a

and hence a

+ b) + x = a +

(b+x) = a + x = x

+ b 5 x. Thus, we have proved a + b = sup{a, b}.

is proved similarly using the product condition. 2.8.

If a

+ x = I and a

'

x = 0.then x = -a.

The second formula

4

CH.1 BOOLEAN ALGEBRAS We have (-a) + I = ((-a)

+ x) = (-a). a + x a + 5 . (-a) + 2 .I = + = x.n +x = 2 (-a) + = z . (.+(-a))

+ x)

= z.a +

‘

(a

’

and hence --a 5 x. On the other hand

hence x 5 --a and consequently x = -a. -(-a) = a

2.9.

This follows immediately from 2.8. 2.10.

-(a

+ b) = (- a).(-b)

and

-(a.b)=(- a)

We apply 2.8 for x = (-a) . (-b) replacing a by a

+ (-b).

+ b.

Equalities 2.10 are De Morgan laws for Boolean algebras. 2.11. 2.12. From a

-I=@

and

-(D=n.

a5b

-b 5 -a.

5 b using 2.10 we have (-b). (-a) = -(a

+ b ) = -b,

i.e. -b 5 -a. Similarly, if - b 5 -a, then a . b = -((-a)

+ ( - b ) ) = -(-a)

= a,

i.e. a 5 b.

2.13.

xla

EE

2-a=(D.

(x - a denotes x . (-a)). From x 5 a, multiplying by -a, we get x - a 5 0.Adding a to x - a = 0 we obtain

a = 2 . (-a) + a = (x

+ a) . ((-a) + a) = x + a ,

i.e. x 5 a. 2.14. If @ < a < 1,then 0 < --a < 1. Because --a = I implies a = 0 and similarly -a = (D implies u = n.

53. ATOMS

5

All the proved above formulas 2.1 - 2.14 are Boolean counterparts of well known laws of algebra of sets. We shall use them in the book without a special reference.

$3. ATOMS.As already mentioned we identify any Boolean algebra A with its universe A . A nonzero element a > 0 of a given algebra A is called an atom, if (D is a unique element smaller than a , i.e. if the following condition holds Vz[z5a-+x=(D v x=a].

For example, in a power-set algebra P ( X ) or in F C ( X ) all the atoms coincide with singletons a = {I}, for z E X . We say that a set D is dense in A, if D consists of nonzero elements of A and for each a > 0 there is a d E D such that d 5 a . In the case A = P ( X )or A = F C ( X ) the set of all atoms is dense in A. Algebras having this property are called atomic. On the other hand there are algebras having no atoms at all (atomless algebras). 3.1. EXAMPLE. Let CI= {0,1}' be the Cantor space of weight K = card1 2 w . Thus, the elements of CI are all functions f : I + {0,1} and basic open-closed sets are of the form

Us = {f

: s

f } = {f : Vi E dm(s)(s(i) = f(i))},

where s is a zero-one function with a finite domain dm(s) C_ I. CI is a compact, zero-dimensional Hausdorff space and the field B(C1)of open-closed sets consists of finite unions Us,U . . . U U s , , of basic sets. Obviously, B(C1)has no atoms, since Us $ Ut holds, if only s is a proper extension of t. Let us note the following simple characterization of atoms: 3.2.

a is an atom

Vx[a 5 x or a . x = 01.

Indeed, we have a . x 5 a and hence, if a is an atom, then a . x = 0 or a . x = a , i.e. a 5 x. Conversely, if a satisfies the right hand side of 3.2. and x 5 a , then either I = a or a . z = (D. In the latter case we have x = 0, since a . x = x.

CH.1

6

BOOLEAN ALGEBRAS

$4. COMPLETE ALGEBRAS. In every Boolean algebra A, for any finite number of elements a l , . . . ,a,, we have

+ . . . + an = SUP{ai ,. . .an} and a1

. ... . a ,

= inf(a1,. . . ,a,},

which follows immediately from 2.7. An algebra A is called complete, if each of its (infinite) subset S has the supremum sup S and the infimum inf S (clearly, it suffices to assume the existence of e.g. sup’s only, since inf S = sup{s : I 5 S}). If this holds for all countable S, then we say that A is a-complete. Any power-set algebra P ( X ) is complete and obviously sup S = U S (similarly, inf S = S). On the other hand, the algebra F C ( X ) is not even a-complete.

x

4.1. ALGEBRA OF OPEN DOMAINS. Let be an arbitrary topological space. A set U X is called an open domain (or regularly open), if U = int[m. For example, for an arbitrary A E X , the set U = int[z] is an open domain. Indeed, U is open, hence U = int[U] in@] and since, U = int[x] C 2,we have

c

whence U = int[u]. Let us note, that

(*I

W n int[V] G i n t [ E T V ]

for open W, v

X.

Indeed, we have

w n int[V] = int[W] n int[V] = int[W n TI s i n t [ W ] , since W n V C Wn. For open sets U , V the following equality holds 4.2.

int[U n V] = int[U] n int[V],

since int[ViT]

s int[D n V] = in@]

n int[V]

54. COMPLETE ALGEBRAS

7

and from (*) we obtain int[U] n int[V] & int[int[U]n V] c int[int[CKV]] = int[U n v], because we know already that the set int[U n V] is a domain. On the set D(X) of all (open) domains we define Boolean operations as follows

U

+ V = int[U u V],

-U = int[X \ U]

U . V = U n V,

and assume 0 = p, 1 = X . From 4.2 it follows that the intersection U n V of domains is again a domain and hence, the definitions of the operations are correct. We have to check that the Boolean axioms are satisfied. Commutativity is obviously true. We check associativity:

(U

+ V) + W = i n t [ i n t [ W ]u W] G i n t [ T Z T u TV] = = int[U u v u W] = int[int[U] u int[V] u W] c int[int[D u V] u W] = (U

+ V) + W

a n d th u s ,( U+V) +W = i n t [ U U V U W ] andweget thesameresultforU+(V+W). Distributivity is proved easily using 4.2.:

U . (V + W )= U n int[V u W] = int[U] n int[V u W] = = int[U n (V u = (U n V)

w)]= int[(TKV) u (

V ) ]=

+ (U n W ) = U . V + U . W.

Similarly, we obtain

U

+ (V.W )= int[(Uu V ) n (U u W ) ]= = i n t [ m ] n i n t [ m ] = (U

Further, we have U

+ V) . (U + W ) .

+ (D = int[UU 81 = U and U . 1 = U n X = U . Finally,

u + (-U)

\ U)] = int[U u (X \ D)] = = int[U u X \ U] = int(X) = X = n = int[U u int(X

and

U .(-U)

= U ni nt ( X

\ U )= @= 0.

CH.1 BOOLEAN ALGEBRAS

8

Thus, D ( X ) is a Boolean algebra. The Boolean ordering on D ( X ) is set inclusion, since U.V = U n V . We prove that the algebra D ( X ) is complete. For any S D ( X ) let V = i n t [ m ] . Then V E D ( X ) and V = s u p s . Indeed, if U E S, then U US and hence U = intU C i n t [ m ] = V, i.e. U 5 V . If W 2 S (i.e. W 2 U ,for every U E S), then U S W and hence

v =int[Dl

int

[W] = W.

Thus V 5 W , which proves V = sup S.

Let us note some formulas valid in any complete algebra A. For example, De Morgan laws can be written in the following form.

4.3.

-supR=inf(-R)

and

- i n f R = sup(-R),

for an arbitrary family R of elements of A (as usual, -R = {-a : a E R}). Indeed, since supR 2 R, hence -supR 5 -R and if x 5 -R, then -x 2 R and thus -x 2 sup R, i.e. x 5 - sup R, which proves the first formula. The second can be proved similarly. We have also the following identities.

4.4.

a + sup R = sup(a + R)

+

and a . inf R = inf ( a . R)

+

for an arbitrary family R, where a R = { a x : x E R} and a . R = { a .I : x E R}. Indeed, we have clearly a sup R 2 a R and if x 2 a R, then x 2 a and x 2 R and hence x 2 a and x 2 sup R, which implies x 2 a sup R. This proves the first equality. The second is proved in a similar manner.

+

+

+

+

$ 5 . HOMOMORPHISMS AND FILTERS

The supremum sup R is often denoted by I . . a , = EL, a i ) and similarly, infR the following formulas: = a1

+ +

4.5

9

C R (because sup (a1 , . . . , a,} -

by

nR. In this notation, we have

a . C R = C ( a . R ) and a + n R = n ( a + R ) .

Indeed, from C R 2 R, multiplying by a , we get a . C R 2 a.R. Suppose x 2 a.R. Hence --r 1: -a (-R), and multiplying by a , we obtain a . (-z) 5 a . (-R) 5 -R. Thus a.(--r) 5 f l ( - R ) = - C R, from 4.3. Passing to complements and multiplying by a we obtain z 2 a . C R, which proves the first formula. The other one is proved similarly.

+

As easily seen from the above proofs, the formulas 4.3, 4.4 and 4.5 are true even in a noncomplete algebra A for all those families R, for which sup R and inf R exist in the algebra A.

$ 5 . HOMOMORPHISM AND FILTERS. Assume that we are given Boolean algebras

We say that A is a subalgebra of B and write A B, if A B (the universe of A is contained in the universe of B), @A = 015and l l = ~ PB(A and B have the same zero and the same unit), and for a , b E A we have

a +pa b = a+B b, a . A b = a.B b, -Aa

= --ga.

(Boolean operations of B are extensions of the corresponding operations of A). 5.1. EXAMPLE. For an arbitrary topological space X we have B ( X ) 2 D ( X ) , i.e. the field of open-closed sets is a subalgebra of the algebra of open domains. Let f : A -+ B be a function mapping (the universe of) A into (the universe of) B. We say that f is a homomorphism (of A into B), if f ( 0 ~ =) 015, PA) = 115

10

CH.1 BOOLEAN ALGEBRAS

If f : A + B is a homomorphism, then the image f[A], of the algebra A, is a subalgebra C B. A one-to-one homomorphism is called an embedding and an embedding "onto" is called an isomorphism.

5.2. EXAMPLE.Suppose that the sets X, Y are of the same cardinality. Thus, there is a function h : X 2 Y, onto Y. If we define f ( a ) = h[a],for a & X, then the function f : P ( X ) + P ( Y ) is an isomorphism of the power-set algebra P ( X ) onto P(Y). Suppose now that cardX 2 cardY. Thus, there is a function h : X "2Y. If we define f ( a ) = h-'[a] for a C Y, then f is an embedding of the algebra P(Y) into

P(X>. Finally, let us suppose that Y X, Y # (3. If we define f(a) = a n Y , for a X , then f is a homomorphism of P ( X ) onto P(Y). The last example above can be generalized as follows. Let A be an arbitrary Boolean algebra and let e be a fixed element of A satisfying 0 < e 5 1. Define a new algebra Ale in the following way: the universe of Ale is the set { a E A : 0 5 a 5 e}, zero is the same as in A and the unit is e. The sum and the product is the same as in A but the complement of a in Ale is e - a. Clearly Ale is a Boolean algebra. If e = 1,then Ale coincides with A. If we define f ( a ) = a . e, for a in A, then f is a homomorphism of A onto Ale.

5.3. SYMMETRIC DIFFERENCE. difference is defined as follows

In any Boolean algebra A, the symmetric

u A b = ( U - b)

+ ( b - a).

Let us note the following property aAb=0

a=b.

Indeed, from a A b = 0 it follows a - b = 0 and b - a = 0, i.e. a 5 b and 6 5 a . Now, if a homomorphism f : A + B satisfies the condition

j-'(0) = {z

:

f(z)= 0)= (01

(i.e. f(z) = 0 holds for I = 0 , only), then f is an embedding. Indeed, from f ( a ) = f ( b ) , it follows f ( a ) A f(b) = 0, hence f ( a A b) = 0. By the assumption, a A b = 0 and thus a = b, which proves that f is one-to-one.

$ 5 . HOMOMORPHISMS A N D FILTERS

11

5.4. PRESERVATION OF ORDERING. Let us note that an arbitrary homomorphism f : A -+ B preserves the Boolean ordering, i.e. the following implication holds

if a 5 b in A, then f ( a ) 5 f ( b ) in B.

+

+

Indeed, a 5 b means a b = b and hence f ( a ) f ( b ) = f ( b ) , i.e. f ( a ) 5 f ( b ) . If f : A -+ B is an embedding then we have an equivalence a5b

=

f(a)

(since f-' is also a homomorphism : f[A]

-+

I f(b),

A).

A subset F of a given algebra A is called a f i l t e r , if F > 0 and the following implications hold b, then b E F , if a E F and a I and if a , b E F , then a . b E F .

For example, if f : A

+B

is a homomorphism, then the set

v(f)= tz :

f(.)

= n)

is a filter in A. On the other hand, the set A(f) = { z : f(z) = @ } has the following properties : A(f) < 1, A(f) is closed under sums and if a E A(f), then b E A(f), for each b 5 a. Such a set is called an ideal. From De Morgan laws, it follows immediately that if F is a filter, then -F(= { - a : a E F } ) is an ideal and conversely, if J is an ideal, then -J is a filter.

5.5. EXAMPLE.For any e > 0 , the set F, = { z : z 2 e} is the smallest filter containing e as its element. Such a filter (i.e. having a single generator) is called principal. We have here F, = V(f), where f : A -+ Ale is the homomorphism described in 5.2 above. 5.6. GENERATING FILTERS. A set R of elements of a given algebra A is called centered, if S > 0, for each finite S E R. Thus, every filter is a centered set. Every centered set R generates the filter F consisting of all elements z satisfying z 2 S, for some finite S C R.

n

n

5.7. MAXIMAL FILTERS. The family of all filters of a given algebra is partially ordered by inclusion. Elements which are maximal in this ordering are called maximal filters or ultrafilters.

CH.1 BOOLEAN ALGEBRAS

12

5.8. LEMMA. For every element e > 0,there is an ultrafilter F containing e. P R O O F . Let Re be family of all the filters containing e. Of course, Re # Psince e.g. the principal filter Fe = { x : x 2 e } is in Re. If a subfamily R Re is linearly ordered by inclusion, then F = U R belongs to Re. Indeed, F is a filter : (D $ F (otherwise we would have (D E H , for some H E R);if a E F and a 5 b, then a E H , for some H E R, and hence b E H C F ; if a , b E F , then there are H I , Hz E R such that a E H I and b E Hz. Since H1 H2 or Hz H1 there is a H E R such that both a , b belong to H and hence a . b E H & F . Obviously e E F and thus we have shown that R is bounded from above. Applying the Kuratowski-Zorn lemma we infer that Re has maximal elements, which finishes the proof. QED

c

5.9. FACTORALGEBRAS. Any filter F in A determines a relation follows a = ~ b

=F

on A as

3x~F[a.x=b.x].

E

This is an equivalence relation : we have a = F a , since x = 1 E F . Symmetry is obvious and if a . x = b . z and b . y = c . y, for some x,y E F , then (multiplying the first equality by y and the second by x ) a z = c . z, for z = x . y E F . Even more, the relation = F is a congruence, i.e. the conditions a = F a' and b = F b' imply the conditions a b = F a' b', a . b = F u' . b' and -a = F -a'. Indeed, for some x , y E F we have a . x = a ' . x and b . y = b ' . y . Hence a . z = a ' . z and b . z = b ' , z , where z = z . y E F and consequently (adding and multiplying these equalities) we obtain ( a b) . z = (a' b') . z and ( a . b) * z = (a' . b') * z. On the other hand, by 9

+

+

+

+

+

+

applying De Morgan laws to a * x = a'. x , we get ( - a ) ( - x ) = (-a' ) ( - x ' ) , which (multiplied by x ) yields ( - a ) . x = ( - a ' ) . x . This shows that the relation = F is in fact a congruence. Thus, we can build up, in a usual way, the factor algebra A / F . The elements of A / F are abstraction classes [ a ]=~ { x E A : x = F a }

and the operations are defined as follows

+

[a] [b]= [a [u] *

+ b]

[b]= [U . b]

-[a] = [ - a ] .

for a in A,

$ 5 . HOMOMORPHISMS A N D FILTERS

13

The zero and the unit of A / F is [O] and [I],respectively. Since = F is a congruence, hence the above definitions are correct and a routine checking shows that A / F is a Boolean algebra. Let us note that the filter F is itself an abstraction class since aE[1]

3xEF(a.x=2)

Z

U = F I

=

3% E F ( a 2 x )

=

a € F,

i.e. [ I ]= F and also [CD] = - F . The factor algebra A / F is a homomorphic image of A and a (canonical) homomorphism f : A '3A / F is given by the equality f ( a ) = [ a l p , for a € A. From F = [I]it follows immediately that V ( f ) = F . Hence, every filter is the kernel of some homomorphism.

5.10. THEOREM. I f f : A + B is a homomorphism of A onto B, then B is isomorphic to the factor algebra A / V (f ) .

PROOF,Define g : A/V(f) + B by setting g ( [ a ] ) = f(a) for a E A. The definition is correct since from a =v b follows a . x = b . I , for some I E V and hence f(a) . f(x) = f ( b ) . f ( x ) and consequently f(a) = f ( b ) (because f ( x ) = I).

+

+

+

+

+

We have g ( [ a ] [a]) = g ( [ a b ] ) = f(a b ) = f(a) f ( b ) = g ( [ a l ) g ( [ b l ) , similarly for the Boolean product. Also g ( - [ a ] ) = g ( [ - a ] ) = f(-a) = -f(a) = - g ( [ a ] ) and g ( [ I ] )= f(1)= I, similarly for 0.Thus , g is a homomorphism and it is onto B, since f is onto B. Finally, g is one-to-one, since g ( [ a ] )= CD

E f(a) = O

and A(f) = - V ( f ) is the zero of A / V ( f ) .

G

a E A(f)

QED

Thus, the factor algebras AIF, where F runs over all filters, exhaust all homomorphic images of a given algebra A. Let J be the ideal adjoint (or dual) to a given filter F , i.e. J = - F . The relation = F can be now denoted by =J and similarly, the factor algebra AIF can be denoted by AIJ. Since we have Z E J -zEF. hence a=jb

U = F ~ G

3xEF(a.z=b.1)

Here is another condition. 5.11.

a=jb

aAbEJ

E

3yEJ(a-y=b-y).

14

CH.1 BOOLEAN ALGEBRAS

Indeed, if a - I = b - I , where I E J , then (multiplying by -b) we shall obtain (a - b ) - I = CD and hence a - b 5 I . Thus, a - b E J and, symmetrically, b - a E J. Hence a A b E J. Conversely, we have always a - a A b = a . b = b - a A b and hence if a A b E J , then a =J b. Now, let us look at the Boolean ordering in A/J. From 5.11 we have [a] 5 [b]

+ [b] = [b]

a

[a]

E

+b

=J

b

(a

+ b ) A b E J.

But (a + b ) A b = a - b and hence we obtain [a] 5 [b]

5.12.

e

a - b E J.

5.13. EXAMPLE.SETS WITH THE BAIRE PROPERTY. Recall that a subset A C X, of a topological space X, is nowhere dense, if int[a] = 9. Countable unions U i c w A i of nowhere dense sets are called meager (or of the first category). Let M = M ( X ) denote the family of all meager sets A C X. We say that a set A c X has the Baire property, if A = M U ,for some U E V ( X ) , (i.e. the symmetric difference A A U is meager). The family B(X) of all sets with Baire property is a a-field. Indeed, if A = M U ,then since ( X \ A ) A ( X \ U )= A A U hence X \ A = M int[X \ U ] = - U , in D ( X ) (because the boundary of a closed set is always nowhere dense). If Ai E B,i.e. A, = M Ui, for i < w , then since

U Ai A IJ

i<w

ui

i<w

hence

U Ai = M

i<w

i<w

i<w

Since U = M int[u] E D(X), for open U ,hence B contains all open and consequently all Bore1 set. Obviously, M(X) C B(X) so that if X satisfies the Baire theorem (e.g. if X is compact or metrically complete), then M is an ideal and even a u-ideal in 1,(since X 4 M). Consider the factor algebra B / M . Each set A E B(X) is congruent to a unique open domain U E D(X) (since U \V # 9implies U\v # 9, for U,V E D ( X ) ) . Let us define

f(A)=U

A

=M

U

for A E B(X) and U E D ( X ) .

$6. ULTRAFILTERS

15

Clearly, f : B(X) + D ( X ) is a homomorphism and

A(f) = { A E B(X) : A

=M

8)=M(X).

It follows from 5.10 that the Baire algebra B ( X ) / M ( X ) is isomorphic to the algebra of domains D ( X ) . In particular, B/M is always complete.

Now, we shall give some properties characterizing ultrafil§6. ULTRAFILTERS. ters (maximal filters). 6.1. THEOREM. A filter F in a given Boolean algebra A is maximal if and only if it satisfies the following condition

(*I

either a E F or -a E F

for each a E A.

PROOF.Assume that F is maximal and let a be an arbitrary element of A. If the family F U { a } is centered (5.6.), then it can be extended to a filter G. Since F is maximal, we have F = G and hence a E F . If FU { a } is not centered, then a.5 = a>, for some z E F and hence z 5 -a. Therefore -a E F and (*) holds. Conversely, assume that (*) holds for an F and suppose F C G, where G is a filter. We have to show that F = G. Let a E G. If a @ F then -a E F by (*) and hence -a E G, which is impossible . QED We say that a filter F is prime if it satisfies the following condition if a

+ b E F , then a E F or b E F .

We shall prove the following characterization:

F is maximal

6.2.

=

F is prime.

+

b) = ( - a ) . (-b) is in F Indeed, if F is an ultrafilter and a, b 4 F , then -(a and hence a b 4 F . Conversely, if F is prime and a is arbitrary, then a (-a) = I is in F and hence a E F or -a E F , i.e. (*) holds. Another characteristic property is the following

+

6.3.

A / F = {@,

+

11,

i.e. the factor algebra has 2 elements, only. Indeed, we have (recall that F is the unit and -F the zero of A / F )

A / F = {a>, I}

--=

VU E A([u]F= F V

[U]F=

V U E A ( U EF V - U E F )

-F) (*).

CH.1 BOOLEAN ALGEBRAS

16

EXAMPLE.The principal filter F = {I : x 2 a } generated by a is maximal if and only if a is an atom. This follows immediately from 3.2. Hence, in every atomless algebra each ultrafilter is nonprincipal. In a finite-cofinite algebra F C ( X ) the family F = { A C_ X : X \ A

is finite)

is a unique nonprincipal ultrafilter. Filters F E P ( X ) in a power-set algebra are called often filters on the set X . Let X be an arbitrary infinite set and denote IE = cardX. We define the Frhchet filter F, as follows A E F, = card(X \ A ) < IE for A C X .

An ultrafilter on X is called uniform, if all its elements are of cardinality IE = w uniform means the same as nonprincipal). It is easy to see that 6.4.

p is uniform

F, C p

tc

(thus, for

for an arbitrary ultrafilter p on X .

Indeed, if there is an A in F, \ p , then X \ A E p and hence p cannot be uniform. Conversely, if there is an A E p , with cardA < I E , then X \ A E F, \ p . Now, let us estimate the number of ultrafilters on a given set X .

IE,

6.5. THEOREM. (Fichtenholz, Kantorovitch). On an infinite set X of Cardinality there is 22* uniform ultrafilters.

PROOF. (Hausdorff [Hz]).A family R C P ( X ) is said to be independent, if the intersection A1 n . . . n An n ( X \ B1) n ... n ( X \ B m ) has cardinality IE, for any distinct sets A1,. . . , A , ; B1,. . . ,B , in R. First, we shall prove that there is always an independent family R cardR = 2K. Let us define an auxiliary set Y as follows (s,H) E Y

s is a finite subset

X

E P ( X ) ,with

and H C P ( s ) .

If we set TA = {(s, H ) : A n s E H}, then the family {TA : A X } has cardinality 2" and is independent. Indeed, we have TA # TB,for A # B since if I E A A B , then ( { I } , ((2))) E TAATB. Now, suppose that A l , . . . , A n ; B I , .. . , B , are all distinct and choose elements xjj E A* A B j , i = 1 , . . . ,m j = 1,. .. ,m.

57. EXTENDING A HOMOMORPHISM If so = { z i j : i = 1,...,n ; j = 1,...,m } , then for any s _> A, n s # Bj n s and

17 SO

we have

Thus, there is an independent family of cardinality 2" of subsets of Y and since cardY = car=, hence there is also such a family R P ( X ) .

c

Now, we can finish the proof of the theorem. For each function f : R define Sf = F, U { f ( A )* A : A E R } ,

+

{ -1, +1}

where we have denoted +1. A = A and -1 . A = X \ A . Each family Sf is centered and hence there is an ultrafilter p f _> Sf.If f ( A ) # g ( A ) , then A E p f A p s . Hence, there are 2*' ultrafilters on X extending the Frkchet filter F,. QED

57. EXTENDING A HOMOMORPHISM. Let s be a subset of a given algebra A. The intersection [S]of all subalgebras A containing S is the smallest subalgebra of A containing S. We say that the set S generates [S]in A.

EXAMPLE.The family S = fin(X) of finite subsets of an infinite set X generates the algebra F C ( X ) in P ( X ) .

c

A be a subalgebra and take an a E A \ Ao. The symbol Ao[a] will Let & denote the subalgebra [& U {a}] C A, i.e. the subalgebra generated by & and a in A. The elements of Ao[a] are of the form 7.1.

I .a

+ y .( - a )

where z,y run over Ao.

Indeed, each element 7.1 is in Ao[a]. On the other hand, all the elements of A0 and a itself have the form 7.1. Hence it suffices to show that the elements 7.1. are closed under Boolean operations. This follows from the following obvious formulas: 21

.a

+ y1(-a) +

22

+

* a yz(-a)

= (21

+

22)

.a

+ (y1 + yz) . ( - a )

- ( . * a + y * ( - a ) ) = (-z).a+(-y)*(-a).

7.2. GAPS. A gap (or a cut) in a given Boolean algebra A is a pair of subsets K = ( S , T ) such that S 5 T , i.e. a 5 b holds for each a E S and b E T . The set S is called the lower and T - the upper class of the gap K .

CH.1 BOOLEAN ALGEBRAS

18

Sometimes we put some additional requirements upon this notion. We say that an element c fills (or separates) a gap K = (S,T ) ,if S 5 c I T (i.e. if the inequalities a 5 c and c 5 b hold for each a E S and b E T ) . Now, we shall see how to extend a homomorphism f : A0 .+ B defined on a subalgebra A0 A into a given algebra B. Any new element a E A \ A0 defines a gap in A0 as follows

Since f preserves the ordering (see 5.4), the image

is a gap in Bo = f[Ao] C B. If b E B fills the gap f[K,] then f[K,] is a subgap (in an obvious sense) of

Using this notation we prove now the following theorem on extending of a homomorphism. (Sikorski [Sz, &I). Let f : A0 .+ B be a homomorphism from a 7.3. THEOREM subalgebra A0 5 A into B. If for an element a E A \ A0 there is a b E B, such that b fills the gap f[K,(&)], then f can be extended to a homomorphism 7 : Ao[a] + B, with T ( u ) = b. In addition, i f f is an embedding and the equality

holds, then f can be extended to an embedding of Ao[a] onto f[Ao][b].

PROOF.The elements of the subalgebra Ao[a] are of the form 7.1. We extend f setting -

f(2. a

+ Y. ( - a ) ) =

f(2)

. b + f(y) . ( - b ) .

In particular, J ( u ) = b. We have to show that the definition is correct i.e. does not depend on the representation of an element in the form 7.1. Suppose that

. a + y1(-a)

= 22 . a

+

Y2

. (-a).

19

$7. EXTENDING A HOMOMORPHISM Hence , from 5.3,we have

and consequently (21

A 12)

*

u

+ (y1 A yz)

*

(-U)

= CD.

This is equivalent to (11

A 1 2 ) . u = CD

and

(YI

A yz)(-u)

An)2 a

and

(Y1

A Y2) 5 a.

=0

and hence to -(21

This means that -(XI A 1 2 ) belong to the upper and y1 A y2 to the lower class of the gap K,, respectively. Hence f(-(11 A 2 2 ) ) = -f(zl A 1 2 ) and f ( y l A y2) belong to the upper and lower class of f [ K , ] , respectively. By the assumption, the element b fills the gap f [ K , ] and hence

Running the argument backward, we obtain

which shows that f : &[a] + f[Ao][b] is a homomorphism. This finishes the proof of the first part of the theorem.

20

CH.1 BOOLEAN ALGEBRAS

To prove the second part, assume that f is an embedding and that (*) holds. If f ( ~a.+ y ( - a ) ) = 0,then f(z)* b = 0 and f ( y ) ( - b ) = @ or, equivalently

-

Thus f(-z) and f ( y ) are in the upper and the lower class of KP,,respectively. By the assumption (*), it follows that -I and y are in the upper and the lower class of K,, respectively. Hence, x . a = 0 and y . ( - a ) = 0,i.e. I . a y ( - a ) = 0.This means that A(7) = {@} and hence f is an embedding. QED

+

The condition (*) (sufficient for extending of an embedding) can be written in the form 7.4.

and

s s a

E

f(z)lb

y 2 a

E

f(y)Lb

for x, y E Ao.

For extending a homomorphism, it is sufficient to have implications (from the left to the right).

7.5. COROLLARY . If B is complete, then any homomorphism fo : from a subalgebra A0 A, can be extended to the whole of A.

A0 -+

B,

PROOF.The family F , of all functions f such that the domain dm(f) = Af is a subalgebra of A containing A0 and f : Af + B is a homomorphism extending fo, is partially ordered by inclusion. If Fo F is a linearly ordered subfamily, then U F Ois an upper bound of Fo. Hence, by the Kuratowski-Zorn lemma, F has a maximal element f . Then the domain Af of f is equal to A for, if a E A \ Af, then b = sup{f(x) : I 5 a } separates the gap f [ K , ] and, by the Sikorski theorem, f is extendable on A r [ a ] ,which contradicts the maximality of f. QED

More generally, Corollary 7.5. holds also for any algebra B, which has no unfilled gaps. Note however, that such an algebra must be complete.

$8. CHAINS A N D ANTICHAINS. A chain in a Boolean algebra A is a subset of A, which is linearly ordered (in the Boolean ordering of A). We shall consider here well ordered or conversely well ordered chains only. Thus, by an increasing chain of length y we mean a sequence ( a , : a < y), of elements of A, where y is an ordinal and a, < u p , for all a < < y. We assume always that a, < 1,for each a < y. Similarly, a decreasing chain is a sequence ( a , : a < y) of nonzero elements of A,

$8. CHAINS A N D ANTICHAINS

21

such that a , > up, for all a < /3 < y. Hence, if ( a , : a < y) is an increasing chain, then the sequence of complements b, = -a, is decreasing and conversely. Elements a, b > 0 are called disjoint, if a . b = 0. An antichain is any set of pairwise disjoint elements. Clearly, an antichain R is maximal if and only if sup R exists and equals 1. Let us note that if a given algebra A has infinite chains, then it has also infinite antichains and conversely. Indeed, if (a, : n < w ) is a decreasing chain, then the elements b, = a , - a,+l form an infinite antichain. Similarly, if (b, : n < w ) is an infinite antichain, then the elements a , = bo . . . b, form an infinite chain.

+ +

8.1. REMARK.Any infinite algebra A has infinite chains and antichains. To prove this remark, let us suppose that each chain (and hence each antichain) is finite. Then the algebra A must be atomic. Indeed, if a > 0 and a is not an atom, then there is an a1 < a and a1 > 0.If a1 is not an atom we find an a2 < a1 and a2 > 0.After a finite number of steps a > a1 > . . . > a , we obtain an atom a,. Now, the set At of all atoms of A is an antichain and hence it is finite. But then, an arbitrary element a can written in the form a = E At : I 5 a } . Since there are finitely many such sums, we infer that A is finite. 8.2. EXAMPLE. In the algebra 'D(X), of open domains of a space X , each antichain has cardinality 5 d ( X ) (= the least cardinality of a dense subset D X ) . Hence, for a separable space X , the algebra D ( X ) has at most countable antichains.

8.3. EXAMPLE. MEASURE ALGEBRA. A nonnegative function m : A (finite) measure if m ( 0 )= 0, m(1) > 0 and m is a-additive, i.e.

+

w is a

m(C R ) = C a E R 4.1 holds for all finite or countable antichains R 5 A (and such that C R exists). As in the case of fields of sets, more general measures (e.g. infinite, signed etc.) can be defined on Boolean algebras.

A measure m on an algebra A is called strictly positive if m ( a ) = 0 holds for a = 0 only. We say that a Boolean algebra A is a measure algebra, if A is a-complete

and there is a strictly positive measure on A. It is easy to see that every measure algebra has the ccc, i.e. that each antichain is at most countable. More generally, assume that m is a strictly positive measure

CH .1 BOOLEAN ALGEBRAS

22

on an arbitrary algebra A (rn can be even finitely additive) and let S antichain. Then S = UnEw S,, where S,={a€S:

1

rn(a)>-} n+l

Clearly, every set S, is finite (has less than (n follows.

2 A be an

forncw.

+ 1) . m(1) elements) and the claim

We already know from 5.4 that each Boolean isomorphism f : A + Iis also an order isomorphism. The converse is also true. 8.4. LEMMA. Each order isomorphism f : A + B is a Boolean isomorphism as

well.

PROOF.We have f(0)= (D and f(1) = 1,since (D and 1 are the smallest and the greatest element, respectively. Generally, it is clear that f(sup S ) = sup f[S]

whenever sup S exists.

since from c = f ( a ) * f ( - a ) > (D it follows b = f-'(c) impossible. Hence, f ( - u ) = -f(a), in view of 2.8.

> 0 and b 5 a,

-a,

which is

QED

Now, we prove the following 8.5. THEOREM. If S & A and T C B are dense subsets of complete algebras A and B, respectively then any order isomorphism f : S + T can be extended to an isomorphism of A onto B.

PROOF.Clearly, we have

(*I

z.y=(D

G

f(z). f ( y ) = 0

for arbitrary z, y E A,

$3. CHAINS A N D ANTICHAINS

23

Let us denote S(a) = {x E S : x 5 a } . From the assumption it follows that a = supS(a), for each a > 0.Indeed, if it was a > supS(a), then for some x E S we would have x 5 a - supS(u) and hence x E S and x . supS(a) = 0 ,which is impossible. Let us define f(@) = 0 and f ( u ) = supf[S(a)]

Since f is an order isomorphism, for every

i.e.

5

for a

> 0.

E S we have

T is an extension o f f .

Assume now that a # b. Thus, there is an x E S such that e.g. x 5 a - b. Hence x E S(u) and x . S(b) = @. It follows that f(z) 5 T ( u ) and, using (*), that f(x) . f[S(b)] = @. The latter implies f ( x ) . T(b) = 0 and consequently f(x) 5 ?(a) - f ( b ) , i.e. f(a) # T(b). Obviously, 7 is onto B because if b E B and

b > 0,then b = supT(b) = supf[S(a)] = f(a), where a = sup f - l [T( b)]. Finally, f preserves the ordering: a 5 b implies S ( a ) C S(b) and hence T ( u ) = supf[S(a)] 5 supf[S(b)] = f(b). Thus, f : A + B is an order isomorphism and by Lemma 8.4.,a Boolean isomorphism as well. QED The binary tree of height w (or the Cantor tree) is the set 2<w = Un<w{O, of all finite zero-one sequences, partially ordered by the reverse inclusion s

5t

E

s

_> t

(i.e. s extends t ) .

Thus, the empty sequence Q, is the greatest element, its immediate predecessors are one-term sequences (0) and (l),on the lower level there are two-term sequences and so on.

CH.1 BOOLEAN ALGEBRAS

24

We say that an element a > 0 (of a given algebra A) splits, if there are elements a o , al > 0 such that a = a. al and a0 . al = 0.If A is atomless, then each element a = ag > 0 has a splitting a ( o ) , ~ ( 1 ) . Repeating this for a(o) and a(1) we obtain four elements a, corresponding to two-terms sequences s. By induction on the length of s E (0, I}<", we define elements a , so that the set { a , : s E {0,1}<"} is order isomorphic to the Cantor tree.

+

Using this construction we prove the following.

8.6. THEOREM. All complete atomless algebras having countable dense subsets axe isomorphic.

PROOF.Let A satisfy the assumptions of the theorem. Thus, A has a countable dense subset {d, : n E w } . We define by induction a new dense subset {a, : s E {0,1}<") order isomorphic to the Cantor tree. Let ag = 1. If do = 1,then let ag split arbitrarily. If do < 1,then define a(o) = do and a(1) = -do. Assume that we already have elements a,, for s E {O,l}, and that a, 5 d,-1 for some s E (0, l}". Now, if a, 5 d,, for some s E (0, l}", then we split all the = n, hence there is an s such that elements a, arbitrarily. Otherwise, since 0 < d, . a , < a, and this a , we split into a, . d, and a , - d,. Let the remaining at split arbitrarily. The resulting set {a, : s E {0,1}<"} is not only order isomorphic to the Cantor tree but also it is dense in A. Now, an application of 8.5 finishes the proof. QED

C,a,

For example, the algebras 'D(X), of open domains, are all isomorphic, for X = Wn, [a,b], (0, 1)" (i.e. for an n-dimensional Euclidean space, an interval, the Cantor space). More generally, D ( X ) is atomless for any regular (because then open domains form a base) space X having no isolated points. If X has countable weight w(X), then D ( X ) has a countable dense subset.

PROBLEMS.

A . FINITE ALGEBRAS 1. Every finite Boolean algebra is atomic.

2. A finite algebra A is isomorphic to the field P ( X ) , where X is the set of all atoms of A. (Define f(u) = { z E X : z 5 a } ) . 3. Finite algebras with the same number of elements are isomorphic.

PROBLEMS

25

4. Each finite algebra can be embedded into any infinite one.

B. POWER-SET ALGEBRA P ( X ) 1. A complete atomic algebra A is isomorphic to P ( X ) , where X is the set of all atoms of A. (Tarski).

c

2. If a field F P ( X ) contains all singletons some R G F , then sup R = U R.

{I}, for

x E X and sup R exists, for

3. The family of finite unions of intervals of the form

where a, b are real numbers (including a = atomless field of sets.

-00,

b =

+00)

is a noncomplete

4. Every infinite a-complete algebra A contains an isomorphic copy of the field P(w). (There is a maximal antichain {ai : i E w } . Define f(x) = supiEza*,for x g w.) 5. Each algebra A having a countable dense subset D can be embedded into P ( w ) . (Let F, be an ultrafilter containing d n , where D = {d, : n E w}. Define f ( ~= ) {n: a E Fn}).

C. FILTERS AND

FACTOR ALGEBRAS

1. Any filter is an intersection of ultrafilters. 2. An infinite algebra A has always a nonprincipal ultrafilter. (The family of complements of atoms is centered).

3. Let I , J be ideals in A and assume that I G J . If J * = { [ a l l : a E J } , then J* is an ideal in A / I and A / J is isomorphic to A / I / J * .

c

4. For a topological space X define N ( X ) = { A X : int[x] = Q, } (nowhere dense sets) and S ( X ) = { A G X : Fr(A) E N ( X ) } (sets with nowhere dense boundary). The family S ( X ) is a field containing all open sets, N ( X ) is an ideal in S ( X ) and the factor algebra S ( X ) / N ( X )is isomorphic to the algebra of domains D ( X ) . 5. If A is a a-complete algebra and I is a a-ideal in A, then the factor algebra A / I is again a-complete.

26

CH.1

BOOLEAN ALGEBRAS

D. COMPLETION

For a partial ordering ( X ,5 ) let us denote U, = {y E X : y 5 x} and regard X as a topological space in which the family {U, : x E X } is an open base. 1. If X is separative, i.e. the following holds for arbitrary elements z and y:

x non 5 y

implies

32 5 x ( U z n U, = (B),

then the sets U, are open domains, for each x E X and the map x order embedding of X onto a dense subset of the algebra ' D ( X ) .

H

U, is an

We say that a complete algebra A* is a completion of A, if there is an embedding f : A .+ A* such that f[A \ {O}]is dense in A*. 2. Each Boolean algebra A has a completion. (Take X = A \ (0) in D.l.) 3. A completion of A is unique up to isomorphism. 4. Every complete algebra A is isomorphic to a D ( X ) , for some space X .

E. GENERATORS AND

HOMOMORPHISM

1. Distributivity law. For an arbitrary matrix

of elements of a given algebra A the following equalities hold

and

where F is the set of all functions f : { 1, . . . ,m} -+ { 1,.. . ,n}. 2. The subalgebra [R],generated by a subset R form

E A, consist of finite sums of the

C ~ l a l ... . .&,,an

where ai E R,

Ei

E {-1, +1} for i = 1,.. . , n (Recall that +1. a is a and - 1 . a

is the complement -u).

PROBLEMS 3. If R

A and a function f : R

+ 1 satisfies the

27 condition

for arbitrary a l , . . . , a , E R and ~ 1 , . .. , e n E {-1, +l}, then f can be extended to a homomorphism 7 : [R]-i B (another form of the Sikorski theorem 7.3.). Also, i f f is one-to-one and the implication (*) is strengthened to an equivalence, then f can be extended to an embedding 7 : [R]+ B.

A family R C A is called a family of free generators, if A = [R]and for each Boolean algebra B any function f : R --+ 5 can be extended to a homomorphism f : A-iB. 4. If A = [R],then R is a family of free generators if and only if R is an independent family (i.e. ~1 'a1 . . . . ' E , ' a , > CD for any distinct a l , . . . a , E R and arbitrary El,

...

,E"

E (-1, +l}).

5. Let X = {0,1}" be the Cantor space of weight K . The open-closed sets U, = { z E X : .(a) = 0 } , for LY < K are free generators of the algebra B ( X ) .

6. Algebras A,

B having the same number of free generators are isomorphic.

F. MEASUREALGEBRA Let m be a finite measure on a a-field F of subsets of a space X . The factor algebra A(m) = F/A(m), where A(m) = { A E F : m(A) = 0 } , is called the measure algebra of m. Let m ( [ A ] = ) m(A) for A E F . 1. For any m the algebra A(m) is a-complete and TTI is a strictly positive measure on A(m) ((2.5.). Hence, A(m) is a measure algebra. (8.3.) 2. If an algebra A is a-complete and has the ccc (each antichain is countable), then A is complete. (sup R = sup Ro, where Ro C R is a maximal antichain in R). Hence, every measure algebra is complete. 3. If a a-complete algebra A has a a-finite strictly positive measure, then A has also a finite strictly positive measure (and hence A is a measure algebra). Let Bo(X) denote the Bore1 sets in X . 4. The algebra A(X,), of an n-dimensional Lebesgue measure, is isomorphic to

Bo(W")/A,, where A, = { A E Bo : X,(A) = 0 ) . 5. The Baire algebraB(X)/M(X) is isomorphic to Bo(X)/M(X)nBo(X) (cf 5.13). 6. Let X = X1 be the Lebesgue measure on the real line W.

ALGEBRAS CH.1 BOOLEAN

28

a) The measure algebra A(X) is isomorphic to A(XJA), where A is any open

segment of reals and XIA is the restriction of X on the measurable subspace A C W.(There is a homeomorphism preserving measure null). b) If A, B are Borel sets and X(A A B ) = 0, then the measure algebras A(X1A) and A(XIB) are isomorphic. c) The function

f

:

C,

+ W,

defined by the formula

f(z) = CiE,2'+' 42)

for z = ( z ( i ): i E w ) E C,,

maps the Cantor space C , = (0, 1}, onto the interval [0,1] of reals. The equality m(A) = X(f[A]), for A E Bo(C,), defines a Borel measure in C, such that for basic open-closed sets U, = {z : s z} we have m(U,) = 2-", where n is the length of the finite sequence s. The measure algebra A(m) is isomorphic to A(Xl[O, 11). (After deleting eventually constant functions ,, f becomes a homeomorphism). from C d) The measure algebra A(m), where m is m in c) above, is isomorphic to the algebra of the n-dimensional measure, for any n 1 1. Thus, for the Lebesgue measure we have a single measure algebra, no matter where the measure is defined: in W" on a segment or in the Cantor space. 7. In a metric space X each finite Borel measure m is regular in the following sense: for each A E Bo(X) there are sets F E F,,(X) and H E G a ( X ) such that F A H and m ( F ) = m(A) = m ( H ) . (The family of sets with this property is a a-field containing all open sets). 8. If m is a finite and atomless (i.e. the measure algebra of m is atomless) Borel measure in a metric separable space X , then there is a partition X = A U B , where A E G a ( X ) with m(A) = 0 and B is meager.

9. The measure algebra A(X), of the Lebesgue measure say - on the interval [0,1], is not separable, i.e. it has no countable dense subset. (If [A,] E A(X), for n E w , then there are sets Bn C An, so that X(Cn<,[Bn]) < 1).

G. COUNTABLE

ALGEBRAS

1. Any two countable atomless Boolean algebras are isomorphic. (There is a dense Cantor tree { a , : s E {0,1}<"} so that each element a > CD is a sum of the form as,

+ .. . +

.)

PROBLEMS

29

H. PRODUCT O F BOOLEAN ALGEBRAS Let {A, : i E I}be a family of Boolean algebras. The product A = defined as follows. The universe A of A is the product

n,,,A, is

A= nA, iEI

of universes, i.e. A consists of all functions a defined on I such that a, E A , , for

each a E I . The operations in A are defined coordinatewise (a, :

2

E I)

+ (b, : z E I ) = (a, + b, :

2

E

I)

(a, : i E I) . (a, : z E I ) = ( a , . b, : z E I ) -(a, :

E

2

I) = ( - a *

:

2

E

I).

Also 1~ = (IA,: z E I) and @A = ( 0 ,: a E I ) . If I = (1,. . . ,n}, then the product A, is usually denoted by A1 x . . . x A,.

nlEI

1. The product A =

nrEI A, of Boolean algebras is a Boolean algebra.

2. Any power-set algebra P ( X ) is (isomorphic to) a product of two element algebras.

3. If nonzero elements e l , . . . , en of a given algebra A satisfy the conditions

1=el+

...+en

then A is isomorphic to A1 lel x

and e , . e j = @

fori#j,

. . . x Alen.

4. If E is a maximal antichain in a complete algebra A, then A is isomorphic to the

product

neEE Ale.

n,,,

5. For any i E I, the function hi : A, --$ A, defined by the formula hi(.) = a, (the i-th projection) is a homomorphism onto the i-th axis A,.

6. If F, is a filter in A, and H is a filter on the set I of indices, then the condition u € F

{ZEI: ajEFi}EH

niEIA,. If H and F,, for each i E I are

defines a filter F in the product A = ultrafilters, then also F is an ultrafilter. I. NUMBER OF FIELDS O F SETS

1. Let A be an arbitrary Boolean algebra. a) If F A is a filter, then [ F ]= F U ( - F ) . Hence if F is an ultrafilter, then [ F ]= A.

c H . 1 BOOLEAN ALGEBRAS

30

b) If F # G are nonmaximal filters, then (F]# [GI. c) If Fi, i = 1,. . . , 4 are distinct ultrafilters, then Fl n F2

# F3 n F4.

2. Let X be an infinite set and A = P ( X ) . a) there are 22' distinct fields of subsets of X , where K. = cardX. (Theorem 6.5. and 1.1) b) There are 22a nonisomorphic fields of subset of X . (The fields [F]are atomic and hence there are 2" fields isomorphic to [ F ] ) .(fieniche).

J. BOOLEAN RINGS

A Boolean ring is an algebraic ring ( R , $, 0, 0, 1) (possibly noncommutative), in which every element a is an indempotent, i.e. a o a = a . 1. Any Boolean ring R is commutative and a = -a, for each a E R.

2. Any Boolean algebra A determines a Boolean ring (with the same universe), if we define a @ b = a A b, a o b = a . b, 0 = 0 and 1 = 1. Conversely, if R is a Boolean ring, then defining a b = a @ b $ a o b, a . b = a o b, -a = a @ 1, (D = 0 and 1 = 1 we obtain a Boolean algebra. The above described correspondence is one-to-one.

+

3. Let R be an arbitrary ring and define Ro as follows. The elements of Ro are all idempotent elements of R (and belonging to the center of R, if R is noncommutative). Multiplication is the same as in R and the new addition is a$b = a+b-2ab. Thus defined Ro is a Boolean ring (called the ring of idempotents of R).

Chapter 2 Gaps and limits

$1. INTRODUCTION. For an infinite set X the symbol fin(X) will denote the family of all finite subsets of X. We write shortly fin, if it is clear what X we have in mind. Obviously, fin(X) is an ideal in the power-set algebra P ( X ) and hence we can form the factor algebra P(X)/fin(X). Clearly, if card X = card Y then P(X)/fin(X) is isomorphic to P(Y)/fin(Y). In a large part of this book we shall be concerned with the algebra P(X)/fin, for a countable X and usually we take X = w as a fixed representative. By definition, any algebra P(X)/fin consists of abstraction classes of subsets A X according to the congruence

c

A=fi,B

=

AABEfin.

The zero of P(X)/fin is fin:

[A]= CD

=

A E fin.

Let us introduce the following notation, which will be extensively used in the sequel. For sets A, B X: A =* B ( A almost equals B ) , if A A B E fin ( A is almost contained in B ) , if A \ B E fin B A if A B and non A =* B , i.e. if A \ B E fin but A (-&B ,

c

c*

c*

B \ A 4 fin, ( A is almost disjoint from B ) , if A n B =* $3, i.e. if AnBEfin. Thus, we have the following equivalences A =* B = [A]= [B] (in P(X)/fin) A IB

1.1.

A C, B

E

[A]5 [B]

(Boolean ordering in P(X)/fin)

A S;. B

E

[A]< [B]

(the strict ordering)

A IB

=

[A]. [ B ]= 0 31

(Boolean disjointness)

CH.2 GAPSA N D LIMITS

32

Let us assume now that X = w. For A w and n E w the set A \ n is a cofinal segment of A : A \ n = { z E A : i 2 n } (since n = ( 0 , . . . ,n - l}, in our notation). If [A]= a, i.e. if the set A is a representative of an a E P(w)/fin, then each cofinal segment A \ n, for n E w is also a representative of a. Let us note also

AC,B

=

3n[A\ncB]

A I B

G

3n[(A\n)nB=p].

and

Now, we prove some simple properties of the algebra P(w)/fin. 1.2. The algebra P(w)/fin has antichains of cardinality c (continuum).' Indeed, from 1.1 we have

[A]>(D

=

A e f i n and

[ A ] . [ B=(D ]

=

A I B

and hence any antichain E of P(w)/fin has a form E = { [ A ] : A E R } , where R is an almost disjoint family (i.e. R consist of infinite subsets of w, any two of them are almost disjoint). It remains to recall a classical result of Sierpiliski - there are almost disjoint families R of cardinality c. To see this let z = (zn: n < w) be a sequence enumerating all the rational numbers. For any irrational number (Y there is an infinite subset A , w such that the sequence slA, = {s, : i E A,} converges to a. Obviously, A , I AD,for (Y # /3 and hence R = { A , : Q E W \ Q} is as required. From 1.2 it follows immediately 1.3. The algebra P(w)/fin has cardinality c. It is almost obvious that 1.4. The algebra P(w)/fin is atomless. Indeed, if [A]> 0,then A 4 fin and hence A splits into two infinite disjoint sets A = B U C. Hence [A]> [ B ]> (D. Any descending chain ( a , : (Y < y) (in an arbitrary Boolean algebra) is called a limit (or a tower) of length (or type) y, if it cannot be further extended, i.e. there is no a > 0 such that a < a,, for each (Y < y. For example, the chain (w \ n : n E w) is a limit of length w in' the power-set algebra P(w).

33

$2. DOMINANCE

1.5. THEOREM (Hausdorff [Hz], [Hs]). In the algebra P(w)/fin there are no limits of length w (and hence of any length y < w1).

PROOF. Let ( a , : n < w ) be a decreasing chain. Choose representatives A , E a,, for each n < w. We have An+1 $ A ,l i.e. *

c*A,,

A,+1

and A ,

\ A,+1

$n!fi

for each n

<w.

Replacing, if necessary, An+l by its far enough cofinal segment we may assume A , , for each n < w . Define a sequence (k, : n < w) by induction that An+l as follows. Let LO be an arbitrary element of Ao: k,+1 is the smallest number in

c

A,+1

\ A,

exceeding all the ko, . . . , k,. Thus we have

{k, : i 2 n } c A , Hence if A = {ki : i

for each n

< w.

< w}, then [ A ]> 0 and [A] 5 [A,] = a,, for each n < w . QED

$2. DOMINANCE. Let us consider two relations defined for functions f,g

:w

4

w

as follows

f =* g

f 5. g

c

{i E {i E

w : f(i)

# g(i)}

E fin

(f almost equals g )

w : f(i)

> g(i)}

E fin

( g dominates

f).

Obviously, the relation =* is an equivalence and 5. is a partial pre-ordering (it means f imply f =* g , the reflexive and transitive) on the set w w . Since f g and g relation 5 , is in fact a partial ordering on the set w w / =* of abstraction classes. The g and non f g) can be described corresponding strict ordering <+(that is f thus

=.

f <, g if and only if f(i) 5 g(i) holds for large enough i and f ( i ) < g ( i ) for infinitely many i E w. Let

XA

denote the characteristic function of a set A . Clearly,we have then

for arbitrary A , B

w,

and

cH.2 GAPSAND

34

LIMITS

s

We say that a set F ww is bounded if it is bounded from above with respect to the relation I,,i.e. if there is a g E ww such that f 5. g holds for each f E F (we write shortly F g in this case).

<*

2.1. Every countable set F = { fj : j < w } is bounded.

+

Indeed, if we define g ( n ) = m a { fj(n) : j 5 n} 1, for n E w then F I*g. The smallest possible cardinality of an unbounded (i.e. not bounded) set will be denoted by lb. Thus

lb = mintcard F : F is unbounded}. We call a set F ww dominating, if it is cofinal in ww with respect to the ordering I* i.e. if, for each function g : w + w there is an f E F such that g f. Of course, any dominating set is unbounded. Hence, if we introduce another cardinal coefficient d = min{card F : F is dominating}, then we have inequalities w1

Ilbldlc.

2.2. THEOREM. The cardinal lb is regular and cf(d) 2 lb.

PROOF.Let F be an unbounded set of cardinality Ib. Suppose that Ib is singular. Then we have F= UF;, i€I

where card I < lb and card F, < Ib,for each i E I . Thus, the sets F, are bounded for each i E w , F; g, for some g; E ww. Since card I < lb, there is a bound g of the set {g, : i E w } . It follows that F <* g, contradicting our supposition. Thus, lb must be regular. Similarly, let D be a dominating set of cardinality d and suppose that cf(dl) < lb. Thus, we have a decomposition D= Dj,

u

j€J

where card J < lb and card Dj < d for each j E J . Thus, no of the sets Dj is dominating and hence there are functions gj such that for all j E J : gj non

I*f

for each f E Dj.

$2. DOMINANCE

35

Now, if g is a bound of the set {g, : j E J } , then we have g f for no f E D, i.e. D is not dominating, contrary to the assumption. Hence cf(d) 2 lb, which finishes the proof.

QED

The next lemma carries more information on unbounded sets.

2.3. LEMMA.Tf F = { f a : a < lb} is an unbounded set of cardinality lb, then there exists an increasing (with respect to 5.) sequence (ga : a < b) dominating the set F (i.e. Va3p(fa 5 g p ) ) . Moreover, we may claim that the terms ga can be taken from a dominating set D, given in advance.

PROOF.The required sequence ga is defined by induction. The first term go is chosen so that go. 2f0 and go E D. If we have already the terms gc E D so that 96. 2{gv : q < 5) and gc+ 2fc, for all [ < a < lb, then we choose ga as any bound of the set {gc :

E < a} U { f a } . Of course, we can find such a bound in D.

QED

Any increasing (in the sense of dominance) unbounded sequence is sometimes called a limit in ww (comp. 1.5). From the lemma it follows immediately

2.4. COROLLARY. The ordinal number lb is the length of a shortest limit with terms in a given dominating set.

A dominating and increasing sequence ( f a : a < y) is called a scale (of length 7). Thus, any scale is also a limit in w w . It is consistent (with set theory) to assert that a scale exists. Indeed, assume CH (the continuum hypothesis). Then we have lb = d = w1. Thus, we may consistently assume that lb = d and then - in the lemma 2.3 - we can take as the F a dominating set of cardinality lb = d. The resulting increasing sequence (ga : a < lb) is dominating (because it dominates a dominating set) and hence - a scale. However, the existence of a scale is independent of set theory (see the theorem 5.11). Let us note that if a sequence ( f a : a < y) is a scale then obviously y 2 (a and there is a subsequence (fca : a < d) of length d, which is a scale too. Indeed, if

D = {ga : a < d} is an arbitrary dominating set, then for each a < d we can find a to < y such that ga 5. fern. The sequence (ft- : a < (a) is dominating, since it dominates a dominating set D and hence it is scale of length 8. 2.5. THEOREM. A scale exists if and only if the equality lb = d holds.

36

c H . 2 GAPSAND LIMITS

PROOF.If there is a scale, then (as remarked above) there is a scale s =

(f, :

(Y

< d)

of length d. Let F be an arbitrary unbounded set of cardinality Ib. Applying the lemma 2.3 to F and D = {fa : (Y < d} we obtain a subsequence t = (fern : (Y < Ib) of length lb, which dominates F and hence t is unbounded, hence cofinal in s, hence dominating, i.e. - a scale. Thus, Ib = d. The converse implication has been proved earlier (after 2.4). QED Now, we shall discus a connection between the cardinal 03 and some gaps in

P( w )/fin. First, let us modify a little the definition of gaps (given in 7.2, Chapter 1) viz. the condition S 5 T will be replaced now by S IT . Thus, a pair of subsets (S,T) is a gap in the new sense, if S I T , i.e. if a . b = 0 for each a E S and b E T . Of course, the difference is completely inessential - replacing the upper class T by -T we pass from one of these definitions to the other. An element c fills (or separates) a gap (S,T ) ,if S 5 c and c I T . A gap K = (S,T) is said to be unfilled, or non-separated, if no c fills K . We often shall be dealing with well ordered gaps, that is gaps K = ( S , T ) where both classes can be represented as increasing chains S = (a, : a! < n) and T = { b p : p < A). Such a gap K is often called a (n, A)-gap. We can always assume that the type (.,A) of K is a pair of regular cardinals, since K contains a cofinal subgap L (in obvious sense) of type (cf(n),cf(A)) (note that K can be filled iff L Can). Finally, if X is an infinite set and S,T

P ( X ) are two families of infinite subsets n B E fin(X) for each A E S and B E T). From 1.1follows that any such K determines (and is determined by) a gap in P(X)/fin

of X , then K = ( S , T ) ,is called a gap on X , if S I T (i.e. A

K* = ( { [ A ]: A E S } ; {[B] : B E 2’)). Clearly, the gap h’* is well ordered in P(X)/fin if and only if A’ is well ordered by the almost-inclusion In this case the types of K* and K are identical. We C and C IT ) if and only note also that a set C X separates the gap K (i.e. S if [C] separates K* in P(X)/fin.

c

c*.

c*

$2. DOMINANCE

37

We shall often make use of the above remarks without specific references. Let us prove now the following characterization of lb.

2.6. THEOREM (Rothberger [RI]). The cardinal Ib is the smallest cardinal K for which there exists a non-separable ( K , w)-gap in P(w)/fin.

PROOF.For an arbitrary function f : w

+w

let us denote

Clearly, we have then

First, we show that there is an unfilled gap of type (lb,w). To see this, let : a < lb} denote an unbounded set of cardinality Ib and let En = { n } x w (a vertical axis). If a set C w x w is almost disjoint from all the axes En, i.e.

F = {f,

c

C n EnE fin(w x w )

for each n E w ,

then we can define a function

g ( n ) = max{j E w

:

(n,j)E C n E n }

for n E w ,

(we assume g ( n ) = 0, if C n En= 9 ). Since F is unbounded, we have f a non <* g , for some a < Ib. From (*) it follows that A(f,) non A(g) and even stronger C, because C C A ( g ) . A(f,)non This means that no set C separates the gap

c*

c*

K = ( { A ( f , ) : a < b} ; {En: n E w } ) of type (lb,w) on the set w x w . Hence, there is an unfilled (K,w)-gap on w and consequently in P(w)/fin as well. Now assume that

L = ( { A , : a < K } ; {B,: n < w } ) is a non-separated (tc,w)-gap on the set w . We have to show that K 2 Ib. We may assume that the sets B, are disjoint (for B, can be always replaced by the sets B,\(BoU...uB,,-l)) and that w = Un<w B, (putting single points from W\U,,<~B,

38

cH.2 GAPSA N D LIMITS

to the B’s) because so modified gap remains unfilled. For each n E w we choose a one-to-one function h, mapping B, onto the vertical axis En = {n} x w . Then h = Un<w h, maps w one-to-one onto w x w so that h[B,] = En, for each n E w and h[A,] is finite on each vertical axis, for all Q < K . Thus, we have an unfilled (K , w)-gap ({h[A,] :

Q


; {h[B,] : n

<w})

on the set w x w . From (*) it follows that the functions

form an unbounded family and hence

K

1 lb.

QED

According to Corollary 2.4 the family F in the above proof can be replaced by a limit (i.e. an unbounded increasing sequence) (fa : Q < Ib). Then we have A(f,) &. A(fp), for all Q < /3. Thus, the lower class of the gap K is an increasing sequence. The same can be said about the upper class {En: n < w } for, it is countable and hence each En can be always replaced by Eo U . . . U En. In other words, the gap K can be claimed to be well ordered. Hence, we obtain the following

2.7. COROLLARY. The cardinal Ib is the smallest cardinal K for which there is a well ordered non-separated ( K , w)-gap. From Ib > w (see 2.1) we get immediately

2.8. COROLLARY. In the algebra P(w)/fin every countable (i.e. of type ( w , w ) ) gap is filled. Nevertheless there are always unfilled (q,wl)-gaps. This famous theorem of HausdorfF will be proved in the next section.

$3. HAUSDORFF GAPS. For almost disjoint sets A, B E w denote p(A, B ) = min{ k E w : (A \ k) n B = 9 }. Thus, for disjoint A, B we havep(A, B ) = 0 and if max AnB = k then p(A, B) = k+l.

$3. HAUSDORFFGAPS

39

For a set A E w and a family R s P(w) we say that A is n e a to R (abbr. A nt R) if the following condition is satisfied 3.1.

the set

{ B E R : p ( A , B ) < k} is finite for each k E w .

Obviously, if R is a finite family, then we have A nt R, for each set A C w . Note that A nt R holds if 3.1. is true for large enough A E w . Hence 3.2.

H A nt R and A C, C, then C nt R as well.

Now, we shall prove the following 3.3. THEOREM (HausdorfT [H3]). The algebra P(w)/fin contains well ordered non-separable gaps of type (w1,wl).

PROOF.We define by induction two increasing, in the sense of

sothatA,nBa=(9,foreacha<w1

c*,sequences

and

is a gap on the set w having the following property

(*)

A, nt {B,

: 9

< a} for each a < w1.

From (*) it follows that the gap L cannot be filled. Indeed, assume that a set C separates L, i.e. A, C, C and C n B, =, Q, , for each a < w1. Then by (*) and 3.2 we have 3.4.

C nt {a,

: 9

< a } for each /3 < w1.

On the other hand, the function f(v) = p ( C , B , ) must be constant on some uncountable set 2 w1. Let us take a p < w1 so large that the set 2 n /? is infinite. Obviously, we have C non nt {B, : 9 E 2 n p } , and hence C non nt {B, : 9

< p } , which contradicts 3.4.

We pass to the construction of the gap L. Let Ao, Bo be arbitrary infinite disjoint sets such that the difference w \ (A0 U Bo) is still infinite. The latter condition will be assumed at each stage of the induction. Assume that we have defined already the sets A t , B, for 6,9 < a,as required. If a is a successor a = B 1, then, since

+

40

c H . 2 GAPSA N D

LIMITS

s1

s*

\ ( A p u Bp) 4 fin, there are disjoint sets A,, B, such that Ap A,, Bo B, and A , U B, has an infinite complement. Obviously, the condition (*) holds too.

w

Now, assume that a is a limit ordinal. Then the sequence @\(At U B t ) : [ < a ) is almost decreasing and, in view of the theorem 1.5, there is an infinite set H C w such that for each [ < a. H ( w I \* ( A t U Bt)

If we denote E = w \ H , then we have At,

B, c*E

for all ( , q < a ,

and w \ E is infinite. Since there are no unfilled countable gaps (see 2.8), there exists a set C separating the family { A t : ( < a } from {B, : 7 < a } , i.e. 3.5.

AcC.C

and C I B ,

C

Of course, we may assume that cn E. According to 3.2 we have

C nt {B,

: 7

forall[,q
E replacing - if necessary

-

the set C by

for each ,f? < a.

< ,f?}

However, this does not mean that C nt {B, : r] larger. In order to do this let us consider the sets

< a } . We have to make C even

and define inductively an increasing sequence

so that

Cn+lnt {B,

: r] E

Wn}

for each n

< w.

Assume that Co,. . . C, are defined. If W , is finite, we can put Cn+l = C,. If W , is infinite, then it must be cofinal in a and its order type is w . Indeed, each initial segment W , n p, for a ,f? < a is finite, since we have C nt {B, : q < p } . Hence, we can write

W , = {q, : i < w }

where

90

< 71 < . . . and supq, = a. i<w

$4.THEPAROVICENKO THEOREM.

41

Clearly, each of the sets

Zi = Bqi \ (BqoU . . . U Bgi-,) intersects the set E on an infinite part. Thus, we can form a diagonal set

D

= { jn : n

<w}

where j , E Zn n E

and j,

> n for each n < w .

Now, let us define Cn+l= C, U D. From the construction it follows that D I B,,,, for each n < w . Hence D I B,, for each 9 < a and consequently we have that Cn+lI B,, for each 9 < a. Since obviously D nt {B,,, : n E w } holds, hence also

Cn+lnt { B , : 9 E W,}, which finishes the inductive step. Similarly as in 3.4we can separate the sets {C, : n < w } from { B E: ( < a} within E: there exists a set A such that C, A , for each n < w and A I BE,for each ( < a. Moreover A E . Additionally, we may demand that CO A (enlarging if necessary A and E by a finite number of points). Now, it is clear that we have

c*

c

c

A nt { B , : 9 < a}. Indeed, suppose that for some k the set

W = (9 < a

:

p(A,B,) < k }

wk

w

wk

is infinite. Then also the set (9 E : p ( A , B , ) < k} is infinite, since C (because CO A). This means that A non nt { B , : 9 E W k } holds, which is impossible, since we have

c

Thus, if we define A , = A and B , = E \ A , then (*) holds, A , n B , = (3 and the complement w \ ( A , u B,) is infinite. This finishes the proof of the theorem. QED

3.6. R E M A R K . Let us note that any two elements a I b of P(w)/fin satisfying a

+ b < 1 are in different classes of some unfilled (w1,w1 )-gap. $4. THEPAROVICENKO THEOREM. Let us write down some basic properties of

the algebra P(w)/fin. It has the cardinality c (1.2) and

c H . 2 GAPSA N D LIMITS

42

1. it is atomless (1.3), 2. it has no countable limits (1.4), 3. it has no countable unfilled gaps (2.8). Any Boolean algebra A having the above properties 1 - 3 is called a Parovitenko algebra. Thus, the algebra P(w)/fin is a ParoviEenko algebra of cardinality c. We prove below that - assuming the continuum hypothesis - it is a unique Parovitenko algebra of cardinality c. First, we show a simple 4.1. LEMMA.Everyfinitefamily {ao,. . .an},ofnon-zeroelementsofan atomless Boolean dgebra A, has a refining antichain (i.e. antichain {bo, ... ,b,} such that bo 5 ao, .... b, 5 a,). In addition, if A has no countable limits, then the same holds for countable families {an : n E w } . PROOF. By induction on n. If we have a family { aC , ~ . . . ~ U , } then , by the inductive hypothesis there is an antichain {bo, .... b n - l } refining {ao, .... u , - l } . Let us define bn = a, - (bo ... b , - i )

+ +

provided b,

> 0.If b, = 0,then we have an

5

+ ... + bn-1

and hence a , . b,

>0

for some i

< n.

Since the algebra A is atomless, we have a splitting

an*bi=c+d

wherec,d>0

and

c.d=0.

We replace the element b, by c and we take d as the n-th element of our antichain. Now assume that A has no limits of type w and let { a , : n E w } be a countable family of positive elements of A. Using the already proved part the lemma we define induction a matrix a0

a1

1 b: 2 by 1 .. 1 ... b: 2 .. . b', 2 ..

... an 1 ... ...

. . . . .. . . . . . . . . . . . b; 1 .. . . . . .. . . . . .

$4.THEPAROVICENKO THEOREM.

43

in which b: = ao, the column { b: , bi } is an antichain refining { b t , a1 } and, generally, for each n E w the column . . . b:} is an antichain refining {b:-,, . . . b,-l, n-1 u,}.

{a:,

In particular, any row a,, b t , b;+,, . . . is a decreasing sequence and - by our assumption - has a lower bound b,. The last column { b , : n < w } , consisting of that bounds, is then a required antichain. QED We can prove now the announced characterization of the algebra P(w)/fin. 4.2. THEOREM (ParoviEenko [PI]). Every Boolean algebra of cardinality 5 w1 be embedded into any ParoviEenko algebra. If CH (the continuum hypothesis) is assumed, then any two ParoviEenko algebras of cardinality c are isomorphic. can

PROOF.Let A be a Boolean algebra with card A 5 algebra. Clearly, A can be represented as a union

w1

and B a ParoviEenko

of an increasing chain of countable subalgebras fulfilling the following conditions =

4.3.

{a,1) for some a E A and each a < w1

A,+1 = A,[4

A, =

u

for each limit ordinal a

Ap

< w1.

B<,

Indeed, the equations 4.3 can be regarded as an inductive definition of the sequence (A, : a < w1). The generator a in the successor case should be defined thus: a = inf(A \ A,), where inf refers to a fixed well ordering of A in type w1. Obviously, we have an embedding fo :

A0 -+

B

viz.

fo((D) = 0,

f o p ) = n.

We shall extend fo successively to A,, for each 01 < w1. Assume that for a limit ordinal a < w1 we have a sequence (fp : p < a),where for each p < a , fp : Ap -i B is an embedding and f p C f-,, for all /3 < y < a. Setting -f, = Up<, f p we get an embedding of A, = Up<, Ap into B.

Thus, it remains to show how to extend an embedding f = f, of A, into B to : A,+1 + B because then the common extension Uo<w,fa embeds A into an I (of course, if A is countable, then the process stops at some stage cy < w1). We

CH.2 GAPSA N D

44

LIMITS

shall apply the Sikorski theorem. According to 7.4 of Chapter 1 we have to find an element b E B such that for each z E A, the following equivalences hold

Consider the gap determined by the new generator a and the embedding f

L

= ((f(5):

5

5 a } ; {f(y)

: y . a = 0)).

For the implications on the right in (S) it suffices to define b as any element filling the gap L. For the converse implications we have to choose b more carefully. In order to do this let us suppose that an element z cuts a , i.e. z - a > 0 and z . a > 0.Then we have

> CD, f(z) - f(z) > 0, f(z) - f ( y )

and

if y . a = CD if z 5 a .

There exists non-zero elements u ( z ) and w(z) such that 4.4.

5 f(z) - f(y) w(z) 5 f(z) - f(z) u(z)

and

for each y Ia for each L 5 a .

Indeed, let {yn : n < w } be an enumeration of the set {y E A, : y l. u } (recall that A, is countable). The elements f(z) - (f(y0) . . . f(yn)) form a decreasing chain and since there are no countable limits in B, we define u ( z ) > 0 as a lower bound of this chain. The element w(z) is defined similarly. Since the family

+ +

is at most countable, we may - in view of the lemma 4.1 - claim that R is disjoint (because for any refinement of R, 4.4still holds). We enlarge now the gap L putting the elements u ( z ) to the lower and w(z) to the upper class of L , respectively. Since the resulting gap L* is still countable, there is an element b 6 B separating L* (and hence also L ) . For each z cutting a we have u(z)

5 b and

b=0

~ ( 2 ) .

and since u ( z ) + W ( Z ) 5 f(z), hence f(z) cuts b: f(z) - b > 0 and f(z). b > 0.

$4.THEPAROVICENKO THEOREM.

45

This means that the element b satisfies the equivalences (S),which finishes the proof of the first part of the theorem. We prove now the second part. Let A and

B be two ParoviEenko algebras both

< w l } and {bp : ,6 < w1} be fixed enumerations of A and B,respectively. We define an isomorphism f : A -+ B as a union f = Ua<wlfa of an increasing chain (fa : a < w1) of partial isomorphism, which is, in turn, defined by induction as follows. Let A0 = {@,I} C A and Bo = {@,I} C B and then fo : A0 + BO is uniquely determined. Assume that we have already defined an isomorphism f, from a countable subalgebra A, A onto a subalgebra B, C B such that a , E A, and b, E B, (we could assume that a 0 and bo of cardinality c = w1, since CH is assumed. Let { a , : a

are the units of A and B, respectively). We extend f, in the following way. Suppose that a,+l E A,. If b,+l E B, then nothing happens but if b,+l @ B, then the inverse isomorphism filhas a n extension (as in the proof of the first part) to an embedding g of B,+1 = B,[b,+1] into A. We define A,+1 = g[B,+1] and f,+l = 9-l. Now suppose that a,+l $! A,. Extend fa to an embedding g : A,[a,+l] + B and if b,+l E g[A,[am+lll then define A,+1 = A,[a,+l], B,+I = s[A,+11 and f,+l = 9; otherwise extend g-' to an embedding h of B,+l = g[A,[a,+l]][b,+l] into A and define A,+1 = h[B,+l] and f,+1 = h-'. In either case we have extended fa to f,+l so that a,+l is in the domain A,+] and b,+l is in the range B,+1 of the isomorphism f,+l, which finishes the successor case of the induction. In the limit case a we are given an increasing sequence ( f p : ,6 < a) of isomorphism such that for each ,6 < a we have u p E Ag = dm (fa) and bp E Bp = rg ( f p ) . We take the elements a,, b, and proceed as in the non-limit case with Up<, f p , Up<, Ap and Up<, Bp in place of fa, A, and B,, respectively.

The sequence

a <

w1,

(fa :

we have a, E

< w l ) (just defined) has the following property: for each dm (f,) and b, E rg (fa). Hence, the common extension a

f = Uo<wl fa is an isomorphism of the whole A onto B,which finishes the proof of the theorem.

QED

We prove later (see Chapter 8) that the Parovitenko characterization of P(w)/fin is in fact equivalent to CH. That is, if c > w1 then there are non-isomorphic ParoviEenko algebras of cardinality c. The first part of the Parovieenko theorem can be reformulated as follows: 4.5.

if CH holds, then each Boolean algebra of cardinality at most c can be embedded into P(w)/fin.

CH.2 GAPS A N D

46

LIMITS

This implication cannot be reversed. In Chapter 8 we prove that there is a model of set theory, in which c > w1 and the conclusion of 4.5 still holds.

$5. TYPESOF GAPS A N D LIMITS. If A is an atomless Boolean algebra, then for each element a > 0 there are limits (towers) (Q( : E < y) in A, in which a occurs as - say - the first term a = ao. Indeed, the family R of all descending sequences f = (f([) : ( < a) of any type a and such that f(0) = a is partially ordered by inclusion f C g (g is an extension of f). If RO C R is a linearly ordered subfamily, then U Ro is an upper bound of Ro. Thus, by the Kuratowski-Zorn lemma, there are maximal elements in R and any such is a limit. For any limit f of length a there is a limit g of length c ~ ( ( Y )which , is cofinal in f. Hence, without loss of generality we may assume that the types of limits under consideration are regular cardinals. A similar remark applies to (well ordered) gaps.

5.1. Assume that CH holds. Then all limits in P(w)/fin are length c = w1 (we restrict ourselves to regular cardinals, cf the remark above). Since Ib = c in this case, the (essential) types of unfilled gaps are ( c , w ) and (c, c), (comp. 2.7 and 3.3). Thus, there exist limits and unfilled gaps of all possible types.

Now we discuss the situation under MA (the Martin axiom) and c we prove 5.2.

LEMMA.

> w l . First,

Martin’s axiom implies the following property called P(c):

if S P ( w ) is uniformly centered (i.e. is infinite, for any finite So C S ) and has cardinality less then c, then there exists an infinite X w such that X A, for each A E S.

s

s*

PROOF.Let P = P(S) consists of elements p of the form

where z p is a zero-one sequence of some length l ( p ) < w and S, is a finite subfamily of S. The ordering on P is defined thus: p 5 q if and only if zp _> zg and S, _> S, and for each i, I(q) 5 i < Z(p), the condition z p ( i )= 1 implies i E S,. It is easy to see that P has the ccc, i.e. each antichain in P is countable. Even stronger, P is a-linked. Indeed, if z p = zP then p is compatible with q , since then the

n

55. TYPES OF

GAPS A N D LIMITS.

47

element (z, S, US,) is below p and q. Since there is only countably many sequences xp, the claim follows. We prove now that the sets and

D , = { p E P : 3i 1 n(zp(i) = 1))

for n E w,

E A = { ~ E PA: E S , }

for A E S

are dense in P. Fix a q = (z,,S,) E P and an n E w. By the assumption, the set S, is infinite and hence there is an i E S, and i 2 max{n, I , ( q ) } . Define now an z E (0,l}i+lby setting

n

n

zIl(q) = 5,

and z ( k ) =

{ il

if k = i if l ( q ) 5 k < i .

Then, clearly, the element (z,S,) is 5 q and is in D,. Now, if an A E S is given, then obviously the element (zg,S, U { A } ) is in EA and below q, which proves our claim. The family { D , : n < w} U { E A : A E S} has cardinality less that c. Thus, the Martin axiom implies the existence of a filter G C P intersecting all the D’s and all the E’s. Let us define

X = { i E w : 3 p E G ( z p ( i )= 1)). Then, X is infinite since G n D , # Q, , for each n < w . Let A E S. There is a q E G n E A . Suppose now that i E X and i > l ( q ) . Then we have zp(i) = 1 for some p E G . Thus, there is an r E G such that r 5 p , q and consequently i E nS,, for zr(i)= 1. Since A E S,, we infer i E A , i.e. X \ l ( q ) A , which proves the lemma. QED The principle P(c) can be also expressed as follows: every centered family R lower bound.

E P(w)/fin of cardinality less that c has a non-zero

Obviously, any descending chain is a centered family. Hence, we obtain the following 5.3. COROLLARY. If MA holds (or a weaker principle P(c)), then m y limit in P(w)/fin has (essential) length c.

48

c H . 2 GAPSA N D

LIMITS

Now let us consider a gap on the set w

L = ({A,

:

< IC}

(Y

; {Bp : /3

< A}),

where IC and X are arbitrary regular cardinals. With any such an L we associate a forcing Q = Q ( L ) ,introduced by K. Kunen, consisting of elements of the form

where u p

IC

and w p

c X are finite sets and xp is a zero-one sequence of some length

I(p) < w . In addition, we assume that

The ordering on Q is defined thus:

5 q if and only if up _> uq and w p _> w q and xp _> x q and for each i, Z(q) 5 i < l ( p ) the following implications hold:

p

if i E

u

A,, then zp(Z)= 1,

a€U,

and

if i E

u

B p , then z p ( i )= 0.

PEW,

It is easy to see that Q ( L ) adds a set X , which separates the gap L . However, a forcing Q ( L ) need not have the ccc. We select below some particular cases when it does.

5.4. I f L is separated, then Q ( L ) has the ccc. Indeed, let a set C separate L. Thus, for any q E Q there is a number k ( q ) E w such that

u

A,

QEU,

Observe that the subset

\qq)cc

and (C \ k(c7)) n

u

PEW,

BB = Q,.

$5. TYPES OF

GAPS A N D LIMITS.

49

is dense in Q,since the sequence x q can be always extended arbitrarily. Now, Q' can be divided into countably many parts

Q' = uu{qE Q' : zq = x z

and

k(q) = k }

k

so that any two p , q in the same part are compatible. In fact, we have ( u pU u q , z, w pU w q ) 5 p , q

(here z = z p = z q ) .

We have shown that Q' is o-linked, which immediately implies the claim. 5.5. If L is a well ordered gap and n

# w1, or (A # w1) then Q(L)has the ccc.

Suppose the opposite, that Q has an antichain E of cardinality w1. If n > wl, then we choose an ordinal y < K so that UqEEu q E y. If we denote

L , = ( { A , : a < 7); {Bg : P < A)), then E is still an antichain in Q(L-,), which is impossible since Q(L7) is separated (by any set A t , with y 5 ( < K ) . Similarly, if K < w1 (i.e. K = w ) , then we can find a y < K for which the inclusion u q y holds for uncountably many q E E (since the function maxuq must be bounded on an uncountable set). Again the gap L, has an uncountable antichain, contradicting 5.4. Now we prove the following

s

5.6. THEOREM (Kunen [K]). If MA and c > w1 hold, then each well ordered gap of type (n,A), where K , X < c and K # w1 (or X # w1)can be separated.

PROOF.From 5.5 we know that the forcing Q = Q(L)has the ccc. Thus, we may apply the Martin axiom: there is a filter G Q,which intersect the following dense subsets for n E w D, = { q E Q : 32 2 n ( x q ( z )= 1)) and for (Y < K and ,B < A. E,,g = { q E Q : (Y E u q A /3 E w q } Clearly, the set X = { z E w : 3q E G ( z q ( i )= 1)) separates then the gap L. QED

Let us define now a so-called dominating forcing. For a family F w w let D = D(F) consists of all elements p = (zp,Fp),where x p is a sequence of nonnegative integers of some length I(p) < w and Fp is a finite subset of F . The ordering on D is defined thus

c H . 2 GAPSAND

50

LIMITS

5 q if and only if x p _> x q and Fp_> Fqand z p ( i )> f ( z ) , whenever f l ( q ) I i < !(PI. p

E

Fq and

It follows immediately that D is a-linked: if z p = z q = z, then (z, Fp U Fq)is below p , q. Clearly, D = D(F) adds a function g , which dominates the set F(i.e. f <* g , for each f E F). Using

ID we can prove easily the following remark.

5.7. MA implies that lb = c.

Indeed, for any F C ww define the sets and

D, = { p E D(F) :

I(p) 2 n}

E j = {PE D(F) : f E F p }

for n E w , for f E F,

which are easily seen to be dense in D(F). Thus, if card F G D(F) intersecting all the D’s and the E’s. If we define

<

c, we have a filter

<*

then g : w + w (since G n D, # $4 for each n E w ) and F g : if f E F then for some q E G, f E Fq (since G n E j # (3 ) and if i > Z(q) then we have g(i) > f(i) (because z p ( i )> f(i) for some p E G, p <_ q). Thus, we have proved that any set F C ww of cardinality < c is bounded, which means that lb = c. Summarizing the above discussion we may say that -- assuming MA plus c > w1 - there are, besides HausdorfFnon-separable (w1, w1)-gaps (which always exist, see 3.3), unfilled Rothberger’s (c,w)-gaps (see 2.7 and 5.7), while all the gaps of type ( K , A) with K , X < c and K # w1 (or X # w1) are separated (theorem 5 . 6 ) . It remains to settle the case of (c,c)- and (c,wl)-gaps, which will be done later in Chapter 8. We prove there that the existence of unfilled gaps of type (c,c) or (c,w1) is both consistent and independent of set theory (with MA plus c > w1). Thus, in the general case (i.e. if no additional assumptions are made) we have two kinds of unfilled gaps only: that of Hausdorff (of type (w1,wl)) and that of Rothberger (of type ( l b , ~ ) ) Note . that, in the latter, the coefficient lb may assume different values, as we have seen above.

$ 5 . TYPES OF

51

GAPS A N D LIMITS.

We may ask now what types of gaps and limits are consistently possible. It turns out that there are here no limitations except that of the theorems 1.5 and 2.8. Actually, we can consistently assert that there is a limit of any type K 5 c, provided c f ( ~ )> w . Similarly, it is consistent that there is a gap of any type (6,A), where K , A 5 c and c f ( ~ )> w (or cf(A) > w ) . In the proof we shall use the Hechler forcing defined as follows. For any partial ordering ( X ,<x)let P = P(X) consist of finite functions p with values 0 or 1 defined on a set of the form d ( p ) x Z(p) = { (z, i) E X x w : z E d ( p ) A

< l(p)}

where d ( p ) is a finite subset of X and l ( p ) E w . The ordering on P(X) is defined as follows: p q if and only if p _> q and p(x,i) 5 p(y, i), whenever Z(q) 5 i < l ( p ) and I s x y.

<

It is easy to see that P(X) has the ccc. Indeed, suppose the opposite and let E be an uncountable antichain. We may assume that Z(p) =const. for p E E . Applying the A-system lemma we find an uncountable EO 5 E and a d, so that dm (p) n dm (4) = d

for any two p, q E Eo.

If p , q are equal on the common part of their domains

then they are compatible, since p U q 5 p , q. On a finite set d there is only finitely many zero-one functions and hence EO must be finite, contrary to the assumption. How does the forcing P = P(X) work ? Let V be a model of set theory,

(X,<x) E V and let G C P be

a generic over V ultrafdter. Then we prove the

following general (Hechler [H4]). In the model V[G] there is an order embedding 5.8. THEOREM of X into P(w)/fin.

PROOF.Define in V[G] the set A,

= {z

Ew

: 3p

E G(p(z,i) = 1))

for x E X .

CH.2 GAPSAND LIMITS

52

Clearly, the family { A , : x E X } is in V [ G ] .We prove that, when partially ordered by C,,it is order isomorphic with X . First, let us note that the sets A , are all infinite. Indeed, the sets

0," = { p E P : 3 i 2 n(p(x, i ) = 1))

for x E X and n E w ,

are in V (since they are definable in V ) and are dense in P : if q E P define a p 5 q by setting pldm ( q ) = q (and p ( x , k ) = 0 for k < l ( q ) , if x # dm ( q ) ) and p(y,k) =

{

0, if l ( q ) 5 k < i 1, if k = i and y 2 x 0, if k = i a n d y n o n 2 x

where i = max{l(q),n}. Obviously, we have then p E 0,". Also the sets

E,, = { p E P : z , y

E dm (p)}

for s , y E X ,

are in V and are dense in P (given a q with x,y # d(q), define a p 5 q setting d ( p ) = d(q) U {x,y} and arbitrary values p(x, k), p(y, k ) , for k < l ( g ) ) . Now, let x < y and choose a p E E,,yflG. We claim that A , \l(p) A,. Indeed, if i 2 Z(p) and i E A , , then we have q(x,i) = 1

for some q E G and q 5 p .

Hence also q(y,i) = 1 and thus i E A,. In order to check that A , to notice that the sets

F,={pEP:

3z>n(p(x,i)=OAp(y,i)=l))

s,

A , it suffices

fornEw,

(which obviously are in V )are dense in P (for a q E P define p 5 q setting d ( p ) = d(q), l ( p ) = i 1, where i = max{l(q), n } and p(z, i) = 1, for z 2 y and = 0, otherwise).

+

Finally, we have to show that if x,y are incomparable (i.e. x non 5 y and 5 x), then A , , A , are also such (i.e. A , non C, A , and A , non A,). This will hold, if we prove that the set

c,

y non

H , = { p E P: 3 2 n(p(~,i)=OAp(y,i) = 1 ) A 3 j 2 n((p(x,j) = 1 Ap(y,j) = 0)}, defined for n E w (obviously in V ) ,are dense in P. To check this, let q E P be arbitrary. w e may assume that x,y E dm ( q ) and define a p 5 q by setting pldm ( q ) = q , d ( p ) = d(q), l ( p ) = j 1, where j = i 1 and i = max{l(q), n } and

+

+

1, if k = i a n d z 2 y 1, if k = j a n d 2 2 x 0, otherwise

53

$ 5 . TYPES OF GAPS A N D LIMITS.

Note that p 5 q holds, since

I

and y are incomparable. Obviously, p is in H, and

QED

the proof is finished.

We shall apply now the above theorem to some specified X. Let e.g. X = K * , i.e. K ordered by the inverse inclusion, where K is a fixed cardinal in V with V I= " c ~ ( K )> w " . We prove that the corresponding decreasing family { A , : a < K } in V[G] is in fact a limit. Let C E V[G] be any infinite subset of w . We have to show that

C non C, A , Let us denote (for y

for some a

< K.


P, = { p E P :

d ( p ) 5 y}

and G, = G n P,.

Clearly, P, is a complete subforcing of the Hechler forcing P = P(X). Hence G, P, is a complete ultrafilter over V . Since P has the ccc and c f ( ~ )> w , the set C has a canonical name C E V(P,) for some large enough y < K and thus C E V[G,]. Now, if an a satisfies y 5 01 < K , then the sets

D, = { p E PIG,

:

3i 2 n(i E C A p ( a , i ) = 0 )

for n E w

are dense in PIG,. Indeed, we have

Let y E PIG, be an arbitrary element. We may assume that a E d(q). Define p 5 q by setting pldm (4) = q , d ( p ) = d(q), Z(p) = i 1 where i 2 max{l(q), n} and i E C and 1, if Z(q) 5 j < l ( p ) and p < y and j E Ap

+

p(P,j) =

{

0 , if Z(q) 5 j < Z ( P ) 0 , if l ( q ) 5 .i< Z ( P )

and and

P < Y and j $ Ap P 2y

Then p E D,, which proves the claim. It follows that C non Now, let X be an ordinal sum

K

+ X*,

where

K,

c*A,.

X are fixed cardinals in V and

c f ( ~> ) w in V . Thus, we have a gap

L = ( { A , : a < ~ } { ;B g : / ? < A } )

c H . 2 GAPSAND

54

LIMITS

in V [ G ]( the lower class of L lies below the upper). In order to show that L is unfilled take any set C E V [ G ]and assume that C Bg,for each p < A. As above, there is a (complete) subforcing Q P such that C E V [ Gn Q] and there is an a E n \ U { d ( q ) : q E Q}. For any q the set Bg \ C is infinite and hence the sets for n E w En= { p E P/G n Q : 3i L n(p(a,i) = 1))

c*

c

nBEd(s)

are dense, which in turn implies that A,non

E t C.

We shall discuss now a problem of a scale. We know already that a scale exists if the continuum hypothesis is assumed (see the remark after 2.4) and as easily seen, it exists also if we assume MA plus c > w1. Indeed, we have then lb = c (see 5.7) and consequently Ib = d = c. The claim follows now from 2.5. Note that in both mentioned above cases there is a scale of length c. Now, we prove a useful technical lemma.

5.9. LEMMA. Let s = (f, : a < n) E V, where IC > w is a regular cardinal of V , be an increasing unbounded sequence in the model V . If P E V has the ccc and V ”card P < tc” holds, then in the extension V [ G ]the sequence s is still unbounded.

PROOF.By a contradiction. Suppose that we have f.

>s, for some function

f E V[G].Thus, for each a < n there is an integer n, E w such that f(i) 1 fa(;)

for each i 2 n,.

Since n is regular and > w , the value na must be constant on a set X cardinality K in V [ G ] there : is an integer N such that

n, = N

K,

of

for each a E X .

The set X need not be in V but it has always a subset Y C X , Y E V , of cardinality K in V . Indeed, let us denote

Yp = {a < n : p IF ”a E X”}

for p E P,

where X is any name of X : X [ G ]= X . Thus,

Y,, E V

for each p E P

and X =

u PEG

Y,,.

55. TYPES OF

55

GAPS A N D LIMITS.

Obviously, we have card G < n, by the assumption of the lemma, and hence card Yp= n for some p E G, which proves the claim. Denote this Yp by Y . Thus, we have

(*I

fu(i) I f(i)

for each i 2 N

and

Q

E Y.

Since card Y = n, the sequence (fu : Q E Y ) E V is still unbounded in V (since, it is cofinal in s). Let A, = {fa(z) : cr E Y } ,for i 2 N . Thus, the family { A , : i N } is in V and from (*) it follows that

>

V [ G ] " V i 2 N ( A i E fin)". Clearly, the same holds with V replacing V [ G ] Therefore, . the function

is in V and obviously g is a bound of s, a contradition.

QED

Note that 5.9 can be restated as follows: if P E V has the ccc and card P < Ib in V , then each unbounded set in V remains unbounded in V [ G ] . A Cohen forcing @ ( X )(where X is an arbitrary set) consist of finite zero-one functions p : d m ( p ) + (0, l}, where the domain dm ( p ) is a finite set X , ordered by the inverse inclusion. @ ( X )has always ccc. If X = A U ( X \ A ) is any partition of X , then @ ( X )is isomorphic to the product @ ( A )x @ ( X \ A ) . If X E V then also @ ( X )E V and card C ( X ) = card X . Models of the form V [ G ]where , G C @(X) is a complete (over V ) ultrafilter and X E V , are called Cohen models (or Cohen extensions). From the lemma 5.9 it follows easily 5.10. COROLLARY. In any Cohen model V [ G ]the set of old functions ww n V is unbounded but not dominating.

PROOF.If f E V [ G ]is any function: w -+w , then there is a countable subset A C X so that f E V [ Gn @ ( A ) ] .Since @ ( A )is countable, we may apply 5.9. to any unbounded sequence s = (fu :

Q

< Ib")

E

v.

Thus, we have fnon, 2 s and consequently fnon, >ww n V . Hence, no function f E V [ G ]is a bound of ww n V, which proves the first assertion.

56

c H . 2 GAPSAND LIMITS

A forcing Q consisting of all functions g : dm ( 9 ) + w , where the domain dm (9) w is finite, ordered by the inverse inclusion, is countable and hence equivalent to a @(w). Thus, if h = U H where H Q is a generic over V ultrafilter, then h is in V[G].Clearly, the sets

c

03 = { q

EQ:

3i 2 n ( q ( i ) > f(i))}

for n E w and f E wW n V ,

are dense in Q (and belong to V ) . This implies that for each f E ww n V we have

h(i) > f ( i )

for infinitely many i E w ,

i.e. h s , f holds for no old function f, which proves the second assertion.

QED

Finally, let us note the following 5.11. THEOREM. Assume that V GCH holds and that IE > w1 is a regular cardinal in V , such that nw = n. Then, in the Cohen extension V [ G ]via C(n) we have lb = w1 and d = c = K . In particular, lb < d and hence there is no scale.

”lb = w l ” , since CH is true in V . It follows from 5.10 PROOF.We have V that lb = w1 holds also in V [ G ]since , the set ww n V of old functions has cardinality w1 and is unbounded. Clearly, V [ G ] ”c = n”, since nw = K . If a set F ww from V[G]is of power < c, then F E V [ Gn @(7)],for some 7 < n. Now, @(n) is equal to @(y)x @(n \ 7) up to isomorphism and V [ G ]= V [ Gn @(y)][G n @(n \ 7 ) ] .Again from 5.10. it follows that the set H = ww n V [ Gn C ( y ) ]cannot be dominating in ww n V[G].Hence, F is not dominating as a subset of H. We have proved that no set F of cardinality < c can be dominating and consequently d = c, which finishes

+

the proof.

QED

PROBLEMS

A. THEMARTINAXIOM 1. Assume MA. If R, S

c P ( w ) are two families of cardinality < c such that

S U {A}is uniformly centered for each A E R,

c

c.

then there is a set Y w such that Y B for each B E S and Y n A is infinite for each A E R. (This is a strengthening of P(c) and the proof can be as that of 5.2)

PROBLEM s

57

2. Let X be a topological space with countable weight and suppose that the sets N,, for 01 < n, where n is a fixed cardinal < c, are nowhere dense in X . a) Let { U , : n < w } be an enumeration of a fixed base, in which each basic set occurs infinitely many times. If Y g w is as in Ex. 1 above with respect to the families

R = {A,

: n Ew}

and S = { B , :

01

< K},

where

A,

=

{i E w

:

Ui

g U,}

and B, = {i E w : U , n N , = q } ,

then the sets 2, =

x\ u{Ui :

iEY

A

i > n)

for n E w ,

are nowhere dense. b) With the notation as in a) we have an inclusion

Hence, if MA holds then the union of less than c nowhere dense sets is meager (in any space with countable weight).

3. Let X be the Lebesgue measure in the real line JR (or in Rn).For an E > 0 let a forcing P,consist of open sets p with X(p) < E , ordered by the inverse inclusion. a) If E & P,is uncountable, then there is an uncountable subset Eo an integer n so that X(p)

+ 1<

E

GE

and

for each p E Eo.

b) For each p there is a finite union of rational intervals w ( p ) G p such that we have X ( p \ w ( p ) ) <

i.

c) If w ( p ) = w ( q ) for some p , g E Eo, then X(p u q )

X(U G) 5 E . (W is hereditarily Lindelof).

d) If G

P, is a filter, then

e) If an A

C JR has measure zero, then

is dense in P,.

< E . Hence P, has the ccc.

the set

CH.2 GAPS A N D LIMITS

58

Hence, MA implies that the union of less than c A-null sets is A-null.

B. GAPS IN P(w)/fin 1. There exists an almost disjoint family R = { A , :

(Y

< w1} such that for each

(Y < w1 the set A , is near to the segment {Ac; ( < a}. (Define R by induction). If S,T C_ R are uncountable and disjoint, then the gap L = (S,T ) is not filled. (Lusin).

2. The separating forcing Q = Q ( L ) , where L is a Lusin gap from Ex. 1 above, never has the ccc.

3. Consider the Cantor space C, = (0,l}". a) There are unfilled well ordered gaps in C of the form

L = ({fa :

a:

<W l }

; {gp :

P <Wl}),

<* fp and go* r g p for a: 5 P < w1 and f, <* g p for any a,P < w1. h <. gp} are all F,-sets. Intervals [fn,gp] = { h E C , : fa

where fa b)

c) The Cantor space C, is an increasing union

U

H,, where each Ha is a

,<Wl

Ga-set. (Hausdorff).

c. ORDERINGS MODULO FILTER For an arbitrary filter F on w , we define two relations on the set w w as follows and

f IFg

{ i E w : f ( i )I g(k)} E F

f =F

{i E W

g

:

f ( Z ) = g(Z)} E F.

1. The relation < F is a partial pre-ordering, i.e. a partial ordering on the set w " / = F of all abstraction classes. 2. The ordering

IFis linear if and only if F is an ultrafilter.

<*

3. The ordering is identical with the < F , where F is the Frkchet filter on the set w , F = { A E w : w \ A E fin}. Sometimes, it is convenient to use the strong ordering defined thus

f

g

3

{ z E w : f(i) < g ( 2 ) ) E

implies the condition f

F.

S F g (i.e. f < F

g and

5. The cardinal lb remains unchanged, when - in its definition - the ordering is replaced by <*. The same applies for the cardinal d.

<*

59

PROBLEMS D. GAPSIN ww Consider the strong ordering

f <* g

{ i E w : f ( i ) 2 g(i)} E fin.

E

We say that f and g are far away, if

).(fI

- g(n)l = +m.

lip 1. If fo

<* go are far away, then there exists a far away pair

f1

<* g1 such that we

have fo <* f1 <* g1 <* go.

2. There exist ordered countable gaps

L = ({frl :

12

< w } ; {gm : m < w } ) (the lower class of L below the upper)

between any far away pair fo

<.

go.

3. Let L be a countable gap as in Ex. 2 above. a) There is an increasing sequence ko i 2 km we have mm{fo(i),

<

Icl

< . . . such that for each m and

. . . ,fm(i)} + m < min{go(i), . . . ,gm(i)}.

b) There is a far away pair f <* g filling the gap L (i.e. fn <* f <* g <* g m for each n,m < w ) . Hence, every countable gap in ww is filled.

4. There exists an ordered gap

of type

(WI

,w~1.

5. There exist non-separated (wl,wl)-gaps L as in Ex. 4 above (Kunen). (This is

a fairly nontrivial construction).

6. It is consistent that there are non-separated gaps L, of a given type ( K , A), where K , X 5 c and c ~ ( K ) ,cf(A) > w . (Modify the Hechler forcing so that the functions p may assume arbitrary values from w ) .

CH.2 GAPSA N D

60

LIMITS

E. CHAINS A N D ANTICHAINS

< n) in ww is regular, then we have n 5 d. 2. It is consistent that there are limits in ww of a given type n I c with cf(n) > w 1. If the length n of a limit ( f a :

(Y

(cf D.6).

3. It is consistent that there are maximal antichains {a, : (Y < n} in P(w)/fin, of a given cardinality n 5 c, cf(n) > w. (Modify the Hechler forcing so that it produces almost disjoint sets). 4. The property P(c) implies that each infinite maximal antichain in P(w)/fin has cardinality c. (If R P(w) is an infinite almost disjoint family, then the family {w \ A : A E R} is uniformly centered). 5. Each infinite maximal antichain in P(w)/fin has always cardinality

2 Ib.

F. DOMINATING FAMILIES 1. The cardinal d remains unchanged when the dominance ordering by the pointwise ordering

f5g

=

I*is replaced

vi E w[f(i) I g(i)].

(Each function f from a dominatingfamily replace by functions fn(i) = f ( n + i ) ) .

2. The cardinal d is the minimum number of compact sets covering the space of irrationals ww.

3. It is consistent that there is a scale of a given length n 5 C, cf(n) > w. (Iterate IE times with finite supports the dominating forcing I D).

G.THECARDINAL t Let t denote the length of a shortest limit in P(w)/fin. 1. The cardinal t is regular and satisfies w1 5 t 2. Let

fA

denote the increasing enumeration of a set A

a) If A C, B, then

b)

I c.

fA+

c w.

2f~.

an arbitrary (infinite) A w and any f : w + w there is an infinite subset B C_ A such that f <* f B . For

Hence, t satisfies w1 5 t 5 Ib.

3. We have the equality 2<' = C. (In P(w)/fin define inductively a binary tree of height t ) .

PROBLEMS

4. It is consistent that t < Ib. (Let V c (see F.3.). Apply Ex. 3 above).

"w1

61

< c < 2w1" and force a scale of length

H.THECARDINAL b A family S P(w)/fin is called splitting, if for each a > (D there is an I E S such that both a . r > (D and a - I > 0.Let s denote the minimal power of a splitting family. 1. Inequality t I s holds. (For a n define a decreasing chain (a, : a for each a < n).

< t and any R = {z, : a < n} P(w)/fin < n) such that either a , 5 I , or a , . I, = (D

2. We have s I d. (If f + > f ~( f = ~ increasing enumeration of A ) then f~ assumes a value in any interval {i E w : f"(h) I i < f " + ' ( k ) } ,for n E w where h is fixed and f ( k ) > 0).

3. It is consistent that s = w1 < c. (Consider a Cohen model).

4. It is consistent that t < s and 2' < 2', simultaneously. (Let V " C = wz < 2,1 < 2,z" and iterate with finite supports wz-times a centered forcing P(F,) determined at a - t h stage by a fixed ultrafilter Fa).

+

5 . Each set X in the Cantor space C, = (0,l}, of cardinality < s has the Lebesgue measure zero. (If g assumes the value 1 infinitely many times, then the set

{f E C ( w ) : I.

g

5. f} is a countable union of null sets).

SEQUENTIALLY COMPACT SPACES

A topological HausdorfF space X is called sequentially compact if any sequence I = { I " : n E w } C X contains a subsequence zlA = { I " : n E A } converging to a point of the space X . 1. Let X be a compact (Hausdorff) space. For any { I " : n E w } C X and any A w denote by SA the set of all cluster points of I ( A= {I,, : n E A } . Assume that I = { I , : n E w } has no convergent subsequences. a) Each subsequence IIA has two subsequences slAo and z)A1 for which we ~8. have S A n~S A = b) There is a binary tree T in P(w)/fin of height t so that the sets S,, corresponding to elements on any level of T , are pairwise disjoint. Hence, each compact space of cardinality less than 2' is sequentially compact.

62

CH.2 GAPSA N D LIMITS

2. A Cantor space C, = {0,1}'( is sequentially compact if and only if the weight K. is less than s. (If R is splitting and card R = s then we have (in ( 0 , l } R )a sequence 2 = {I,,: n E w } , where zn(A) = 1 iff n E A ) .

Chapter 3 Stone Spaces

$1. THESTONE REPRESENTATION. By a compact space we mean always a compact Hausdorff space. Thus, a topological space X is compact, if any two points have disjoint neighbourhoods and each open cover X = U R contains a finite subcover Ro C R. In the above, we can assert that the covers R consist of basic sets (sets taken from a fixed base). The cover condition can be replaced by the following: each centered family F consisting of (basic) closed sets has a nonernpty intersection.

A space X is called zero-dimensional, if the family B ( X ) of all the open-closed sets is a base. It appears, as M. H. Stone had proved, that each Boolean algebra is of the form B ( X ) , for some compact X . More precisely, we have the following representation theorem. 1.1. THEOREM (Stone [Sl],[Ss]). For any Boolean algebra A there is a compact zero-dimensional space X , such that A is isomorphic to B ( X ) .

PROOF.Let X be the set of all ultrafilters in the algebra A. Let us denote

U, = { F E X

:

a E F}

for a E A.

Thus, we have U1 = X and Uo = 8 and from characterizations given in Sec. 6 of Chapter 1 we obtain immediately the following identities

Moreover, the sets U,, ub are distinct for distinct indices a,b. Indeed, if a # b, i.e. a A b (D, then there is an F E X containing a A b as an element. This in turn implies F E U, A Ub, i.e. Uu # ub. It follows that the formula for a E A

f(u) = U,

63

cH.3 STONE SPACES

64

defines an isomorphism of A onto the field R = {U, : a E A} of subsets of X . We regard X as a topological space with the topology generated by the family R. Clearly, X is zero-dimensional (since, R is closed under complementation) and R itself is a base (since, it is multiplicative). Let us check that X is Hausdorff. If F, G are distinct, then there is an element a E A such that a E F and -a E G. This means that U, and U - , are neighbourhoods of F and G, respectively and are disjoint: U, r) U-, = UQ= (3. Now, let us show that X is compact. Let S = {U, : a E T} be a centered family of basic sets. Thus, for any finite number of elements a1 , . . . ,a,, E T we have

and hence a1 . . . . a , > 0,i.e. the family T c A is centered as well. Hence, there is an ultrafilter F E X containing T as a subset. But the condition T c F implies FE U,,i.e. F E S, which proves the claim. 3

naET

n

It remains to prove that R = B ( X ) . We know already that R c B ( X ) so let U E B ( X ) . Thus, for some Ro c R we have U = URo, since U is open. Since it is also closed and hence compact we have U = U S for some finite S c Ro and therefore U E R, because R is additive. This finishes the proof. QED Any zero-dimensional compact space X such that B ( X ) is isomorphic to a given algebra A is called a Stone space of A. Obviously, for homeomorphic spaces X and Y the fields B ( X ) and B ( Y ) are isomorphic. Conversely, we shall prove the following

1.2. LEMMA.& for compact zero-dimensional spaces X and Y , the fields B ( X ) and B(Y) are isomorphic then X and Y are homeomorphic.

PROOF. Let f : B ( X ) determines an ultrafilter

+

B ( Y ) be an isomorphism. Every point z of X

B ( z ) = {U E B ( X ) : x E U } of its open-closed neighbourhoods and conversely, each ultrafilter F form: if x E F then F c B ( z ) and hence F = B ( z ) . Let h : X by the formula f[B(z)] = B ( h ( r ) ) for z E X ,

n

c B ( X )has that +

Y be defined

which is correct, because the image f [ B ( z ) ]of an ultrafilter is an ultrafilter. If z # y then there are disjoint neighbourhoods U , V of x and y, respectively. It follows that

$1. THESTONEREPRESENTATION

65

f [ B ( z )# ] f [ B ( y ) ]and hence n B ( h ( z ) )# B ( h ( y ) ) .If y is an arbitrary point of Y then f ( z ) = y, where { z } = f-'[B(y)]. Thus, h is one-to-one.

n

Finally, let V E B ( Y ) and denote U = f - ' ( V ) . Then we have z E h-"V]

= h ( z )E v = = u E B(z) =

V E f[B(.)] 5

E

=

u,

which proves that h is continuous and hence a homeomorphism, as X is compact. QED

From the above lemma we obtain immediately the required uniqueness viz. the Stone space X = S[A] of a Boolean algebra A is defined uniquely (up to a homeomorphism). In a similar way we prove the following 1.3. THEOREM. Any compact zero-dimensional space X is the Stone space of the field B ( X ) , i.e. X = S [ B ( X ) ] .

PROOF.We have remarked already that the family B ( z ) = {U E B ( X ) : z E U} of neighbourhoods of a point 2 is an ultrafilter in B ( X ) and each ultrafilter in B ( X ) has this form. Thus, the mapping z H B ( z ) is one-to-one from X onto S [ B ( X ) ]and it is a homeomorphism, since for any a E B ( X ) we have

i.e. B-'[U,] = a , which proves the theorem.

QED

It is easy to see that a principal ultrafilter becomes an isolated point in the Stone topology. Thus, we have 1.4. THEOREM. The principal ultrafilters of A coincide with the isolated points of the space X = S[A]. PROOF. If F is a principal ultrafilter, i.e. F = {z E A : z 2 a } , where a is an atom, then U, = { F } which means that F is isolated. Conversely, if F is non-principal and U, is any neighbourhood of F , then a E F and hence a cannot be an atom. Thus, a has a splitting a = b c , where b, c > CD and b . c = 0 and consequently U, = ub U U, where ub, U, are nonempty and disjoint. Thus, uais never a singleton and hence the point F is non-isolated. QED

+

cH.3 S T O N E

66

SPACES

It follows immediately from the above theorem that an algebra A is atomic if and only if the space S[A] has a dense set of isolated points. 1 . 5 . EXAMPLE. The atomic algebra F C ( X ) , for an infinite set X , has a single non-principal ultrafilter consisting of cofinite sets. Hence, the Stone space Y = S[FC(X)]has n = card X many isolated points and a single cluster point. Therefore Y is a one-point compactification of the discrete X.

$2. SUBALGEBRAS A N D HOMOMORPHISMS. The Stone representation theorem clarifies also the relationship between the subalgebras or homomorphic images of a given algebra A and the corresponding Stone spaces. First, let us consider the case of subalgebras.

THEOREM. Let X = S[A] and Y = S[B]. Then B is (isomorphic to) a subalgebra of A if and only if the space Y is a continuous image of X . 2.1.

PROOF.If B setting

A is a subalgebra, then the function f : X f(F)= F n B

+

Y defined by

for F E X ,

is onto Y and continuous, since for b E B we have

f-'[U[]= { F E X

:

F n B E V,'} = { F E X

: bE

F}=Ut

(U,"and U[ denote the basic Stone neighbourhoods in X and Y , respectively). Conversely, if Y = f[X] for some continuous f , then the function g ( V ) = f-'[V], for V E B(Y) is an embedding of the field B ( Y ) into B ( X ) . Hence, 1 is isomorphic to a subalgebra of A. QED Now, let us consider the homomorphic images.

THEOREM. Let X = S[A] and Y = S[B].Then B is a homomorphic image of A if and only if the space Y is (homeomorphic to) a subspace of X . 2.2.

PROOF.If h : A defined by setting

-+

B is a homomorphism onto B, then the function f

f(F)= h-'[F]

for F E Y,

:

Y

--*

X

67

$3. ZERO-SETS.

maps Y into X in a one-to-one manner: if F # G then for some b E B we have b E F and -b E G, whence a E f ( F ) and -a E f ( G ) , for any a E h-'(b). The function f is continuous, since

f-'[u,x] = {F EY :

h-'[F] E U,"} =

= { F E Y : h ( a ) E F } = UGa,.

c

Thus, we may identify the space Y with a closed subspace f[Y] X . Conversely, suppose that Y C X is a closed subset. We have then a homomorphism h : B ( X ) + B(Y) defined by the formula

h(U)= U nY

for U E B ( X ) .

It remains to prove that h is in fact onto B(Y),i.e. that each open-closed set V E B ( Y ) has the form V = U fl Y , for some U E B ( X ) . Since V is open in Y , there is an open W & X such that V = W n Y . Now, W is a union of some basic sets

W=

u

U,

where Ui E B ( X ) , for each i E J.

iEJ

Hence, we have an open cover V = u i E J ( U in Y ) of a compact set V whence V = UiEJo U,n Y , for some finite set Jo J. If we denote U = then we obtain V = U n Y for a U E B ( X ) , as required. QED

c Y,

uiEJo U,

c

n

Any closed subset of X = S[A]has the form F , where F is a filter in B ( X ) (because B ( X ) is a base for closed sets). On the other hand, the homomorphic images of A can be identified with the factor algebras B ( X ) / F . Let us note that the function h(u)= u n

nF

for

u E qx),

is a homomorphism of B ( X ) onto B ( n F ) with the kernel V ( h )= F and consequently B ( X ) / F is isomorphic to B ( n F ) . In other words, we have proved the following identity

SPWFI =

nF,

for any filter F in B ( X ) .

s

X §3. ZERO-SETS. Let X be a topological space. Two disjoint sets A , B are called completely (or functionally) separated, if there is a continuous real-valued function f : X + W such that A

s f-'(a)

and B C f - ' ( b )

for some a

# b in R g ( f ) .

68

cH.3 STONE

SPACES

The function f is said then to separate A and B . Clearly, if f separates A and B then adding to f (or multiplying f by) a suitable real number we may assume that f l A = 0 and f l B = 1. Additionally, we may assert that f is bounded - say 0 5 f 5 1 (replacing f , if necessary, by min{max{f, 0}, 1)). We recall that a HausdorfF space X is called completely regular (or Tychonoff), if for any closed B and an arbitrary I 4 B the sets {I} and B , are completely separated.

A subspace of a Tychonoff space is Tychonoff. Also a product of an arbitrary family of Tychonoff spaces is again Tychonoff.

A HausdorE space X is normal, if any two disjoint closed sets can be separated by open sets or - equivalently - completely separated, by a well known Urysohn lemma. Also, the following property known as the Tietze extension theorem characterizes normal spaces (in HausdorE spaces): for each dosed set F X , any continuous function f : F -t W can be extended to the whole of X (in a continuous way). Every compact space is normal. For any topological space X , let C ( X ) consist of all continuous real-valued functions. We regard C ( X ) as a linear ring with pointwise operations. The family of all bounded functions in C ( X ) is then a subring denoted by C * ( X ) . Both C ( X ) and C * ( X )are closed under the operations min and max. Now, let us assume the following notation z(f) = {z E X : f(z) = 0)

for f E C ( X ) .

Sets of this form are called zero-sets. Let

Z ( X ) = (4.f) :

f

E C(X)}

be the family of all zero-sets in the space X . If A = z(f) is a zero-set then we may always assume that f is nonnegative f 2 0, since z(f) = z( Ifl). We can also claim that f E C * ( X )since z(f) = z(min{f, a } ) , where a is any real > 0. Each zero-set A = z(f) is closed and a Gs, since

$3. ZERO-SETS.

69

If X is normal then the converse is also true: each closed Gs set A is a zero-set. Indeed, we have A = n n < w A n ,where the sets A , are open and form a decreasing sequence. By normality, there are functions fn E C ( X ) such that fnlA = 0, fnl(X\An)= 1and 0 5 fn 5 1,for each n < w . Then, the function f = 2-" fn is in C ( X ) (as a uniform limit) and we have z(f) = A , which proves the claim. The following theorem contains some basic informations on the family Z ( X ) .

3.1. THEOREM. The family Z ( X ) of all zero-sets contains B ( X ) and is closed under finite unions and countable intersections. If X is a Tychonoff space, then Z ( X ) is a base for closed sets.

PROOF.The characteristic function X A of an open-closed set A is continuous. Hence B ( X ) Z ( X ) . The second assertion follows from the obvious identities

c

4f)u z(g) = z(f . 9 )

n

r ( f , ) = z(Z, min{fn,2-("+')})

where

fn

2 0.

n<w

Finally, if X is a Tychonoff space and F X is closed then for any z E X \ F there is an f E C ( X ) such that f l F = 0 and f(z) = 1, i.e. F C z(f) and z 4 z ( f ) . It follows that F = n { A E Z ( X ): F A } , which finishes the proof. QED Let us observe that continuous counter images of closed initial and cofinal segments of the real line f-'[(-m,a]]

and

f-'([ b,+m)]

for a, b E

W,

are always zero-sets, since we have clearly

f(z) 5 a

=

rnax{f(z) - a,O} = 0

and

f(z)2 b

min{f(z) - b,O} = 0.

Thus, from the theorem 3.1 it follows that continuous counter images of closed subsets of W (so called functionally closed sets) coincide with all zero-sets. Consequently, functionally open sets, i.e. continuous counter images of open sets R are precisely the complements of zero-sets, (called also cozero-sets).

70

cH.3 STONE SPACES

Disjoint zero-sets are completely separated and also can be 3.2. COROLLARY. separated by functionally open neighbourhoods. PROOF.If A = z(f) and B = z ( g ) where f,g 2 0, then h = f / f + g is in C ( X ) and we have 0 5 h 5 1 and A = h-'(O), B = h - l ( l ) . Hence, h separates A and B . If 0 < a < b < 1 then the functionally open sets h-'[ [0, a)] and h-'[(b, 111 separate the sets A, B .

QED

54. THES T O N E - ~ E C COMPATIFICATION. H

Let X be a Tychonoff space. Although the family Z ( X ) of zero-sets need not be a field, the notion of a filter can be defined in a usual way. Specifically, a centered family F E. Z ( X ) is a filter, if from A E F and A B E Z ( X ) follows B E F . As in the case of Boolean algebras, maximal filters in Z ( X ) will be called ultrafilters. Any centered family S Z ( X ) generates a filter

Therefore, using the Kuratowski-Zorn lemma, we can prove (as in the lemma 5.8, Chapter 1 ) that each centered family S C Z ( X ) can be extended to an ultrafilter.

A counterpart of the characterization 6.6 in Chapter 1 can be stated as follows. Z ( X ) is an ultrafilter, then for each A E Z ( X ) either A E p or there is a disjoint from A set B E p . 4.1. I f p

Indeed, for each B E p , the condition A n B # (3 means that p U { A } is centered and hence p U { A } = p . In particular, if p # q are distinct ultrafilters, then there are disjoint sets A , B E Z ( X ) such that A E p and B E q. It is easy to check that each ultrafilter p Z ( X ) is prime:

s

4.2. I f A , B E Z ( X ) and AU B E p , then A E p or B E p .

4p

then there is an A1 E p such that A n A1 = (3. Hence A1 n ( A U B ) = A1 n B and A1 n B is in p . Consequently B E p , as A1 n B G B . Indeed, if A

The family of all ultrafilters in Z ( X ) is denoted by @[XI.We regard p [ X ] as a topological space with basic open sets of the form U A = {p E p[X] : A

4p}

for A E Z ( X ) .

$4,THES T O N E - ~ E CCOMPATIFICATION H

71

Note that from 4.2, it follows that UA n U B = U A ~ and B hence the family { U A : A E Z ( X ) } is multiplicative. We recall that a compatification of a space X is any compact space having a dense subset homeomorphic to X .

4.3. THEOREM (Cech-Stone [Ss], [C]). If X is a Tychonoff space, then @[XIis a compatification of X . PROOF.

Let us define

F(I)= { A E Z ( X ):

I

for

E A}

I

E X.

If B E Z ( X )\ F ( I ) then the Tychonoff property implies that B is disjoint from some A E F ( z ) . This means that F ( z ) is a maximal filter in Z ( X ) , for I E X . Thus, F is a function from X into /3[X]. Clearly, F is one-to-one since distinct points are completely separated. It is also continuous since F-'[UA] = {I E X : A and X

\ A is open.

4 F(I)}= {I E X

: I

4 A} = X \ A

On the other hand we have

F[A] = {F(I): A E F ( I ) } = { p E @[XI: A E p}

n F[X] = (@[XI\ UA) n F[X],

which proves the continuity of the inverse function. Now, we check that the image F[X] is dense in @[XI. Let UA be a nonempty neighbourhood. Then, we must have A 5 X and if I E X \ A then A 4 F ( I ) , whence F ( I )E UA. It remains to show that @[XIis a compact space. First, let us prove the Hausdorff property. If p # q then there are disjoint zero-sets A E p and B E q. From the corollary 3.2 it follows that there exist functionally open disjoint sets A1 and B1 such that A 2 A1 and B B1. Thus, the zero-sets a = X \ A1 and b = X \ B1 satisfy p E U, and q E ub and a U b = X . The latter property yields

c

u, n ub = UaUb = U X

= (8,

which proves the claim. Finally, let R = {@[XI\ UA : A E S} be a centered family of basic closed sets. For any finite subfamily So = {Al, . . . ,An} S there is an ultrafilter q such that

c

72

cH.3 STONE

SPACES

whence A l , . . .A,, E q. Thus, A1 n ... n A,, E q which shows that the family S is centered in Z ( X ) . Thus, there is an ultrafilter p E Z ( X ) extending S. Obviously, we have then p E ( - ) { @ [ X I UA \ : A E S} =n R and the proof is complete.

QED

As we have seen above, the space X is homeomorphic to the set of fixed ultrafilters { F ( I ) : I € X } , where F ( I ) =-{B E Z ( X ) : I E B}. Developing further the Cech-Stone theory (in which we follow [GJ] and [CN]) we prove the formula (valid in any @ [ X I ) :

4.4.

-

F [ A ]= {p E p [ X ]: A E p}, for each A E Z ( X ) .

Indeed,let A be a zero-set. If I E A , then A E F ( x ) , i.e. F[A] C { p : A E p} and hence F [ A ] {p : A E p} as the latter set is closed. On the other hand, if p 4 F[A] then we have UB n F[A] = $9, for some B E Z ( X ) , B $ p . It follows that A G B, since x E A implies F ( x ) 4 UB i.e. B E F ( z ) , whence I E B. Therefore A 4 p, i.e. p E U A , which proves 4.4. Identifying each point x E X with the corresponding fixed ultrafilter F, = {A E Z ( X ) : I E A } we may assume for simplicity that always X E @[XI. The formula 4.4 can be written now as follows

Let us note also the following property 4.5. In @[XI, we have

= 2 n B for arbitrary A , B E Z ( X ) .

Indeed, from 4.4 we get directly

Any compatification K , of a given Tychonoff space X , determines (and is determined by) the subring SK G C * ( X )of all functions in C * ( X )which are (continuously) extendable to A'. For K = @[XI,the ring SK is equal to the whole C * ( X ) . Thus, the characteristic property of P [ X ]is that each bounded continuous function on

$4.THESTONE-CECH

COMPATIFICATION

73

X can be extended onto a [ X ] .This follows from the general theorem on a continuous extending proved below. 4.6. THEOREM (Taimanov [TI). Let X be a dense subset of a (Tychonoff) space T . The following conditions are equivalent. 1. For any compact space Y , each continuous function f : X -t Y has a continuous extension fT : T -t Y . -T -T 2. Disjoint zero-sets A, B E Z ( X ) have disjoint closures in T : A n B = 9.

PROOF.Assume 1 and let A , B E Z ( X ) be two disjoint zero-sets. By the Corollary 3.2, there is an h E C * ( X ) separating A and B . By assumption, the function h has an extension h~ E C ( T )which in turn separates the closures ; Iand B in T,. Thus, 2 follows. Now, we prove the implication 2

1. Let a continuous f : X -t Y be given. For any point p E T let B(p) denotes the family of all (open) neighbourhoods of p in T . Let us observe that the family -t

is centered. Indeed, if U1,.. . ,Un E B(p) then we have U1 n... n Unn X X = T ) . On the other hand, the inclusion

-

f[u1n . . . n Unn XI

# 9 (since

C f [Ul n XI n . . . n f [Un n XI

holds always, whence the claim follows.

n

Thus, the intersection R(p) is nonempty and we claim that it has one point only. For proof, suppose the opposite - let y1 # yz be in n R ( p ) . There are _ _ neighborhoods & and Vz of y1 and yz, respectively such that Vl n Vz = (B. Applying the Urysohn lemma we find a function g E C * ( Y )so that g l F = 0 and g l E = 1. Now, if we set

A = (gf)-'(0)

and B = (gf)-'(l),

then A, B E Z ( X ) and A n B = 9. In addition, we have

4.7.

f

-'[F]C A

and f-'

[El

B

By assumption, the closures 2 and B are disjoint and hence T = (T \ 2)U (T \ B). Therefore p E U ,where U = T \ 3 or U = T \ B.

74

cH.3 STONE SPACES

Consequently both y1 and claim follows.

y2

are in f [ U n X I , which contradicts 4.7 and the

n

Now, let us define f ~ ( p )as the unique point of R(p). Clearly, f~ is an extension of f. It remains to show that f~ is continuous. Let V be a neighborhood of a point f~(p).Thus, n R ( p ) S V and by compactness we have

f[u,n X ] n . . . n f[unn XI

C V,

for some U1,.. . ,U,, E B ( p ) . The left hand side of this inclusion contains the set f [ U n X I , where U = U1 n . . . n U,,E B(p). Hence, f ~ [ U Cl V holds (because f~[Ul C f[U fl X I ) , which proves the continuity. QED Let us note that the property 1 can weakened to the following 1'. Each f E C * ( X )has an extension 7 E C*(T). Indeed, 1 implies l'and 1' implies 2. From 4.5 it follows that the statement 2 of the Taimanov theorem holds for m y Tychonoff X and T = /?[XI. Hence, each continuous f : X + Y (into any compact Y ) has a continuous extension fa : /?[XI+ Y . Note that the extension fp (denoted also by 7) is unique, since X is dense in /?[XI. As we have remarked earlier, the property 1 (and hence also 1' or 2) of the theorem 4.6 characterizes p [ X ] .Indeed, assume that 1 holds for a Tychonoff X and its compatification T (we may assume that X S T ) . Thus, the identity function id: X + X has two continuous extensions: f : /?[XI"5 T" and g : T " 2 /?[XI. Hence, g = f-' (since it is so on a dense subset) and therefore f is a homeomorphism of @[XIonto T. Hence, we have proved 4.8. If the property 1 or 1' or 2 holds for a Tychonoff space X and its compatification T , then T = /?[XI.

$5. SPACESOF UNIFORM ULTRAFILTERS.. Let X be an infinite set and consider the power-set algebra P ( X ) . It is easy to see that the Stone space of P ( X ) is p [ X ] , (where X is treated as a discrete space). Indeed, we have Z ( X ) = P(X) (each subset of a discrete space is a zero-set) and consequently ultrafilters in Z ( X ) is the same as ultrafilters on the set X . Any basic neighbourhood U A = { p : A E p } , determined by a set A X , in the Stone representation coincides with the basic set

55. SPACESO F

75

UNIFORM ULTRAFILTERS

Ux\, = { p : X \ A 6 p } in the topology of @ [ X I .When A runs over all subsets of X , then so does X \ A and hence the both bases coincide. Thus, in view of the theorem 1.1 we obtain immediately the following 5.1. T H E O R E M . The Stone space of a power-set algebra P ( X ) is p [ X ] (with discrete X ) . In addition, the function

f ( A ) = 2 (= { p E p [ X ]: A E p } )

for A

CX,

is an isomorphism of P ( X ) onto L?(P[X])

Now, let X be any infinite set. For every infinite cardinal n 5 card X , the family

X<" = { A C X

:

card A

< n}

is clearly an ideal on the set X . In particular, X<" = fin(X). Generalizing slightly the notion of a uniform ultrafilter we say that an ultrafilter p on X is n-uniform, if each set A in p has cardinality 2 n. The family of all n-uniform ultrafilters on X is denoted by U , ( X ) . Denote also (a n-Frechet filter)

F, = { A C X

:

card ( X \ A )

< n}

and say that a family R C P ( X ) is n-centered, if card (A1 f l . . . n A , ) 2 n for any finite number of sets A1 , . . . ,A , from R. With this terminology, we check easily the following two properties (cf 6.4, Chapter 1):

1. p E U , ( X ) F, C p , for any ultrafilter p on X . 2. Every n-centered family R C P ( X ) extends to an ultrafilter p in U,(X). For any discrete X the set U , ( X ) p [ X ]will be regarded as a topological space with the topology inherited from p [ X ] .The forthcoming theorem carries some basic informations on U,(X).

5.2. T H E O R E M . For any infinite discrete set X the space U K ( X )is closed in p [ X ] and hence compact. The function f ( A ) = 2 n Uk (the relative closure) for A C X is a homomorphism from P ( X ) onto B ( U , ( X ) ) with kernel A(f) = X < " . Hence the Stone space of the factor algebra P ( X ) / X < ' is U , ( X ) . If p 6 U, then there is a set A E p , with card A < open-closed neighbourhood PROOF.

-

A = {QE

"1

:

A E ql,

K.

The basic

76

c H . 3 STONE SPACES

of the point p, is then disjoint from U,. Thus, the set U, G @[XIis closed. Now, let A C X be an arbitrary set. We haveX = AU(X\A) and An(X\A) = (3, from which we infer

This implies f ( X \ A ) = U, \ f ( A ) . Obviously, f ( A U B ) = f(A) U f(B), f ( X ) = U, and f(9)= 9. Therefore, the function f is a homomorphism of P ( X ) into B(U,). We have to show yet that f is onto. Let V be an arbitrary open-closed set in U,. Thus, we have V = U n U, for some U E B ( @ [ X ] )If. we set A = U n X then we get A = U n X = = U,since X is dense in @[XI,and the equality f(A) = V follows. Thus, the homomorphism f : P ( X ) + B(U,)is onto. Finally, let us note that the condition f ( A ) = 9, i.e. 2 n U, = $3 is equivalent to card A < x. Indeed, if card A < x then no element of the set = { p : A E p } is x-uniform. Conversely, if card A 2 x, then the family { A } is K-centered and hence there is a p in 2 n U,. Thus, we have shown

A(f) = { A C_ X : f ( A ) = $3) = X',, QED

which finishes the proof.

For x = w we have X<" = fin(X) and therefore the Stone space of an algebra P(X)/fin is U w ( X )= @[XI\ X consisting of nonprincipal ultrafilters on X . Generally, any space of the form @[XI\ X is called a remainder (or a growth) space of X . Let us denote A*=X\X For example, we have X * =

\X

forAC_X.

= @[XI\ X . From the theorem 5.2 we obtain

5.3. COROLLARY. i Y X is an infinite discrete space, then X * is the Stone space of P(X)/fin, B ( X * )= {A* : A X } and the following equivalences hold

s

A* = B *

=

A=, B

(i.e. A n B ~ f i n )

A'SB'

G

Ac,B

(i.e. A \ B E f i n )

=

A I B

(i.e. A n B ~ f i n )

A*nB*=Q)

A'

# (3

<E

A is infinite.

$5. SPACESOF

77

UNIFORM ULTRAFILTERS

Now, we can list some basic properties of the remainder w* = P [ w ] \ w . Thus, the space w* is compact and zero-dimensional (since w* = S[P(w)/fin]),it is dense in itself (since P(w)/fin has no atoms), nonempty Gs-sets in w* have nonempty interiors (since there are no w-limits in P(w)/fin). The weight w (w *) is c (since card P(w)/fin = c), while the cardinality of w* is 2' (the theorem 6.5, Chapter 1). If A C X is an infinite subset of a discrete space X then the basic neighbourhood A in /?[XI is homeomorphic to /?[A](since e.g., the property 2 of 4.6 holds with A and 3 in place of X and T , respectively). In particular, if card A = card X then the closure 2 is homeomorphic to /?[XI and also A* is homeomorphic to X * . Let us recall that each infinite Hausdorff space has a countably infinite discrete subset.

-

5.4. THEOREM (Cech [C]). If D C w * is a countable,discrete set, then the closure is homeomorphic to /3[w]. Hence, the space @[w]can be embedded into the remainder w* (as a nowhere dense set). P R O O F . Let D = {q, : n E w } . Thus, there exist basic neighbourhoods A;, so that A: n D = {q,}, for each n E w . The sets A , are infinite and pairwise almost disjoint (Corollary 5.3). We may assume that they are in fact pairwise disjoint (replacing A , by A , \ (A0 U ... U An-l)). We may also assert that w = UnEw A, (adding at most one element to each particular A , ) . Let f E C*(D). We show that f has an extension 7 E C ( n ) and then apply 4.7. Let us define a function g : w -t W setting glA, = f(q,), for each IZ E w and let ij E C ( / ? [ w ] be ) the Stone extension of 9 . Then

ij[A,l= -[A,] = S[Anl=

{f(qn))

for any n E w ,

and hence the function f = ?jlDis as required. Thus, the set p[D]and consequently to / ? [ w ] . QED

is homeomorphic to

From the theorem 5.4 it follows at once that each infinite closed set F C w* contains (a homeomorph of) / ? [ w ] , since it contains a countable discrete set. In particular, the cardinality of F is that of / ? [ w ] ,i.e. 2'. Finally, let us note that each closed set E in w* has the form

E=n{A*: AEF} where F is a filter on w (cf. the remark following 2.3).

78

cH.3

STONE SPACES

$6. STRONGLY ZERO-DIMENSIONAL SPACES. Let X be a Tychonoff space. A standard open base of p [ X ] ,defined in the preceding section, is additive (i.e. it is closed under finite unions). Indeed, we have

It follows that each open-closed set U E B ( p [ X ] )is a basic set, U = U A for some A E Z(X)(since U ,as compact, is a finite union of basic sets). On the other hand, we have the following equivalence 6.1.

A E B(X)

U A E B ( B [ X ] ) for A E Z(X).

Indeed, if A E B ( X ) then also X

\ A E B ( X ) Z ( X ) and obviously

X = A U ( X \ A ) and A n ( X \ A ) = (3. Therefore

p [ X ]= X = ; IU ( X \ A ) and 2 n X

\ A ) = A n ( X \ A ) = Q,.

Thus, 2 is open-closed and so is U A = @[XI\ A. Conversely, if U A E B ( P [ X ] )then, as remarked above, p [ X ]\ U A = UB for some B E Z(X), and we have and

p [ X ]= U A U U B = UAnB,

whence A n B = Q,,

U A n UB = 9 = U A U B ,

whence A U B = X .

Thus, A must be open-closed in X , which proves 6.1. A(Tychonoff) space X is called strongly zero-dimensional, if disjoint zero-sets can be separated by open-closed sets: if A , B E Z ( X ) and A n B = Q, then there is an a E B ( X ) so that A aOB=q.

a and

6.2. THEOREM. The space p [ X ]is zero-dimensional if and only if X is strongly zero-dimensional.

PROOF.If p [ X ]is zero-dimensional, then from 6.1 we have

(*I

B(p[X]= ) {Z

: aE

B(X)}.

$6. STRONGLY ZERO-DIMENSIONAL SPACES

79

Let A, B E Z ( X ) be two disjoint zero-sets. Then also 2 nB = $3, by 4.5. Note that for a E D ( X ) we have -

ACii

z

Xn(p[X]\??)=Q, z -

-

G

AnX\a=$3

G

Asa.

E

An(X\a)=$3

E

Hence, from (*) it follows -

A = n { E : a E D ( X ) and A

E a}.

By compactness, there is an a E D ( X ) containing A and disjoint from B , since -

A n B = n { m : U E D ( X ) andAca}.

Conversely, assume that X is strongly zero-dimensional. Let U A be a basic set and take a p E U A . Since A 4 p , there must be a B E p so that A n B = $3. By assumption, there is an a E D ( X ) separating A and B, i.e. A a and a n B = $3. It follows that

c

p E U, U A , which proves that the family of open-closed sets {U, : a E B ( X ) } is a base in @[XI. QED

A point x of a space X is said to be a P-point, if for any countable family {U,, : n E w } of open neighbourhoods of x we have

i.e. the intersection of countably many neighbourhoods of x contains a neighbourhood of I. A (Tychonoff) space X is called a P-space, if each point of X is a P-point.

6.3. EXAMPLE. Let X be an uncountable discrete space and let Y = X U { p } where a local base at the new point p consist of sets { p } u ( X \ A) for a finite or countable A C_ X. Clearly, Y is a P-space with a single limit point p . There is also a dense in itself P-space. Let ( X , < ) be an 71-ordering, i.e. a dense linear ordering having no countable cofinal or coinitial subset and no unfilled countable gaps. The simplest example of such an X is that of Sierpiriski: X consists of all functions x : w1 + {0,1} for which the set x-'(l) = { a < w1 : .(a) = 1) has a greatest element, with the lexicographical ordering, (see [GJ]). Clearly, X with the order topology is a P-space and has no isolated points. 6.4. THEOREM (Rudin [Rz]). Assume CH (or MA plus c > w1). Then the remainder space w* has a dense subset of P-points.

CH.3

80

STONE SPACES

PROOF.Let ( A , : a < c) be an enumeration of all infinite subsets of w . We define by induction a decreasing (in the relation chain (B, : a < c) so that

c*)

(*)

either B,

c*A ,

or

B, IA ,

for each a

< c.

Assume that we have already defined a part (Be: 5 < a) satisfying (*). Both CH and MA imply that there are no limits of length < c (see 5.3, Chapter 2), so let /3 be the smallest index for which

Ap

c* BF

Let us define

B,

=

{

for each ( < a. if Ap IA , otherwise.

Ap 7 Ap n A,,

Obviously, (*) holds now for B,,with 5 5 a. By compactness and (*) the intersection BZ has a single point p and the family {BZ : o < c} is an ordered base at p. Since under our assumptions the cardinal c is regular, the point p is a P-point and, even stronger a P(c)-point (i.e. p E int[n R ] ,for any family R of neighbourhoods of p , with card R < c). Indeed, if R = { A : : a E S} is any family of neighbourhoods of p and card S < c, then 7 = sup S is < c and p E B1; int R, as required. Of course, we may claim that p E A*, where A is any set given in advance (it suffices to define A0 = A). QED.

no,,

c n

A point x in any Hausdo& space X is called a weak P-point, if x E Dl for no countable set D X \ {x}. Any P-point x is a weak P-point. Indeed, if D E X \ {x} is countable, then we have x E

U , where

u = int n{u, : and U, is a neighbourhood of x such that y disjoint from D.

y E D)

4 U,.

Thus, U is a neighbourhood of x

It follows that every compact (infinite) space X always contains non-P-points. Indeed, no points of D,for any countable discrete D X , is even a weak P-point.

n\

c

Now, we shall prove a simple characterization of P-spaces. 6.5. THEOREM (Gillman, Henriksen [GHl]). For a Tychonoff space X the following conditions are equivalent. 1. X is a P-space (i.e. each point is a P-point).

$7. EXTREMALLY DISCONNECTED

81

SPACES

2. Each function f E C ( X ) is constant in a neighbourhood of each point. 3. Z ( X ) = B ( X ) (i.e. any zero-set is open). PROOF.

Let f E C ( X ) and take a point z E X . Define neighbourhoods

4y E X :

u

-

a-

where a = f(z)is the value of f at z. If z is a P-point then for some open U we have z E u n{u. : n E W )

c

and therefore flu = a. Thus, 1 implies 2. Now, assume 2 and let A E Z ( X ) . Thus, A = z( f) for some f E C ( X ) . If z E A , then f is constant on a neighbourhood U of z, whence U A. Hence, A is open and 3 follows. Finally, from 3 it follows at once QED that B ( X ) is a base and a a-field, which in turn immediately implies 1.

c

Clearly, every P-space is zero-dimensional and even strongly zero-dimensional, which follows easily from 6.1 and 6.2.

$7. EXTREMALLY DISCONNECTED SPACES. Any Boolean algebra A can be identified with the field B ( X ) , where X is the Stone space of A. The Boolean ordering in B ( X ) is the inclusion. Let us consider the suprema of infinite subsets C B ( X ) . It is easy to see that for any family R B ( X ) and V E B ( X ) we have 7.1.

V=supR

E

V=UR.

c

c

V for each U E R, whence U R V and also Indeed, if V = supR, then U is open, it must be empty (see V, as V is closed. Since the set V \ e.g. 4.5, Chapter 1). Conversely, if V = then V is a bound of R and clearly, V V' holds for any bound V' of R.

c

c

m,

A topological space X is extremally disconnected, if the closure of any open set U is open (and hence E B ( X ) . ) Clearly, in any extremally disconnected space, the closures of disjoint open sets are still disjoint. The property 7.1 yields immediately the following characterization of complete algebras: 7.2. A Boolean algebra A is complete if and only if its Stone space X = S[A]is extremally disconnected.

cH.3 STONE

82

SPACES

In the case of a-complete algebras, the unions U R occurring in 7.1 will be countable, i.e. open F,-sets or even cozero-sets if we restrict ourselves to normal spaces only. Hence, we call a space X a-disconnected (or basically disconnected) if each cozero-set U E X has an open closure. Thus, the Stone space of a cT-complete Boolean algebra is a-disconnected and vice versa. In the general case any Tychonoff extremally disconnected (or a-disconnected) space is always zero-dimensional. Indeed, the cozero-sets is an open base and hence 7C U . for any open U and a point p E U there is a cozero V so that p E V Therefore, U(X) is a base. Now, we shall prove the following

7.3. THEOREM. For any Tychonoff space X we have the following equivalence X is a-disconnected

p [ X ]is a-disconnected.

Similarly, for extreme disconnectedness.

PROOF.Assume that X is a-disconnected and let U be a cozero-set in p [ X ] . Obviously, U n X is then a cozero-set in X and hence

is open-closed in p [ X ] ,as the closure of an open-closed set in X. Conversely, if U is

a cozero-set in X then U = V n X , for some cozero-set V E p[X]. Indeed, if U = { z E X : f(z) # 0) for an f E C * ( X ) ,then let us define V = {PE @[XI:f(p) # 0 ) . Hence, the set = 8 n X = V n X n X = V n X is in

vx

W).

QED

Recall that a continuous map f : X " 5 Y" is irreducible, if the equality f[F] =Y holds for no closed F 5 X . Let us note the following useful remark: if X, Y are compact then for any continuous f : X "3 Y there is a closed E E X so that f[E]= Y and the restriction flE is irreducible. Indeed, the family R consisting of all closed F X such that f[F]= Y is closed under the intersections of descending chains. By the Kuratowski-Zorn principle, there are minimal sets in R, which proves the claim. 7.4. LEMMA.Let X and Y be compact spaces. If Y is extremalJy disconnected then any continuous irreducible map f : X "3 Y is a homeomorphism.

$7. EXTREMALLY DISCONNECTED PROOF.

If U

SPACES

83

X is an open set, then

Indeed, if z E U and V is a neighbourhood of the point y = f(z), then we certainly have X \ (U n f-'[V]) X , since z E U n f - ' [ V ] . Thus, the image f[X\ (U n f - l [V])is a proper subset of the space Y . Let w = f(u) be a point of the set

y

\ f [ X \ (U n f - l

[Vlll.

Then, we have u E U n f - l [ V ] which implies w E V \ f [ X \ U ] and hence V n (Y \ f [ X \ U]) # $3, i.e. y E Y \ f [ X \ U], which proves (*). Since X is compact, it suffices to show that f is one-to-one. Thus, assume that 21 # 2 2 and let Ul and Uz be disjoint neighbourhoods of X I and 1 2 , respectively. Obviously, we have

(Y \ fP \ Ull) n (Y \fix \ U21) = $3 and hence also

y

\ f[X\ Ull n y \ f[x\ U2l = $3,

since Y is extremally disconnected. From (*) we obtain f[U1]n f [ U z ] = proves that f is one-to-one. QED

q, which

For any compact space X let us denote G ( X ) = S [ D ( X ) ] i.e. , G ( X ) is the Stone space of the algebra of open domains of the space X . We know already that the algebra D ( X ) is always complete (see 4.1, Chapter 1). Consequently G ( X ) ,the Gleason space of X , is a compact extremally disconnected space. The space X is always an irreducible image of G ( X ) . To see this, let us observe that the intersection njj =

:

A Ep}

where p E ~

(x),

has exactly one point. Indeed, njj is nonempty since X is compact. Let x E njj. We claim that D ( z ) C p , where D ( z ) = {U E D ( X ) :

2

E

U}.

Indeed, if U E D ( s ) then z E U n 2,for each A E p . Hence also z E U n A , since U is open. Thus, the family p U { U } is centered, which implies U E p and the claim follows. Clearly, for distinct x 1 , z 2 E X , the filters D ( z 1 ) , D ( z 2 ) contain disjoint members (since D ( X ) is a base) and therefore njj = {z}.

84

c H . 3 STONE

SPACES

Let us define f : G(X) + X by the equivalence

It is easy to see that f is onto X . Indeed, if I E X, then the filter D ( x ) can be extended to an ultrafdter p E Q(X) and, as shown above, we have then f(p) = 2. It remains to prove that

7.5. The function f : G(X)

"3 X is continuous and irreducible.

PROOF.The continuity o f f is equivalent to the condition

f[E] E f[El

for any set E

c G(X).

f[z]

So let z E and take a neighborhood B E D ( X ) of the point A E D(X) so that x E A c 2 E B. We have

x = f(p)

for some p E

2.

There exist an

E,

and we know that A , B E p . Therefore UA n E # (3, where U A = { q E G ( X ) : A E q } is a basic Stone neighbourhood. If q E UA n E then f ( q ) E 2 B , whence x E f[El. Thus, the function f is continuous.

c

Now, let F 5 Q(X) be a closed proper subset. The complement G(X) \ F contains a basic open set U A . Take a domain B , so that B B A. Then we have - B f l ( - A ) = (Band consequently B C X \ f[F], since f[F] C ~ [ U - A ] -A. Thus, f

c c

is irreducible.

c

QED

For an arbitrary topological space X the density d(X) is defined as the minimum cardinality of dense subsets:

d(X) = min{card D : D

. It is not 7.6.

cX

and

=X }

difficult to check that d(G(X)) = d(X)

for any compact X.

Indeed, if D is a dense subset of G ( X ) , then we have

PROBLEMS

85

which means that the image f[D] is dense in X , whence d(X)
f[E] = f[El = D = x. This proves d(G(X)) 5 d(X) and 7.6 follows. Finally, we prove the following

7.7. THEOREM (Efimov [El]). For any compact separable X , the space G ( X ) , can be embedded into the remainder w'.

PROOF.From 7.6 it follows that the space G ( X ) is separable as well. Let D C G ( X ) be a countable dense set and choose an f : w 2 D. The function f : w + G(X), as continuous on the discrete w , has a continuous extension f : P[w] --t G ( X ) . We have -

f [ P [ X ]= ] f[G] = f[wl=

D = G(X),

i.e. f is onto G(X).Thus, for some closed F P[w], the restriction g = TIF is an irreducible map from F onto G ( X ) and hence g is a homeomorphism, in view of the lemma 7.4. We have proved that G(X) can be embedded into P[w]. An application of 5.4 finishes the proof. QED

A well known Marczewski theorem says that the product of at most continuum many separable spaces is still separable. In particular, the Cantor cubes {0,1}'( and Tychonoff cubes [0, 1]'(for K 5 c, are all separable. Therefore, the Gleason spaces

G({O,l}n)and G([O,l]'() for

K

5 c,

are all embeddable into the remainder space w'.

PROBLEMS A . PRODUCT O F BOOLEANALGEBRAS 1. Let A = A1 x

. . . x A,

be a finite product of Boolean algebras.

86

cH.3 STONE

SPACES

a) Each ultrafilter p in A has the from p = { a E A : ai E F }

where F is an ultrafilter in A, . b) The Stone space of the product A1 x Stone spaces of A1 , . . . ,A,:

for some i,

. . . x A,

15 i 5 n,

is the disjoint union of the

2. If {Ai : i E J } is an arbitrary family of Boolean algebras, then we have ~ [ n i G , ~ i= ]

B[c~,JsIA~I].

B. SUPERATOMIC BOOLEANALGEBRAS 1. An algebra A is atomic if and only if its Stone space has a dense set of isolated points.

2. If J is the ideal of A generated by all the atoms, then we have

S[A/J] = S[AId

(the set of limit points).

A Boolean algebra A is superatomic, if every homomorphic image of A is atomic. 3. An algebra A is superatomic if and only if its Stone space S[A] is scattered (has no perfect subsets). Thus, e.g. the algebra F C ( X ) is superatomic. 4. Let X be a 0-dimensional compact space. a) The space X is scattered if and only if some Cantor-Bendixon derivative is empty. b) If Y = f[X] is a continuous image of X then for each ordinal a! we have Y ( a )C f[X(")]. Hence, if X is scattered, so is Y. (For y E Y d , f-l(y) cannot be open). l},} in c) If X is non-scattered, then there is a family { X , : s E UnEw{O, B ( X ) , order isomorphic to the Cantor tree (i.e. Xo = X and each X, splits X , = X(,o)UX(,l)). Hence the Cantor space C, is a continuous image of X . (Define f : X + C , so that I E Xf(=)ln,for each I E X ) . (Rudin).

nnEw

Therefore, a Boolean algebra A is superatomic if and only if each subalgebra B A is atomic.

PROBLEMS

87

C. IDEALSIN C ( X ) 1. If J is an ideal in a ring C ( X )then the image z [ J ]= { ~ ( f:)f E J} is a filter in

Z(W 2. If F is a filter in Z ( X ) then the counter image t - ' [ F ] = {f E C ( X ) : z ( f ) E F } is an ideal in C ( X ) . Every maximal ideal J has this form. Let M ( X ) denote the family of all maximal ideals in the ring C ( X ) ,where X is a Tychonoff space.

3. The function J : @[XI+ M ( X ) , where

maps p [ X ]in a one-to-one way onto M ( X ) so that points z E X (i.e. fixed ultrafilters in Z(X))correspond to fixed ideals J ( z ) = {f E C ( X ) : f(z) = 0). For each f E C ( X ) , the set V j = { J E M ( X ) : f @ J} is the image of a basic set J[UZ(/)l. Hence, the space M ( X ) , with the topology generated by the base {Vj : f E C ( X ) } ,is the same as @[X].(Stone). 4. If X is compact then X is homeomorphic to M ( X ) (in particular, each maximal ideal in C ( X ) is fixed). Hence, if for compact X and Y the rings C ( X ) and C ( Y ) are isomorphic, then the spaces X and Y are homeomorphic. (Gelfand, Kolmogoroff).

D. THE RING

OF BOUNDED FUNCTIONS

Let M * ( X ) denote the family of all maximal ideals in the ring C * ( X ) ,where X is a Tychonoff space. 1. The map f ++ 7 is an isomorphism of C * ( X )onto C ( p [ X ] ) Moreover, .

Thus, C * ( X )is isometrically isomorphic to C ( @ [ X ] ) . 2. Each ideal J E M * ( X ) has the form J * ( p ) = {f E C * ( X ): 7 ( p ) = 0}, for some P E

a".

88

cH.3 STONE

SPACES

3. The function J' : p [ X ]-+ M ' , defined in Ex. 2 above, maps p [ X ]in a one-to-one way onto M * ( X ) . The set V; = { J E M * ( X ) : f !$ J } is the image under J* of the cozero-set p [ X ]\ z(7), for each f E C * ( X ) . Hence, the space M * ( X ) with the topology determined by the base {V; : f E C * ( X ) }is the same as p [ X ] .(We identify points of X with fixed ideals of

C*(X)).

E. SOME PROPERTIES O F P [ X ] Let X be an arbitrary Tychonoff space. 1. If Y & X and each function f E C * ( Y )can be continuously extended to X , then Y = @[XI(the closure y is taken in @[XI).

2. If X

Y & @[XIthen p[Y] = p[X].

3. For any f E C * ( X )we have -

f(p) = n{f[Al: A Ep}

F. THE SPACE U ( x )

FOR A DISCRETE

for each P E @[XI.

x

Let X be an infinite discrete space. 1. The space @[XIcan be embedded into the space U ( X ) of uniform ultrafilters.

2. Let rc = card X be a regular cardinal. The space U ( X ) contains discrete subsets of cardinality rc and for any such set D we have B = p [ X ] .

3. If Y is a compact space and d ( Y ) 5 card X , then the Gleason space G(Y)can be embedded into U ( X ) .

4. Let n = card X be a regular cardinal. Any sets A , B C X are called n-disjoint, if card A = card B = rc and card ( A n B) < rc. a) If a family { A , : a that

< r c } is rc-disjoint, then there is a function f : n -iX

f(a)E A ,

b) There exist

rc+

\

u

A@

for each

Q

< n.

many r;-disjoint subsets of X .

Hence, the space U ( X ) has always antichains of cardinality

> K.

so

PROBLEMS G . THE RING C ( X * )FOR

A DISCRETE

89

X

In C ' ( X ) = Z"(X) we have the uniform norm

Let us denote

c o ( X ) = {f E P ( X ) : VE > O{z E X

: If(z)l

2

E}

E fin(X)}

1. The set c o ( X ) is a closed ideal in P ( X ) . 2. The map f H f*(= T l X * ) is a linear homomorphism from P ( X ) onto C ( X * )

with the kernel co(X). Hence, the ring C ( X * )is linearly isometric to P ( X ) / c o ( X ) . H. LOCALLYCOMPACT

SPACES

A Hausdorff space X is locally compact, if each point z E X has a neighbourhood U , so that the closure is compact. Such a space is always Tychonoff. Each open subset of a compact (and even locally compact) space is locally compact. 1. Any locally compact space X is open in /3[X]. Hence, for any Tychonoff space X we have the remainder X * = @[XI\ X is compact if and only if X is locally compact. The symbol K ( X ) denotes the family of all compact subset of X 2. If X is a zero-dimensional locally compact space, then the mapping A H A* (= Ti n X * ) is a homomomorphism from the field B ( X ) onto B ( X * ) with the kernel K ( X ) . Hence, the field B ( X * ) is isomorphic to the factor field B ( X ) / K ( X ) . 3. Let X be a locally compact space. a) We have the following equivalence p E X * if and only if p n K ( X ) = $3, for p E P [ X ] . b) Each noncompact zero-set A E Z(X)belongs to some ultrafilter p E X * . c) The set of all functions vanishing at infinity

c o ( X ) = {f E C * ( X ): V& > 0{2 E x is a closed ideal in C ' ( X ) .

:

If(2)l

2

E}

E K(X)}

cH.3 STONE SPACES

90

d) I f f * denotes flX' then we have

f * = 0 if and only if

f E Q(X)

Hence, C ( X * )is isomorphic to C * ( X ) / c o ( X ) .

I. THESPACE p[W] 1. For an arbitrary Tychonoff space X we have:

X

is connected

=

p [ X ] is connected.

2. The space P[W]is a continuous image of p [ w ] and, on the other hand the space p[w] can be embedded into p[W](cf E.l). Hence, card p[W]= 2'. Let us denote

W+ = { u E W :

a 2 0) and

W- = { u E W :

u

5 0).

3. We have a splitting R* = W; U W?,where W; n W? = 0. Hence, the remainder space W* = p[W]\ W is disconnected. 4. The spaces W; and W*_are connected. (Each point p E W; contains an unbounded interval [a,+m)).

J . P-SPACES 1. 2. 3. 4.

Any set of P-points of a given Tychonoff space is a P-space. An infinite P-space is never compact. A Tychonoff space is P-space if and only if each Gs-set is open. Let X be a P-space. a) If z ( f ) = z ( g ) then f = h . g, for some h E C ( X ) . b) Each ideal J C C ( X )has the form J = z - ' [ z [ J ] ] ,that is

fEJ

3g E J ( z ( f ) = z ( g ) )

for f E C ( X ) .

c) If J is a prime ideal in C ( X )then F = 4.71 is a prime filter in Z ( X ) . Hence,

F is maximal. ( Z ( X ) = B ( X ) is a field). 5. Let X be a Tychonoff space.

a) If x E X is a non-P-point, then there exists a zero-set A , E Z ( X ) such that x E A , \ intA,. b) Any maximal extension G, of the filter F, = { B E Z ( X ) : x E intB} omitting the set A, is a prime filter in Z ( X ) .

PROBLEMS

91

c) If F is a prime filter in Z ( X ) then J = z-'[F] is a prime ideal in C ( X ) . Therefore the prime ideal J = 2-l [G,] is not maximal. From Exs. 4 and 5 above we infer immediately:

A Tychonoff space X is a P-space if and only if each prime ideal in C ( X ) is maximal. (Gillman, Henriksen [GHz]).

K. EXTREMALLY DISCONNECTED

SPACES

1. The following condition characterizes extremally disconnected spaces (in all Tychonoff spaces):

(*)

if U n V = (?then V n V = $9,

for any open U,V

X.

If, in addition at least one of the U , V is assumed to be functionally open, then the property (*) characterizes a-disconnected spaces. 2. Every P-space is o-disconnected. 3. Each dense set and each open set in an extremally disconnected space is also extremally disconnected. L. NORMALMEASURES Let B(X) denote the a-field of Bore1 sets or sets with the Bake property of a space X . A finite measure j i defined on B(X) is normalif A@) = M ( X ) , i.e. measure null sets coincide with the meager sets. 1. Let A be any measure algebra (carrying a finite measure). a) The Stone space X = S[A] is extremally disconnected. (F.1, Chapter 1). b) On B ( X ) there is a strictly positive measure m satisfying m

(@)= C U E R m ( U )

for every antichain R C B ( X ) .

c) B ( X ) = D(X)(for every extremally disconnected space). d) There is a homomorphism h : B(X) '3B ( X ) with the kernel

A(h) = M ( X ) (5.13, Chapter 1). e) The measure m can be extended to a normal measure p by setting

p ( A ) = rn(h(A))

for A E B(X).

cH.3 STONE

92

SPACES

Hence, an arbitrary measure algebra induces a normal measure in its Stone space. (Halmos (HI]). 2. If a (compact extremally disconnected) space X carries a strictly positive normal measure p then M ( X ) = N ( X ) , i.e. every meager set is nowhere dense. ( p is regular). 3. Let m be a finite measure defined on a a-field F of subsets of a space S. Denote by X the Stone space of the measure algebra A(m). a) The subring R C C ( X ) of simple functions (i.e. assuming finitely many values) is dense in C ( X ) . b) The ring L"(S, F, m ) of measurable essentially bounded functions is linearly isometric to C ( X ) . c) There is a normal measure p on X extending m. (See Ex. 1 above or alternatively extend the linear functional I(f) = CnEre,f) (Y . m ( f - ' ( a ) ) from R to C ( X ) and apply the Ftiesz representation theorem) (Dixmier). Let & ( X ) denote the smallest a-field of subsets of a space X containing the field B ( X ) of open-closed subsets of X . 4. If X is compact and 0-disconnected then the algebra B ( X ) is isomorphic to

BO ( X ) / M ( X ) . Hence, every a-complete algebra A has the form F / A , where F is a a-field of sets and A is a a-ideal. (Loomis-Sikorski). 5. Any measure algebra A is of the form A = A(m), i.e. A is the measure algebra of some measure m (defined on a a-field of sets). (Exs 1 and 4 above).

Chapter 4

F-Spaces

$1. EXTENDING A FUNCTION. Let X be a Tychonoff space. We say that a subset E C X is C*-embedded in X , if every function from C * ( E )has an extension in C * ( X ) . Similarly, a set E 2 X is said to be C-embedded in X , if any function in C ( E )can be continuously extended to the whole of X . Clearly, any C-embedded set is also (?*-embedded. In this terminology each closed set E in a normal space X is C*-embedded in X (and C-embedded as well). In view of the Taimanov theorem (Theorem 4.6, Chapter 3) any Tychonoff space X is C*-embedded in @[XI. Moreover, the space @[XIis a unique compatification of X in which X is C*-embedded. Furthermore, from the theorem 5.4 of Chapter 3 follows that any countable discrete set D C w* is C*-embedded in the space w * . 1.1. THEOREM (Urysohn [GJ]). Let E be an arbitrary subset of a Tychonoff space X . If any two disjoint zero-sets A, B E Z ( E ) axe completely separated in X , then E is C*-embedded in X .

PROOF.Let S = F be the closure of E in @ [ X I . By assumption, there is a function f E C * ( X )separating given disjoint zero-sets A, B E Z ( E ) . Since f has a continuous extension defined on @ [ X I we , obtain 2 n B = $3 in @[XIand hence also in S. Thus, by the Taimanov theorem, E is C*-embedded in S and hence in @[XI (because S is C*-embedded in p [ X ] as a closed set). Since E g X g @ [ X I ,E is C*-embedded in X as well. QED Now, using the Urysohn theorem, we prove the following 1.2. THEOREM. In a a-disconnected Tychonoff space X every functionally open (and in an extremally disconnected X every open) set is C*-embedded.

93

CH.4 F-SPACES

94

PROOF.Let U = X \ z(f) be a functionally open set and take any disjoint A, B E Z(U).There are disjoint functionally open in U sets V,W separating A and B (Corollary 3.2, Chapter 3). Thus, we have e.g. V = U \ z ( g ) , for some g E C*(U) and A G V . Define f(z) . g(z), for I E U h(x) = 0, for x 4 U. Clearly, h is in C * ( X ) and V = X \ z ( h ) , i.e. V is a cozero-set in X , too. By E B ( X ) and hence the characteristic function xv of the set is assumption, continuous on X and separates A and B . From the Uryhsohn theorem, it follows that U is P e m b e d d e d in X . In the case of an extremally disconnected X the proof is even simpler, since the set V as open in U is open in X , whence E B ( X ) . QED

{

v

v

A Tychonoff space X is called an F-space (Gillman, Henriksen), if every cozero (=functionally open) set G X is C*-embedded in X . Theorem 1.2 says that any extremally disconnected (and even u-disconnected) Tychonoff space is an F-space. In particular, @[XIfor a discrete X is always an F-space. In the next section we give a Boolean interpretation of F-spaces.

8'2. CHARACTERIZATION OF COUNTABLE GAPS. w e prove now that in the class of Stone spaces (i.e. compact zero-dimensional spaces) the F-spaces correspond to Boolean algebras in which all countable gaps are filled. First, we show a well known 2.1. LEMMA.Any zero-dimensional Lindelof space X is strongly zero-dimensional.

PROOF.Let A, B E Z ( X )be two disjoint zero sets. Any Lindelof space is normal and therefore the family R={UEB(X): UnA=(2VUnB=qj} is an open cover of X . We may assume that R is countable (since X is Lindelof) and disjoint (since R G B ( X ) ) . Now, if we define

then we have A cover we obtain

U and U n B = (2. Obviously, U is open and since R is a disjoint X \ U = ~ { V E R :V n A = ( 2 } ,

$2. CHARACTERIZATION OF COUNTABLE GAPS

95

whence U E B ( X ) . Thus, the sets A and B are separated by an open-closed set. QED Recall that countable gaps in a Boolean algebra are of the form

L = ({ui where a, . bj = 0 ,for all i, j

:

i

<w} ;

{bj

:

j <w}),

< w.

2.2. THEOREM. If X is a compact zero-dimensional space, then X is an F-space if and only if all countable gaps in the algebra B ( X ) axe filled.

PROOF.Let L = ({Ui : i < w } ; {V, : j < w } ) be a gap in B ( X ) . Thus, the sets U=UUi i<w

andV=UV, j<w

are open F, and hence U U V is a cozero-set. Now, if X is an F-space, there is a function f E C ( X ) such that flu = 0 and flV = 1. Consequently, the sets U and V are contained in disjoint zero-sets. By the lemma 2.1 there is a set W E B ( X ) separating U and V , i.e. filling the gap L . Conversely, let us assume that the algebra B ( X ) has no unfilled countable gaps and take a cozero-set U X . Then we have U = Ui<w U,,for some disjoint family {Ui : i < w } of open-closed sets. Indeed, we can write U = U R, where R c B ( X ) and also U = Un<wF,, where the F's are closed. Hence, for each n E w there is a finite subfamily S, c R such that F,, c US,, for all n < w . The family S = Un<wS, is at most countable and we have U = US. Clearly, we may assume that the family S = { U, : n

< w } is disjoint and our claim follows.

We shall apply to the set U the Urysohn theorem on a continuous extending. Let A , B E Z ( U ) be two disjoint zero sets in U. Using again the lemma 2.1, we find for each n E w a set W, E B(U,,), which separates the sets A n U , and B n U,. Since U, E B ( X ) , we have also W, E B ( X ) , for all n < w . Thus, we have a countable gap

L = ({Wn : n < w } ; {Urn\ W,,, : m < w } ) . By assumption, there is a set W E B ( X ) separating the gap L. The characteristic function xw E C ( X ) separates then the sets A and B in X . An application of the theorem 1.1 shows that U is C*-embedded in X . QED

96

CH .4 F-SPACES

A compact zero-dimensional space X is called a Parovicenko space if it has the following properties: 1. X has no isolated points. 2. Nonempty Ga-sets have nonempty interiors. 3. X is an F-space. Property 1 means that the algebra B ( X ) has no atoms, property 2 - that it has no limits of length w and property 3 - that all countable gaps are filled. Hence, the remainder space W* = p [ w ]\ w is a Parovieenko space a weight c as the Stone space of the algebra P(w)/fin. Theorem 4.2 from Chapter 2 can be reformulated as follows

2.3. THEOREM. (Parovitenko [PI]) Every compact space of weight 5 w1 is a continuous image of the remainder w*. If CH is assumed, then every ParoviEenko space of weight E is homeomorphic t o w * . In the first part of the theorem, we have omitted the assumption of being zerodimensional. We can do this because each compact space X is a continuous image of a compact and zero-dimensional space S of the same weight as X . To see this, choose a family R c D(X)such that card R = w(X) and R is a base of X (clearly, any base contains such an R). Let A = [R] be the subalgebra of D ( X ) generated by R and define S as the Stone space of A. We have w(S)= cardA = cardR = w(X). We can now define a function f : S + X in a similar way as for the Gleason space G ( X ) and prove its continuity (cf 7.5, Chapter 3). In the recent case, the space S need not be only extremally disconnected.

53. CONSTRUCTION OF PAROVICENKO SPACES. A topological space X is called a-compact if it is a union X = U i < w X , of countably many compact subspaces. Obviously, any a-compact space is Lindelof and hence normal. Recall that a space X is locally compact, if each point I E X has an open It is well known that any locally compact neighbourhood U with a compact closure space is Tychonoff. Moreover, the remainder X * = p[X]\X is closed (in P [ X ] )and hence compact. Indeed, if z E X is an arbitrary point and U is a neighbourhood of I with a compact closure, then for some open in p [ X ]set V we have U = V n X and

v.

z€vcV=vnx=Vcx since the closure of U in p [ X ]is the same as in X (because is open in @[XIand the claim follows.

is compact). Hence X

$3. CONSTRUCTION OF PAROVICENKO SPACES

97

It is easy to see that the map U H U*(= \ X ) , for U E B ( X ) is a homomorphism of the algebra B ( X ) onto B ( X * )with kernel K n B ( X ) , (i.e. the compact elements of B ( X ) ) . 3.1. THEOREM. (Gillman, Henriksen [GHl,GHz]). If X is a non-compact but locally compact and a-compact space, then the remainder X * = p[X]\X is a compact F-space.

PROOF.Let U c X * be a cozero-set. Since X is open in p [ X ] ,it is also open in X U U . Hence, U is closed in X U U . We note that X U U is a normal space. Indeed, X is a-compact (by assumption) and so is U as an F,-set. Therefore, X U U is also a-compact and hence normal. It follows that U is C*-embedded in X U U (by the Tietze theorem). Since X is C*-embedded and dense in p [ X ] ,we infer that U is C*-embedded in @[XIand hence also in X * . QED 3.2. EXAMPLE.The remainder W; = p[W+]\ W+ of the non negative real numbers is an F-space (in view of 3.1). Consequently, there are connected F-spaces (comp. Exs G.3 and 4, Chapter 3.).

For any Tychonoff space X we define a space T = T ( X ) as follows. First, we form the product w x X (with the discrete w ) and note that it is identical with the disjoint sum of w copies of X

x = Ui<,{i} x x = Ci<,x. Let us denote T ( X )= (w x X ) * = p[w x x]\ (w X x). w x

3.3. THEOREM. If X is a zero-dimensional compact space, then T ( X ) is a Pa.rovii.enko space.

PROOF.Clearly, w x X is a non-compact but a-compact and locally compact space. In addition, it is strongly zero-dimensional (the lemma 2.1). Applying the theorem 3.1 and the theorem 6.2 from Chapter 3, we infer that the remainder T ( X ) is a compact and zero-dimensional F-space. It is easy to see that there are no isolated points in T ( X ) . Indeed, if U* # (3 for a U E B(w x X ) , then U is a non-compact set and therefore the set A={iEw : Un({i}xX)#Q)}

CH.4 F-SPACES

98

must be infinite. Let us split A into two infinite disjoint parts A = B U C and set and Thus, the open-closed sets V and W are non-compact, disjoint and U = V U W . It follows that Z* and W* are nonempty, disjoint and U* = V *U W * ,which proves the claim. We have to show yet that the algebra D(T) has no limits of type w . Let

(U;: n < w ) be a decreasing chain in B(T). Thus, the sets U, are non-compact while U, \ Un+l are compact, for each n < w. Hence, the sets A , = {i E w :

U, n ({i} x X )# 8 )

are all infinite and we have An+l GI A,, for each n A,, for each n < w . A such that A Now, if we define

< w . There exists an infinite set

u = ~ { u , n ( { ~ } x nxE)A:} then U is a non-compact open-closed set and clearly U* C the proof.

nnCw U:, which finishes

QED

3.4. REMARK. If the weight w(X)= K is 5 c, then the space T ( X ) has weight c. Indeed, since the cardinality of B(w x X ) is non-greater than rcw 5 c" = c, the weight w(T(X)) is also 5 c. To prove the converse inequality, let us take an almost disjoint family R (of subsets of w ) of cardinality c. Setting V, = a x X for a E R, we = c. see that the sets V: E B(T) are disjoint each other. Hence w(T(X)) 3.5. SPACESOF ORDINALS. Any ordinal number y > 0 can be regarded a topological space, in which the set ($9) and the intervals

are basic open-closed sets. Thus, the ordinal space y is zero-dimensional. It is always

+

normal and it is compact exactly for successors y = 6 1. The point (3 and the successors are all isolated in y and limit ordinals > 0 are precisely the cluster points of y. Let K 2 w be an infinite cardinal. The point K of the ordinal space rc 1 has

+

$3. CONSTRUCTION OF PAROVICENKO SPACES

99

an ordered base of neighbourhoods (the intervals ( a ,K], for a < K) of cardinality Thus, if K is a regular cardinal > w , then K is a P(rc)-point in the space K + 1.

K.

In an arbitrary topological space X , a set A X is called a P(n)-set, if any intersection of less than K neighbourhoods of A contains a neighbourhood of A:

A

c int nR

for any family R, of neighbourhoods of A, with cardR

< K.

Hence, a point z E X is a P(n)-point if and only if the one element set {z} is a P(n)-set. A P(wl)-set is called also a P-set. Now, let us fix a regular cardinal S = w x (K 1). Clearly, the set

K

+

A =w

X {K}

> w and consider the disjoint sum = { ( z , ~ ) :Z Ew}

is a P(n)-set in the space S. Moreover, the set A has an ordered base of neighbourhoods, viz. the sets V, = w x ( a ,K ] , for a < K . Indeed, if U is any neighbourhhood of A, then for each i < w there is an ordinal ai < K such that {i} x ( a i , ~ ] U . Hence, if a = ai then a < K and V, U. Note also that A = V, is closed and nowhere dense.

c

3.6. THEOREM. Let K > w be a regular cardinal. The ultrafilter space w* is (homeomorphic to) a closed nowher dense P( 6)-set in the ParoviEenkospace T ( K + ~ ) . -

PROOF.Clearly, the remainder A* = A \ S, where S = w x ( K + 1) and A = w x {K}, is closed in T(n + 1). It is easy to see that A* is also a nowhere dense set. Indeed, if U E B ( S ) ,then there is a V E B(S)such that V U and V n A = $3, (because A is nowhere dense in S). It follows that V * C U* and V' n A* = $3 (since , proves the claim. In addition, we have A* ( S \ V ) * ) which

(*I

A* =

n

where Va = w x ( a , ~ ] .

V:,

a
Indeed, if V E B ( S ) is any basic set, then

vnA=$3 Therefore, for any p E T(K

=

3a
+ l),we have

~ $ G2 3 V E B ( S ) ( p E V A V n A = $ 3 ]

= =

3 v E B(S)3a < n [ p E V A V n V , = $31 3a<

K[P$V,],

=

CH.4 F-SPACES

100 -

-

V,,which implies (*). The family {V: : a < n} is then an ordered i.e. A = base of open-closed neighbourhoods of the set A*. Consequently, A* is a P(n)-set in the space T . It remains to show that the space w* is homeomorphic to A*. First, we note that any function f E C * ( A ) can be continuously extended to the whole space S = w x (n 1) by setting f ( i , a ) = f ( i , n ) for i < w and a < n. Since S is C*-embedded in p [ S ] ,the function f can be further extended to @[S],in particular to 2. It follows that A is C*-embedded in 2,which means 2 = @ [ A ] .But A is homeomorphic to w (as countable and discrete) and hence we have 3 = ,B[w], which implies A* = w * , as required. QED

+

For any topological space X and any continuous function f : A --f Y , which maps a closed set A X onto mother space Y , define a factor space X / f as follows. The underlying set is X / f = ( X \ A ) U Y and a set V X / f is open if and only if V n Y is open in Y and (V \ Y ) U f - ' [ V ] is open in X . From this definition it follows at once that a set F C X / f is closed if and only if F n Y is closed in Y and ( F \ Y ) U f-' [ F ] is closed in X . The topology on Y inherited from X / f coincides with the original one and hence Y is a closed subspace of X / f and the complement X / f \ Y is identical with X \ A .

c

c

It is well known and easy to check that if X and Y are compact and zerodimensional, then so is X / f .

Now,we prove the following useful 3.7. THEOREM. Let X be a ParoviEenko space and A E X a closed nowhere dense P(K)-set (for some regular tc > w). I f f : A + Y is a continuous map onto a compact zero-dimensional F-space Y , then X / f is a Parovitenko space, in which Y is closed and nowhere dense P(n)-set.

PROOF.We already know that X / f is a compact and zero-dimensional space. Suppose now, that X / f has an isolated point z. If z E X / f \ Y , then z would be A would be open in X . Since neither case is isolated in X and if t E Y , than f-'(z) possible (the latter because A is nowhere dense), the space X / f is dense in itself. Let : i < w ) be a decreasing sequence in B ( X / j ) . Then the sets U, = (V, \Y)U f -l [V,] form a decreasing sequence in B ( X ) and hence there is a nonempty set U E B ( X ) such that U U,, for all i < w . Since A is nowhere dense, the set U \ A must contain

(x

53, CONSTRUCTION OF PAROVICENKO SPACES a nonempty set V E B ( X ) and then V E B ( X / f ) and V there are no countable limits in B ( X / f ) .

101

V,, for all i < w. Thus,

Now, let us consider a countable gap

L = ({U,

:

i

< w } ; {Vj : j < w } )

in the algebra B(X/f). Of course, we may assume that the sequences (Ui : i < w) and (Vj : j < w ) are both increasing. If at least one of these sequences lies outside of Y , i.e. if U , n Y = Q, for all i < w, then

K = ( { U i : i < w } ; {(V,\Y)uf-'[Vj]:j < w } ) is a gap in B ( X ) and therefore it can be separated by a set U E B ( X ) . Since A E n,<,(X \ V i ) , there is a set W E B ( X ) such that A g W and Vi n W = Q,, for every i < w (since A is a P-set). The intersection U n ( X \ W) is open-closed in X / f and separates the gap L. Finally, let us assume that both the U's and the V's intersect Y . Denote ui = (U, \ Y )U f-'[Vi] and v j = (V, \ Y )u f-'[Vj]

for i, j E w .

Clearly, L x = ( { u , : i < w} ; {vj : j < w}) is a gap in B ( X ) and L y = ( { U , n Y : i < w } ; {V, n Y : j < w}) is a gap in B(Y). By assumptions and the theorem 2.2, there is a set v E B(Y) separating the gap L y . The counter image w = f-'[v] separates that part of the gap LX which lies in A. Simultaneously, we have w = W n A for some W E B ( X ) and hence ui n A W and W n A X \ v j for all i, j < w. Since A = n { H E B ( X ) : A C H}, there are open-closed neighbourhoods H i of A for which

s

ui n Hi G W

s

and W n Hi ( X \ v i ) for all i

< W.

By assumption, A is a P-set and hence there is a neighbourhood H E B ( X ) of A contained in all Hi, i < w. We have then

u,nHsW

andWn(vjnH)=Q,

fori,j<w,

i.e. the set W separates that part of Lx which lies in H. Now, if a set U E B ( X ) separates Lx on X \ H, then the set

U n ( X \ H)u [(W n H )\ A1

102

CH.4 F-SPACES

is open-closed in X / f and separates the gap L . We have shown that any countable

gap in X / f (that is - in B ( X / f ) ) is filled and consequently X / f is a ParoviEenko space. Clearly, Y is a closed and nowhere dense P(n)-set in X / f , which follows directly from the same properties imposed on A . QED Let us note additionally the following inequalities

which follow immediately from w(Y)5 w(A) 5 w ( X ) .

x

c w* be a closed set. It follows $4. CLOSED SETS IN THE SPACE w * . Let immediately that X is a compact zero-dimensional space and the weight w ( X ) 5 c = w(w*).If is also not difficult to prove the following 4.1 LEMMA. Every closed set X

c w* is an F-space.

PROOF.It is sufficient to show that in the algebra B ( X ) any countable gap L = ( { U i : i < w } ; {Vj

: j

<w})

is filled. Each set U E B ( X ) is of the form U = u n X for some open-closed u E B ( w * ) . Hence, we can define a gap K in w* as follows. Let u0,vo E B ( w * ) be such that uo n X = Uo and vo n X = Vo. We may assume also that uo n vo = Q, (replacing, if necessary, uo by uo \ vo). Assume now that we are already given { u ; : i 5 n } and {v, : i L n} so that ui n X = U,and v, n X = vi for all i L. n, and ui n LJ, = (3 for all i , j 5 n. There is a un+l E B(w*) such that un+l n X = U n + 1 . If un+l intersect some of the v’s, then replace un+l by un+l \ (vo U . . . U vn). The set vn+l is defined symmetrically. Thus, we have defined a gap K in f?(w*) whose trace on X is L . Now, if a set W E B(w*) fills K , then W n X fills L and the proof is complete.

QED

Clearly, the same argument proves a little more general theorem:

If, in a given Boolean algebra A, every countablegap is filled, then the same property has any homomorphic image of A. As we have shown above, any closed set X G W * is a compact zero-dimensional F-space of weight L: C. In the presence of the continuum hypothesis, the converse theorem is also true (Louveau [L]). Even stronger, we prove the following

$5. O N T H E PAROVICENKO THEOREM

103

4.2. THEOREM. (Balcar, Frankiewicz, Mills [BFM]). Assuming CH, every compact zero-dimensional F-space X of weight 5 c can be embedded into w* as a closed nowhere dense P-set.

PROOF.In view of the theorem 2.3, the space X is a continuous image of the space w*: there is a continuous f : w* + X , onto. Substitute K = w1 = c in the theorem 3.6.The ParoviEenko space T = T ( w l + 1)has weight c (by Remark 3.4) and contains the space w* as a closed nowhere dense P-set. Moreover, T is (homeomorphic to) the space w* (again by the ParoviEenko theorem 2.3). Applying the theorem 3.7, we infer that T l f is a ParoviEenko space containing X as a closed nowhere P-set. Clearly, we have w ( T / f ) = c, since T l f contains e.g. open-closed sets E w* = T , disjoint from X . Hence, by the ParoviEenko theorem, we have T f f = w*,which QED finishes the proof.

$5. ON T H E PAROVICENKO THEOREM. From the theorem 3.6, we already know that the remainder w* is contained in any ParoviEenko space T ( K 1) as a P(rc)-set. It is, in fact, contained as a regular P(K)-set, i.e. it has an ordered base of open-closed neighbourhoods.

+

c

w* be a closed regular 5.1. LEMMA.(Frankiewicz, Zbierski, [FZl]). Let X P(K)-set. If the space p [ w ]is a continuous open image of X , then there exists a scale

of length

K.

PROOF.Let g : X + P[w]be a continuous open map (onto p [ w ] ) . The sets X , = g-l(n), for n E w , are open-closed in X and hence we have X , = A; n X for some infinite A , C_ w and n E w. Since the sets X,, n E w , are disjoint each other, we may assume that the relative closures A; are disjoint as well (because each A; can be replaced by A; \(A$ U . . . U A;-l)). Therefore, the sets A , can be also chosen mutually disjoint. Let {Vz : a < K} be an ordered open-closed base of X . In particular, X = V,l and hence the sets V, n A , are infinite, for all (Y < K and n E w . Let us define f , = min(V, n A , )

If a < p, then Vp

c* V,

for a < K and n < w.

and hence f , <* f a . Moreover, the sequence (f, :

OL

< K)

104

(3.4F-SPACES

dominates. Indeed, let f : w

-+ w

be an arbitrary function. Let us denote

S, = A , n {i E w : i 5 f(n)} and S =

U S,. n<w

Since the sets A , , n E w , are mutually disjoint, we have S n A , = S , E fin, for all n E w . Hence S' n A: = $3 for all n < w . On the other hand, the set

UAEnX= U X , nEw

nEw

is dense in X , since the map g : X -+ P[w] is open. It follows that S' is disjoint from X and consequently S' n Vz = $3, for some a < K . Thus, S n V, E fin, which implies S, n V, = 8 for sufficiently large n. This means that f <. fa and the claim follows. Now, the dominating sequence (fa : a < K ) contains an increasing (in the QED relation 5') cofinal subsequence, which is obviously a scale of length K . 5.2. EXAMPLE. If X is a a-disconnected compact space, then @[w]is a continuous open image of X (obviously, X is assumed to be infinite). Indeed, the (infinite) algebra B ( X ) is then a-complete and contains a countably infinite antichain C = { X , : n E w } . We may assume that C is maximal, for if

supx, = n<w

u

X,

is less than X in B ( X ) ,

nEw

then we can add an element X as a subalgebra: if we set

\ UnEv X,

g ( A ) = supX, = nEA

to C . It follows that B ( X ) contains P ( w )

uX,

for A C w ,

n€A

then a simple checking shows that g : P ( w ) -+ B ( X ) is an isomorphic embedding. Identifying points z E X with fixed ultrafilters F, = {V E B ( X ) : z E U } we see immediately that the induced map f : X -+ @[w]of the corresponding Stone Spaces has a form f ( ~= ) g-'[F, n ~ [ P ( w ) = ]] { Ac w : z E

U x,}.

n€A

In order to check that f is open, it suffices to show that for any U E B ( X ) we have

f [ U ]= 2

where A = {n E w : U n X ,

# $3).

$ 5 . ON THE PAROVICENKO THEOREM

105

Clearly, we have A C_ f[U]and hence also 2 f [ U ] . On the other hand, the condition f(z) $ implies w \ A E f(z),whence I E g(w \ A). But, since U g ( A ) and g(A)n g(w \ A ) = (3, we obtain f(z) 4 f [ U ] ,and the claim follows. QED

Let P denote the following property

(P) Every ParoviEenkospace of weight equal c is homeomorphic to the ultrafilter space w* . Thus, the ParoviEenko theorem 2.3 says that CH implies P. Conversely, we have the following 5.3. THEOREM. (van Douwen, van Mill [vDvM]). The property P implies CH and hence P is equivalent to the continuum hypothesis.

PROOF.Let w1 5

K:

5 c be a regular cardinal. The space T(K: + 1) contains w*

as a closed nowher dense P(/c)-set (the theorem 3.6). The assumption P implies that

T(K+ 1) is itself homeomorphic to w * . Let us note now that the following holds (*)

any separable compact space X is a continuous image of w * .

Indeed, X is a continuous image of p [ w ] , since there is a map g : w X onto a dense subset X and the extension 5 : B[w]-t X is then onto X . On the other hand, the space p[w] is a continuous image of w * , since the algebra P ( w ) is easily seen to be embeddable into P(w)/fin: if w = UnEw2, is a division into countably many infinite parts, then the map for A

w,

nEA

is clearly an embedding. Thus, we have proved (*). Now let us specify X as e.g. X = G((0,l } w )(the Gleason of the Cantor space). X . By the theorem 3.7, the space T(n+ l)/f Hence, there is a continuous f : w* is still a ParoviEenko space of weight c and contains X as a closed regular P(K:)-set. We may also write T ( K l)/f = w*, in view of the assumption P. Hence, we can apply the lemma 5.1, since X - as extremally disconnected - can be mapped onto p [ w ] with a continuous open map (see example 5.2). It follows that there is a scale of length K . Thus so far we proved that the assumption P implies the existence of a scale of any regular length K:, w1 5 K: 5 c. This implies CH for if c > w1, then we would have two scales of length w1 and w2 which is obviously impossible. QED

oz

+

cH.4 F-SPACES

106

We have mentioned earlier that the first part of ParoviEenko theorem is not equivalent to CH. The proof will be given in Chapter 8.

56. ON P - S E T S IN THE SPACE w * . If the cardinal c is regular, then each

P(c)-set X in the space w* is a regular P(c)-set, i.e. X has an ordered base {V, : a < c } of neighbourhoods. Indeed, if {U, : (Y < c } is the family of all open-closed neighbourhoods of X , then the V's are defined by induction so that

From theorems of the sec. 4 we know that, assuming the continuum hypothesis, the following holds (BFM) Any closed set X

w*

is homeomorphic to a closed P(c)-set Y

w*.

Now,we prove that the converse is also true: the property BFM implies CH. Before the proof let us express the property BFM in Boolean terms. By a P(K)-filter we mean a filter F on the set w having the following property: if R G F is a family of cardinality less than such that A GI B, for all B E R.

K,

then there is a set A E F

Clearly, if F is a P(K)-filter, then n { A * : A E F } is a closed P(K)-set and vice versa. The property BFM can be now stated as follows: For any filter F on w there is a P(c)-filter G such that the factor algebra

P ( w ) / F is isomorphic to P(w)/G. First, we prove the following 6.1. LEMMA. There exists a closed set X G w* for which the algebra B ( X ) has a non-separable ordered gap of type (w1 ,LO).

PROOF.Let {H,: n E w } be an arbitrary disjoint family of open-closed sets. Since B(w*) has no countable limits, there exist decreasing chains of length w1:

H, = H t 2 HA 2 . . . 2 H," 2 . . , We choose points q: E H,"\

for

(Y

< w1 and n < w .

and denote

Q={q:: a<wl

and n < w } .

$6. ON P-SETS I N THE SPACE

107

W*

Clearly, Q is a strongly discrete set. Define a
Qy={q::

and

and

nEw}

&;I= { q g : a < y and ZLn}.

Let C, be any open-closed set separating the countable gap

({Hn\Ht:

n<w};

{H,":

a<w}).

Hence, the set C, separates also Q" from Q \ Q* and the set 0," = C, n separates QE from Q \ QE. If we set

X=V\

u UE=W n

a<wl n<w

Ui<, Hi -

n(w*\D:),

a<wl n<w

then X is a nonempty closed nowhere dense subset of w * . Let us define and

nX = Hn nX

< w1

a, = C ,

for a

bn

for n E w .

From this definition, it is clear that the sequence (a, : a < w1) is increasing and the sets b, are disjoint each other. Of course, the a's and the b's are both relatively

open-closed in X . Thus, we have a gap in B ( X )

L = ({a, :

<WI} ; {b,

< w}).

:

In order to show that L is unfilled, take an arbitrary u E B ( X ) and suppose that a, 2 u for all a < w1. Since u = U n X for some U E B ( w * ) , we have obviously

c, 0

n

nn

(w*

\ D,") C_ U

for every

<wl.

a<wl n<w

By compactness, there is a number n ,

c,nQn

< w , such that

UH~EU. i>na

The value n, must be constant on an uncountable subset 2 w1 : n, = n for arbitrarily large a < w1. Hence, we have

108

CH.4 F-SPACES

which implies bi C u, for i 2 n. Thus, the set u does not separate L.

Q ED 6.2. THEOREM. (Frankiewicz, Zbierski [FZI]). The property BFM implies (and hence is equivalent to) the continuum hypothesis.

PROOF.From BFM it follows immediately that there are P(c)-points in w * . In particular, the cardinal c is regular. We know already that the space @[w] can be embedded into w* (the theorem 5.4, Chapter 3). Therefore, we may assume that P [ w ] is a regular closed P(c)-set in the space w*. Hence, there exists a scale of length c (the lemma 5.1). From the theorems 2.5 and 2.7 of Chapter 2 it follows that every ( ~ , w ) - g a pin B(w*), where K < c, is separated (because Ib = dl = c). Let us suppose that c > w1. Thus, any (w1,w)-gap is separated. We shall derive a contradition. By the lemma 6.1 there is a non-separated (w1, w)-gap

L = ({u,

: a

< w1} ; {v,

: n<w})

in B ( X ) , for some closed X w * . By the assumption BFM, the set X can be regarded as a P(c)-set. But if so, then the gap L determines a gap K in B(w*), whose trace on X is L. Indeed , let {W, : CY < c} be an ordered open-closed base of the set X . We choose open-closed sets U,,V, E B ( w * ) such that U,

= U, n X

and v, = V, n X

for all a

< w1 and n < W .

Since L is a gap, we have

(U, n V,) n X = Q, Now, using the equality X =

n,,,

for all a

< w1 and n < w .

W-, and the compactness of X , we obtain for some y = $a,n).

(Urnn V,) n W, = Q, Thus, if 6 = sup{y(a,n) : a < w1 and n

U, n V, n w6 = 8

< w } , then 6 < c and we have

for some a

< w1 and n < w .

In other words li.=({u,nWb:

a<w1};

{V,:

n<w})

$6. ON P-SETS I N THE SPACE

W*

109

is a gap in B(w*) and the trace of K on X is the gap L. Since K is a separable, so must be L which contradicts the choice of L. QED We shall discuss now the Louveau theorem mentioned earlier. Consider the following property.

(L) Every Boolean algebra of cardinality at most c, in which all countable gaps are separated, is a homomorphic image of P(w)/fin. Thus,the Louveau theorem says that CH implies L. We prove below that the converse is not true, i.e. that L is not equivalent to CH. For the proof, it is sufficient to construct a model (of set theory) in which c > w1 and L holds true. Actually, a Cohen model is good for this purpose. Let V be a ground model in which CH holds and let V1 = V[G] be an extension obtained by adding w1 Cohen reals. Denote A = P(w)/fin. Obviously, we may assumethat A n V G A n V l . First, we prove the following

6.3. LEMMA.Let B E Vi be an algebra of cardinality c in which all countable B gaps are separated. Every homomorphism h E V, of A' onto a subalgebra Bo can be extended to a homomorphism H E Vl, of A' onto B. PROOF.We may assume that h is defined on P(w) and h ( s ) = 0 for s E fin. Let us consider a countably generated (over V) subalgebra A of the form

A = AV[a,

: a

< y]

i.e. A is a subalgebra generated (in V l ) by A' where a < y.

where y < w1, and countably many new sets a,,

Every new set a E V1 determines a gap La = ({u E A : u 5 a } ; {v E A : a 5 v}). We claim that any such La is countably generated. Indeed, any element u E A is a finite sum of elements of the form 2 n a ( & ) ,where z E V and a ( & )= &(a1). a,,

n . . . n €(a,) . a,,

cH.4 F-SPACES

110

and & is a function from a finite subset C w1 into {-1, +l}. There is a countable subforcing C, of C(wl) such that A, a E V [ G n @,I. For any X E V(@l) and p E @, denote X , = {i E w : IF "i E X " } . From this definition, it follows immediately that

z E V and p It. "z C X" imply Assume now, that

I

I

X,,.

n a(&) c*a and hence that

p It. "z n a(&)

a u k"

It follows at once

znU(E)

for some p E Caand k E w .

c x ( p , E , k) n a(&) c u u k,

where X ( p ,E , k) = [ ( a U k) U ( p \ In particular, the sets X ( p , & k) , are all in the ground model V. Hence, any set u E A such that u 5 a, is contained in a finite union of sets of the from X ( p , e , k) ( I a ( € ) . There is only countably many such sets. Consequently, the lower class of the gap La is countably generated. The proof for the upper class is similar and the claim follows. Now we can describe how to extend a given homomorphism h : AV*a Bo. Since CH holds in V,, we have card AV1= c a r d B = w1, in V1. Let Avl \Av = {a, : a < w1} and 1 \ Bo = { b , : a < w 1 } be some fixed enumerations. Define by induction (in Vl) continuously increasing chains (A, : a < w l ) and (B, : (Y < w l ) of subalgebras of A' and B, respectively and a chain ( h , : a < w1) of homomorphisms such that 1. A0 = A', lo is already given and ho = h 2. h, maps A, onto B,, a, E A,+l and b, E B,+l, for all a 3. A, is countably generated over AV,for all (Y < w1.

< w1.

Assume inductively that the terms A€, Be and h~ are already defined for all ( < a. If a is a limit ordinal, then define

Now let a = p

+ 1 and suppose that a = up # Ap. As proved earlier, the gap

56. ON P-SETS IN THE SPACE

111

w*

where a = ap, is countably generated (because Ap is countably generated over A0 = A'). Let u n and v,,, for n E w , be generators of the lower and the upper class of L a , respectively. Thus, for arbitrary 2,y E Ap, we have x < a

and

3 n ~ w ( s < u ~ )

=

y>a

3mEw(y>vm)

By assumption, there exists an element b E B separating the countable gap

Consequently, the element b separates also the gap

By the Sikorski theorem (Theorem 7.3, Chapter l),there is a homomorphism f : Ap[a]" 2 Bp[ b ] , extending hp. If bp E Bp[b],then we define A, = Ap[ap], B, = Bp[b]and h, = f. If bp 6 Bp[b] then we extend f further as follows. There is a countable subforcing such that Ap and ap are in V [ Gn @a]. Let A w , A E Vl be a Cohen real over V [ Gn Cp]. Clearly, for any subsets u , v C w from V [ Gn Cp] we have the following:

s

if u

A

v , then u E fin a n d w \ v E fin.

Again by the Sikorski theorem, f has an extension g : Ap[a,A]+ B with an arbitrarily chosen value g(A). Of course, we set g(A) = bp and define A, = Ap[ap,A ] ,B, = Bp[b,b p ] and h, = g. In the case of a0 E Ap and bp

4 Bp we proceed in a similar way.

Now, since A' = A, and B = U,<wl B,, the function H = Uu<wlh, is a homomorphism of AVLonto B, which finishes the proof. QED Let us make now a more general remark. Suppose that K E V is a regular uncountable cardinal in a model V and P, = C,
tcm

s

K

(for such that --t

K

CH.4 F-SPACES

112

Indeed, choose a canonical name f E V@*)of the operation f f = { ( ( c u , ~ ( Eq ),) @ * ) , q ) :

cu < K and q E A @ ) } ,

where z = ((YO,. . .( ~ ~ - 1A)@~ ) C P, is an antichain and q5 : is a function satisfying the condition q Ik "f@) = ~ ( c uq)" , -

U,<,{Tii}

x A@} + K

for every q E A @ ) .

It is easy to check that the set

is a cub in

K.

Let us denote

Clearly, fly is a P-,-name and (fly)[F n P,] = fly, for all y E the proof of 6.4.

Nj.

This completes

Let V, denote the forcing extension of V via @(wl x y), i.e. the model obtained from V by adding y times w1 Cohen reals.

A subset N C w2 is called an w1-cub, if N is unbounded in w2 and closed under wl-limits (Le. if supE<,, f(E) E N , for any increasing function f : w1 -+ N , see Section 2, Chapter 5 ) . Using the remark 6.4, we obtain easily the following 6.5. COROLLARY. For every Boolean algebra B E V,,, if card B = c = w2 and all countable gaps in B are separated, then there is an wl-cub N _C w2 and a decomposition B = UYEN B-, such that 1. the sequence (B, : y E N ) is increasing and By = U{B, : $, < y A E N } for any wl-limit y E N . 2. For all y E N , By is in V, and (in V-,) has cardinality c = w1 and all countable gaps in B, are separated.

PROOF.In view of 6.4 there is a cub NO E w2 and a decomposition B = UyENo B, fulfilling all the requirements except possibly that on separating. Let y E No and L be a countable gap in 1,. By assumption, L is filled in B and hence in some

I , , Y I Y L and Y L

E No.

$6. ON

P-SETS IN THE SPACE w*

113

Now, if we set

y* = sup{yr, : L

is a countable gap in

By},

then y* E NO and all countable gaps of By are separated in By*.Define by induction an increasing function f : w2 + NO by the equations f(0) = y, f(a 1) = (f(a))* and f(a)= f(P),for limit a. Finally, let l(y) = SUP^<,^ f(a).Clearly, l(y) E NO and Bl(.,)has all its countable gaps filled. Thus, we have proved that the subset N 5 NOconsisting of all y E NO such that By has all its countable gaps separated is unbounded in w2. Obviously, N is also closed under wl-limits, which finishes the proof. QED

+

Now, we can prove the announced theorem. (Dow, Frankiewicz, Zbierski [DFZ]). In the Cohen model V,, 6.6. THEOREM. every compact zero-dimensional F-space of weight 5 c can be embedded into the remainder space w* .

PROOF.Let B E VWzbe a Boolean algebra of cardinality c having no unfilled countable gaps. We have to show that B is a homomorphic image of AVwz, (where A = P(w)/fin). Take a set N C w2 as in 6.5 and assume inductively that we have already a homomorphism h, : AV7 %o

By where y E N ,

and h, belongs to V, (if y =inf N , then the existence of h, follows from the Louveau theorem, since V, k CH). Let 6 E N be the successor of y in N . Applying the lemma 6.3, we obtain a homomorphism hg : Av6 "3 Bg, extending h,, (since is obtained from V., by adding w1 Cohen reals). If y E N is an wl-limit, i.e. y = S U ~ ~ y
h.,, then h, = for a < w l .

..

AVO Onto + Bye

for all a

< w1,

h,, is a homomorphism of AV7 onto By extending all the h.,,,

Indeed, we have AV7 = forcing and Corollary 6.5).

uu<wl Av7- and By = Ua<wlBy,, (by the ccc of Cohen

CH.4 F-SPACES

114

Finally, let {y, : n < w } C_ N be an increasing sequence of length w and let y = ~ u p , < ~ y , in , N and let h,, : Av7n + B,, be an increasing sequence of homomorphism onto, for n < w. Then h = h,, maps A = Un<wA v y n onto Un
Un+

a],

Hence, every gap La is countably generated in A [ a , : a < p] over V,, where 6 = y, (since La is countably generated over v 6 in any A v T n , n < w ) . Therefore, the lemma 6.3 applies also in this case and consequently h = Un<wh , can be extended to a homomorphism h, : AV7 By,which finishes the proof.

oz

QED

$7. CHARACTER OF POINTS. For any topological space X the character x(z) of a point 2 E X is defined as the smallest possible cardinality of a local base of neighbourhoods at the point z, (the character x ( A ) of a subset A C_ X is defined similarly). Clearly, the condition x ( z ) 2 w characterizes the limit points 5 E X , (for a regular space X ) . For a discrete space X the character x(p) of a point p E p [ X ]is the smallest possible cardinality of a family R p, generating the ultrafilter p . 7.1. THEOREM. For everyp E w* we have x(p) > w . PROOF.Obviously, we have x(p) 2 w . Suppose that some p has a countable base A0 1 A1 _> . . ., i.e.

A Ep

3n E w ( A ,

A)

for any A

w.

Since the sets A , are infinite, there is a Bernstein set: by induction, at each step n choose two points zn,yn E A , \ {zi,yi : i < n}. Then, the set B = (2, : n E w } is in p (since p u {B} is a centered family) and also w \ B is in p (by the same reason), which is impossible.

QED

Clearly, we have always x ( p ) 5 c , since p

P(u),

$7. CHARACTER OF

115

POINTS

7.2. THEOREM. (Pospil;il [Pz]). There exist pointsp E W * of character c.

PROOF.Let R be an independent family of subset of proof of the theorem 6.5, Chapter 1). Then, the family

R U {w \

7.3.

0S : S c R

w,

card R = c (see the

and S is infinite}

is uniformly centered. Indeed, for arbitrary sets A l , . . . A , S1,.. . , S, R we have

E

R and infinite

c

n . . . n A, n (w \ n s l ) n . . . n (w \ ns,)

=

= ~ ~ n . . - n ~U, (W\Bl)n...n n =

u ... u BiETi

BiESi

U

(w\B,)=

BmESm

Aln...nA,n(w\B,)n...n(w\B,),

BmETm

where Ti = S, \ {Al, . . . ,A,} for i = 1,.. . ,rn. There exists an ultrafilter p E w* containing the family 7.3. Suppose that p has a base K of cardinality less than c. Now, if for b E K we set Rb = {A E R : b C A}, then R = UbEKRb, and hence some Rb must be infinite, whence Rb E p (because b C Rb). On the other hand, from 7.3 it follows that w \ Rb E p , a contradiction.

n

Hence, x ( p ) = c.

n

QED

An arbitrary open-closed set U = A' in w* is homeomorphic to w *. It follows immediately that the set { p E w* : x(p) = C} is dense in w'. Assuming CH, we have x ( p ) = c for every p E w * . The same holds, if we assume MA - or even weaker - the principle P ( c ) , (cf 5.2, Chapter 2). Indeed, if a family S p has cardinality less than c, then P ( c ) implies int n { A * : A E S} # $?J and hence S cannot be a base. However, it appears that - consistently - there can be points of character < C. 7.4. THEOREM (Kunen). There is a model of set theory, in which x ( p ) < c holds for some p E w'.

PROOF.Let V be a model in which c > w1. Fix a regular cardinal IE E V , w1 5 IE < c. We shall iterate with finite supports an almost disjoint forcing Q = Q ( F ) , i.e. the same forcing as in the proof of 5.2 in Chapter 2. Thus, Q ( F )for an ultrafilter F is always a-linked and adds a set X = X ( F ) which is almost contained in each A, A E F .

CH.4 F-SPACES

116

The iteration is defined by induction. Assume that we have already defined P, and an F, E V@Q) such that P, It ”F, E w * ” . Define P,+l = P, * Q(F,), where Q is a P,-name of Q(F,). Since we have

P,+1 Ik ”FaU {X(F,)}

is centered”,

we can choose a P,+l-name F,+1 such that

At limit stages a > 0 we define P, as the direct sum P, = CB<,Pp.It is easy to check that for all (Y < K P, IF ”&(Fa)has the ccc and card Q(F,) 5 d’. Hence, the finite support iteration P, = C,<, P, has the ccc and in the generic extension V[G]we have w1 5 K < c = c”. Moreover, any set A w in V[G]occurs at some stage (Y < K of the iteration. It follows that p = Urn<.F,[G]is an ultrafilter and the sequence (X, : a < K ) is a base of p (because, if A E p occurs in an a-th stage, then Xa+l C_, A ) . Hence, x(p) = K and the proof is complete. QED

c

It is also possible to construct a model in which, for a given in advance sequence ( K C : t < y) E V of regular cardinals < c, there exist ultrafilters p~ satisfying x(p6) = K C , for all < y. The proof of this is much more difficult. We shall discuss yet some simple properties of arbitrary F-spaces. In m y F-space X every countable discrete subset D 7.5. THEOREM. C*-embedded.

c X is

PROOF. Clearly, a discrete set D = {zn : n E w } is strongly discrete - there are mutually disjoint basic neighbourhoods U,,such that zn E U,,for all n < w (this is true for any regular space). We may assume that the U’s are all cozero-sets, since X is a Tychonoff space. Hence, the set U = Un<wU, is still a cozero and consequently it is C*-embedded. Now, i f f E C*(D),then first extend f to U ,setting flu,, = f(z,,), and then to the

whole of X.

QED

We say that a sequence (I,,: n < w ) has the property disjoint neighbourhood U,,of z,, for all n E w , such that

--

U

rEA

~i

n

U U, = Q,

3gA

for every set A

(p), if there are mutually E w.

57. CHARACTER OF POINTS

117

7.6. COROLLARY. Every (countable) discrete sequence in an F-space X has the property ( P ).

PROOF.We choose U, as in the proof of 7.5. The sets = U j q A U j are then disjoint. By 7.5 there is a function

flu = 0 and flv = 1, whence I n v = (2.

7.7. COROLLARY. N o point x points # x).

f

2~

= U i E A U i and

E C ( X ) such that

QED

E X of an F-space X is a limit of a sequence (of

Every sequence contains a discrete subsequence (zn: n < w ) . From the property ( p ) it follows that z has a neighbourhood omitting infinitely many terms of PROOF.

the sequence I,.

QED

7.8. COROLLARY. Everynonisolatedpoint x ofan F-spaceX has an uncountable character.

PROOF.The condition x ( x ) = w implies that x is a limit of a sequence of points, which contradicts 7.7.

QED

7.9. COROLLARY. Every (infinite) compact F-space X contains the space /?[a].

PROOF.Any infinite space X contains a countable discrete set D. Then is a compatification in which D is C*-embedded. Hence, = S[D] = P[w] (up to a homeomorphism).

QED

Finally, let us recall the notion of a convergence with respect to a filter F on w . X ( X is assumed to be Hausdofi), we say that a point Given a sequence f : w I E X is the F-limit of the sequence f, and write x = F-limi f(i), if

{i

E w : f(i) E

V }E F

for every neighbourhood V of x.

Clearly, every sequence may have at most one F-limit. The ordinary convergence is an F-convergence, where F is the Frkchet filter. Hence, if z = limi f(i) in the usual sense, then I = F-limif(i), for every nonprincipal ultrafilter F. Now, examine plimits for ultrafilters p E w * , in more details. 7.10. If X is compact, then for every sequence f : w there is a p-limit off and p-limif (2) = f(p).

+

X and every p E w*

118

CH.4 F-SPACES

This follows immediately from the continuity of the extension 7 : p [ w ] + X at the point p . 7.11. In any space X , we have {p-lirnif(i) : p E p [ w ] }= { f ( i ): i

<w}.

Indeed, if z E { f ( z ) : i < w } , then the family {f-'[V] : V is open A z E V } is centered. 7.12. If Y is compact and g : X .+ Y is a continuous map, then the condition z = p-lirnif(i) implies g(z) = p-limig(f(i)). 7.13. EXAMPLE. Let I : w we have

--t

p [ w ] be the identity

p = p-limI(n) n

7.14. If D =

(2, :

I ( n ) = n, for n E

w.

Then

for all p E w * .

n E w } is a discrete set in a compact F-space X , then

plim, z,

# q-limn 2,

for p

# q.

This follows from 7.10.

PROBLEMS

A . F-SPACES 1. A Tychonoff space X is an F-space if and only if any two disjoint cozero-sets are completely separated. 2. Let X be an F-space. If Y C X is C*-embedded in X , then Y is an F-space.

3. For any Tychonoff space, we have

X is an F-space

=

p [ X ]is an F-space.

4. If an F-space X is locally compact, then the remainder X * = p [ X ]\ X is an F-space. In particular, for any discrete space X , the remainder X * is an F-space. 5. The remainder Y * of a cozero subspace Y F-space.

s

X of a compact space X is an

6. For an infinite, compact F-space X , we have always w ( X ) 2 c.

119

PROBLEMS B. P-SPACES 1. Any countable set D C X of a P-space X is closed and discrete.

2. Let X be a P-space. a) If D X is countable, then there exist mutually disjoint open-closed neighbourhoods U, of points x E D. b) Every cozero-set in X is open-closed. Hence, any countable subset of a P-space X is C-embedded in X .

3. If a cozero-set U 2 X , where X is a Tychonoff space, is C-embedded in X , then U is open-closed. Hence, a Tychonoff space X is a P-space if and only if any cozero-set X is C-embedded in X , (Ex. 2.b, above).

C. POINTWISE ORDERING

IN

C(X)

Any ring C ( X )is partially ordered by the relation

We say that (the ordering I on) C ( X ) is complete, if every subset F C C ( X ) bounded from above (i.e. F 5 h, for some h E C ( X ) ) has a supremum g = sup F in the ordering 5 (equivalently, any set bounded from below has an infimum). If, in the above definition, F runs over countable sets only, then we obtain the notion of a-completeness. We assume that X is a Tychonoff space. 1. Let F C C ( X )be bounded from above. a) If g ( x ) = supfEFf(x), for x E X (pointwise supremum), then g is lower semicontinuous (i.e. g-'[(a, m)] is open for all a E W). b) There is an F C C(p[w])such that s u p F exists and is distinct than the pointwise supremum of F. 2. Assume that g = sup F in C ( X )exists for the following sets F: a) F = {f E C ( X ) : f I xu} for an open set U X . Then, we have g = xu, whence E B ( X ) . b) F = {min{l,n . f} : n E w } where f 2 0. Then, g = X \ ~ ( f ) whence , U E B(X).

xu,

where U =

Hence, if the pointwise ordering on C ( X ) is complete (a-complete), then the space X is extremally disconnected (a-disconnected).

C H .F-SPACES ~

120

3. If {U, : a E Q} is a family of open sets C_ X (indexed by rational numbers), satisfying U, _> Ub for a < b and UaEQU, = X , Ua = 9, then the equality h ( z ) = sup{u E Q : z E Ua}defines a continuous function h : X + R.

naEQ

4. Let X be an extremally disconnected (a-disconnected) space and F 2 C ( X ) a bounded from above set of functions (and countable, in the case of a-disconnected X ) . The sets U, = U f E F{z E X : f(z) > a} satisfy the conditions of Ex. 3 above and we have h = sup F. Hence, from Exs 2 and 4 we obtain: X is extremdly disconnected (a-disconnected) if and only if (the pointwise ordering on) C ( X ) is complete (a-complete). (Stone, Nakano). 5. Let U and V be two disjoint cozero-sets: U = X \ z(f) and V = X \ z ( g ) , where f,g E C ( X ) and 0 5 f , g 5 1. If F = { m i n { l , n . f} : n E w } and G = ((1- g)n : n E w } , then L = (F;G) is an ordered countable gap in C ( X ) .

If a function h E C ( X )separates L (i.e. F 5 h 5 G), then hlU = 0 and hlV = 1, i.e. h separates U and V. Hence, if every countable gap in C ( X ) is separated, then X is an F-space (A. 1, above).

6. Let X be an F-space and L = (F, G) - a countable gap in C ( X ) . If, for rational a,b we denote the sets

U, =

u

{z :

f(z) > u } and fi =

f EF

u

{x : g(z) < b } ,

g€G

then there are open sets W,, satisfying the conditions of the Ex. 3, and such that U a 5 Wa and Fa n Wa = 9 for all a E Q. The function h, determined by the sets W,, separates then the gap L . (Define W, by induction). Hence, X is an F-space if and only if every countable gap in C ( X )is separated. (Seever).

D. EMBEDDINGS OF c ( X ) Let X and Y be Tychonoff spaces. Any continuous function h : X an adjoint map H : C * ( Y )-+ C * ( X )defined by H ( g ) ( z )= g ( h ( z ) )

+Y

induces

for g E C * ( Y )and z E X .

1. The function H is a continuous (in the uniform norm) linear homomorphism of the linear ring C * ( Y )into C * ( X ) .

PROBLEMS

121

2. H is an isomorphic embedding if and only if the image h [ X ]is dense in Y . Hence, if X and Y are compact, then H is an embedding if and only if h is onto Y .

3. H is onto C * ( X ) if and only if h is a homeomorphism onto a C*-embedded subset Y . Hence, if X and Y are compact, then H is onto C * ( X )if and only if, h is a homeomorphic embedding of X into Y.

4. Let X and Y be compact zero-dimensional spaces. The algebra B ( X ) can be embedded into B ( Y ) if and only if the ring C ( X ) can be embedded into C ( Y ) . 5. Assuming CH, if X is compact and w ( X ) = C, then C ( X ) can be embedded into

C ( w * ) = ZW/co. (For the same with non CH see the theorem 3.1, Chapter 8)

E. CHARACTER OF ULTRAFILTERS 1. Let card X = K > w . Every uniform ultrafilter p E U ( X ) has character greater than K . More generally, if p E U x ( X ) , then x(p) > A.

2. There are always points p E U ( X ) ,for which x(p) = 2".

F. REMAINDERS 1. Let X be an infinite compact space. a) If K > cardx is a regular cardinal, then the set the space ( K 1) x X .

K

x X is C*-embedded in

+

b) The space X is the remainder

P[K x X ] \ ( K

x X ) , i.e. X = ( K x X ) * .

2. Every Tychonoff space X is a remainder space ( X = Y * ,where Y = K x P [ X ]x X * ) .

G. ZERO SETS 1. Let X be a discrete space.

a) Each nonempty zero set A E Z ( X * ) has the nonempty interior. b) Each nonempty zero set in X * is a closed domain. 2. Let X be a strongly zero-dimensional space such that card C ( X ) = c. a) If A

X * is a zero set, then we have A = Y * ,where Y = a [ X ]\ A .

b) Each zero set A

X* is a ParoviEenko space of weight c.

Hence, assuming CH, we have A = w * .

This Page Intentionally Left Blank

Chapter 5

r-Base Matrix $ 1 . BASETREE. Let M = M ( w * ) denote the family of all maximal antichains in the algebra B(w*). Thus, the elements of M are all disjoint families A B(w*) consisting of (nonempty) open-closed sets and such that

c

U A = U { a : a E A} = w * . The family M is partially ordered by the refinement relation

A5B

V a E A3b E B[a b]

( A refines B).

If A 5 B holds and for some a E A and b E B we have a n b # Cp, then a C b. Moreover, for an arbitrary b E B we have b = U { a E A : a C b } . Let lh denote the smallest cardinal, for which there exists an unbounded from below family R M , of cardinality Ih:

h = min{n

:

VR E M[cardR < K

-t

3B E MVA E R(B 5 A ) ] } .

1.1. LEMMA.If R = { A , : a < K} 2 M is a family of maximal antichains and 5 lh, then there is a decreasing (in the sense of the refinement relation) sequence S = (B, : (Y < K ) such that B, 5 A,, for all a < K. K

PROOF.The sequence S is defined by induction. Let Bo

= A0 and if for some we have already defined antichains BC for E < a , then we take as B, an arbitrary antichain B E M , which refines all Bc,for [ < a and A, (such a B exists,

(Y

<

K

since a < Ih).

QED

1.2. THEOREM. The cardinal Ih is regular and satisfies w1 5 h 5 123

t.

CH.5 X-BASEMATRIX

124

PROOF.The family of all antichains of the form {U,w* \ U},where U runs over B(w*)and U # $9 , w* has no lower bound. Hence, h 5 c. If A l , . . . , A , are arbitrary elements of M , then

is also in M and obviously A 5 A1,. . . ,A,. Hence, we have h 2 w . Now, if R = { A , : n E w } C M is a countable family, then we may assume that A0 2 A1 2 . . ., by Lemma 1.1. Consider a decreasing sequence g(0) 2 g(1) _> . . . such that g(n) E A,, for all n < w . The intersection H , = n n < w g , is then a nonempty Ga-set, hence int H g # $9 (since f?(w*) = P(w)/fin has no countable limits). Let A, be a maximal antichain in int H, and put A = U, A,. Clearly, we have A 5 A,, for each n < w . If U E B ( w * ) is a nonempty basic set, then there is an a0 E A0 such that U n a0 # $9, since U A o is dense. Similarly, we have U a. a1 # 8 for some a1 E A l . Hence a0 _> a l . In this way we find a decreasing chain g(n) = a , E A , such that U n g(i) # 8 for all n < w . Thus, U n Hg # $9 and hence also U n intHg # $9. Consequently U n a # $9 for some a E A, A, i.e. A is a maximal antichain below each A,, n E w . It follows that h 2 w1.

n

nil,

Finally, the regularity of Ih follows directly from Lemma 1.1. Indeed, there exists a descending sequence s = (A, : a < h) having no lower bound. Therefore, any cofinal in s subsequence has no lower bound either and hence Ih must be regular.

QED In the definition of Ih, we can replace arbitrary families R C M by decreasing sequences (by Lemma 1.1). Any decreasing sequence s = (A, : tree

T, = (

u

A,,

2)

cy

< K ) of antichains from M determines a

(the inverse inclusion),

m
whose vertices are some open-closed sets of w * , ordered by the inverse inclusion. Thus, the elements a E A0 (the 0-th level) are minimal in the tree T,; for every a E A , (the a-th level) the set of predecessors { b E T,: b _> a } has the order type (Y (with respect to 2 ) and any a E A, has immediate successors on the next level (viz.

{ b E A,+1 : b C a } is the set immediate successors of a ) . Any (maximal) linearly ordered set g Tb is called (as in any tree) a branch of T,. The ordinal /c is here the height of T,.

$1. BASETREE

125

We are looking for a tree of the form T = T,, which is simultaneously a r-base of the space w* (i.e. each open set U of w* contains a set a from T ) . B ( w * ) is splitting (or that K splits), if for each Recall that a family K ' such that U n a # Q, and U \ a # Q,, nonempty u E B(w*) there is an a E h (see Exs. H, Chapter 2).

We also call a family R

M(w*) splitting, if

UR is splitting.

1.3. THEOREM. (Balcar, Pelant, Simon [BPS]). No tree T, of height less than h can be splitting and there is a splitting tree T, of height lh. PROOF.If the height K of T, is less than lh, then the sequence s = ( A , : (Y < K ) , of the levels of T,, has a lower bound - there is an A E M such that A 5 A,, for all (Y < K . Clearly, no U E A is splitted by s. of

K

Now, let K be the minimum power of a splitting family R is correct since, e.g. the family

R = {{U,w f

\ U} : U

E B(o*) and U

c M . The definition

# Q,,w*}

splits. We show that K = Ih. Clearly we have K 2 lh, as shown above. On the M be any family with card R < K . We claim that each other hand, let R nonempty U E B(w*) contains a subset u U, u E B(w*) such that the formula VA E R3u E A(. u ) holds. Indeed, suppose the opposite, i.e. there is a set U E B(w*), U # Q, such that for every subset u U, u E B(w*) we have A ) . This means that R splits u , since there must be 3A E RVa E A ( . non u n a # Q, for some a E A (by maximality). In other words, the trace of R on U

c

c

c

c

c

RIU = { { a n U : a E A }

:

A E R}

c M(U),

is a splitting family in the subspace U. But this is impossible since card RIU < and the subspace U is homeomorphic to w*. This proves the claim.

K

Now, it follows at once that if

K = { u E B(w*) : VA E R3a E A ( . 5 a ) } , then the union U K is dense in w* and hence any maximal antichain B C K is also maximal in B(w*), i.e. B E M ( w * ) . Simultaneously, we have B 5 A , for all A E R. Therefore, card R < h and consequently K 5 lh.

cH.5 R-BASEMATRIX

126

Thus, we have proved that K = Ih. It follows that there is a splitting family R = { A , : a < Ih} of cardinality Ih. Applying Lemma 1.1, we find a decreasing sequence s = (B, : Q < Ih) such that B, 5 A,, for all a < Ih. Thus, the corresponding tree T, splits and its height is lh, which finishes the proof. QED Clearly, any 7r-base is a splitting family. On the other hand, any splitting tree obtained as in 1.3 can be modified a little so that it will be also a R-base. This is done in the following theorem. 1.4. THEOREM. (Balcar, Pelant, Simon [BPS]).There exists a tree T of height Ih, which is a a-base.

PROOF. Let T, be a splitting tree determined by a decreasing sequence = ( A , : a < Ih), as in Theorem 1.3. Clearly, the family T, = U{AB : a < p < Ih}, for arbitrary a < Ih,is a splitting family as well. First, we show that each nonempty set U E B(w*)intersects c many elements of some level A,. Indeed, there is an a0 < Ih such that some a0 E A , splits U. In particular, we have U \ a0 # 9 and hence there is an al E A,, such that (U \ ao) n a1 # 9 and a1 # ao. Thus, U intersects at least two members of A m 0 .Now, let us repeat the same argument for the sets U n a0 and U n al replacing the tree T, by T,,. It follows that there is an a1 > a0 such that U n a0 has nonempty intersection with some distinct a00, a01 E A,, and similarly, U n a1 intersects some a10,all E Au1. Hence, a00 and a01 are subsets of a0 and a l . Thus, we can define by induction an increasing sequence similarly, a10, all a0 < a1 < a2 < . . . of ordinals < Ih and a family {a, : E E Un
1. a, E A,,, for E E (0, l}n+' and a, 2. a, C_ a,, if 3.

E

# a,,

whenever E

# 71;

C v;

U n a, # 9, for all E .

Thus, by 1 and 2, the sets a, form a binary Cantor tree (of height w ) . Since the cardinal Ih is regular and > w , the ordinal a = sup{a, : n < w } less than < Ih. By compactness, for every function g : w + {0,1} we have

U n int H, #

8,

where H, =

agln. n<w

Hence, there is a set a, E A, intersecting U n intHg, in particular ag n U # 9, and a, n ah = 9, for g # h. Thus, we have U n a # 9 for c many elements a E A,. Let

$1. BASETREE

127

us denote

K , = {U E B(w*) : card{a E A ,

:

U na

# (D}

= c}

for a

< h.

We have proved above that each nonempty set U E B(w*)belongs to a K O ,for some a < Ih. Now, we can define the required tree T . First, we replace some of antichains A , by their suitable refinements Ab, as follows. If A ' , = (D, then let Ab, = A,. Suppose now that K , # $3. Since K , has at most c elements, we have an enumeration X, = (Ug : ,!? < c}. Fix also an enumeration A , = {aT : y < c } and define a function f : c + c, by induction on ordinals ,!? < c, setting

Let bg = Up n a f ( p ) . Thus, the sets b,, for a < c are all distinct. There is a maximal antichain Bg E M ( a f ( p ) )containing the set bg. Replace now the set a f ( g )by all the members of Bg,for each ,!? < c. The remaining elements a E A , are left without changes. In this way, we obtain a new antichain Ab, 5 A , such that every set U E K , contains an element a E Ab,. In view of Lemma 1.1, we may assert that the sequence s' = (Ab, : a < Ih) is decreasing. Clearly, the corresponding tree T = Tst is a n-base. QED We know (from the proof of 1.4) that each of the levels A , of any n-base tree has c many elements, for sufficiently large a. Since any cofinal segment of a n-base tree is also a n-base tree, we may assume that this holds for all a < h and hence, we can represent a n-base tree T in a matrix form 1.5.

T = {t(a,,!?):a
where the levels A , = { t ( a , P ) : ,!? refines A,, whenever a < ,!?.

<

c } are maximal antichains in B(w*) and Ap

It will be useful to have a better estimation of the cardinal coefficient Ih. In the forthcoming proof, the symbol gx denotes the increasing enumeration of an infinite w . It is easy to check that gx set X g y , whenever Y X (the strict almost inclusion) and that for a given infinite X E w and any f : w + w there is an infinite subset Y X , for which f <* g y holds. 1.6.

We have the inequality h 5 lb.

CH.5 T-BASEMATRIX

128

Indeed, let K < lb and suppose that F = {fa : a < K } is a family of functions from w to w . Recall that the cardinal lb is the minimum power of an unbounded family of functions. Hence, it suffices to show that the family F is bounded (because then the inequality K < lb and hence lh 5 lb follows). Let R = { A , : a < K } be a family of antichains in w * , (i.e. A , E M(w*), for a < K ) . For a nonempty set a E B(w*) we fix an infinite X , w , such that the relative closure X l = a. We may gx., for all 01 < K and a E A , (k.that R can be chosen so assert then, that f, that the above holds). Indeed, if a E A , is such that fa gx. does not hold, then we can choose a maximal almost disjoint family

<*

and replace the set a by the antichain {Y* : Y E Sa}. Now, if B is a lower bound of the family R (it exists, since K < h) and b E B is an arbitrary element, then we have

gb.

whence F

QED

$2. STATIONARY SETS. Let K > w be a regular cardinal. A nonempty set C 5 K is said to be cub (closed and unbounded), if supC = K and for all a < K we have supC n a E C, (the latter condition means that C is closed in the order topology of the space K ) . The closedness condition can be also written thus: if X < K and (a{ : ( < A) is an increasing sequence in C,then S U ~ ole ~ ,E ~C. 2.1. The intersection of less than

K

many cub sets is cub.

PROOF.Let C, C IF be cub sets, for a < A, where X < K . Obviously, the intersection C = C, is closed. We shall prove that C is unbounded. Let p < K be given - we shall find a y 2 p such that y E C. Define by induction a function f : K x w --f K with help of the following equations

n,

f(0,O) = min{( E Co : f ( a , n )= min{( E

62 P }

C, : [ > supf(r],n)}

for a > 0,

rl
f(0,n

+ 1) = min{[

E CO:

6 > sup{f(a,n) :

a < X and n < w } } .

$2.

129

STATIONARY SETS

Clearly, the sequence s, = ( f ( 0 , n ): n < w ) is increasing and all its terms are in C,. Hence, the ordinal yo = sup{f(cu,n) : n < w } belongs to C,, for all a < A. But we have f(a,n ) < f ( f f ' , n ) < f ( a ,n

+ 1)

whenever a

< a' < X and n < w .

It follows that Y~ = 7,' and the common value y = y,, for a

< X is in C. QED

2.2. The diagonal intersection

of K many cub sets is cub.

PROOF.First, we show that the set C is unbounded. Let define inductively a sequence ( a n: n < w ) as follows a,, =/? and an+]=min{y>cu,:

/? < K be given and

Cc}.

ye €
a,. If [ < a , then we have

By 2.1, the definition is correct. Let a = (

forsome n < w ,

Cc, the ordinal a = sup{a; : n < i < w } belongs and since { a , : n < i < w } to the set Cc. It follows that a E C (and a > /?,by definition). In order to check that C is closed, take an increasing sequence ( f ( E ) : < A), of length X < IE, with terms in C and let y = supc<xf(0.If 17 < y, then 17 < f ( 5 ) for all 5 > 17 and hence f(E) E C,. Consequently, the ordinal y = sup{f([) : 71 < E < A } is in C,, whence yE

c. QED

2.3. EXAMPLE. Each cofinal segment C, = diagonal intersection

{t E K

n

c = { ~ < ~ : c,} C 4 - i

consists precisely of limit ordinals y

< K.

: LY

< < K } is cub. The

c H . 5 A-BASEMATRIX

130

c

c

A subset S K is said to be stationary, if S n C # $3, for every cub set C K . From 2.3 it follows immediately that any stationary set S is unbounded and all the cofinal segments St = {a E S : a > <} of the set S are also stationary. Therefore, the intersection S n C, where C is an arbitrary cub set, is always unbounded. 2.4. LEMMA.(Fodor (Fl]). If S K is a stationary set, then any regressive function f : S + K (i.e. f(a)< a,for all Q! > 0), must be constant on a stationary subset T S.

c

PROOF.Suppose that no counter image Tp = {a E S : f ( a ) = p } is stationary. Thus, there are cub sets Cp, such that Cp n Tp = $3, for all p < K . The diagonal intersection C of the sets Cp is then a cub set, by 2.2. Hence C n S # $3, but for a E C n S we have f ( a ) 2 a,which contradicts the regressivity of f . QED Obviously, the set S = {a < K : c f ( a ) = w } of limit ordinals a cofinality, is Stationary.

< K , of countable

Every stationary set S K can be divided into K many stationary parts (Ulam, Solovay). We prove here a weaker version of this theorem.

2.5. THEOREM. The set S = {a < K : c f ( a ) = w } can be divided into K many mutually disjoint stationary subsets.

PROOF. For every a E S we choose an increasing sequence gu = ( g l * ( n ): n < w ) of ordinals satisfying the condition a = supn<,, ga(n). Let us denote Tnp = { a E S : go(n) 2

p}

for n

< w and p < K .

It is not difficult to see that for some no, the sets Tnopare stationary, for all (*)

p < K.

Indeed, in the opposite case, there are ordinals P,,, such that Trip, is nonstationary, for all n < w . Thus, for some cub sets C, C K , we have C, n Trip, = $3 and hence

2.6.

C n Un<wTrip, = $3

where C =

on<,, C,.

Obviously, the set C is a cub set, in view of 2.1. We know already that the set S n C is unbounded in K . But, if p = SUP,,<,, Pn and a E S n C, then 2.6 implies a 5 p, a contradiction. This proves (*). Now, let us define a function f : S + K by setting for a E S. f ( a )= g,(no)

$3. C-POINTS

131

Clearly, f is regressive on S. Applying the Fodor lemma (2.4) for the restrictions f I Tn,p, where /3 < K , we find stationary sets Sp C Tnop,on which f is constant: for some yp

f ISp = y p

< n and p < n.

Obviously, Sp n Sp, = (3, if only y p # 7 p 1 . Assume inductively that we have already defined an increasing sequence ( 6 ~: [ < A) of distinct values and take

<

+

/3 = sup(6e : < A } 1. By definition of Trip, we have yp 2 /3 so we may set 6~ = yp and then 6x > 6 ~for , all [ < A. Hence, the stationary sets Sa, S are disjoint each other, for [ < ti and the proof is finished. QED An unbounded set C 5 n: will be called w-cub, if C is w-closed, i.e. E C, for every increasing sequence (an : n < w) with terms in C. In other words, C is w-cub, if C is unbounded and contains all its limit points of countable cofinality. Hence, we obtain immediately

2.7. THEOREM. If S c {a < n : c f ( a ) = w } is a stationary set, then S n C # Q, for every w-cub set C K . Now, let n be regular and > stationary. An unbounded set C points of cofinality w1:

Clearly, the set {a E n : c f ( a ) = wl}, is n will be called q - c u b , if C contains its limit

w1.

sup{ac : for any increasing sequence (at : of 2.7:

2.8.

t < WI} E c,

5 < w1) with terms in C. We have then an analogon

if S C {a < K : c f ( a ) = w 1 } is stationary, then S n C # (3 for any

Indeed, we have S n S n C f Q,.

w1

-cub set C c

K.

c # (3, where c is the closure in the order topology, whence

$3. C-POINTS. A point x E X (where X is an arbitrary topological space) is said to be a n-point, if there is a disjoint family {Ua: a < n} of open sets, such that for d l a < n. xE

132

c H . 5 T-BASEMATRIX

Here, the cardinal coefficient K may assume also finite values, e.g. no point of an extremally disconnected space can be even a 2-point, since disjoint open sets have disjoint closures. In particular, the space P[w] has this property. In the remainder space w* the situation is completely different. Solving a problem posed by R. Pierce, A. Szymaliski proved that every point p E w* is an w-point. This result has been strengthened by Balcar and Vojt&, who proved that every point p E w* is a c-point, i.e. p belongs to the closures of continuum many disjoint open subsets of w*. The proof given below is due to R. Frankiewicz. Let B ( p ) denote a local base of open-closed neighbourhoods at a point p E w*. Consider the following property:

3.1.

for every U E B ( p ) there is an open-closed subset w(U)C U , such that the family {w(U): U E B ( p ) } is disjoint.

We claim that 3.1 implies that p is a c-point. Indeed, each of the sets w ( U ) contains c many mutually disjoint, open-closed subsets w , ( U ) , Q < c. If we denote W , = U{w,(U) : U E B ( p ) } ,for Q < c, then the family of open sets {Wa : (Y < c} is disjoint and U n W , # (3, for all Q < c and any U E B(p), (since w a ( U ) U n Wa). Hence, we have p E for all Q < c, which proves the claim. It follows that p E w* is a c-point if and only if there is an antichain in B(w*) refining p ( = {A* : A E p } ) . Now, we shall prove two auxiliary lemmas.

w,,

3.2. LEMMA.U ( b , : n < w) is a decreasing chain in B(w*), then there exists a disjoint family {V, : Q < Ib} of open sets, such that Ua
nncw

U n V,

# (3,

whenever U n ( b ,

\ bn+l)

+ Q, for infinitely many n < w.

PROOF.Let us choose a decreasing sequence { B , : n < w} of infinite subsets B, C w such that BG = b,, for all n < w. We may assume that B, \ Bn+l is infinite, B, = (3. It will be more convenient to regard the sets for all n < w and that B, as subsets of w x w. To this end we fix arbitrary functions

nn<w

gn :

Bn \ Bn+l 2 { n }x w , onto, for n E w,

and let g = U n < w g n . Any set A

w can be now identified with its image

53. C-POINTS g[A] C w X U . In particular, the sets B, for all n < w.

133

become now the vertical axes { n }x w ,

Let us choose an increasing and unbounded (in the ordering <*) sequence (fa :

a

< Ib) such that the functions f, are increasing (Corollary 2.4, Chapter 2).

Let us denote

W, = {(n,j)E w x w :

fa(.)

for a < Ib.

<j }

The sequence (w, : a < Ib), where w, = W: is a decreasing chain in B(w*). We claim that for any ordinal a E S, where

s = {a < lb :

c f ( a ) = w},

the set v, defined by setting

3.3. is nonempty. Indeed, if (Y = sup{&, : n < w } , where sequence of ordinals < a,then obviously

and hence v,

# $9,

&s

(tn:

n

< w)is an increasing

a nonempty Ga-set.

According to the theorem 2.5, we have S = Ua
V, =

U vy

for

(Y

< b.

-ES,

Thus, the open sets V, are nonempty and mutually disjoint. If

(W, \ W,) n ( E n\ Therefore(w,\w,g)nU,,,(bn\bn+,) and consequently

v, c

is finite, for all n

(Y

< /3 then the set

< w.

= $9, whichimpliesat once w,\w~

n

n<w

b,

for all a

< ~b.

c nn<w b,

cH.5 T-BASEMATRIX

134

Now, we show that the sets V, fulfill the other required condition. Let a set B(w*) be given, such that U n ( b , \ b,+l) # 0 for all n E K , where K is some infinite set w . Choose an A w such that A' = U. We have A n ( B , \ B,+1) $ fin A and hence, for a fixed cv < b, the function for all n 6 '

U

E

h ( n ) = min{j > fa(.) is well defined. If n

:

for n E K ,

( n , j )E A }

< m are two consecutive elements of the set h', then we set h ( i )= h ( m )

Since the sequence (f, : a < y fp non h, i.e.

fa(.)

> h(n)

for all i, where n

< i < rn.

< b) is unbounded, there for infinitely many n

is a /3

> a such that

< w.

It follows immediately that the set A n (W, \ Wp) must be infinite. Thus, we have proved that for every Q < b there is a /3 > a such that U n (w, \ w p ) # 8. Hence, there exists an unbounded in b set 2 E b, such that the sequence (Un w, : Q E 2 ) is decreasing. Moreover, we may assert that 2 is w-closed (this follows by the same argument as in 3.3). According to the theorem 2.7, we have S, n 2 # 0 for all Q < b. For any y E S, n 2 there is in increasing sequence (y, : n < w ) such that y = s ~ p , < ~ y and , , y, < y, for all n < w . Then, we have v,

n u = (inti

w,]

\ w,) n u = inti

,<-l

This yields immediately U

n

( w m n U ] \ (wrn u)]#

8.

n<w

n V, # 8, for all a < lb and the lemma is proved. QED

For a given descending chain s = (b, : n

U E B(w*) intersecting the boundary of

nn<w b,

Y, = { u E qw*) : u n Fr[

<

n

w ) , let Y, be the family of all

b,]

# 8).

n<w

3.4. LEMMA.Let s = (b, : n < w ) be a decreasing chain in B(w') and let bE b, be an open-closed set. For every U E Y, there exist open-closed sets w(U) U n \ b ) such that the family {w(U): U E Y,} is disjoint.

nn<w

nn<w(bn

53. E-POINTS

135

s

PROOF.As in the preceding proof, let Bn w be such that B; = b,, for all n < w . Similarly, for any U E Y, we choose an A = A ( U ) such that A* = U . Obviously we have

n<w

n<w

n<w

Hence, for an arbitrary set U E Y,, the set

is infinite. Let T be a .rr-base tree. We know already that for any nonempty set U E B(w*), there is a level 1 of T such that U n a # $3 for continuum many elements P ( w ) having the following two a E 1. Hence, there exists a function t : Y, properties: 1. t*(U) E T and

t * ( U ) K*(U),for any U E Y,.

2. t * ( U )# t*(V),whenever U

# V (even if K * ( U )= K * ( V ) ) .

In particular, we have t ( K ) G* K ( U ) ,for U E Y,. Let us denote

L ( U ) = U{Bn \ ~

nE

n + l :

t(u)).

It follows that the set A ( U ) n L(U) is infinite on infinitely many vertical axes, i.e.

U n L* n ( b ,

\ b,+l) # $3

for infinitely many n E w .

Applying the preceding lemma to the chain ( b , disjoint open sets for a V, C ( b , \ b)

n

\b:

n

< w ) we find mutually

< lb,

n<w

such that U n L * ( U )n V, # $3, for all a < lb. For every set U E Y, we choose an ordinal a = a ( U ) such that the set t * ( U ) belongs to the a - t h level of the tree T. We claim that the sets U n L * ( U ) n V,(u) are then disjoint each other (note that h 2 lb, by 1.6). Indeed, if a ( U ) # a("), then V , ( u )n Vo(w)= 8. If a ( U ) = a ( W ) ,then t * ( U ) n t * ( W )= $3 and hence t ( U )n t ( W )=* $3. Consequently

L * ( U )n L*(W)n Vp = $3

for each

p < lb,

136

CH.5 T-BASEMATRIX

which proves the claim. Now, if we take as w = w(U)any open-closed subset of U n L * ( U )n Vu(u),then the family {w(U): U E Y }has the required properties. This finishes the proof.

QED Now, we can easily prove the announced theorem.

3.5. THEOREM. (Balcar, VojtG [BV]).Every point p in the remainder space w*

is a c-point.

PROOF. For a given point p E w', let us choose a decreasing chain s = ( b , : a < K ) , of open-closed neighbourhoods of p , such that p E Fr[n,,, a,]. If p is a non-P-point, then there is such a chain s of length K = w . Applying the lemma 3.4 we obtain immediately the condition 3.1 for p . If p is a P-point, then there is an s with a regular K > w . Let S = {a < K : cf(a) = w } . For every Q E S fix an increasing sequence (an: n < w ) , with SUP^<^ an = Q and apply the lemma 3.4 to each subchain sLI = (ban : n < w ) , substituting b = b,. This yields a disjoint family

{w(U): u E Y&}. Each neighbourhood U of p intersects some boundary Fr[nn<, ban], for some a E S. Hence, the set w(U)is defined for each U E B(p) and the condition 3.1 holds

for p in this case as well.

QED

Now, we shall show an interesting application of Theorem 3.5.First, let us prove the following easy lemma.

3.6. LEMMA.Let X be a compact space and f : G ( X ) X - the canonical mapping of Gleason space of X onto X . If, for a K 2 w , an element x E X is a &-point, then the counter image f-l(z) has at least 2'= many points. PROOF.By assumption, there is a disjoint family R of open sets in X such that x E n { T i : a E R } and cardR=K.

We may assert that the sets a E R are open domains (replacing a by i n t [ ~ ] ) . Let U, be a basic neighbourhood in Gleason space

u, = { p E G ( X ) :

a E p}.

PROBLEMS

137

By definition of the canonical mapping f (see Sec 7, Chapter 3), it follows immediately that f[Ua] = a, (for any open domain a). Hence, there is a point qa E U,, such that f(qa) = x, for every a E R. Since, for a, b E R we have Ua n ub = Uanb = 8, the set S = {qa : a E R } is strongly discrete and has power K . It follows from the Taimanov theorem that S is C*-embedded in G ( X ) (because G(X)is extremally disconnected). Therefore, 3 = p[S]= P[.] and hence card 3 = 2**. On the other hand, we have obviously f [ S ]= f[sl = {x} and the proof is finished. QED From the lemma 3.6 we obtain immediately

3.7. COROLLARY. Every counter image f-'(p), of the canonical mapping C f : G(w*) w', has power 22 (and contains B[c]).

PROBLEMS A. THECARDINAL h. 1. We have always t 5 h 5 s (Exs G and H, Chapter 2). 2. The Martin Axiom implies h = c. 3. Assume V k CH and let K E V be a regular cardinal, V " K > w1". If V[G]is obtained from V by adjoining K Cohen reds, then we have V[G] "lh = w1 A K = c". Hence, it is consistent that lh = w1 < C.

4. Let

+

E V be a regular cardinal in V such that V "w1 < finite support iteration (P, : a < K ) by induction as follows

Pa

K

= Cp<,PO

Pa+l= P a

*Q

Pa+l = Pa * D

K

< c". Define a

for limit a (direct sum) for even a,where Q is a P,-name of a ccc forcing Q such that Q It- MA. for odd a,where D is a Pa-name of the dominating forcing D.

If P = Ca
2

K.

It follows that V[G]k "h= K " . Hence, it is consistent that

w1

< h < C.

138

CH.5 a-BASEMATRIX

B. COVERSWITH

NOWHERE DENSE SETS

The Nov$k number n (of the space w*) is defined as the minimum cardinality of a cover of w* consisting of nowhere dense sets n = min{n : 1. We have w2 5 n

3R

N(w*)[cardR= n and

uR

= w*]}

5 2'.

2. Let R C N ( w * ) be a cover of cardinality n. If, for N E R, A N is a maximal antichain in w* \ N , then the family { A N : N E R } splits. Hence n 2 Ih. 3. Let A,, for Q < Ih, denote the levels of a given n-base tree T of height Ih. a) The sets N , = w*\U A , are nowhere dense and form an increasing sequence b) The equality w* = Ua
4. The equality n = Ih holds if and only if there exists a n-base tree of height h without long branches (i.e. every branch has length< Ih). 5. Let T be a r-base tree of height

Ih and K

-a

regular cardinal 5 c.

D(w*)is an antichain such that U,(a) a) If {U,(a) : y < n} a E T and y < K , then the sets N , = w*\ u { U , ( a )

: a E

T

and

Q

a , for all

< y < n}

are closed and nowhere dense.

N , = w* and n 5 K . Hence, the inequality Ih < c b) If Ih < X, then implies Ih 5 n 5 'hI 5 C. In addition, if the cardinal c is singular, then there must be n < c.

C. CONSISTENCY PROOFS 1. The Martin Axiom implies Ih = c < n. 2. It is consistent that Ih < n

< c. (A Cohen model).

PROBLEMS

139

D. P-POINTS 1. Any dense-in-itself regular space X satisfying the ccc (every disjoint family of open sets is at most countable) has no P-points. In particular, the Stone space of an atomless measure algebra has no P-points. Given a set A, we say that cardA is a measurable cardinal, if the power-set algebra P ( A ) contains a nonprincipal a-complete ultrafilter F . Note that the a-completeness of F is equivalent to the following condition:

If A =

(*I

u

A , is a disjoint partition of A ,

n<w

then A,, E F, for some n < w . Measurable cardinals (if exist) form a cofinal segment in the class of all cardinals.

2. Let X be an extremally disconnected (Tychonoff) space and assume that 2 E X is a nonisolated point. a) There exists a maximal antichain A C B ( X ) , such that 2 4 U A . b) The family F, = {R A : x E is a nonprincipal ultrafilter (on the set A ) . c) If card A is nonmeasurable, then there is a function f E C ( X ) nonconstant on every neighbourhood of the point x. (If A,, are as in (*) above, set f I U A,, = i, for n E w and note that U A is C*-embedded in X). Hence, there are no P-points (except isolated) in any extremally disconnected space with nonmeasurable Suslin number (= the smallest upper bound of powers of antichains).

m}

3. The following condition characterizes P-points p E w * : for any disjoint partition w =

(*I

u

n<w

A,, such that

A,, E p for no n < w , there is a set A E p such that A n A , is finite, for all n < w .

4. A point p E w* is a P-point if and only if for any compact lst-countable space

X , any sequence f : w + X and any x E X the condition x = p-limi f(i) implies x = limicA f(i), for some A E p.

140

c H . 5 T-BASEMATRIX

E. P-POINTS

FROM A TREE

1. Assume that n > c (then we must have Ih = c, see Ex. B,5). a) For every partition 2 = (2, : n < w } of the set w , there is a maximal antichain A z in L?(w*) containing as elements all sets Zi, n E w . b) There exists a decreasing sequence s = ( A , : Q < Ih) such that for an arbitrary partition 2 of w , there is an Q < Ih such that A , 5 A z . c) The tree T = T,, corresponding to the sequence s, has a branch g = (g(a) : a

< Ih) of length Ih (i.e. the sequence g is decreasing and < Ih).

g ( a ) E A,, for all a

d) The intersection of any such a long branch g is a singleton g(a)= { p } and the point p is a P-point. The family {g(cr) : a < Ih} is an ordered base of neighbourhoods of the point p . Hence, if n

> c then there exist at least n many P(c)-points in w *.

If the condition (*) in Ex. D 3 is replaced by a stronger one: A n A , has at most one point, for all n < w , then we obtain a definition of a selective P-point. 2. If n > c, then there exist at least n many selective P-points. (Choose suitably antichains A z , in Ex. 2a).

F. FORCING WITH

THE ALGEBRA P(w)/fin

Let P = B(w*) \ ((3) or P = T, where T is a fixed T-base tree of height Ih and let G P denote a generic (over V ) ultrafilter. 1. The forcing P is t-closed. In particular, the cardinals unchanged.

IE

E V,

IE

5

t remain

2. We have P ( w ) n V = P(w) n V [ G ]and ww n V = ww n V [ G ] .

3. The cardinal coefficient t calculated in V is the same as in V [ G ] .

4. There exists an order embedding of the tree T into a tree of sequences U,
nG is a P(c)-point and, actually, a selective P(c)-point.

PROBLEMS

G. SPACESOF

141

ORDINALS

Any ordinal y > 0 is a topological space, with order topology, which subbase consists of sets {( : [ < a} and {[ : ct < ( < y} for an arbitrary a < y. 1. Every space 7 is a 0-dimensional normal space. 2. If an ordinal y is a successor (y = p+1, for some p), then the space y is compact. If y is limit then y is not compact.

3. For any nonempty A C y the ordinal inf A is never a limit point of A . Therefore, y is always a scattered space. 4. Let K > w be a regular cardinal and f : K -+ W a continuous function. a) The set L C K , of limit ordinals < K , is cub in K . b) For every E > 0 there is a regressive function h : L -+ K such that for all a E L and any [ the condition h ( a ) < [ < ct implies

If([)

- f ( a ) l< E.

Hence, each function f E C ( K )is constant on a cofinal segment of C ( K )= C * ( K ) (EX. , 2).

K.

Therefore,

+

5. If K > w is a regular cardinal then the set K is C*-embedded in the space IC 1. In addition, we have C ( K = ) C ( K 1) and P [ K ]= K 1. On the other hand, the set w is not C*-embedded in w 1.

+

+

+

6. For any compact scattered space X , every function f E C ( X ) takes at most countably many values only.

H . THEALGEBRA O ( K+ 1) Let K: 2 w be an arbitrary cardinal. We consider the compact space the order topology).

K

+ 1 (with

+ 1) is superatomic. (Exs B, Chapter 3) 2. The algebra B ( K + 1) contains a chain of length Every chain in O ( K + 1) of 1. The algebra O ( K

K.

length

K

generates the whole of O ( K

+

+ 1).

3. The space K 1 is a continuous image of the remainder w* if and only if the algebra P(w)/fin contains a chain of length K .

This Page Intentionally Left Blank

Chapter 6 Inhomogeneity

$1. KUNEN'SPOINTS. A topological space X is called (point) homogeneous, if for any two points x,y E X there is an autohomeomorphism f : X + X such that f(x) = Y. The ultrafilter space w' is not homogeneous. This can be easily seen if the continuum hypothesis or - more generally - the existence of P-points is assumed. Indeed, any homeomorphism f : X + X sends P-points into P-points and there are always non-P-points in w'. S. Shelah in [Sl]has constructed a model of set theory in which the space w' has no P-points at all. Thus, the existence of P-points in w* cannot be proved in set theory but the inhomogeneity itself can be proved using a slightly different type of ultrafilters introduced by K. Kunen. A nonisolated point x E X is called an OK-point, if for every descending chain of neighbourhoods UO2 U1 2 . . . of x there is a family {V, : a < c} of neighbourhoods of x, such that

Valn . . . n V,,

c U,,

for every n E w and every sequence (YO,.

. . ,a ,

of indices.

The neighbourhoods V, in the above definition need not be different. For example, each P-point x of a given space X is OK: we set V, = int U,] , for each

[ nnEw

a

< c. 1.1. THEOREM. Each OK-point x E

any countable set D

X

is a weak P-point (i.e. x $ D, for

c X \ {x}).

PROOF.Let a set D = {z, : n E w } be given, D C X \ {z}. For every n E w , there is a neighbourhood U, of x omitting 2,. We may assume that the sequence

143

144

c H . 6 INHOMOGENEITY

{U, : n < w ) is decreasing and let {V, : a < c } be a corresponding family of neighbourhoods. Clearly, the sets

are finite. In fact, A , has at most n elements, since V,, n...nV,, Thus, if a $ Un<w A,, then V, is disjoint from D. QED

& Un and x,

6 U,.

Now, it is easy to see that the existence of an OK-point p E w* implies the inhomogeneity of w * . Indeed, any homeomorphism sends OK-points into OK-points but there exist always non OK-points in w * . For example, no limit point of a countable discrete set E w* is even a weak P-point.

$2. A form

MATRIX OF INDEPENDENT SETS.

K = { A n ( a , @:) n E w

Consider an arbitrary family K of the and a , p < c},

where the sets A n ( a ,p ) are infinite subsets of w . For finite sets t

c x c denote

r(t) = { p < c : 3a((cY,/?)E t ) } tg = {a < c :

(a,p) E t }

ng = the number of elements of t g .

Let F be an arbitrary filter on the set w extending the Frkchet filter Fo = { X C w : w \ X E fin}. We say that the family K is linked with respect to F , if the following conditions are satisfied. 1. For every t the intersection A ( t ) is centered with F , i.e. A ( t )nx is infinite, for all X E F .

2. A,(a.,P)

A m ( a , P ) ,for n 5 m and arbitrary a , ,9

3. For any p < c and i < np, the intersection

< c.

nuEtp A i ( a ,p ) is finite.

2.1. LEMMA. There is a family h’ linked with respect to the Frkchet filter Fo.

$2. A

145

MATRIX O F INDEPENDENT SETS

PROOF.(P. Simon). We define K on a countable set S = { ( k , f ) : k ~w as follows. For any n E w and X , Y

A,(X,Y) = { ( k , f )E A

:

and f : P ( k ) + P ( k ) } w let

card f ( Y

n k) I n

and X n k E f ( Y n k ) } .

Let K = { A , ( X , Y ) : n E w and X , Y C w } . The property 2 of linkedness is obviously satisfied. To check 3, let us take pairwise distinct sets X I , .. . ,X , E w . For sufficiently large ko E w , the sets X1 n ko,. . . ,X , n ko are all different. Thus, if we have

(k,f ) E Aj(X1, Y )n . . . n A i ( X , , Y ) then clearly k

where i

< n,

< ko and consequently the intersection above is finite, which proves 3.

Finally, for any finite t C P ( w ) x P(w), the set A ( t ) contains all the elements

( k , f ) with an arbitrary k E w and f such that f ( Y n k) = { X n k : X E t y } , for Y E r ( t ) and f ( y ) = Q, for the remaining y C k. Hence, the set A ( t ) is always infinite.

QED

2.2. THEOREM (Kunen [K]). There are OK-points in the space w * .

PROOF.We shall define by induction an increasing sequence (Fy: y < c) of filters and, simultaneously, a n auxiliary decreasing sequence ( K y : y < c) of families of the form K y = ( A , ( ( Y , ~: )n E w and (Y < c and p E I-,} linked with respect to F-,, for all y < c. The sets of indices Iy will form a decreasing chain, I0 = c and always card I-, = c. We begin with the FrCchet filter F = FO and a fixed family K = KO linked with respect to Fo,(which exists, in view of the lemma 2.1). At limit stages X > 0 we take FA = F-, and I A = I-,.Now, we describe the inductive step from y to y 1.

+

n-,,,

Let z = (2, : y < c} be a fixed enumeration of all infinite subsets of w . Similarly, let b = {(BZ : n E w ) : y < C} be an enumeration with repetitions of all B, = Q, and all the decreasing sequences (B, : n E w ) , such that Bo = w and B’s are infinite (hence, the family {B, \ B,+1 : R. E w } is a partition of w ) . By the

146

c H . 6 INHOMOGENEITY

inductive assumption we have already defined filter F, and a family I<,. Now, we take Z = 2, and distinguish the following cases. 1. The set Z is centered with F, and K , is linked with respect to the filter [F,, 21, generated by F, and 2. We define F = [F,, Z ] and K = K,.

2. The set Z is centered with F, but K , is not linked with respect to [F,, 21. Then for a certain t c XI, and X E F, the set A ( t ) n X n Z is finite, whence w \ Z E [Fyr A ( t ) ] . In this case assume F = [F,, A(t)]and I = I, \ .(t). The family K (obtained from K, by limiting the range of the index p to I) is then linked with respect to F , since for any finite s c x I and Y E F the set A ( S )n ( A ( t )n Y ) = A ( . u t ) n Y is infinite.

3. The set Z is not centered with F,. Thus, we have w \ Z E Fy and we assume F = F, and K = K,. In all the cases the family K is linked with respect to the filter F . Now, consider the sequence ( B , : n < w ) occurring as the y-th term of the enumeration b. If some of the B's are not in F , then we define

Suppose now that B, E F , for all n E w . In this case we set

C, =

u

Am((I:,6)n ( B ,

\ B,+1)

for a fixed 6 E I.

m<w

It is easy to see that

2.3.

A , ( ( I : , ~n) B,

C,

for any m E w and

(I:

< C.

Indeed, let i E A m ( a , 6 )n B,. For a certain n < w we have i E B, hence m 5 n, since the sequence B, is decreasing. Thus, we have A,((I:,6) A,((I:,6 ) and therefore

which proves the claim 2.3.

\ B,+1

and

$2. A

147

MATRIX OF INDEPENDENT SETS

Let X E F be an arbitrary element. Using 2.3 we obtain easily

C,, n . . . n C,, 2 A , ( a l , 6 ) n.. . n A , ( a m r 6 ) n B , n X = A ( t ) n ( B mn X). Since B, n X E F and h ' is linked with respect to F we infer that all the sets C, can be adjoined to F. Let us define

Fy+i= [F,{C, : a < c}]

and 1,+1 = I \ ( 6 ) .

Clearly, is linked with respect to F-,+l, since for any s 2 c x I,+l and any X E F we have A ( s ) n x n C,, n . .. n C,,

2 A(s u t ) n ( B , n X ) $ fin,

where t is as above. We have to show yet the inclusion 2.4.

C,

n . . . n C,,

g* B,

for n 2 1.

This follows immediately from

since the right-hand side is a finite set. The condition i E C,

i E &(a, 6 ) n ( B , \ B,+1)

\ B,

implies

for some m E w .

Thus, we must have m < n and hence A m ( a , 6 ) A,-1(a16). It follows that i E A n - l ( a ,6 ) , which proves (*) and 2.4. The inductive step has been now completely described.

u,
F, is easily seen to be an OK-point. Indeed, The resulting ultrafilter p = let { V , : n < w } be a decreasing family of basic neighbourhoods. We have U, = B: and the sequence B, can be assumed to be decreasing, Bo = w and B, = (?. There is a y < c such that {B, : n < w } F, and also { B , : n < w } is the y-th term of the enumeration b. By the construction, the sets V, = C:, for a < c are then by 2.4. This finishes the proof. neighbourhoods of p and V , , n . . . n V,, C U,,,

nnEw

QED

148

CH.6

INHOMOGENEITY

Modifying a little the above proof we obtain the following stronger form of 2.2.

2.5. THEOREM. (Kunen [K]). There exists 2‘ many OK-points, none of which is a P-point.

PROOF.Any division of the set of indices c = I U J into two disjoint parts, the both of cardinality c, determines a division K = K I U K J of the family h ‘ = { A , ( a , p ) : n < w A a , p E c} linked with respect to the Frkchet filter Fo, where Kr = {A,( a , p ) : n < w A a < c A ,B E I} and the family K J is defined similarly. Now, fix an a < c and define

Clearly, the family R is independent and moreover for an arbitrary finite t the set

c x I,

is infinite (here, as usual, E ( P ) = 5 1 , + A = A and - A = w \ A). Indeed, if a then Al(a,p) n A l ( a ’ ,p ) E fin, i.e.

# a‘

Consequently, the set 2.6 contains the intersection

A ( t ) n A ( s ) = A(t U s)

for some s C {a’} x

J.

It follows that the family K I is linked with respect to each of the filters

F E = [ F o , { & ( P ) . A l ( c r , pP) :E J}]

where&: c * { - l , + l } ,

and hence every F, can be extended to an OK-point p , E w * . Obviously, we have p , # p , for E # 17 and therefore there are 2c many OK-points of the form p,. In order to ensure that no of the p , is a P-point it suffices to adjoin to the filter F, all sets of the form w \ Y where

2.7.

Y

G ~ ( n. A)

I (n ~)

for every n E w .

$3. COUNTABLE SETS IN F-SPACES We have to notice only, that for any finite t

149

c x I , the set

where Y1,. . . ,Y k satisfy 2.7, must be infinite. Since we have inclusions

hence for any n # PI,.. . ,PI the set 2.8 contains some intersection A ( t ) n A(s) = A(t U s) from the family K , similarly as in 2.6. QED

$3. COUNTABLE SETS IN F-SPACES. We already know (Theorem 7.5, Chapter 4) that every countable discrete subset of an F-space X is C*-embedded in X . Actually, this is true for arbitrary countable sets (not necessarily discrete), which follows easily from the following lemma.

3.1. LEMMA.(Comfort, Negrepontis [CN]). Let A, B be countablesubsets of a Tychonoff space X . I f A n B = $4 and Zn B = $4, then the sets A, B can be separated by cozero sets.

PROOF.Let us write A = {xn :

n E w } and B = {ym : m E w } . By assumption,

10 4 B and hence there is a neighbourhood UO of x o such that V O B = $4. Then we have B n (2U VO)= $4 and hence there is a neighbourhood & of yo such that V On(AU UO)= $4. It follows that A n (BUVo) = $4. Assume inductively that we are given neighbourhoods Uo,.. . U,; VO,.. . ,V, (of 2 0 , . . . , x n ; yo,.. . ,y,, respectively) so that U, n V, = pfor i, j 5 n and in addition

B n ( Z u U V i ) =$4

and

isn

An(BU

Uvj)=@. 3

9

It follows that there is a neighbourhood Un+l of zn+l such that

Then we have also

Bn(Zu

u v,)=q,

i
150

cH.6 INHOMOGENEITY

and hence there is a neighbourhood Vn+l of yn+l such that

We have then also

An(BU

U

Vj)=(B.

jln+l

Thus, we have defined some neighbourhoods U,,, V, of points x,,,ym respectively such that u,, n V, = (3 for all n,m E w . Moreover we may assume that the U’s and V’s are cozero sets (because X is a Tychonoff space). But then the cozero sets U = UnEwUnand V = UmEwVm separate A and B . This finishes the proof. QED 3.2. COROLLARY. Every countable set in an F-space X is C*-embedded.

PROOF.The preceding lemma implies that the assumptions of the Urysohn theorem (Theorem 1.1, Chapter 4) are satisfied. QED 3.3. COROLLARY. Let A,B be countable subsets of an F-space X . If AnB = 8 and

7in B = (D, then ;inB

= 8.

PROOF.In any F-space X disjoint cozero-sets are functionally separated. QED -

f

Recall that every function f : w --+ w has a unique continuous extension : p [ w ] + p[w] (sometimes, we do not distinguish strictly between f and its Stone

extension 7). The relation pLq

=

3f: w+w(f(q)=p)

forp,qEw*,

is called the Rudin-Keisler ordering (on w*). 3.4. THEOREM. If C E q and for some function g : C 4 w an ultrafilter p contains the family R = {g[A]: A E q A A C}, then there must be p 5 q (Rudin- Keisler).

PROOF.The family R is obviously centered, since

151

$3. COUNTABLE SETS IN F-SPACES g[Al n . . . n A,]

g[Al] n . . . n [An]. Therefore the family R generates the filter

F ={B

cw:

3 A E q[A

cC

and g[A]

c B]}.

By assumption, we have F C_ p . Moreover F is an ultrafilter. Indeed, let us suppose that w \ B is not in F,i.e. that g[A] n B # 9, for all A E q and A C. This means that the set g-'[B] is centered with the base {A E q : A C} of the ultrafilter q and hence g-'[B] n C E q. But since g[g-l[B] n C ] B, we obtain B E F and the claim follows.

c

c

c

Thus we have F = p and R is a base of the ultrafilter p . Let f be a continuous extension of the function g to w and hence to the whole of /3[w]. For an arbitrary set A E q we have q E ; Iand hence f ( q ) E f[Z]= f[Al,i.e. f[A] E f(q). It follows that R f(q) and consequently F = p = f ( q ) . QED

c

Now we are ready to prove a technical lemma which will be applied to homogeneity problems. We use the notion of convergence with respect to an ultrafilter (Sec. 7, Chapter 4). 3.5. LEMMA.Let D = {zn : n E w } be a discrete and E = {ym : m E w } an arbitrary set in a compact F-space X . If p-limn zn = 5 = q-lim,,, ym for some p , q E w * , then one of the following possibilities holds: 1. the sequence E is q-constant, i.e. { m E w : ym = x} E q,

2. p is Rudin-Keisler 5 q,

3. p is not a weak P-point.

PROOF.Let us assume that the property 1 does not hold, i.e. that the set { m E w : y,,, # z} E q. Without loss of generality we may assert that ym # z for all m E w . Denote B = { m : ym E D} and consider the case of B E q. Suppose that the set C = { m E B : ym E D } is in q. The condition g(m)=n

=

ym =x,

for mE C ,

defines a function g : C + w . We claim that g[A] E p for each A E q, A Indeed, we have a decomposition

into disjoint parts (for D is a discrete set in an F-space). On the other hand

c C.

152

cH.6 INHOMOGENEITY t

= q-limy, E {ym : m E A } = {z, : n E g [ A ] } m

and consequently g[A] must be in p (since t = plim, t, as well). This proves the claim. Applying the theorem 3.4, we infer p 5 q in the Rudin-Keisler ordering, i.e. the property 2 of the conclusion. Let us suppose now that the set B \ C = \ D} is in q. In view of the corollary 7.9 of Chapter 4 there exists {m : y, E a homeomorphism f : D + p [ w ] , such that f(z,) = n for all n E w . By the assumption t = plim,t,, we obtain f(t)= p-limf(z,) = plim,n = p. On the other hand q = q-lim, ym, hence p = f(t)= q-lim, f(ym). This yields

since B \ C E q and f(ym) # f(t)= p. It follows that p cannot be a weak P-point (since it is accessible from a countable set). Thus, the property 3 of the conclusion holds. Now, let us consider the remaining case of w \ B = { m E w : y, 4 D} E q. First, we choose mutually disjoint cozero neighbourhoods U, of the points t,, all n E w. Let U = UnEw U n and C = { m $! B : ym E U}.Clearly, we have {y,,,: m E ( w \

mE(w\B)\C}=(ZJ.

)\C}nD=(ZJ andDn{y,:

According to the corollary 3.3 we obtain

n,

and since I E we see that (w \ B) \ C $! q, whence C E q. Define a function g : C -+ w by the condition

In view of the theorem 3.4 it suffices to show that g[A]E p

for every A E q , A C C .

To prove this let us notice that the condition ( p ) holds for the sequence (I, : n E w) (Corollary 7.6, Chapter 4) and hence ~

U

nEgIA1

_

unn

U

n4dAI

_

_

un = g .

$4. INHOMOGENEITY OF PRODUCTS OF COMPACT SPACES

153

But we have obviously

Therefore z 4 {zn : n 4 g[A]}, which implies w \ g[A] p, whence g[A] E p. We have proved that p is Rudin-Keisler 5 q, in the case in question. This finishes the proof.

QED

$4. INHOMOGENEITY OF PRODUCTS O F COMPACT SPACES. Recall that a sequence (zn : n E w) of points of a space X has the property ( p ) , if there exists mutually disjoint neighbourhoods Un of the points zn, respectively such that - U, n U, = Q,

U

U

n€A

n@A

for every A

c w.

If the space X under consideration is a product X = Hi,, X i , then the property ( p ) depends on these coordinates only, on which the open sets Un depend. Any basic open neighbourhood depends on finitely many coordinates only. Therefore, we obtain the following corollary A sequence (xn : n E w) has the property(p) in the product X = n i E r X i if and only if there exists

4.1.

a countable subset (znlIo : n

product

n

I0

I such that the sequence

E w)has the property(@)in the sub€

Xi.

Obviously, no sequence with the property

(a) is convergent (in ordinary sense).

The cardinal number x(Y) = sup{x(y) : y E Y} is called the character of the space Y. Thus, a countable character x(Y) 5 w means that there is a countable base of neighbourhoods at every point y E Y. It is easy to see that the lemma 3.5 implies immediately the inhomogeneity of any compact (infinite) F-space. More generally, we prove here the following 4.2. T H E O R E M . (Frankiewicz, Kunen, Zbierski [FKZ]). Let X be a product of compact spaces

X = nieixi x

nj€jYj,

154

cH.6

INHOMOGENEITY

where Xi is an infinite F-space for all i E I and ~ ( 5 w) for all j E J. If I # @, (i.e. at least one F-space occurs), then X is point inhomogeneous.

PROOF.Let us choose a sequence (zn : n E w ) of points of X having the following two properties 1. On each axe X i the sequence (zn(i): n E w ) is discrete.

2. On each axe Y, the sequence (zn(j) : for any n,m E w .

R

E w ) is constant: z,(j) =

Zm(j)

We fix also two weak P-points p , q E w* such that p non 5 q (in the RudinKeisler ordering). For example, let q be an arbitrary OK-point. Since there are 2' many OK-points (Theorem 2.5) and only c many Rudin-Keisler predecessors 5 q, there is always a p, as required. Denoting z = plim,zn and y = q-limz, let us suppose that there exists an autohomeomorphism f : X + X sending y to z : f(y) = I. Let ym = f(zm) for m E w. We have then q-limy, = q-limf(z,) m

m

= f ( y ) = z = plimz,

Hence, plim, z,(i) = z ( i ) = q-lim, y,(Z), infer that

n

for all i E I. Applying the lemma 3.5, we

Ai = { m E w : y,(i) = z ( i ) }E q

for all i E I.

By assumptions, at least one axe Xi of the product X is an F-space. The sequence (z,(i) : n E w ) in this axe has then the property (p) (since it is discrete). From 4.1 it follows that the sequence (zn : n E w ) has the property (p) in the whole product X. The same holds also for the sequence (y, : m E w ) (as a homeomorphic image).

Applying again 4.1 in the opposite direction we conclude that for some countable set of indices K I U J the sequence (ymlK : m E w ) has the property ( p ) in the countable subproduct X K = Xi x Y,.

ni,,,,

njEKnJ

Now let { V i : n E w } be countable local base at the point z ( j ) E Since we have q-lFy,(j) = z ( j ) = plimz,(j) n the sets BA = {m E w : y,(j)

E

for any j E J ,

V j } are all in q.

Thus, the countable family

{A,: i E K n i } u { & :

nEwAjEKnJ}

5 ,for j

E J.

PROBLEMS

155

is centered. Therefore there is an infinite set C C w such that

C C* Ai

for all i E K n I

C c * BL

for all n E w and j E ' h

and

nJ.

It follows that the sequence (ymlK: m E C) converges to the point s l K , in the usual sense, i.e. each neighbourhood U of zIK contains all the terms ymlK, m E C except possibly finitely many. But this is impossible, since the sequence (ymlK: m E C ) has the property (8) in X K . Thus, there is no homeomorphism f of X with f(y) = I, which proves the theorem. QED

PROBLEMS A. AUTOMORPHISMS The Stone theorem determines a one-to-one correspondence between automorphisms of a given Boolean algebra and autohomeomorphism of its Stone space. 1. I f f : w + w is a permutation, then 7 : p[w] + P [ w ] is an autohomeomorphism and each automorphism of P[w] is of this form. 2. Every permutation f : w + w induces an autohomeomorphism

f' = &*

of

the remainder space w * .

3. If f', g' are automeomorphism induced by permutations f,g : w have

f'

= g'

=

{i

E w : f(i)

+w,

then we

# g ( i ) } E fin.

4. Permutations inducing autohomeorphisms can be replaced by one-to-one functions f : w \ s + w \ t , w h e r e s , t ~ f i n .

5 . There exists exactly c many induced and a t most 2c many arbitrary autohomeomorphisms of w'. 6. If CH is assumed, then there is 2' many autohomeomorphisms of w'. remainder of w x {0,1}( is a ParoviEenko space).

(The

7. Assume the continuum hypothesis. If p , q E w* are two P-points, then there is an autohomeomorphism of w' sending p to q. (The proof of the theorem 1.1, Chapter 9).

156

CH.6 INHOMOGENEITY

B. CHARACTER O F POINTS 1. If p E w* is an OK-point and x(p) < c, then p must be a P-point. Therefore, there are 2c many points p E w * , with x(p) = C. 2. There exists a family R = { B ( a ,@): a,@ < c}, consisting of infinite subsets of w , satisfying the following conditions: a) B(a,/31)n B(a,Pz) E fin, for @1 # /3z and an arbitrary cr < c, b)

naEaB(a,

g(a))

$ fin, for any finite s c c and any function g

: s + c.

(Let S = { ( k , f ) : E E w A f : P(k)+P(k)}anddefine B ( X , Y ) = { ( k , f ) E S : j ( X n k) = Y n k}, for X , Y w ) .

c

3. Let { A , : a < c} be an enumeration of all infinite subsets E w . a) There is a function f : c + c such that A,non B ( a , f ( a ) ) ,for all a < c and any y < a,where the B’s are taken from the family R of Ex. 2.

c*

b) If an ultrafilter p contains the sets B(a,f(cr)), for all cr have x-x(p) 2 cf(c).

< c,

then we must

Hence, if the cardinality c is regular, then there exists a point p E w’, such that ?r-x(p)= c (Bell).

4. (Kunen, Bell). It is consistent that c > w1 and ~ - x ( p )= w1, for every p E w’. To see this, prove the following:

P, as follows a) Let V I=GCH and define a finite support iteration P = Co<wl P,+, = P, * Q,, where Q , IF “MA AC = w , + ~ ” , and P, = Pa (direct sum) for limit a > 0 Then, the resulting P has the ccc and P It- ”c = uwl”.

b) There is a sequence f = (fa : cr < w1) E V ( 9 such that the terms fa are -+P(w)”,for all cr < w1. functions, j, E V(Pa+l)and Ba+1 It- ”fa : wu+l onto Moreover, P It- ”fa C fa,’, for all a < /3 < w1. c) If F is an ultrafilter in

V(@, P It- ”F E w’”, then we define

L ( F ) = { t < wu+1 :

3 E Pu+l(q It- ”fu(t) E F”)}.

There is a PU+1-nameF, of the set F have P It- ” F = Ua<wlF,”.

n {fa(<) : t E Z,(F)} such that we

c,

d) For every a < w1 there is a P,+z-name A , E V(’,+,) such that A , B, for all B E ,(use the P(c)-princ$le in V(pn+l)). Therefore, the family {A: : a < w 1 ) is a local x-base at the point F .

PROBLEMS

157

C. PRODUCT SPACES 1. If a product X x Y is an F-space then X or Y is a P-space (Henriksen, Curtis).

(If zo E X and yo E Y are non-P-points, choose an f E C ( X ) and a g E C(Y), which are nonconstant on neighbourhoods of 10 and yo, respectively. Define h ( X , Y ) = If(z)l - IS(Y)I).

2. Any product X x Y of infinite compact spaces is not an F-space. Also an arbitrary product Xi of infinite compact spaces is never an F-space. (A compact P-space is finite).

ni,,

D. LINEAR-MULTIPLICATIVE FUNCTIONALS Let X be a Tychonoff space. The linear ring C * ( X )is a Banach algebra with the uniform norm llfll = suprEx

.[).(fI

1. Every (algebraic) homomorphism h : C * ( X ) -+ h ( a . f) = (Y . h ( f ) , for (Y E W and f E C * ( X ) .

W

is homogeneous:

2. Let A be an ideal in C * ( X ) . If f - (Y E A, for an f E C * ( X )and

(Y

E W, then

ffEf[Xl. 3. Every homomorphism h : C * ( X ) + W is continuous. In fact, h as a linear functional is bounded with norm one: Ih(f)l 5 Ilfll, for all f E C * ( X ) . ( h - h ( f ) E A ( h ) and EX. 2). Let H ( X ) denote the set of all linear-multiplicative functionals on C * ( X ) (i.e. homomorphism from C * ( X )into W).

4. For any p E @[XIwe set h p ( f ) = f ( p ) , for f E C * ( X ) . The function F ( p ) = h, is then a one-to-one map from p [ X ]onto H ( X ) . Now using the function F , we can transfer the topology from @[XIto H ( X ) , and thus obtain a new representation of the Stone-Cech compatification p [ X ] .The points z E X correspond (in this representation) to the point functionals h , ( f ) = f(z). 5. Denote f(h) = h ( f ) for f E C * ( X )and h E H ( X ) . The map f isometric isomorphism of C * ( X )onto C ( H ( X ) ) .

H

f

is then an

6. The sets of the form

n

{ h E H(W : I h ( f i ) - ho(fi)I

< &I,

lsiln

where fi,. . . ,fn E C * ( X ) , constitute a functionally open base at the point ho E H ( X ) . Hence, the topology of H ( X ) transferred from @[XIcoincides with

cH.6 INHOMOGENEITY

158

the weak topology inherited from the space of all linear (bounded) functionals on C * ( X ) .

7. If X is compact, then each functional h E H ( X ) has the form h = h, for an x E X, and vice versa. E. F-LIMITS For a closed set K C w* let FK = { A

s w : K C A*}

1. Each filter FK extends the FrCchet filter {w \ s : s E fin} and K = n{A* : A E FK}.

2. The filter FK is the intersection of all ultrafilters p E K :

3. For any z E C*(w) = l m the limit FK-lim, x ( i ) exists if and only if x*lK =const.

F. MULTIPLICATIVE SUMMABILITY

A linear-multiplicative functional 4 defined on a closed (in the uniform norm) subalgebra E s C * ( w ) is called regular, if the domain E contains all convergent sequences and $(z) = limn z, holds, for each convergent I = (z,, : n E w ) . 1. Let K C w* be a closed set and FK the corresponding filter of neighbourhood (see Ex. E).

a) The functional q 5 ~ ( f ) = FK-lim, f(i) defined on

E K = {f E C * ( w ) : F*lK = const} is regular. b) If K = w*, then E K = the ring of convergent sequences and 4 = lim. If K = A*, then $(x) = lim,eA x ( i ) (the limit of the subsequence ( z ( i ): i E A )).

2. (Henriksen). E C*(w).

Let q5 be a regular functional defined on a closed subring

a) The domain E consists of all f E C * ( w ) such that the Stone extension 7 is constant on abstraction classes of the relation

PROBLEMS

159

b) The factor space X = /3[w]/ = E is compact and E is isometrically isomorphic to C ( X ) . c)

4(f)= FK-lim, f(i),

for some abstraction class K

Hence, every regular functional

w*.

4 is of the form 4 = F-lim for some filter F .

3. If 4 = F-lim then F = { A G w : F-limXA(i) = 1). I

4. For any closed sets K , L G w* the corresponding functionals 4~ = FK-lim, 4~ = FL-lim coincide on EK n E L if and only if h' n L # Q, and then C $ K ~ Lis their common extension. 5. For any p E w* the family R, = { 4 ~ :. A E p } is a maximal family of functionals having a common extension. Moreover, if '&f, = 0, where fi E EA; for some A ] , .. . , A , E p , then we have CEI~ $ ~ ; ( f = i ) 0.

6. A family Rp = { 4 ~ :- A E p } , where p E w * , summs every bounded sequence (i.e. ~ { E A: . A E p } = C * ( w ) ) if and only if p is a P-point.

G. HOMOMORPHISM OF c ( X ) Let X , Y be a compact spaces and T : C ( X ) + C ( Y )a nonzero homomorphism. 1. a) For every y E Y the map a H T ( a ) ( y )is a homomorphism of the field W into itself. Hence, there is a set U E B(Y) such that

b) T is a homogeneous: T ( a . f ) = a . T ( f )for all a E

W and f

E

C(X).

Hence, every homomorphism T : C ( X )H C ( Y )is linear. 2. a) T preserves the (pointwise) ordering: if b) sup{IIT(f)ll : IlTll = 1.

f Ig, then T ( f )I T(g).

llfll 5

1) = 1, i.e. T as a linear operator is bounded and

Hence, every homomorphism of C ( X ) into a C ( Y ) is continuous.

This Page Intentionally Left Blank

Chapter 7

Extending of continuous functions

$1. WEAKLINDELOF PROPERTY. We say that a subset S E X, of a space X is weakly Lindelof, if for every open cover R of S there exists a countable subfamily & R whose union & is dense in S, i.e. S = S n &. We know already (Corollary 3.2, Chapter 6) that all countable subsets of the remainder space W * are C*-embedded. The same is true also for arbitrary weakly Lindelof sets S C w * . In fact, we have the following

u

u

1.1. THEOREM. (Comfort, Hindman, Negrepontis [CHN]). Let X be a conipact zero-dimensional F-space. Every subset S C X having the we& Liiidelof property is C*-embedded in X.

PROOF.Let us take two disjoint zero sets A, B S. In view of the Taiinanov and - Tietze theorenis it is sufficient to show A n B = $9 in X. First, we find (functionally open in S) sets u , w separating A and B: A C u, B

w

and i i n z = P i n S.

Each point z E S has a neiglil~ourhood V , E B ( X ) , clisjoint either froin w (if z E v) or froin u (if z E E ) or from both u and w (for z 4 ii U m).The cover {Vz : z E S} has a countable subfamily R covering a dense subset of S. We may assuine that R is a disjoint family. Let 11s define

R ( u ) = { V E R : V n u # $9) iUld

R ( w ) = { V E R : V n w # $9}. Clearly, we have then

R(u) n ( R \ R ( u ) )= $9 and u n u { V n S : V E R \ R ( u ) } = $9, 161

162

cH.7 EXTENDING OF CONTINUOUS

FUNCTIONS

u

u,

and hence A where U = U R(u) is a cozero set in X. Similarly, if W = R(w), - then B c and W is a cozero set in X. Since X is an F-space, we have U n W = 9 Therefore iin B = 9, which finishes the proof. QED

w

In the presence of the continuum hypothesis the weak Lindelof property characterizes C*-embedded subsets S c w * . More generally, the characterization mentioned above is true in any compact F-space X , for which the ring C * ( X )has carclinality c.

[w]).x

1.1. THEOREM. (woods Let be a compact zero-di~neiisio~id F-space, for which card C ( X ) = C. If CH is assiuned, then C*-embedded subsets S C X coincide with weakly Liiidelof sets c X.

PROOF.The mapping f I-+ flS, for f E C ( X ) is a homomorphism of the ring C ( X )into C*(S). If S is C*-embedded in X, then the lioniomoq~l~sni is onto C*(S) and then card C*(S) = C. Hence, it is sufficient to prove that card C*(S) > c = w1 for any S So let S 2

X without the weak Lindelof property.

X be such that for some covering R -

R we have S \ U& #

B ( X ) and any countable

9. Since obviously w ( X ) 5 C, we obtain card

subfamily & R = W I , by CH. Therefore we can write R = {U, : a

< W I } and

1.3. Now, let us define by induction ordinals 7, < W I and sets V, U,, BS follows: 70 = 0 and VO = UO;if ye a i d Ve for ( < a are already clefined, then applying 1.3 for 6 = sup{-ye : ( < a} we can find a 7, > 6 and an open-closed set V, C_ U,, such that V, n s # 9 and n u, = 9.

v,

U

€
Let us denote 6, = U, ns and V, = V, nS. The sets U , and ? , are relatively open-closed in S and, in addition, the sets ? , are disjoint each other. It is easy to see that the set V = Un
u , ~ ~ ~

$2. A LONG CONVERGENT SEQUENCE

163

unions of increasing chains at the limit stages, we obtain an extension g = Un<wlga of the function f to the set 0, which proves the claim. Clearly, there is at least 2w1> w1 = c continuous bounded functions on the set P (because, consists of w1 mutually disjoint pieces). By the claim above there is also 2w1 bounded continuous functions on 6. But the set S is dense in U , since we have

It follows that also card C * ( S )2 2w1 > c, which proves the theorem.

QED

Let us note (it is clear from the above proof) that the theorem 1.2 holds also when the fanlily of weakly Lindelof sets is replaced by the family {S X : C * ( S )= c}.

$2. A LONG CONVERGENT SEQUENCE. No uncountable discrete set S c w* has the weak Lindelof property. Also, it is immediate to see that card C * ( S )2 2wl. Thus, from the theorem 1.2 we obtain at once the following 2.1. COROLLARY. If CH is assumed, then w1

NO

discrete set

S C w * , of cardinality

is C*-embedded in w * . On the other hand it is not difficult to prove the following remark. 2.2. If c = 2"'1, then there is a discrete set S C w * , of cardhiahty w1, which is

C*-embedded in w * . Indeed, by the assumption, the weight w ( / 3 [ w l ] )is 2"' = c (w1 is taken with the discrete topology). Thus, the space / j [ w l ] can be embedded into the Cantor cube (0, l } c which, in turn, is a continuous image of @[w](asa separable compact space). Therefore /3[wl] itself is a continuous image of some closed set F /3[w], i.e. there is a continuous f : F + /3[wl], onto. The mapping f can be further restricted to an irreducible one and hence a homeomorphism (since /3[wl] is extrenially disconnected cf the lemma 7.4, Chapter 3). Since g [ w ] is embeddable into w * , we conclude that, under our assumption, /3[wl] is a closed subspace of w * . Now, the discrete set w1 is C*-embedded in /3[wl] and therefore also in W * (since / j [ w l ] is closed in w * ) . Balcar, Simon and Vojth constructed some discrete sets as in 2.1 (i.e. not C*-embedded) using no aclditional assumptions. This will follow from the following more general result.

164

cH.7 EXTENDING O F CONTINUOUS

FUNCTIONS

2.3. THEOREM (van Douwen [DvM]). Every P-space X of weight than c can be embedded into the remainder w * .

11011 greater

PROOF.Since every P-space is zero-dimensional, we niay assume, as in the proof of 2.2, that X {0,1}'. But now X is no longer closed (because no i i h i t e P-space is compact) and hence we have to use a different argument. We know that the Gleason space G = G({O,l}c)can be embedded into w* (Sec. 7, Chapter 3). Hence, it is sufficient to embedded X into G. We have an irreducible cauonical mapping f : G + (0, 1)' defined by the equality

f ( ~=)n F =

n { x : A E F}

G such that the restriction

We shall determine a subset I' phism of r onto X. First, let us define for a E (0,

ha(z)= z

for F E G.

+a

fir is a homeonior-

for z E (0, I}',

(addition mod 2, coordinatewise). The iiiappings ha, for a E (0, l}c, are authomeorphisms of the cube (0, 1)' and induce authomorphisnLs

A

A

+ a(= {z + a : z E A } )

of the algebra D({O,l}')of open domains, which in turii give a rise to autohomeorphisms F + F a of the Gleason space G (of course F a = { A a : A E F}). It is easy to see that

+

f(F

+

+

0)

=f(F)

+a

+

for arbitrary F E G a d a E {0,1}'.

Indeed, we have

f ( ~ + = n{B: B E F + =

n{x+

=

n{z:

a : A E F} = n { Z: A E

A E F} =

F} + a = f ( F ) + a.

Let FO be a point in G sent by f to the null function of the cube {O,l}c (i.e. define r = { F o + a :a ~ ~ }

f ( F 0 ) = 0 ) and

+

+

Since f(F0 a) = f ( F 0 ) a = a, f is one-to-one on the set r mid maps is open for onto X. It remains to show that (flI')-' is continuous, i.e. that f[U]

$2. A

165

LONG (.:ONVERGENT SEQUEN(!E

any relatively open U r. Let A E D({O,l}c) be an arbitrary domain. Then A determines a basic neiglibourhoocl in the Gleasoii space

Hence, the relatively open set U in

r has the form

and then we have the iniage

f[uAn r] = { a E x :

A E F~

+a}.

Since the Cantor cube { O , l } ' is separable, the closure of the open domain A depends only on countably many coordinates (if R is a maxiiiial disjoint family of basic neighbourhoods contained in A, then R is countable and =

x

m).

+

Let a E ~ [ U nAr],i.e. A E FO a. We have to show that a is an interior point. Let E c be the set of those coordinates on which depends the closure 71 and take any 6 E X such that 6lE = alE. Then a 6 vanishes on E and hence

+ A = 2 + ( a + 6 ) = A + (a + 6 )

and

+ (a + 6 ) ] = A + ( a + 6 ) , which implies A + a = A + 6. Since A E FO + a, i.e. A + a E Fo,hence also A E FO+ 6, which yields 6 E ~ [ U nAr].As we have remarked already, the set E is at A = int[x] = int[A

most countable, and we conclucle that the set ( 6 E X : 6lE = alE} is a relative G6 in X and hence open in X (because X is a P-space). T~LIS, the iniage ~ [ Unr] A contains with every point a soiiie neighbourhood of a, which finishes the proof. QED

As a particular case of 2.3 let us define

x = {zn : a < w1} u {z}, where the points zn are all isolated and a local base at the point z consists of the sets {zn : a E Z}u {z},where w \ 2 is at most countable. Thus, z is a P-point and therefore the space X is a P-space of weight w1, which can be called a convergent sequence of length w1. Applying the theorem 2.3 we a1)tain the following

166

cH.7 EXTENDING OF CONTINUOUS

FUNCTIONS

2.4. THEOREM. (Balcar, Simon, Vojta+ [BSV]). The illtrirfilter space W* cozitiiills a convergent sequence of length

w1.

It is easy to notice that the discrete set X \ {z} = (3, : (Y < w1} cannot by C*-embedded, for otherwise we would have X \ {z} = /3[wl], which is impossible, since no ultrafilter p E p[wl] \ w1 is a limit in the sense as z is.

$3. STRONGLY DISCRETE SETS. From the theorem 2.4 we know that there are always discrete sets of cardinality w1 which are not C*-embedded in the space w*. On the other hand we may assert that - consistently - every strongly discrete set of cardinality w1 is C*-eml~eddedin w*. Recall that a given set X is called strongly discrete, if each point p E X has a neiglibourlioocl U, such that tlie family {Up : p E X} is disjoint. Of course, the ultrafilter space W* contains strongly discrete sets of cardinality continuum, since the Boolean algebra P(w)/fin contains anticliains of cardinality c.

In this section we shall prove the following

3.1. THEOREM. (F'rankiewicz, Zbierski [FZz]). There is a model of set theory in which the Martin axioni plus c = w2 hold and each strozigly discrete set X C_ W* of Cardinality w1 is C*-embedded in w*. For the proof we need two auxiliary lemmaq. Let R = {A, : Q < w1} be an almost disjoint family (of subsets of w) and let 2 w1. We shall use a kind of a separating forcing Q = Q( R, Z),depending on R and 2,whose elements are of the form p = (s,, t,) where s p : dm(sp) -+ w

and t, : dm(tp) -t w

are functions with finite domains, dm(sP) C Z ancl clm(t,) C w1

p

5q

8,

2 sq ancl

tp

\Z

2 tq.

$3.

167

STRONGLY DISCRETE SETS

It is immediate to see that tlie sets

E,p = { p E Q :

E clni(sp)

cy

ancl [I E clni(t,)}

where cy E Z ant1 /I E w1 \ 2,are dense in Q . Therefore, any generic filter G determines two functions

Q

where fc = U { s Y : p E G} and gc; = U { t p: p E G}. If we denote

then A n B = (Band we have

A,

ant1

(Z* A

Ap C, B

for all

cy

E Z

for all [j E w1 \ 2.

Tlius, the forcing Q = Q( R, 2)separates the division

R = {A,

:

E Z } U { A p : /I€

w1\

Z},

of the family R into two parts.

Now, let 11s suppose that we iterate sonie of the separating forcing with finite supports w2 times so that for sonie families R ancl every Z w1 in the final nioclel the corresponding forcing Q = Q ( R ,2 ) will occiir in the iteration. Let F = { F , : cy < w 1 } he an arbitrary strongly discrete set of ultrafilters in so obtained model. In view of the Tainianov theorem it is sufficient to show that 3.2.

{F, :

cy

E

2)n {Fp :

[j

E w1 \ Z } = (3

for each Z C w1

There exists open-closet1 neiglilmurlioocl U , of the point F, such that the family (17, : cy < w 1 } is disjoint. If A, (Z w are such that A: = U , for all cy < w1, then tlie family R = {A, : cy < w1 } is almost disjoint ancl if tlie forcing Q(R, 2 ) occurs at sonie stages of tlie iteration for each set Z w l , then 3.2 is satisfied, and X is C*-enibedclecl. We note however that in general the separating forcings do not have the ccc. Thus, the main difficulty of the proof is to find, for a given strongly discrete

s

cH.7 EXTENDING OF CONTINUOUS

168

FUNCTIONS

set F, a suitable family R (representing disjoint neiglibourhoods of points of F) such that the corresponding forcing Q = Q(R,2) has the ccc. Now, we shall examine antichains in a forcing Q = Q(R,2).Let us denote

K(P)=

U

Aa \ s p ( a )

and L ( P ) =

aEdm( a,,)

U

~ a \ t p ( ~ )

BEdm(tp)

Hence, we have K ( p )r l L(p) = 8, for each p E Q. For p , q E Q clefine the sets

Thus, e(p, q) is always a finite subset of w. Two elements p , q E Q are incompatible (p I q), if at least one of the following conditions hold 1. s p U sq or t, U t , is not a function. 2. The set e(p, q ) is non-empty.

Let C = { p n : a < w1) be an uncountalde antichain in Q. Clearly, the property 1 may hold for at most countably many elements of C only (which follows immediately from the A-system lemma). T~IIIR, we may assume that the property 2, (i.e. e ( p , , p p ) # 0 for all a < @ < w1) liolcls for all elements of C. In addition, performing some obvious operations, we may assume that the cloniains of s p and t, have the same number of elements: for some n, 712 E w we have dm(s,,

= {7:, . . . , 7 ~ }mid dm(tp, = {tf,.

Moreover sp,(7,P) = s p s ( - y f ) for any a < 19 < tp,(aq)= tpp(6!)for all a < /j ancl j = 1,. . . ,772.

w1

. . ,b{t}

for a,/ j

< w1.

and i = 1, ...,72. Similarly,

Finally, we may claim that

max dm(s,,) < niin d m ( s p p )

for all a

< P < w1

and similarly for the t’s. Thus, if Q has mi uncountable antichain, then it has a regiilar one (i.e. satisfying all the properties listed above). The forthcoming lemma expresses a fuudamental property of (regular) antichailis in Q.

3.3. LEMMA. Let C = { p a : a < w1) be a regular zuiticJiahi in Q. There exists a tree T,consisting of finite sequence (eo, . . . ,e,,) wliere e i E fin, mi o r d i d 7 < w1

$3. STRONGLY DISCRETE SETS

169

mid a function q : T + {pa : a < y } such that for every a 2 y, (a< w l ) , there is a brmicli g = (e,, : ti E w ) of the tree T fulfilliiig the following coiiditioiis: 1. e,, = e ( p a , q ( g l n ) ) for eadi 71 E w .

2. the sets e,, are pairwise disjoiiit.

PROOF.For an arbitrary iionempty finite set eo

w clefiiie

C ( e o ) = {a < W I : a > 0 and e(po,pn) = eo}

and

S = {eo E fin : card C ( e 0 ) = w l } .

Some of the sets C ( e 0 ) can he finite or countable, so let us define 70 = s u p ( s u p C ( e 0 ) : eo 4 S } and denote Y ( e 0 ) = C ( e 0 ) \ 70. Then we have

Set q(e0) = pa, where a =inf Y ( e 0 ) . Next, we repeat the above construction with replaced by q(eo), for eo E S. Tlius, we define

C ( e o , e l ) = {a E Y ( e 0 ) : e(q(eo),pa) = e l )

and

S ( e 0 ) = {el E fin : card C ( e 0 , e l ) = w l } .

71 = s u p { s u p C ( e o , e l ) : eo E C(e0, e l ) \ 71, then we have

Now, if we set

s and el 4

S ( e o ) } and Y ( e 0 , e l ) =

We put q ( e 0 , e l ) = p a , where a = infY(e0,el). Continuing in this way we obtaiii an increasing sequence 7 0 < 71 < . . . of countable ordinals, and uncouiitable subsets

Y = w l , Y ( e o ) , Y ( e 0 ,e l ) , . . . of w1

mid sets

S, S ( e o ) , S ( e 0 , e l ) , . . . such that

3.4. a d ,for all n E w

Now, let

US

define y = sup{y,, :

T = {(eo,.. . ,e,,)

71

Ew}

: eo E

aid

S,. . . ,e,, E S ( e 0 , . .. , e I I - l ) } .

cH.7 EXTENDING OF CONTINUOUS

170

FUNCTIONS

We check that y and T are us required. If a 2 y, then from 3.4 a i d 3.5 we can find elements eo E S, el E S(eo), e2 E S(e0, el), . . . such that for each 71. < w we have Q E Y(e0,. . . ,eI,). Therefore, the sequence g = (eI, : 11 < w ) is a branch of T and en+] = e ( p , , q ( e o , . . . ,eI,)). Also eo = e ( p o , p , ) , so if we clefiiie aclditioiially q(9)= 9,then the equality e,, = e(p,,q(gln)) holds for 71 = 0 its well. It remains to show that the sets e,, are disjoint each other. First of all, observe that

Assume inductively that the family {eo,.

. . ,eI,-l} is disjoint. Then we have

fiom the obvious inclusions

it follows that

since K(p,) n L(p,) = 9. On the other hand, froin the inclusions K(q(gln))

n L ( P ~C) K(d!7b))~

1 Ui
that K((I(SI72))

n L(Ym) n

u

K(q(gl4)

n ~ ( p , )c

~ ( q ( g I 7 1 ) )it

follows

n L(Y,) = 9,

i
since K(q(g1n))n L(q(g1n)))= (D. Hence, the set K(q(gln))n L(p,) is disjoint from eo U .. . u en and, syninietricdly, the set L(q(gln)))fl K(p,) has the same property. This implies at once e,, n (eo U . . . u eI,-l) = 9, which finishes the proof. QED 3.6. REMARK. Let 9 exceed the clomaii~sc l m ( ~ ~and ~ ) cliii(tpc) , for all 6 < 7.It follows froin the proof that, the branch g determinecl by an a 2 y, is definable from the following parameters: T, q, { A ( : ( < T} ,p a ancl

{A,

:

71

E clm(sp,)U clm(tpm)}.

As mentioned earlier, a separating forcing Q = Q(R,Z),wliere

53. STRONGLY DISCRETE

SETS

171

R = { A , : a < w l } , need not have the ccc. Let S = {B, : a < w 1 ) be another almost disjoint family. We say that the forcing Q ( S ,Z ) is an iiiiprovenieiit of Q ( R , Z ) , if B, C A, for all a < w1 and Q(R,Z ) has the ccc. We prove now that every forcing Q(R,2)has an improvement possibly after at most w1 stages of the iteration. 3.7. LEMMA. Let (P, : a < w1) be a finite support iteration of nontrivial forcings having the ccc. If R, Z E V, then for every family of iiltrafZters {F, : a < w l } E V[G] such that A, E F, for all a < w1 there is a11 almost disjoint family S = {B, : a < w 1 ) E V[G], such that Q ( S ,Z ) is xi i~nproveme~it in V[G] of Q(R,Z ) mid B, E F,, for all a < w1. PROOF.

<

Let (Am : a < w1) be au increasing enunieration of all limit ordinals

w1. It is well known that there are Coheii reals, (over V[G[A,]), c, E V[G(X,+i],

w (we may also ass,ssumedirectly that PA+1 = P A* c, for h i t where c, Let us define A, n c,, if c , E F, B, = A m \ c,, if C, F,

< w1).

{

It follows that B, C A, a i d B, F,, for all a < w1. We shall prove that Q ( S ,Z ) , where S = {B, : a < w l } , has the ccc in V[G]. Suppose that it is not the case ancl let C = { p , : a < w1} E V[G] lie an uncountable mitichain in Q ( S ,Z ) . We may assume that C is regular. In particular, iiiax dm(sy,)

< iiiiii clni(s,,)

for a < /I < w l ,

and the same holds for the t ’ s . Clearly, an eleiiieiit p E Q ( S , Z ) need not be in

Q(R,Z ) . But the just meiitionecl property of C implies iiiimediately the following: for every a < w1 there is a k, E w such that if

and

A,

= B,

for y

U

c~m(s,,) u clm(tp,),

,<W,

the11 C C_ Q(k, Z ) where R = {A, : y < w l } . Moreover, C is in fact an antichain in ~ ( k , since the set e ( p , , p P ) , calculatecl in Q ( S ,z ) ,is coiitainacl in e ( p , , p a ) , calculatecl in Q(R, z ) ,for any a < /j < w l . NOW, we apply the leiiiiiia 3.3 to C in Q(k,Z ) and fix a < w1 such that the paranieters T, q {At : ( < T} \)elo~>gto the submodel V[Gl/3] (cf. the remark 3.6). The existence of such a /9 follows froin the

z),

172

cH.7 EXTENDING OF (:ONTINUOUS

FUN(:TIONS

ccc property of the iteration. Fix an N > [j and let y1 < . . . < ya be an eiiuiiieratioii of tlie set dm(s,,) U dm(t,,). Since p p and A,, ,. . . ,A,, are all in V, the lxaiich g = (eIl : n < w ) , corresponding to a,belongs also to V(G(/j]. Let us denote =

Since p < (Y 5 y1 5 A,,

{ :i

, the set E,,

c,,

if c, $ F, if cy E F,

is a Cohen over V[Glp]and heiice tlie set

mist be infinite. Repeating this operation r times we obtain an infinite sul)family

Consider an arbitrary element definition of B, it follows that

err

of E. Thus, e,, E EYnf

Byn e,, = (3

l

. . . n E,,

ancl froin the

for y = y l , . . . ,yr.

Consequently, the set wI, = e(y,, q(g1n))) calculated iii Q ( S ,2)is disjoint from eIl. On the other hand, wtl is a subset of the set e(yp,q(g1n)))calculated in Q(k,Z),i.e. we have w,, G en. It follows that e(y,, q(g1n))) = wIl = 0,whence y n and q(gln)) are QED compatible, a contradiction, since they are members of the anticliaim C.

REMARK.The improvement Q ( S ,Z),of the forcing Q(R,Z ) , coiistructecl in the above proof has an additional property, viz. B, E F, for all N < w1, where F = {Fa: N < w1) is a given family of ultrafiltars. Thus, Q ( S , Z ) will be referred to as an F-improveiiient of Q(R,2). Now, we can finish the proof of tlie theorem 3.1. Let H = u { H , : N < w 2 ) be the family of all sets of hereditary power less than w2, in V. We shall use the diamond principle for H: there is a seq~ie~ice (T,: a < w2 and c f ( a ) = w1) SIICJIthat for every X C H, the set {a E w2 : ~ f ( a=) w1

is statioiiaxy.

x n H, = T,}

$3.

173

STRONGLY DIS(:RETE SETS

We assume that the grouiid motlel V satisfies the allove principle as well as CH and 2w1 = w 2 , (e.g. as V we can take the constructible iiioclel). In the iiiodel V we define by induction a finite support iteration (P, : cy < w z ) , with card P, 5 w1 mid therefore all the stages P, can be regarded as eleinelits of H: (the direct limit) for liiiiit a Pa = Ca<,Pp Po+l = P, * T,, if cf(cy) = w1 and T, is a P,-iianie of a forcing such that P, It- “T, has the ccc”

* Q, if c f ( a ) = w1 and T, = R x

2 x D for soiiie P,-iiaiies R , 2,D (for a family, a subset of w1 and a strongly cliscrste set, respectively) and for some Q E vP- sucli that P, It- ”Qis an D-improvement of Q(R,2)’’ Po+l = P, * C, in all reiiiaiiiing cases (C is a P,-iiitliie of the siiiiple Colieii

Po+l = P,

forcing). Thus, the forcing P = Cn<wz P, has the ccc and card P = w2. Let G P be a geiieric filter. By tlie diamond principle every ccc forcing of cardinality < w2 occurs in the iteration as some T,. Heiice, we have V[G]

”MA ant1 c = ~

2

”

Now, let F = {F, : a < wl} lie a strongly discrete subset of w* in V[G]. We R = {A, : cy < w 1 } such that A, E F,, for all a < w1. If 2 E w1 is a fixed set, then choose an arbitrary aliiiost disjoint family

since P has tlie ccc. If we denote Fly = {Fan V[G/y]: a

< wl}

for y

< w2,

then the set {y E w2 : Fly E V[Gly]} is an w l - c u b (Sec. 2, Chapter 9). Indeed, the set F = { F, : a < w 1 } cau be represented as

F = { ( a , z )E w1 x P ( w ) : The restrictions Fly have then a form

cy

< w1 aiicl z E F,}.

174

cH.7 EXTENDING OF CONTINUOUS

FUNCTIONS

Let F be a canoiiical nmiie of F coiisistingof pairs ( ( ( ~ , g ) ( ~ ) , p where ) , g is a canonical name of the set z w. Thus, the set

are then antichailis in P and hence they are all countable. Let as defiiie subnames

Clearly, each Fly can be regarded as a P,-name. Let us coiisicler the following set

N1 = {y < w2 : Vz,(Y[z E V(P,)

+ F((Y,Z)

c P,]}.

Obviously, Nl is closed under wl-limits. If 6 < w2, then since z mid (Y run over some sets of cardiiiality w1 (note that P, Il-CH) mid siiice all F(a,z)are countable, F(a,x) 5 Pa*. Define by induction an increasing there is a 6* < w2 such that sequence: 60 = 6, 6(+1 = 6; aid 6~ = sup(6, : < A } , for limit A. Now, let y = 6,, .

u,,

Thus cf(7) = w1 mid hence, by the ccc, any z E V(',) is already ill some V(Pst), whence ~ ( c Yx), Pa,,, C_ P,. It follows that y E iV1 (and y > 6). Consequently, the set N1 is an wl-cub. Note however that

and hence Fly E V[Gly], for y E N1 , which proves the claim. Clearly, the family Fly is strongly discrete in w* in the model V[Gly] (whenever it belongs to V[Gly].). Since F C H and H = Un<w2 H, is a decomposition of H into a continuously increasing chain, it can be easily checked that the set

is an wl-cub. Thus, N = Nl n Nz is an wl-cub (see 2.1, Chapter 5 ) . Applying the diamond principle we find an ordinal y E N , y > /" a i d such that T, = From the lemma 3.7 it follows that for some P,-nanie Q we have

x

zx Fly.

P, Il- " Q is aii (F1y)-iiiiProvemeiitof Q ( R ,Z),',

mid hence, the definition of the iteration iiiiplies P,+1 = P, ately that the parts

* Q.

It follows iiiiiiiedi-

P HOBLEM s

175

+

are separated in V(G1-y 11 with sonic open-closed sets A* and B*.The also separate the parts

{F, :

a E Z}

and {Fp :

in V [ G ] wliicli , finishes the proof.

same

sets

B E w1 \Z}

QED

PROBLEMS A . C*-EWBEDDED SETS

\ {p} is C*-embedded. be an independent family of subsets of w and F c (0, a set

1. If CH is assumed, then for no p E w* the complement w* 2. Let {A, : a

< C}

of cardinality w1 of zero-one functions. Assume c = 2”1 am1 fix an enumeration

{X,

: a

< c } of P ( F ) and denote

T(f)= {fin : a) The family

ti E w }

mid T,(f)= { f l t i :

ti

E A,}

for a

< c.

Sf,for f E F,consisting of T(f)and either of T,(f),if f T,(f),if f $ X,, for all a E c, is centerecl.

E X,

or of T ( f )\

b) Let Ff be an ultrafilter (on the countal,le set UtrEw{O, 1)”) extending The set { F j : f E 7 )is strongly discrete an C*-eiiil~eclded.

Sf.

Hence, assuming c = 2 w l , there exist strongly discrete a i d C*-embedded sets of cardinality w1 in the remainder w * .

B. FORCING Q(R,Z) 1. Let R = {A, : a < w 1 } be an aliiiost disjoint family. If, for a given set 2 E w1, the gap ({A,: a E 2 ) ; {Ap : [j E w1 \ 2 ) ) is filled, then the forcing Q(R,2) has tlie ccc. 2. Let R = { A , : a < w l } be mi aliiiost clisjoint faiiiily satisfying tlie following coiidition the set {( < a : niax{At n A.} < k} is finite for all a < w1 and k < w . (It means that A, is near to {A( : ( < a}).If both 2 w1 micl w1 \ Z are uncountal)le, tlieii Q(R,2)has not the ccc. Even stronger, we have then

O F CONTINUOUS cH.7 EXTENDING

176

FUNCTIONS

C. FATIDEALS An ideal I on the set w is called fat, if fin c I and for every family {A,, : n E w } of subsets of w , if lim,,ininA,, = +oo, then there is an iiifiiiite set 2 w such that

U{A,, :

nE

Z} E I .

1. If I is a fat ideal, then the dual filter F = { w \ A : A E I}is a P-filter (i.e. the set n { B * : B E F} is a P-set in w*). 2. If p E w* is a P-point, then the dual icleal I = { w \ A : A E p } is fat.

3. Assume CH. a)

The remainder W * contains a nowhever dense P-set X homeomorphic to the Gleason space G({O,l}w).

b) There are no P-points in X . Therefore, the P-filter F = { B cannot be extended to a P-point. c) The dual icleal I = {A

w : w

w :

X E B*}

\ A E F} is fat.

Hence, assuiiliiig CH, there are fat ideals, which cannot be extended to P-points.

D. EXTREMALLY DISCONNECTED SPACES 1. Any open subset of an extreiiially discoiuiected space is C*-embedcled. Conversely, if every open subset of a space X is C*-eiiil~eddecl,then X is extreiiially clisconiiectecl. 2. A Tychnoff space X is extreiiially disconnected if and only if every cleiise subset D c X is C*-eiiil~edcled in X. 3. If X is an extremally cliscoiinectecl compact space, then X = / j [ D ] ,for every dense subset D C X. 4. Let X and

Y be coiiipact spaces and f : X ' 3Y be an irreducible map.

a) If z is an isolated point in X , then f ( z ) is isolated in Y. b) The Glearoii space G({O,l}w)is separalde a i d cleiise in itself. c) The ultrafilter space w* contains countable dense in itself subsets. Hence, there are dense in itself spaces X, in which every subset is C*-embeddecl.

177

PROBLEMS

E./ j [ ~ AS ] For A

A SEMIGROUP

w mid

ti

E w clefine

A - T I = {z' E w

:

i + ti E A }

and

A / n = {i E w : i . n E A } . 1. Let p , q E /3[w]. Define p

+ q and p . q as follows

and

a) For any ultrafilters p , q E /I[w], p

+ q asid p . q are ultrafilters on w as well.

+

H p q and p , q I+ p . q extend the ordinary addition mid iiiultiplicatioii mi w , respectively (identifying the priiiciple ultrafilters with integers).

11) The operations p, q

c) The associativity law holcls for

+ a i d - on /I[w].

+

d) The both operations awl . are left-continuous (i.e. p p . z) are coiitiiiuous functions of z,for every p E /I[w]).

+ z (mid siixiilarly

e) The ultrafilter p + q is principal if mid only if the hoth p and q are principal.

Siiiiilaly, for the procluct p . q . Heiice ( P [ w ] ; + ) and ( / l [ w ] ; . ) (mid continuous semigroups. (Glazer).

also ( w * ; + )

asid ( w * ; * ) ) are compact left-

2. Let (G; .) be a compact left-continuous semigroup. a) The family R = {X C_ G : X miiiiiiial eleiiieiit A E R.

# (D mid

x =X

mid X

.X

X}

ha5

a

1)) If a E A , then n A = A. ( a . A is a continuous image of A ) .

c) A = { L E A : a . 8 = a } mid therefore a . a = a, for any n E A . Heiice, every coiiipact left-continuous semigroup has an idempotent. makura).

+

+

(Na-

For any set X C w let u ( X ) cleiiote the set of all finite s u m s z1 ... z,,of distinct elements of X. Siixiilarly, r ( X ) denotes the set of all 2 1 . . . . * z,,,where

..

Zl,. ,Zit E

x.

3. Let q E w* be an idempotent with respect to (say) the siiiii

+.

178

cH.7 EXTENDING OF CONTINUOUS

FUNCTIONS

a) If A E q , then there is a clecreasing sequence A,, E q and an increasing sequence k,, E A,, such that Ao = A and A,,+, = A,, n (A,, - k,,), for all n E w.

b) a({lc,, :

E w}) C A . (Induction on the nunilier of smunancls).

Hence, for any A E q there is a11 iiifinite X r ( Y )E A for some infinite Y . (Galvin).

A such that u ( X ) C A. Similarly,

4. (Hindman). For any division of w : w=

z, u . . . u z,,,

into finitely many parts, at least one part contains a set c ( X ) ,for some infinite X w. Also, oiie of 21,. . . Z,,contains r ( Y ) ,for some infinite Y.

Chapter 8

T h e M a r t i n Axiom IMAGES.From the ParoviEenko theorem we know that, as$1. CONTINUOUS suming CH, every compact space of weight c is a continuous image of the remainder space a*.In the case of c > w l , the situation is more compound even if the Martin axiom is assumed.

<

First, we prove the following

1.1. THEOREM (van Douwen, Przymusixiski [vDP]). If MA is assumed, then every compact space of weight less than c is a continuous image of the space w*.

PROOF.We know already (see 2.3, Chapter 4) that every compact space is a continuous image of a zero-dimensional compact space of the same weight. Hence, it is sufficient to prove that every Boolean algebra A of cardinality less than c can be embedded into the algebra P(w)/fin. The algebra A determines a forcing P, as follows. The elements p of P are the from p = ( f p , m p )where mp E w and fp : dm(fp) -t P ( m p ) ,where the domain dm(fp) is a finite subset of A. Thus, for any a E dm(fp), the value !,(a) is a finite subset of w bounded by m p from above. The ordering on P is defined thus: p 5 q if and only if 1. m, 5 mp and dm(f,) C dm(fp), 2. for each a E dm(f,), f p ( u )n m, = f,(a), 3. for arbitrary a , b,c E dm(f,), if c = u + b, then fp(c) \ m, = fp(a)U f p ( b ) \ m,. Also, if c = --a then f p ( c )\ m, = ( m p\ fp(a))\ m,.

It follows immediately that if mp = m,, then p Iq holds if and only if fp(a)# f,(u) for some a E dm(fp) n (f,). An obvious application of the A-system lemma shows that the forcing P has the ccc. Now, let us check that the sets &(a, b) = { p E P : a , b E dm(fp) and 32'

179

1 n [i E fp(u) A f p ( b ) ] } ,

180

cH.8 THEMARTINAXIOM

where a # b, are dense in P. Take an arbitrary q E P. The finite set dm(fp) U { a , b} generates a finite subalgebra Ao 5 A. Let 2 1 , . . . , z k be the atoms of Ao. Thus, A0 = P ( { z l , . .. ,l k } ) (up to isomorphism). Let us divide the set w into k infnite parts W = z 1 u * * * u z k .

.

Clearly, the mapping zi H Zi, i = 1,.. . ,k can be extended to an isomorphic embedding g : A0 -t P(w), so that all symetrical differences g(z) A g(y), for # y, are infinite. Choose an i E g(a) A g ( b ) such that i 2 max { n , m q }and an arbitrary m = mp > i. Define

Obviously, the condition p = (fp, m p )is in E,,(a, b ) and we have p 5 q, which proves the claim. Since card A < c we can apply the Martin axiom. Thus, there is a filter G C P intersecting all sets En(u,b), for n E w and a, b E A. It follows immediately that the equality f(a)= u { f p ( a ) : p E G} mod fin

defines a function from the whole of A into P(w)/fin. Of course, f is one-to-one and the property 3 of the definition of the ordering of P ensures that f preserves the Boolean operations. Thus, f is an embedding. QED The Martin axiom does not settle the case of w ( X ) = c. Suitable models will be constructed in the following sections.

$2. THE SPACE p [ w l ] . Martin’s axiom together with c > w1 imply the equality 2w1 = c. Hence, the compatifiction P [ w l ] has weight c. We are going to construct a model with MA plus c > w 1 in which the space P [ q ] is not a continuous image of the remainder w*. In other words, we show that - consistently with MA plus c > w 1 the power-set algebra P ( q ) is not embeddable into the algebra P(w)/fin. Let us consider an increasingly ordered gap L , of type ( w I , w I ) in the algebra P(w)/fin. We know that L can be represented in P ( w ) in the form

52. THE SPACE p[w,]

181

where A,, Bp C w are all infinite and have the following properties: 1. The sequences ( A , : (Y < w l ) and (Bp : relation C, of almost inclusion.

p < w1) are both increasing in the

2. A , n Bp E fin, for all cu,p < w1. Moreover, replacing A, by A , \ ( A , n B,), we may assert also

3. A , n B , = $9, for each a < w1. Each gap L determines a separating forcing Q = Q ( L ) , whose elements are of the form q = ( u q , o g w , q ) ,where u g ,wq C w1 are finite sets, zgis a zero-one sequence of some finite length l ( q ) and

u

A, n

&up

u

Bg C l ( q ) .

PEW,

From the theorem 5.4 of Chapter 2 we know that

2.2. If the gap L is separated, then the forcing Q(L)has the ccc. The gap 2.1 determines also another forcing E = E(L), (due to Kunen), whose elements are finite sets e C w1 such that

A,nBp#$9

or A @ n B , # $ 9

foranyar#pEe.

The forcing E is ordered by the inverse inclusion. Its action is clarified by the following

2.3. LEMMA.If the a gap L is unfilled, then E = E(L) has the ccc and E IF "Q(L)has an uncountable antichain".

PROOF.Suppose, tending to a contradition, that {e, : a < w1) is an uncountable antichain in E. Applying the A-system lemma, it is clear that we may assert that max e, < min ep for all o < /3 < W I . Since e, I ep, we have At S, r l Tg = $9, where S, = the property 3 of the gap L.

n B, =

$9, for some f E e, and 77

E e p . Therefore,

ncEea A€ and Tg = nqEep B,. Also, S, n T, = $9, from

Now, let a < w1 be given. There is a y < w1 such that a < mine-,. Then we have A, S, and hence A, G* S, where S = S,. Similarly, for every

cH.8 THEMARTINAXIOM

182

up<wl

p < 01 we have Bp C, T , where T = Tp. But S n T = 8 which means that the set S separates the gap L,contradicting our assumption. This proves that the forcing E has the ccc. Now, we shall prove the other assertion. Define a subset Eo C E consisting of all e E E with an uncountable splitting, i.e. E

e E Eo

e

E E and card {a E w1 :

e U {a}E

E} = w1.

We claim that the set E\Eo is at most countable. Indeed, suppose that card (E\Eo) = and choose an arbitrary eo E E \ Eo. Thus, there is an ordinal yo < w1 such that e U {a}4 E for every a 1 yo. But, there is an el E E \ Eo containing some elements 2 70 and then we have eo I e l . Therefore, there is an ordinal y1 < w1 such that eo U {a}4 E for each a 2 7 1 etc. Continuing in this way we obtain an uncountable antichain, a contradition. This proves the claim.

w1

From the claim it follows easily that Eo also has the ccc (since the set {z E E : e 5 z} is finite, for all e E E). Hence, deleting countably many elements e 4 Eo, we may assume that E = Eo. Thus, the sets

are dense in E. Consequently, any generic filter G so is the set

c E is uncountable and therefore

Z = {a E ~1 : {a}E G}. Now, if we denote q, = ({a},@{a}), then we have q, I q p for a # /? E 2,in Q(L) (since {a}and { p } are compatible in E). Thus, {q, : a E Z } is an uncountable antichain in Q(L).

QED

Consider the subalgebra A

P(w1) generated by at most countable sets, i.e.

A = {z C w1 : card z 5 w

or card

(w1

\ z) 5 w } .

Obviously, card A = c. It is easy to see that A has 2w1 unfilled gaps. For example, let w1 = U{Z, : a < w 1 ) be a division of w1 into w1 sets Z,, each of cardinality

Lx

= ({A, : a

< w l } ; {Bp : P < WI})

w1,

Define for X G w1,

$2. THE

183

SPACE p[wl]

where A , = UtEX2, r l a and Bp =

2, n p.

Clearly, each Lx is a non-separated gap in A and the L X ’ Sare all distinct for distinct X C_ w l . Moreover, if f : A + P(w)/fin is an arbitrary embedding, then the images

f [ L x I = ( { f ( A a ): a < W } ; { f ( B p ): P < ~

1 1 )

have the following property if, for distinct X and Y , the gaps f[Lx] and f [ L y ] are separated in P(w)/fin, then they are separated by distinct elements (of P(w)/fin). Thus, we have proved the following 2.4. REMARK. If c < 2”1, then for an arbitrary embedding f : A * P(w)/fin remains unfilled in P(w)/fin. there is an unfilled gap L in A, such that the image f[L] In the proof of the following theorem we use (as we often did) P(w) rather then P(w)/fin (identyfying sets equal mod fin). Also, embeddings f : A + P(w)/fin should be understood as relations f A x P ( w ) sastysfying an obvious condition.

c

2.5. THEOREM. (Frankiewicz [Fz]). There is a model of the Martin axiom plus a continuous image of the remainder w’.

> w1,in which the space P [ q ] is not

We may assume that in the ground model V we have c = w1,the cardinal 2y1 = IC = I C
There is a sequence (T, : a H ( K ) the set

< IC and c f ( a ) = w1) such that for every set

X

{ a < K : cf(a) = w1 and X n H, = T,} is stationary in

K.

Here H = H ( K )denotes as usual the family of all sets of hereditiary power < IC and H = Ua

as follows:

(the direct sum) for limit a if cf(a) = w1 and T, is a Pa-name of a forcing such that P, IF “T, has the ccc”;

184

cH.8 THEMARTIN AXIOM

P,+l = P,

*E

P,+1 = P,

*C

if c f ( a ) = w1 and for some gap L of A in V('a) we have P, IF "T, : A + P(w)/fin and T,[L] is unfilled" and E is a Pa-name of E(T,[L]). in all remaining cases (C is a name of a Cohen forcing). P, has the ccc. Clearly,

From the lemma 2.3 it follows that the forcing P = in the model V[GJ the Martin axiom and c = IC hold true.

It remains to show that there is no embedding of the power-set algebra P(w1) into P(w)/fin. Suppose conversely that F : P(w1) + P(w)/fin in V[G] is such an embedding and let f = F ( A . Let us consider restrictions fa

= f n V[Gla]

for a < IC and c f ( a )= w1.

f C H of the embedding Since V[G] "c = IC", we can find a canonical name f, consisitng of some pairs ((z,u)('),p) where z E A, a E P(w)/fin in V(') and the set

f(z,a) = {P E P : ((Z,4'p',P)

E f)

is an antichain (and hence is countable). It follows easily that the set

is an wl-cub in IC. Indeed, N1 is obviously wl-closed. Let a < IC be an arbitrary ordinal. Since P has the ccc and c < K holds in the submodel V[Gla], we can find : Z , U E V(pa)} C Pa*. Define by induction an an a* < IC such that U{f(z,u) increasing sequence (7t : ( < w1) as follows 70 = a, 7,+1 = 7; and 7, = supB<, 78 for limit a < w1. Clearly, the ordinal 7 = S U ~ 7~ ~ is< in~ N1, ~ which proves the claim.

f =f n ( V ( p ~x) P,) is a P,-name Obviously, for any y E IVl the restriction -7 and we have

f

-7

[GI71 = f, = f

n V[GI71.

Therefore, the embedding f, belongs to V[Gl-y], for all y E N1.

A similar argument shows that the set N 2 = {a < K : cf(a)= w1

and f n H,

=fa>

$3. ON THE

PAROVICENKO

THEOREM

185

is an wl-cub as well. Let N = N1 n NZ. From the diamond principle applied to the f it follows that for some y E N we have T, = f, and f, E V[Glr]. Since name c < 2"' in V [ G l r ] we , may apply the remark 2.4. Thus, we have P,+1 = P, * E , where E is a P,-name of a forcing E(f,[L]) for some unfilled gap L in A n V [ G l r ] , such that f,[L]remains unfilled in P(w)/fin. An application of the lemma 2.3 shows ) $ ( F [ L ] )has an uncountable antichain in that the forcing Q = Q(f,[L])= Q ( f [ L ]= V[Gly 11 and hence in V [ G ]as , well. On the other hand, the gap L is separated in P(wl), in V [ G ] .Therefore, the image F [ L ]is separated in P(w)/fin, whence Q ( F [ L ] ) has the ccc, a contradition. QED

+

Thus, if the cardinality c is > w1, then the conclusion of the statement 4.5 of Chapter 2 is no longer true, even if the Martin axiom is assumed.

53. ON THE PAROVICENKO THEOREM. In this section we prove that the first part of the Parovitenko theorem, (cf the statement 4.5, Chapter 2), is not equivalent to the continuum hypotesis. In fact, we shall construct here a model in which the continuum is large and every Boolean algebra of cardinality 5 c can be embedded into P(w)/fin (equivalently, every compact space of weight 5 c is a continuous image of the space w * ) . The construction can be so arranged that, additionally, the Martin axiom will hold also in the model. To simplify the proof we shall limit ourselves to a slightly weaker version of Martin's axiom, viz. MA (a-linked): for every a-linked forcing P and any family R, card R < C, consisting of dense subsets of P there exists a filter G C P intersecting all sets D E R.

Thus, we prove the following 3.1. THEOREM. (Baumgartner, Frankiewicz, Zbierski [BFZ]). There is a model in which MA (a-linked) plus c > w1 holds and every Boolean algebra A of cardinality 5 c can be embedded into P(w)/fin.

PROOF. Fix any regular cardinal K > w1 in a ground model V ,fulfilling 2<" = n. We intend to have c = K in the final model V [ G ] . We define inductively a finite support iteration (Pa : a < K), with the ccc, which will be a combination of two iterations: at some steps we take care of embeddability into P(w)/fin and, at the others we treat (in a known way) the a-linked forcings. To be more precise, let us

186

cH.8 THEMARTIN AXIOM

fix a pairing function j :

K

x

K

5 K , onto K and such that

j(a,p) > a

for all a < K ,

and assume that at a given stage a we have already defined a Pa, algebras Bp P ( w ) for p 5 a, card Bp < K and functions ( h p : p < a) defined on K , such that hp enumerates all gaps L E V('@) of the algebra B p and all forsings Q in V('@), with card Q < IE. We assume that each hp enumerates with repetitions (that is s u p h j ' ( ~ )= K , for every value a of hp). Note that our assumption 2<" = K ensures that there is, at every stage, at most K (canonical names of) objects to be enumerated. Now, define Pa+l = Pa

* Q,(L)

if a = j ( p , ( ) and L = hp(<)is a gap in Bp and Q,(L) is a Pa-name of a separating forcing determined by L,

P,+l = P,

* Q,

if a = j ( p , () and Q, = h p ( ( ) is a Pp-name of a forcing such that Pa IF " Q , is a-linked",

P,+1 = Pa

*C

in all the remaining cases (C is a Cohen forcing).

Define also X, as the set added by Q,(L), if we force with a separating forcing, and let B,+1 = B,[X,]. Finally, let h, be an enumeration (with repetitions) of all gaps in Ba+l and all forcings of power < /c in V('Q).

At all limit stages Q we take direct limits and h, is defined as above. K , cofinal in K , such that It follows from this definition that there is a set r at every stage LY E r we force with a Q,(L) and add a set X, separating a gap L in the algebra 8, = [ X p : /3 < a and p E r],generated by all earlier generically added P, has the ccc, every K-generated gap (i.e. sets X p , ,d < (Y.Hence, if only P = a gap generated by less than K sets) in the algebra B = Urn<" B, = [X,: a E r] is separated. (More exactly, the above remarks refer to algebras B,/fin and B/fin, respectively, since gaps are understood in the sense of almost inclusion El).

Ca<"

It follows also that if C = K \ I', then also supC = K and at every stage a E C we force with a a-linked forcing and each a-linked forcing of cardinality < K in V[G] occurs at some stage a E C. Hence, if P = C P, has the ccc, the axiom MA(a-linked) and c = K > w1 hold true in the model V[G]. Assume, for the moment, that P has the ccc and let us prove that every Boolean algebra A in V[G], of cardinality 5 c = K , is einbeddable into B/fin C P(w)/fin.

$3.

O N T H E PAROVICENKO

THEOREM

187

There is a decomposition A = U a < r A u , into a continuously increasing chain of subalgebras, such that A0 = {0,1} and Au+l = A,[a,] for some generators a,, all Q < IE. Assume, that we already have an embedding

f : A, such that f(ac) = X,,

for all {

4

B

for s o m e a

< IE,

< a. It suffices to extend f to A,+1 (we take unions

of increasing chains at limit stages). To do this we shall use the Sikorski theorem (see 7.4, Chapter 1). Clearly, the image L , under the mapping f, of the gap

has cardinality less than

2 sup{yt : ( equivelences

y

<

K

and hence L is separated by an X , for some y E

a}. According to the Sikorski theorem we check the following

5 a, 3.2.

r,

=

f ( z ) c*

x, for x , y E A,.

and

f(y)nX,=,Q,

y.aa=O

The implications to the right are obvious (X, fills the gap L ) . The converse implications can easily be obtained from the following remark on a forcing Q(L) determined by a gap L = ( S , T ) . Recall that the elements q E Q(L) are of the form q = ( u q , z q , w q )where , uq S and wq T are finite sets and zq : I ( q ) 4 ( 0 , l ) is a finite zero-one sequence (see Sec. 5, Chapter 2).

c

c

Let A c w be in the ground model, n E w

3.3

and X

c w be

a generic set added by Q(L). If q It- ” A \ n C X ” , for some q E Q(L), then A c , u u q . Similarly,if q I t - ” ( A \ n ) n X = ( P t h e n A C , U w , .

Indeed, if i 2 I(q) and i 4 U u q then there is a p 5 q, with z p ( i )= 0 and hence p It- ” i 4 X ” . This yields non q It- ” A \ n X”, if only i 2 n. The second assertion is proved in a similar way. Now, to finish the proof of 3.2, suppose that f ( z ) El X , holds in V[G]. Thus, we have V[Gly+l] ”f(z)\n X,” for some n E w , and hence q It- ” f ( z ) \ n X,”

c

c

188

cH.8 THEMARTINAXIOM

for some q E Q , ( L ) . An application of 3.3 yields f(z) C, f(11) U ... U f(z,), for some 11,. ..,2, 5 a,, which implies z 5 a,. The second implication is proved in a similar way. P, has the ccc. For the proof It remains to show that the iteration P = CaCr of this property we need some auxiliary lemmas. For any finite functions s : dm(s) + {-1, l},where dm(s)

~ ( =4

n

s ( ~ ) . ~=c

€Edm(a)

n xCn n

c l?, denote

( w W.

a(v)=-l

.9(()=+1

Thus, the algebra B, consists of finite unions X(sl)U...UX(s,), where dm(si) C Q for i = 1,.. . ,n. Hence, we may assume that all gaps under consideration consist of sets of the form X ( s ) or, even simpler, of sequences s. Let E,

c P,

consist of all p E P,, for which:

r n supp (p) there are Pa-names uc, ze,we such that

for each ( E

PI (

It- ” p ( ( ) = (u$,zf,wf)” and dm(s)

c

supp (p) for each s E uc U wEl

there is a number I = I(p) such that each sequence

6E

I$

has length I, for all

SUPP ( P ) n r. Since for every p 6 C(= n \ I’) we have Pg IF ” Q g is a a-linked forcing”, we can Kg,of a function, such that

choose a Pg-name

Pb It- ”Kg : Q g Let

E:

c E,

w A Vn E w[ICa’(n) is linked]”

be the set of all p E E, with the property:

for every ( E C n supp (p) there is an n E w , such that pl( IF

”I<&(())

= 12”

3.4. LEMMA.For e v e r y p E P, and m E w there is a q 5 p such that

q

E Et

and I(q) 2 m.

PROOF.By induction on a. If Q is a limit ordinal, we apply the lemma to a PIP, where p < o and supp (p) c p, to obtain an r 5 PIP, P E Ei and I(r) 2 m. Clearly,

$3. ON THE PAROVICENKO THEOREM

189

the element q = r * 1 (i.e. a unique q E P, such that qI,B = r and q ( q ) = 1, for all q , p < q < a) is as required. Thus, the only essential case is that of a = p 1 and p E supp ( p ) . Suppose that p E C. There is an r1 5 p i p such that

+

r1 It- "h'p(p(p)) = n"

for somen E w .

By the induction hypothesis, there is an r2 5 rl such that r2 E E; and l(r2) 2 m. Hence, the element q = r2 * p ( p ) E P, has all the required properties. Finally, let p E r. Since, the relation It- "p(,B) E Qp" holds, we can find an r1 5 pIp such that r1 11- " p ( p ) = ( u p , z p , wp)" for some PB -names u p , 2, wp. Furthermore, we may assume that U{ dm(s) : s E up U wp} supp ( r l ) (setting segments of the X's on suitable places). By the inductive assumption, there is an r2 5 r1 such that r2 E E; and l(r2) 2 max { m , l ( p ( p ) ) }We . have then r2

It- " X ( s )n X ( t )

l(p(p))"

for each s E u p and t E w B ,

and 1-2 determines all the segments X e n l ( r z ) ,for E dm(s), s E u b u w p . Therefore, we can define an extension 2 of length I(r2), such that r2

1- " ( u p , I,wp)5 (up, 2, wfl)".

Thus, the element q = 7-2 * ( u f l , z , w f l is ) 5 p , and obviously q E EZ and we have I(q) = I ( f - 2 ) 1 m.

QED

3.5. LEMMA.If p IF " X ( s )E fin", where p E E , and the ordinal y = max dm(s) is < a, then there is a q 5 p , q E E , and I(q) = I(p) and such that if qIy II- "q(7) = (u:, z:,w:) then we have q It- "sly E ui", if s(y) = -1 and we have q It- sly E w;", if s(y) = +1.

PROOF.By induction on a. The case of a limit a is obvious, so let us assume

+

+

that a is successor and even a = y 1, since we have ply 1 II- " X ( s ) E fin". Suppose e.g. s(y) = -1, (the case of s(y) = +1 is treated symetrically). Thus, in V(',+l) we have X(sly) C, X , and therefore X ( s l y ) n X ( t ) =* $!j, for any t E w;. Let us define w = { t E w: : sly U t is a function}. For t E w: \ w , we have P, II- " X ( S ~ Y n)x ( t ) = p.

If t E

w , then X(sly) n X ( t ) = X(sly U t ) and yt = max dm(sly u t ) is

< 7.

CH.8 THEMARTINAXIOM

190

Thus, ply It- " X ( s l 7 U t) E fin" and, by the induction hypotesis, there is an

I ply, rt

E,, I(rt) = [(ply) = l ( p ) and rt forces that (sly U t)lyt is in the left or right part of rt(y1) depending on wether the value of SIT u t at 7t is -1 or +1. In rt

E

either case

Repeating this for each t E w we find an r I ply, r E Ea, 1(r) = l ( p ) such that r forces (*) for each t E w . It follows that

and consequently, q = r * (up' U {sly}, xa,wa) is well defined and has all the required properties. This finishes the proof. QED The next (and the last) lemma states a useful sufficient condition for compatibility in P,.

3.6. LEMMA.Let p l q E EZ be such that l ( p ) = l ( q ) and for each ( E supp (p) n supp ( q ) the following hold

) n" and ql( It- "KE(q(t))= m" then n = m, if ( E C and pIt It- " K € ( p ( ( )= and

if t E = XE P g*

r and p l t It-

" p ( ( ) = ( u $ , x $ , w $ )and ~ ~ q l t IF " q ( 0 = ( u $ , x $ , w $ ) "then ,

XE

Then p and q are compatible and we may assert that there is an r r E Ea and I(r) = I(p) = I(q).

5 p , q such

PROOF.By induction on a. Clearly, the only essential case is (Y = p + 1, where p E supp (p) n supp (q). Let 1 denote the common value I = l ( p ) = l ( q ) . Assume that p E C. By the inductive hypothesis, there is an r1 5 p i p , qIp such that rl E Eg and l ( q ) = 1. By assumption, we have

whence there is a Pg-name T such that rl It- " T 5 p ( p ) A T 5 q(/?)", (because every counter image K i ' ( n ) is linked). Thus, if r = r1 * T , then r 5 p , q , and r E E, and I(r) = l ( q ) = 1 (it may happend that r 4 E:, since r1 need not determine the value

K s ( 7 ) )*

$4. GAPS Now, suppose that

P

E

191

r and that PIP

g,

It- " P ( P ) = pa.(,

Wpa)"

and

QIP IF " d P ) = (u$xpa, By the inductive assumption, there is an =

I(T1)

Wpa)".

E Ep such that

rl

rl

5 PIP,

qlP and

M P ) = 1.

Clearly, we have rl

It- " X ( s )n X ( t ) =*

p

for any s E

U uf and

16;

t E wi uw f .

Applying the lemma 3.5 for rl and each X ( s U t ) such that s U t is a function (for, otherwise rl It- " X ( s )n X ( t ) = p)we obtain an r2 5 r l , with r2 E Ep and l(r2) = l(r1) = 1 and such that (s u t ) l y , where y = y ( s , t ) = max dm(s u t ) , is in uT2 or in w&, depending on the value of (s U t ) ( y ) . In any case, this implies that

X ( S )n X ( t ) = X ( S u t ) c and therefore

r2

I(r2) = 1

It- " r E QD", where r denotes (u,"u u f , 28,wpa u wf),

where zp =

z t = zt.

Hence, if we set r = r2

* r , then r has all required properties.

QED Now, we can finish the proof of the theorem. Suppose that some P, has an uncountable antichain C. Since E: & P, is dense (Lemma 3.4), we may assume that C EZ and that 1(p) = 1(q), for all p , q E C. Applying the A-system lemma to the family { supp ( p ) : p E C} we find a common root, on which all assumptions of the lemma 3.6 are satisfied for uncountably many p E C. This yields a contradiction and the theorem is proved.

54. GAPS. We know already (Sec. 5 , Chapter 2) that, assuming the Martin axiom, every ordered gap in P(w)/fin of type (.,A), where w 5 n, X < c and cf(6) # w1 or cf(A) # w l , is separated. Here, we shall discuss the remaining case of

types (c,c) and ( c , ~ ) .

cH.8 THEMARTINAXIOM

192

In fact, we prove that the existence of non-separated ordered gaps, of types mentioned above, is both consistent and independent of MA. Thus, we consider here ordered gaps in P(w)/fin of the form

L = ({A, : where S = { A , : a < K}and sequences such that A , n l 3 p

=.T

a < K)}; { B , : r]

< A}),

= {B,: q < A } are increasing (in the ordering Qfor any a < n and p < A.

If c f ( L ) = w1 (i.e. c f ( ~ )= cf(A) = subgap K of type (w1 ,w1), e.g.

wl),

then the gap

c*)

L contains a cofinal

where (at: [ < w1) and (a, : 7 < w1) are arbitrary increasing sequences of ordinals satisfying sup{ac : ( < w l } = n and sup{B, : 17 < w l } = A. In the proof below we use forcing of the form E(L) adding uncountable antichains to a separate forcing Q(L),(see the lemma 2.3). 4.1. THEOREM. (Kunen). There is a model of the Martin axiom plus a large continuum in which every ordered gap of type (E, E) or (c,w1) is separated.

PROOF.Let V be a ground model such that V K

CH and, for some regular

> w1, the diamond principle for H(n) = Ua
H(K) the set

For example, V can be taken as the contructible model and K - as any regular cardinal > w1. We define a finite support iteration (Po : a < K ) with the ccc, by induction as follows: P, = Ca
= Pa * E ( L )

(direct limit) for any limit a > 0 if cf(a) = w1 and Ta E V(‘0) is such that P, Il- ”T,has the ccc” if ~ f ( a = ) ~1 and Pa ll- ”T, is an unfilled gap and cf(T,) = w1” and L is a cofinal subgap of T, of type (w1, w l )

193

$4. GAPS

P,+1 = P,

*C

in all remaining cases (C is a simple Cohen forcing).

From this definition and the lemma 2.3 it follows that the forcing P = satisfies the ccc. Let G C P be a generic ultrafilter. Clearly, we have

P,

V[G] It- "MA and c = K " . Suppose that there is in V[G] an unfilled gap L = ( { A p : a < c} ; {Bg : p < c}). We shall derive a contradition. Let us fix a pair 1 = ( a ,b) of canonical P-names of the lower and the upper class of L,respectively. Thus, a consist of some triples ((a,A)('),p), where A is a P-name of a subset of w, and

4 % A ) = { P E p : ( ( a , A ) , p )E a } is an antichain (and hence is countable). Clearly, (cf the proof of the theorem 2.5), the set

N , = (7 < K : cf(7) = K is an wl-cub in

K.

A V ( a ,A )

E dm(a) n V ( p ~ ) [ ~ A ( a) , P,]}

Hence, the restrictions

are P,-names such that

.,[GI71 = 4 G I n V[Glrl = = { A , : a < K } n V[Glr]

for all 7 E N,.

Doing the same for b we infer that for any 7 E N1 = N , n Nb the part L n V[Gl7], of the gap L in the submodel V[Gly], belongs to V[Glr]. Moreover, the set

Nz = (7 < K : cf(r) = w1

and P, IF "1, is unfilled"},

where I , = (a,, b,), is also an wl-cub. Indeed, no set z C w in V(',), where a < K , separates all Zg, p > a (since Up,,Lg is a cofinal subgap in I ) . Thus, there is a /3 = P(z) < K such that z does not separate l a ( = ) . Since c < K in V('*), we have

a* = sup{/3(1) : z E w A I E v'p")} < 6.

+

Iterating this w1 times, i.e. defining $0) = a , -y([ 1) = 7([)* and 7(X) = ~ u p ~ , ~ r ( [ for ) , limit X we obtain an ordinal y = sup{r([) : [ < wl}, which is

cH.8 THEMARTIN AXIOM

194

> a and belongs to Nz. Obviously, Nz is wl-closed and the claim is proved. Also, we have seen earlier, that the set

is an wl-cub (by a similar argument as for N1). Thus, the set N = N1 n Nz n N3 is an wl-cub and by the diamond principle applied to the names a,b H(n), we can find an arbitrarily large 7 < n such that T., = I, and P , IF , 'Z"

is unfilled A cf(T,) = wlll.

It follows that at the 7-th stage, we force with an E ( K ) ,where K is a cofinal subgap of T,[Gl7] = L n V[Glr]. Thus, the forcing Q ( K ) has an uncountable antichain in V[Gl7 1) and hence in V[G], as well. On the other hand, some A, separates K in V[G] (e.g. a = sup{@: AD E V[Glr]} l),and consequently, Q ( K )has the ccc (see the remark 2.2), a contradition.

+

+

The case of gaps L of type (c,w1) is even simpler - we can leave unchanged the upper class of L,since it has the order type w1. This finishes the proof. QED Now, we want to prove that, consistently, there may exist unfilled gaps of types 4.1 and 4.2 below are due to K. Kunen (unpublished), (c,c) and ( c , w ~ )Theorems . but the proofs are different from that of the autor's. 4.2. THEOREM. (Kunen). There is a model of the Martin axiom plus a large

continuum, in which there exist non-separated gaps of types (c, c) and (c, w l ) .

PROOF.In a ground model V satisfying CH we choose a regular cardinal n > w1 fulfilling 2<" = n. We apply a Hechler forcing (Sec. 5 , Chapter 2) to obtain a generic extension V1 of V, in which there exist unfilled (tc,tc)-gaps (or (n,wl)-gaps or both). Clearly, we have V1 "c = n: and 2'" = 6". Now, we use an ordinary iteration ( P a : a < n) which forces MA. At each stage a we choose an enumeration e, = { e m ( ( ): ( < n} of all (representatives of) forcings in V('*) such that P a IF "V( < tc[card e,(() < K ] " . Let j :

IC

xn

n be a fixed pairing function satysfying j(a,p)> a

for each a

< n.

§4.GAPS

195

We define Pa by induction as follows: (the direct limit) if a is a limit Pa = Pp Pa+l = Pa * eC(9) if Q = j ( [ , 9) and Pa IF "eC(9) has the ccc" Pa+l = P a * Cohen in all remaining cases.

Pa be a generic It follows that card Pa < n, for all a < n. Let G c P = Ca<s filter. We know that Vi[G] "MA and c = K " . We assert that any unfilled gap L E V1 of type (n,n) or (n,w1)remains unfilled in Vl[G].Indeed, let e.g.

L = ({Aa :

< K } ; {Bp : p < K } )

and suppose that a set S c w separates L in Vl [GI.Since P satisfies the ccc, there is Vl[Glr].There are integers n,,mp E w such that

a 7 < n such that S E

Aa \ na C S and S n (Bp \ mp) = $9 for all a,p < n. Hence, for some sets X, Y E Vl [GI?] of cardinaly n: in V1 [GI71and some n, m E w we have n, = n and mp = m for every a E X and p E Y. The sets X and Y need not be in V1 but we can write

X , = { a < n : p IF " a E X " }

for p E P,,

and then X = UpEGl,X p and X p E V1, for each p E P. Since card GI7 5 card P, < 7, there is a p E G1-ysuch that card X , = n and then the set A = UaEX,A , is in V1 and A c S. Similarly, we can find a g E GI7 such that the set Yq = { p < n : g It- " p E Y"} is in V1 and has cardinality n. Then we have

(Bp \ m ) n A = $9

for each p E Yq,

since A S. Thus, the set A separates in Vl the gap ( { A a : ct E X,}; {Bp : p E Yq}), which is cofinal in L and therefore, A separates L as well. This yields a contradition and the proof is finished. QED

cH.8 THEMARTINAXIOM

196

OF C ( X ) . Let X be a compact space. The linear ring 55. HOMOMORPHISMS C ( X ) ,of all continuous functions, is a Banach algebra with the uniform norm

for f E C ( X ) . Suppose that h : C ( X )+ B is a linear homomorphism from C ( X ) into a Banach algebra B. We know (Ex.1.2, Chapter 6) that in the case of B = R or B = C(Y) any such a homomorphism is automatically continuous. Kaplansky posed a problem, if it is so in the general case (i.e. for an arbitrary B). Thus, let us formulate the following statement:

(K)

For any compact space Xand a Banach algebra B every linear homomorphism h : C ( X ) + B is continuous.

Assuming CH, Dales with X = p[w].

[D]and Esterle [Ez]constructed counterexamples for K,

On the other hand, Solovay proved the existence of a model in which K holds. Thus, the property K is both consistent and independent (of set theory). Here, we present (cf Jech [J])the Solovay's construction simplified by Woodin. We shall use the (strict) dominance ordering

f <* g

=

{ i E w : f ( i ) 2 g ( i ) } E fin

and orderings determined by ultrafilters cp E W *

f

<+,g

=

{ i E w : f ( i )< g ( i ) } E $2.

More exactly the relation <+,is a linear ordering on the set 3+, = wwf =+,,of equivalence classes of the relation

f

=vg

I

{ i E w : f ( i ) = g ( i ) } E cp.

For any function g : w -+ w let 3+,lg denotes the initial segment 3 v l g = {f : f We shall make use of the following theorem which we state without proof.

<+,g } .

(Woodin [DW]). If, for some compact space X, the algebra 5.1. THEOREM. C ( X ) has a discontinuous homomorphism (into some Banach B), then there exists

$5. HOMOMORPHISMS OF c(.x) an increasing function g : w -t w and an ultrafilter segment Fvlg can be order embedded into (w", <*).

Q

197 E

W*

such that the initial

In fact, g can be taken as the identity g = id. Let T be the set (0, l } w lordered lexicografically: for a , b E T let a


a7

< b.,

where 7 = min{a < w1 :

a,

# b,}.

We shall prove first that, assumnig MA, the set T can be order embedded into any segment F+,Ig, (for arbitrary Q E w* and any increasing g : w + w). Then, to prove the consistency of the hypothesis I< it is sufficient, in view of Woodin's theorem, to construct a model in which MA holds and there is no order embedding of T into (w", <*). Suppose that we are given an ordered gap L , of type (w1 ,w1), in ( w w ,<*):

Thus, the sequence (f, : a < w1) is increasing and (gp : p the ordering <, and we have also

< w1) is decreasing in

5.3. LEMMA.Assume MA and c > w1. Every gap L of the form 5.2 is 9-separated, for each Q E w * , i.e. there is an h such that

PROOF.Let P consists of elements of the form p = ( u p ;xp, yp), where u p w1 is a finite set and x p r y pare finite sequences x p r y p: I ( p ) -t w such that for every i 2 I(p) we have

where a1,< . . . < a,,are all elements of u p , in increasing order. The ordering on P is defined thus p 5 Q if and only if u,, u p ,i pC_ i p , y g yp and we have f,(z) < xp(i) < g,(z)

c

or f,(i)

c

< yp(i) < g a ( i ) for all i, l ( q ) 5 i < I(p) and Q E u,,.

It is easy to see that P satisfies the ccc and even it is u-linked. Indeed,

cH.8 THEMARTINAXIOM

198

is a partition into countably many linked parts. For, suppose p = ( u p ; z,y)and q = (up; z,y) are taken from one part and let 1 = l ( r ) be so large, that for i 2 1 the property (*) holds, where a1 < . . . < an is now an enumeration of u p U up. Then, the sequence x has an extension xr, of length 1 such that

fu(i) < zr(i) < g u ( i ) for all i , l ( p ) 5 i < 1, and a E u p . Similarly, y has an extension yr of length 1 such that fu(i)

< yr(i) < g u ( i )

for all i , l ( q ) 5 i

< I , and

(Y

E up.

Hence, the elemnt r = (up U u q , zr,yr) is in P and lies below p and q , which proves the claim. It is also easy to check that the sets

D: = { p E P :

a E u p and l ( p ) 2 n}

for n E w and a

< w1,

are dense in P (for a given q E P we extend first xp to have a suitable length, depending on a,and then adjoin a to u q ) . By the assumption there is a filter G intersecting all the sets DI. Therefore, we obtain functions

P

h, = U{zp : p E G} and h, = u { y p : p E G}. Clearly, for arbitrary 5.4.

Q

< w1 we have

f d i ) < h z ( i ) < g u ( i ) or fu(i) < h,(i) < g a ( i ) ,

for large enough i E w (i 2 min{l(p) : p E G and a E up}). Now, if Q E w* is any ultrafilter, then one of the above inequalities 5.4 holds on a set from Q. Thus, we have (for a given a) 5.5.

fa
h=
or

fu


Let h = h, or h = h, depending on which of the relations 5.5 holds for uncountably many a. Obviously, we have then fu
h
for all a,P < q ,

$5. HOMOMORPI[ISMS since ft

OF

c(x)

199

h implies ft <,+,h. Thus, the function h 9-separates the gap L.


QED

Now, we can describe an order embedding of T into a segment .Fvlg, using the results of Exs. D, Chapter 2. Let S = (0,l}clbe the binary Cantor tree of height w1. To each s E S we assign a pair of far away functions fa <. gal (i.e. limi If 8 ( i ) - gs(i)l = 00 by induction on the length of a sequence s, as follows. For s = 9, we take any pair of far away functions below g: fv <. gg <. g. It fa <. gs are given, then for the extensions s * 0, s * 1 we find four functions between fa and gs in the following order

and such that any two consecutive terms are far away. Thus, the splitting goes according to the following scheme

If s has a limit length a,then we have an ordered gap

of countable cofinality. We know that L is filled and as the pair fs <. gs we take any far away pair separating L. Thus, the family { (fs, gs) : s E S} has been completely defined. Suppose now, that a E T, i.e. a is a branch of the tree S. Thus, a determines a gap L a = (tfalt : €

<W l l

; {Salt, :

17

< wl)

in (ww,<.), of type (w1,wl). By the lemma 5.3, there is a function ha 9-separating the gap La,(for any given E w * ) . If a < b, in T (i.e. lexicografically) then we have a y = 0 and b, = 1 where -y = min{a < w1 : a , # b,}, and consequently

cH.8 THEMARTINAXIOM

200

Therefore, the mapping a c-) ha is an order embedding of T into Fvlg. We introduce yet a counterpart of the forcing E (cf Lemma 2.3). Let us notice first, that if a given gap L = ({f,}; {gp}) satisfies the condition for every a (HI

a#

p, then for some i E w eitherf,(i) 2 y p ( i )

fp(4

or

< wland i E w , f , ( i ) < g,(i) and if

2 gdi),

then L cannot be separated. Indeed, suppose that h separates L. Thus, there are integers ka, for a < w1 such that

The function a c-, k, must be constant, k, = k on an uncountable subset Y Hence, we have

E w1.

" , is an Since Y = UI{a < w1 : f,lk = s}, where s runs over U n E w ~ there uncountable YOC_ Y such that the initial segments f,lk are identical each other for a E Yo. Similarly, we find an uncountable Y1 Yo so that gplk is constant for p E Y1. But then, for any i < k we have f , ( i ) < ga(i) = gp(i) and fp(i) < gp(i) = g,(i), for arbitrary a,p E Y1. This contradicts the property H, which proves the claim. Secondly, let us note that the property H is absolute: if a gap L satisfies H in a given model, then L satisfies H in any bigger model (and hence, L remains unfilled). Now, let us define E = E(L) as the set of all elements of the form ' w1 is a finite set and the functions f:,gb, satisfy H restricted to a E U, and are equivalent to the corresponding functions in L, (i.e. f: =* fa and gb, gal for each a E U e ) . The ordering on (E(L) is the inverse inclusion. e = (f' @,g, : a E .Ye),where Ue

=.

Let L' = ({f:};{gb}) gb, =* g a l for each a

be another gap. We write L'

=.

L , if f:

=.

fa

and

< w1.

5.6. LEMMA.If L = ({f,}; {gp}) is an unfilled gap, then the forcing E = E(L) has the ccc and E Il- 113L'(L' =* L and L'satisfies H).

c(x)

$ 5 . HOMOMORPHISMS OF

201

PROOF.If e l I ez, then there exist a,/3 E U,, A Ue, such that fA(i) < g b ( i ) and fb(i) < g & ( i ) , for all i E w . Therefore, if {e, : a < w1} is an uncountable antichain, then - in view of the A-system lemma - we may assert that the sets U,, are pairwise disjoint and in addition, that max U,, < min Ue,

whenever a

< p < w1.

Let us define

fa(i) = max EEu., f{(i)

and i p ( i ) = rnin g : ) ( i ) rlEU,,

for i E w .

Then, we have

f,(i)


for all i E w and a , @< w l ,

and hence the function h ( i ) = f,(i) separates the gap L , contrary to the assumption. Thus, we have proved that E ( L ) has the ccc. To prove the second assertion, let G c E(L) be a generic filter. Clearly, the sets

are dense in E. Hence, the gap L' = ({f;}; {gb}), determined by G in a n obvious way, has the property H and also L' =* L. QED Now, we can prove the announced theorem.

5.7. THEOREM. (Solovay). There is a model of MA plus c > w1, in which the hypothesis K is true. PROOF.According to the remark following 5.1 it is sufficient to construct a model of MA and c > w1, in which there is no order embedding of T = (0,l}w* into 7 = (ww,<*). We choose a ground model V so that CH holds in V and the cardinal n = 2w' is large and n
c H(K)

(3.8TIIEMARTINAXIOM

202

Such a model V can be obtained from the constructible model by lifting the cardinality 2"' up to a regular K 2 2w' and then forcing the diamond for H ( K ) , (comp. the proof of the theorem 2.5).

s

In V, let D T be the set of all a E T, which are constant from some point on. Thus, card D = c = w1 and D has 2"' = tc gaps, since every element a E T determines a gap La = ({z E D : I < a } ;{y E D : a < y}) and 2' = 2"' = K . Define by induction a finite support iteration (P,, : a < K ) as follows:

pa = CB<,PB Pu+l = Pu * Tu Pu+l = P a

* E(T,[L])

P,+l = P,

*c

(the direct limit) if cy is a limit ordinal, if cf(cy) = w1 and Ta is a P,-name of a forcing and P, It- "T, has the ccc", if c f ( a ) = w1 and T, is a P,-name such that P, It- 'IT, : D + 7 is an order embedding" and L is a P,-name such that P, IF " L is an unfilled gap in D and T,[L] is unfilled in F" (where C is a Cohen forcing) in all remainig cases.

Let G 5 P, where P = C,
and

E

= K".

Suppose now that in V[G] there is an order embedding of T into 7 ,and consider its restriction f on the subset D. We shall derive a contradition. Since V[G] /= "card D = card 3 = K " , we can choose a canonical name f

H ( K )of , the mapping f. Clearly, the set

N1 = {a < K : cf(a) = w1 and Vz E dm(f) - n V'P"'[f(z)

E Pm])

is an wl-cub and for each (Y E N1, the set f, = f n (V('0) x P,) is a P,-name of the restriction f, = f n V[GIcy], (cf the proof of 2.5.) In particular, fa belongs to the submodel V[Gla],for all a E N1. We also check easily that the set

N 2 = { a < t c : cf(a)=wl

fnH,=L} and -

PROBLEMS

203

is an wl-cub. Let N = N1 n Nz and apply the diamond principle for X = f. Thus, there are ordinals a such that -f = T, and a E N , i.e. fa is an embedding in V[Gla] of D into 3. Since at the a-th stage of the iteration we have card.F = c


= card {L

D : L is an unfilled gap },

there must exist a non separated gap L such that the image fa[L] remains nonseparated in 3 (note that the gaps of the form f,[L,] are separated by distinct elements, if at all). This means that we force with the E(f,[L]) at this stage of the iteration. By the lemma 5.6, the gap fa[L] remains nonseparated in the final model V [ G ] On . the other hand, the gap L is separated in V[G] by some a E T ( T is in fact a complete ordering) and consequently, j a [ 3 ]is separated in 3 as well, (because we have an embedding of the whole of T into F ) , which yields a contradiction. QED

PROBLEMS

A. LIMITSAND ANTICHAINS 1. Let R be an almost disjoint family (of subsets of w ) . The alsmost disjoint forcing S = S(R) consists of elements of the form p = (R,,,xp), where R, C R is a finite subset and z p : Z(p) -t (0, l},for an Z(p) E w . The ordering on S is defined as follows x p and for each i, I(q) p 5 q if and only if R, 2 R,,, x g condition i E U R, implies z p ( i )= 0.

5i <

I(p), the

a) S is o-linked. b) If G

S is a generic ultrafilter, then the set Xc

= {z E w :

3 p E G ( x p ( i= ) 1))

is infinite and almost disjoint from each A E R. c) Let S, be tc-fold iteration (with finite supports) of S. Thus, at each stage a < K of the iteration we add a new set almost disjoint from all sets added at earlier stages. If c f ( k ) > w , then we have S, IF ” There is a maximal antichain of cardinality

K

in P(w)/fin”.

CH.8 THEMARTINAXIOM

204

Hence, maximal antichains, of a given cardinality, can be added step by step. 2. Let L = ( A , : (Y < y) E V be a descending chain (of subset of w ) . Define a forcing Q = Q(L)as the set of all elements p = ( u p ,z p ) ,where u p y is a finite set and z p : I(p) + (0, I}, for some I(p) E w . The ordering on Q is defined thus p 5 q if and only if uq E u p r z qC z p and for each i, I(q) A,. condition z p ( i )= 1 implies i E

naEup

a)

5 i < I@), the

Q is a-linked Q is a generic ultrafilter, then the set

b) If G

X G = {i E w : 3 p E G(zp(Z) = 1)) is infinite and Xc C, A,, for all n

< y.

c) If Q 6 , where 6 E V is an infinite ordinal and c f ( 6 ) > w , is a &fold iteration of Q with finite supports, then we have &a It-

”

There is a limit of length 6 in P(w)/fin”.

Hence, limits in P(w)/fin, of a given length can be added step by step.

3. Let Db be &fold iteration of a dominating forcing, so that at each stage we add a function dominating all previously added functions. If c f ( 6 ) > w then we have D6 It-

B. PRODUCT

OF

”

There is a limit of length 6 in (ww,I*)’’

FORCINGS

ni,,

1. Let { P i : i E I}be a family of forcings. The forcing product Pi is defined as the subset of the ordinary product, consisting of all p = (pi : i E I),for which the support {i E I : pi c 1) is finite. a) A forcing product Pi,

x

x Pi,

niE,

has the ccc if and only if each finite subproduct has the ccc. Pi

b) For each i E I the map z H p z ( z E P i ) , where

PROBLEMS

205

is a complete embedding of Pi into P. c) If G P is a generic ultrafilter, then Gi = { p i : p E G} is a generic ultrafilter in Pi and V('i)[Gi] V(')[G], (for any i E I).

2. Let K > w1 be a regular cardinal of a ground model V and W C_ K a set of cardinals of cofinality > w , W E V . If G C_ P = n x e w P , is a generic filter, then we have in V[G]: a) If PA= Sx (Ex. A l , c), then there are maximal antichains of each cardinality

XEW b) If PA = &A, (Ex. A.2, c), then there are limits in P(w)/fin of each length X<

w.

c) If P A= Dx, (Ex. A.3), then there are limits in (w", 5.) of each lenth X E W .

3. There is a model, in which c > w1 is regular and for every cardinal X 5 c, with cf(X) > w , there exist limits in P(w)/fin (and in w " ) of length X and maximal antichains of cardinality A.

c. ADDINGCOHEN REALS 1. Let K > c be a regular cardinal in a ground model V . If C ( K )is the &-foldCohen forcing and G C C ( K )a generic ultrafilter, then we have: a) If in V[G]there exists an increasing chain F = (fa : a < K ) in ww or in P(w)/fin (for fa : w + {O,l}), then there exists a 7 < K and a subchain (fa( : E < w1) E V[Glr], which dominates the set

(Define f,( by induction). b) There is a function g E V[Gl-y],such that 9

fa,

I* g , for all < w1, and also

I* f,.

Hence, in V[G] there are no chains of length

K.

c) Every limit in the model V remains a limit in V[G].

+

2. Let V CH and let & , A be two regular cardinals in V such that There exists a generic extension V[G] of V such that:

w1

< IE < A.

206

c H . 8 TIIE

a) For every n' 5 n, with w w ) of length n'.

c~(K')

MARTINAXIOM

> w , there exist limits (in P(w)/fin and in

b) There are no limits (chains) of any length n' satisfying n < n' 5 A.

3. Let CH hold in a ground model V and choose some regular X > n > w l , in V . There is a generic extension V [ G ]such that V [ G] "c = X" and:

+

a) There are limits in P(w)/fin and in ww of length n: and A.

b) Every limit (of a regular length) has length either n or A. 4. Let V be an arbitrary ground model. Then in the model V ( c ( w l ) ) [ G(i.e. ] , after

adding w1 Cohen reals), there is a limit of length

D. COVERINGS

w1.

WITH NOWHERE DENSE P-SETS

Let B C w be an infinite set. If the set {i E B : f(i) 2 g ( i ) } is finite, then we g , on B. say that f 1. Any unbounded family F E w w , consisting of nondecreasing functions, is unbounded on every infinite set B w . (If g : w --+ w , then the function gB, defined by setting g B ( i ) = max { g ( j ) : j 5 min{lc E B : k 2 j}}, does not dominate the family F ) . 2. Let X

> IE > w be regular cardinals. If the rows of a matrix {U,p :

(Y

< n and /3 < A } ,

where the U's are open-closed in w * , are decreasing and the columns -increasing limits, then the intersections

are closed nowhere dense P-sets, such that

3. Let (fa : a < n) and (go : /3 < A) be two limits in ww and assume that the lengths n, X are regular and K < A.

P no B L EM s a) We may assert that f, non

g p , for

207

all a

< IE

and p

< A.

b) If A,@ = { z E w : fa(;) 2 g p ( z ) } , then the sets U,p = A:p form a matrix as in Ex. 2. (Use Ex. 1).

Hence, it is consistent with set theory that c > w1 and there is a covering of the space w* consisting of closed nowhere dense P-sets. (Kunen proved that there is no such a covering, if CH holds).

E.

SEPARATING FORCING

1. Let L = ({A,};{Bp})be an ordered nonseparated gap of type (w1,wl) in a ground model V. If Q = Q ( L ) is the separating forcing determined by L and E = E(L)is as in Sec. 2, then we have E * Q IF " w y is countable".

2. Let L = ({A,}; { B p } )be an unfilled gap of type (w1,w1) in a ground model V. If Q is an arbitrary forcing satisfying Q IF " L is separated",

then we have: a) There is a Q-name f such that

b) The function g ( a ) = min{m : 3q E Q ( q It- "f(a)5 m")}is in V and there is an rn E w such that the set { a < w1 : ( A , \ m )n Bp = (B} is uncountable in V . c) There is a forcing P E V, with the ccc, such that P It- "Q has an uncountable antichain".

d) We have P * Q I!- " w y is countable". Hence, the separating forcing Q = Q ( L ) is in a sense unique.

cH.8 THEMARTINAXIOM

208

F. DISTINGUISH SOME

REMAINDERS

If 2w = 2w1, then the remainders w* and w; have equal powers and weigths. 1. If IE 2 w is non-measurable (Ex. D , Chapter 5), then every nonprincipal ultrafilter on IC contains a decreasing chain K = A0 _> A1 _> . . ., with A , = 9. Hence, every P-point p E K* contains a countable set.

nnEw

2. No cardinal IE 5 c is measurable. (If c is measurable, then there is a branch in the Cautor tree of basic sets {Us: s E UnEw{O, l}n},each member of which has measure 1).

3. Assume MA and c > w1. Let homeomorphism. a)

n,

K

<

c and suppose that h :

The intersection h[V,'],where V, = {( P-point (Theorem 6.4, Chapter 3)


: a

K*

"3 w*

is a

< ( < I E } , contains a

b) There is a P-point in IE* containing all cofinal segments V,. This contradicts Ex. 1 and 2, if IC > w . w

Hence, MA plus c > w1 implies that the remainders < IE < c, are not homeomorphic.

w* and

K*,

for any

K,

Chapter 9

Partitions of Antichains

51. PARTITION ALGEBRAS. Let E be a maximal antichain in the algebra B(w*), (or in P(w)/fin). A set U E B(w*) is called a partitioner of the antichain E , if we have e

U

or

e n U = (3

for each e E E .

Clearly, unions U U V , intersections U n V and complements w* \ U of partitioners of E are still partitioners of E. Therefore, the family P = P ( E ) , of all partitioners of a given antichain E , is a Boolean subfield of B(w*). Obviously, the elements e E E , finite unions el U . . . U en, ( e l , . . . en E E ) , and their complements are partitioners of E . We call them trivial partitioners. The ideal [El, generated by the family E in the algebra P ( E ) , consists of finite unions el U . . . U en, where e l , . . .en 6 E . Thus, the factor algebra P ( E ) / [ E ]is called the partition algebra of E. The unit 1 (or a>) represent all trivial partitioners. Other elements (if any) correspond to nontrivial partitioners of E . In an equivalent approach we are dealing with maximal almost disjoint families (mad - for short) R, of subsets of w , rather than antichains. A partitioner of R is then a set A w,such that e

C, A or

e

n A =* (i3 for every e E R.

The algebra P ( R ) , of all partitioners of R, is, in this formulation, a subalgebra of P ( w ) and in order to obtain the partition algebra of R we factorize P ( R )by the ideal [fin,R],generated by R and all finite sets.

Of course, the notion of a partitioner is meaningful for an arbitrary Boolean algebra A and a maximal antichain E A. Partition algebras P ( E ) / [ E give ] us some 209

cH.9 PARTITIONS OF ANTICHAINS

210

insight into the structure of maximal antichains (maximal almost disjoint families), since they describe the separable divisions E = Eo U El, Eo n El = Q, . 1.1. THEOREM. (Baumgartner, Weese [BW]). If CH holds, then every Boolean algebra A of cardinality 5 c is a partition algebra, (i.e. for some maximal antichain E f?(w*), A is isomorphic to P(E)/[E]).

s

PROOF. We shall construct a Boolean algebra B and an antichain E the following properties will be satisfied:

B so that

1. B is a ParoviEenko algebra of power c extending A. 2. Every nonzero element a E A is a nontrivial partitioner of the antichain E. 3. If b E B is a partitioner of E, then b = E a, for some a E A, ( b = E a means

b A a E [El). F’rom the condition 1and the assumption, it follows that B is (up to isomorphism) the algebra P(w)/fin. It follows from 2 that the element a in 3 is uniquely determined. Thus, the condition 2 and 3 define a map h : P ( E ) + A, onto A, which is easily seen to a be homomorphism with the kernel A(h) = [El. Thus, A is isomorphic to P(E)/[E]. It remains to make sure that E is in fact a maximal antichain. Suppose that b E B and b . E = 0, then b is a trivial partitioner and (by 3) so is h(b), whence h(b) = 0 (by 2), i.e. b E [El,which implies b = 0. An algebra B and an antichain E C B satisfying the above conditions can be defined as unions B= B, a n d E = E,,

U

U

a<wt

a<wt

where (B, : a < w1) and (E, : (Y < w1) are increasing chains defined by induction as follows. Bo = A and EO= Q, and B, =

u

Bp

and E, =

u Ea

for every limit

Q

> 0.

@
We describe now the successor step from inductive assumption):

(Y

to

(Y

+ 1.

Suppose that we have (the

(W,) card B, 5 w1 and every element b > 0 of A is a partitioner of the antichain Ea and b 4 [&I. Assume also that at each stage 0 we choose an enumeration (zf : ( < wl) of all

31. PARTITION ALGEBRAS a) nonzero elements of B p , b)decreasing chains bo > bl > . . . in B p , c) countable ordered gaps a0 < a1 < . . .

. . . bl < bo

211

in Bp.

(Note that, by CH, we have at most w1 objects to enumerate) We fix also a pairing function j : w1 x w1 + w1 (one-to-one, onto) such that j((, p) > p, for all E, p < w1. Thus, at the given stage a we have two ordinals E, /3 < a such that j ( [ , p ) = a,which determine an object x = x!. Let X , be the Stone space of the algebra B,. If z is a chain or a gap, we define E,+l = E, and B,+1 = [B(X,),n,,, b,], i.e. 8,+1 is the subalgebra of P(X,) generated by B , (after identification with B(X,)) and the set fin<, b, (the intersection of the chain or the upper class of the gap x). In other words, B,+l is obtained from B , by adding a new element under the chain x or separating the gap z. Obviously, W,+l holds. Now, suppose that z is a nonzero element of 8,. Consider first the case of z E A = Bo. We may claim that there is an atom a 5 z, disjoint from E,. Indeed, the inductive assumption W, implies that the family

F, = { z \

U s : s E fin(E,)}

c B(x,)

is centered. Hence, we can always replace B , by the subalgebra of P(X,), generated by 6, = B ( X , ) and a point p E F,. Then a = { p } proves the claim.

n

+ +

Define Xa+l = ( X , \ a ) (w l),i.e. delete the isolated point (corresponding to the atom a ) and put into its place a one-point compactification of the discrete w , and let B,+l = B(X,+l). Clearly, B, C B,+l (since X , is a continuous image of X,+1) and in B,+1 there are infinitely many new atoms eo, e l , . . . lying under z and disjoint from E,. Define E,+1 = E, U {en : n E w } . Obviously, W,+1 holds true and the element z is from now on a nontrivial partitioner of Finally, suppose that x is in B,

(*I

5

\A

=E a

and x is a partitioner of E, such that for no a E A.

(If z is not as above, then we add only to B , a new atom a under I, a . E , = 0, and set Ea+l = Em). Let us notice that the Boolean complement -z has similar properties and (*) implies that the ideal J E A, generated by the set { a E A : a . z E [E,] or

a - x E [E,]}

cH.9

212

PARTITIONS OF

ANTICHAINS

is proper. Let p be any ultrafilter in A extending the dual filter - J = { - a : a E J } . Then the set n p \ U E, intersects both 2 and X , \z (in the Stone space X , ) . Hence,

we can add to B, two new atoms that eg

+ el 5 a

eo

or eo

5 z and el 5

+ el 5 - a

-2,

disjoint from E , and such

for each a E A.

+

Thus, if we define Ea+l = E, U {eo e l } , then W,+l holds and the element z splits eo + el (and hence z is no longer a partitioner). This finishes the description of the inductive process. Obviously, every element, countable chain and countable gap of B = B, occurs at some (uncountably many) stages. Therefore, the conditions 1, 2 and 3, imposed upon B and E , are satisfied. QED

u,<wl

The proof given above is different from the authors' original one (see [FZ1]).

$2. COMPLETE ALGEBRAS. The assumption of CH in the theorem 1.1is essential - we show here that, consistently with non CH, there exist Boolean algebras of cardinality C, which are not represented as partition algebras. More exactly, we prove that in a Cohen model with a large continuum no complete (infinite) algebra is a partition algebra. Let C = C(w2) be a multiple Cohen forcing consisting of finite functions p : dm(p) + (0, l},where dm(p) G w2.

2.1. THEOREM (Baumgartner, Weese [BW]). If V CH, then in the generic is not a partition algebra. extension V(',)[G] the field B ( ( 0 , l}"z)

PROOF.Suppose that the theorem is false. We shall derive a contradiction. Thus, the field B({O, 1 } w 2 )is isomorphic to some partition algebra P ( R ) / [ R ]where , R is a mad. The basic sets

U, = {z

E (0,

: .(a) = 1) for a

< w2,

form an independent family in B({O,l}"~). Let us choose the corresponding partitioners A, C w of the mad such that

R. By the ccc there are countable sets X ,

A , E V[GIX,] for every a < w2.

w2, X

,

E

V

52. COMPLETE ALGEBRAS

213

By the generalized A-system lemma (Ex.A . l ) we may assume that for some countable d C w2 we have

X, n Xg = d for all a < /3 < w2.

2.2.

Any one-to-one function h : X,

H : C(Xa)

+ C ( X s ) such

-t

Xp, onto Xp, determines an order isomorphism

that

By 2.2, we may require that h ( ( ) = (, whenever ( E d. We claim that we have then H , ( A a ) = Ag, where H. : V(C(xa)) -t V(c(xa)) is the induced isomorphism of names. Indeed, choosing an h, : X , -t X (where X 2 d is fixed) we must have H:(Aa) = H t ( A g ) for w2 indices a,p (since there is only w1 canonical C(X)-names). But then h j ' h , : X , -t X p determines an isomorphism (in the sense described above) and the values p((), for ( E d are invariant.

For every sequence t E

UnEw{O,l}", the set

n

=

t(i)=l

A, n

(W

\Aj)

t(j)=O

is a nontrivial partitioner of R. Every function g : w -t {0,1}, (g E V[G]), determines a decreasing chain (A,ln : n € w ) . Hence, there is an infinite set B, C w such that

B, C, ABln for every n E w . Since R is a mad, we have

B, n e is infinite, for some e E R, and therefore e C, Aglnrfor all n E w , i.e. e

L A,

if t(i) = 1

and

e n A j =, (3

if t ( j ) = 0.

Let us choose such an e corresponding to the function g defined as follows g(i) =

u

G(y

+ i)

where y = sup { sup Xi : i E w }

+ 1.

214

c H . 9 PARTITIONS OF ANTICHAINS

There is a countable set X, (Y < w2 so large that

E

min(X,

w2

such that e E V[GlX,]. We fix also an ordinal

\ d ) > m u { sup X,,

7 +w}.

Such an CY exists, since the family {X, \ d : (Y < w 2 ) is disjoint (and hence the ordinals min(X, \ d ) are arbitrary large). We prove that A, splits e. This will give a contradiction, since e is a partitioner of R. Thus, it suffices to show that the sets

D, = {g E C : 3i 1 n(q It- ” i E e n A,”)) and

En= { q E C : 3i 2 n(q It- ” i E e \ A,”)} are dense in C. Let p E C and n E w be given. Since dm(p) is finite, there is an i E w , such that 7 i 4 dm(p) and Xi n dm(p) d. Let H : C(X,) + C(Xi) be an isomorphism sending A, to A; (and invariant on d). Then, H(plX,) and p are compatible and hence p u H(plX,) E C. Since y 1 4 dm(p u H ( p l X , ) ) we can define an r 5 p U H(plX,) satisfying r ( i ) = 1 (or r ( i ) = 0 in the proof of density of En).Thus, we have

+

+

r It- ”e

c*A,”

and hence r It- “e n A,

It follows that for some r1 5 r and j r1 I(X,

4 fin”

1 n we have rl It- ” j E e n A;”, whence

U Xi) ll- ” j E e n Ai”.

Since r1 IX, = PIX,, we infer that q = rlUH-l(r1 IXi) is in C. Obviously, q It-: j E e”, since q 5 r1. Also H-’(rI IXi) It- ” j E A,”, since rl IXi IF: j E Ai”. Hence, we obtain q IF ” j E e n A,”, which proves the density of D,. In the case of the sets En the proof is similar.

QED

From the proof it follows also that no algebra (in V[G]) containing the field

B({O, l } w z ) is a partition algebra. Indeed, we have used only the fact the algebra under consideration contains an independent subset of power w2.

53. PARTITION ALGEBRA

UNDER MA

215

Let A be a complete (or a-complete) infinite Boolean algebra. Thus, A contains an infinite antichain {ai : i E w } . If S is an independent family (of subsets of w ) , then the elements a A = C{u;: i E A } for A E S form an independent subset of A. In other words, the algebra A contains B({O,l}c) as a subalgebra. Therefore, the theorem 2.1 can be formulated as follows.

w2

2.3. COROLLARY. If V CH and V[G]is obtained from V by adding at least Cohen reds, then no a-complete infinite algebra A E V[G]is a partition algebra.

$3. PARTITION ALGEBRA UNDER MA. In connection with partitioner representality a problem appears if the assumption of CH in the theorem 1.1 can be replaced by MA (plus c > w l ) . Here, we answer this question negatively. In fact, we shall prove that, consistently with MA, the power-set algebra P(w1) need not be a partition algebra. More exactly, we prove the following 3.1. T H E O R E M . (Frankiewicz, Zbierski [FZ3]). There is a model of the Martin axiom with arbitrarily large c, in which P(w1) is not a partition algebra.

The proof of this theorem will occupy the rest of this section. The general idea is similar to that used in the proof of the theorem 2.5, Chapter 8. Suppose that the power-set algebra P ( w l )is a partition algebra, i.e. there exists an isomorphism

f : P(wl)

P(R)/[fin, R]

for some mad R.

Thus, for any I C w1 the value f ( z ) consists of (equivalent mod [fin,R ] )partitioners of R. Let A E f ( z ) and B E f(y). If z C y, then A \ B E [fin,R],i.e. A \ B Ro for some finite Ro R. Similarly, if I n y = 0,then

=.u

A nB

=+

uRo

for some finite

Let us denote

R(A)= { e E R

: e

E* A}.

RO G R.

C H .PARTITIONS ~

216

OF

ANTICHAINS

For trivial partitioners A, the family R ( A ) is either finite or co-finite. Using the above notation we prove 3.2. LEMMA.If 2 E P ( w 1 ) is infinite, then card R(A) = C, for each A E f(z).

PROOF.Let us divide z = $0 into two infinite parts zo = ~ the division process up to obtain a binary tree

Tz= {z,

: s

E

u

( 0U ) z(1)

and repeat

(0,l)")

nEw

such that for arbitrary sequences s, t: sCt

_> z t

implies

z,

implies

z , ~ i +n~z + + ~= $3.

and s(i)

# t(i)

Let us choose partitioners A , E f(z,),for

s

# $3 and Ap, = A .

Hence, for any

8,t :

sCt

implies

At C A, mod [fin, R]

implies

A++l n Atli+l = $3 mod [fin,R].

and s(i)

# t(i)

Modifying the sets A , by trivial partitioners (i.e. replacing A , by A , \ R, for some finite R, R) inductively, along the levels of T,,we may assume that for any indices s, t : sCt

implies

At C A ,

s(i)# t(i)

implies

n

and = $3.

Every branch g E (0, determines a descending chain (Agln : n E w ) . We choose an infinite set B, C w such that

Bg c* Agln

E R such that eg n B, is infinite. If follows that (since the sets Agln are partitioners). In particular, if

Since R is a mad, there exists an eg

C* Agln for each

g1

# g2, then we have

E

w,

for every n E w .

egi

e,

n eg2 C* Ag, In n Agzln

= $3,

for a large enough n, which proves that the eg's are disjoint, for distinct g's. QED

$3, PARTITION ALGEBRA

UNDER

MA

217

Now, we generalize the notion of a gap as follows. Let D be an almost disjoint family (not necessarily maximal). A system

is called a D-gap if the following conditions are satisfied: 1. The sets a,, bp are uncountable partitioners of D, i.e. the families D(a,) (= { d E D : d E* a , } ) and D(bp) are uncountable for a,P < w1. 2. a, ilbp =* 8, for d l a,@< w1 and a , n b, = (3, for each a < w1. 3. a , ED up and b, G D bp, whenever a 5 P < w1, where a ED b means that the family D(a \ b) is finite or countably infinite. Besides the ordinary separation of a D-gap 7-1 we can also consider a more general notion. We say that a set S C w D-separates (or D-fills) the gap 7-1, if a , C _ D S and bp G D w \ S, for all 0,P < wi. Let Q = Q(7-l) be a separating forcing as in Sec. 3, Chapter 7. Thus, Q consists of pairs p = (sp,tP) of finite functions from w1 into w such that

n

a,

u

\Sp(a)n

oEdm(s,)

bp \ t p ( ~ = ) 8.

BEdm(t,)

The ordering on Q is defined as follows p

5 q if and only if

sq

s p and t , C t,.

3.3. LEMMA.If 3.1 c m be filled, then Q ( H ) has the ccc.

PROOF.Suppose, tending to a contradiction, that Q has an uncountable antichain {qa : a < w l } . Using the A-system lemma and simple cardinal arguments we may assume that max dm( spa) < min din( sqp) for a < p and similarly, for dm(t,,). Let us denote

Incompatibility of q,,qp means that the set

218

cH.9

PARTITIONS OF

ANTICHAINS

is nonempty (cf the proof of the lemma 3.1, Chapter 7). Since some set S separates 7-1, we can find an integer m E w such that

k, \ m E S and (S \ m ) n I , = (3 for every a E Xo, where X O C w1 is some uncountable set. Let a0 = minxo. There exists an uncountable set X I C_ X O such that qua * q p is constant for P E X I . Again, if a1 = minx1 then we find an uncountable set X2 X I , such that qc*, * q p is constant for P E X p , etc. Hence, for every n E w there is an uncountable set X , 5 w1 such that, if an = minx,, then the nonempty sets Quo

* Q a , ,. . . ,qu,-, * qo,

are disjoint each other and bounded by m, which is impossible, (for n > m). QED e

We use also the forcing E = E('H), associated with 3-1, consisting of finite sets w1, such that au n bp

# $9

or ap

n bu # (3

for distinct a,P E e.

The ordering on E is the reverse inclusion (Sec. 2, Chapter 8). Any function g : w1 "subgap" 3-1, of 'H, where

+

w1, satisfying g ( a ) 2 a (all a

< w l ) , determines

Let E'(3-1) be the finite support product

E*(3-1) = rI,E(3-1,) (Ex. B . l , Chapter 8). We prove that the forcing E* freezes up all the 3-1, (as could be expected). 3.4. LEMMA.If no set D-separates 3-1, then E* = E*('H) has the E* It- "Q(3-1,)

has an uncountable antichain"

ccc

and

for every g E V.

a

219

$3. PARTITION ALGEBRA UNDER MA

PROOF.For the countable chain condition it is sufficient to prove that finite subproduct E('H,,) x . - .x E(X,,) has this property (Ex.B.l a, Chapter 8). Suppose, by way of contradiction, that the sequence e, = ( e i , ...,e:) (a < w 1 ) is an uncountable antichain in E('H,,) x . . . x E('H,,,). Applying the A-system lemma n times, we may assume that the families {eq : a < w1} are disjoint, for i = 1,.. . ,n, and hence that also whenever a < /3 and i = 1,... ,n.

maxep <mine! Since eu, e@are incompatible em I ep

for all a < p < w1 ,

we have eq I e f , for some i, and hence there are [ E ey and

n a g i ( o ) n (bq n ' g i ( q ) )

=8

and

r] E ef

such that

n a g i ( q ) ) n (4n b g i ( t ) ) = 8.

Thus, if we denote

S, = n { a t n u ~ ~: ( E~ e:)

a n d i = l , ..., n}

Ta = n { b , nbsicq,: q E e!

and i = l,.. .,n},

and then we have S, n Tp = 8, for any a # a < w1, we have S, n T, = 8, as well.

p. Since, by definition, a, n b,

= (3, for all

For 7 5 mine:, ...,mine: we have a, G D S, and hence a , C D S, where S = Ua<wlS,. Similarly, if T = U{Tp : P < w l } then bp C D T , for each p < w1. Since S n T = 8, we derive that the set S D-separates the gap X,which contradict E('H,) has the ccc. the assumption. Thus, we have proved that E*(H) =

n,

In particular, each E('H,) has the ccc and consequently, (see the proof of the lemma 2.3, Chapter 8), we may assume that for every e E E(N,) the set {a < w1 : e U {a}E E(H,)}

is uncountable. Let G E* be a generic ultrafilter. The projection G, onto the g-th coordinate is then a generic filter in E('H,) and the sets { e E E('H,) : 3P(P

1 a and P E e)}

220

c H . 9 PARTITIONS OF

ANTICHAINS

are dense. It follows that G, contains uncountably many singletons {a}. For any such a,define qa = ( S a , t a ) E Q(7-t,) as follows

Now, q, and qp are incompatible in Q('H,), since {a}and { P } are compatible in E('H,), as elements of G,. Hence, the sequence (q, : {a}E G,) is an uncountable antichain in Q('H,), which finishes the proof. QED

Now, we shall construct a particular gap 3c from a given "partial" representation of P(w1) as a partition algebra. It is more convenient to deal with P(w1 x w1) rather than P(w1). w1, let (X), denote its initial segment of order type For an uncountable set X a,a < w1. Suppose that the complement w1 \ X is also uncountable and define

and and Let B(X) be a subalgebra of P(wl x w1) containing the sets z,, yp and the axes {a}x w1, for all a,P < w1. Let us assume the following

3.5. DEFINITION. A D-representation (where D is an almost disjoint family) of a gap k ( X ) = ({z,}; {yp}) is a function r : B(X) + P ( R ) such that: 1. For every z E B(X), r(z) is a partitioner of D such that set

D(r(z)) = { d E D : d c* r ( z ) } is uncountable, for each I # 8. 2. r is congruent with Boolean operations (in obvious sense). In particular, the conditions z E y and z n y = 8 imply that the sets D(r(z) \ r(y)} and D ( r ( z )n r(y)), respectively are finite.

A gap k ( X ) and its D-representation r determine the corresponding system

K ( X )= ({Aa :

< W I } ; {Bp : P < w i } )

$3. PARTITION ALGEBRA UNDER MA

221

where A , = r(z,) and Bg = r ( y p ) , for a,p < w1. However, IC(X) need not be a D-gap, since the intersections A , n Bp are, in general, infinite.

3.6. DEFINITION. We say that a D-gap 3-1 = ( { a , } ; { b p } ) is containedin K ( X ) , if

A , C D a , C. A,,

a,

n Bp

8

Bp G D bp Ea Bp,

bp n A , =.

8

and

for all cr,p < w l ; 3-1 is said to be regular, if

for all a < p

and b, n b,

< 7 < w1

Gs b, n bp.

We prove below that if there are enough Cohen reals and dominating functions, then we can always find a regular, unfilled D-gap 3-1 contained in K ( X ) , for some X . Recall the dominating forcing D. The elements are pairs p = ( s p ,Fp), where s p : I(p) + (0,l},for some I(p) E w , and Fp is a finite set w w ; p 5 q holds if and only if sq s p and Fq C Fp and f ( i ) < s p ( i )for each i,, I(q) 5 i < I(p), and f E Fq. If G C D is a generic filter, then the function fc = U { s p: p E G } } dominates every function f from the ground model V : fc(i) > f ( i ) for all but finitely many i E w. Finally, D has the ccc, (Sec. 5 , Chapter 2).

3.7. LEMMA.Let D E V and let P, = C , < 7 P o be a finite support iteration with the ccc, such that there axe two increasing sequences (at : < w1) and

<

*

(& : < w l ) , both cofinal in 7 and such that Pa,+l = P,, * C and Ppc+l = Pg, D, for all ( < w1 (C denotes the Cohen forcing). Then, in V [ G ]there , is an X 7 such that for an arbitrary D-representation r of k ( X ) there exists a regular D-unfilled D-gap ‘H contained in Ii’(X). P R O O F . Let ct be the Cohen real added at the stage at. Define a set

as the Cohen subset determined by the sequence (cc :

X = {w * ( + i

:

q ( i ) = 1 and

C$

< wi):

< < w1 and i E w } .

X C w1

CH.9 PARTITIONS OF ANTICHAINS

222

Thus, X is in V [ G ]but for no a

<

y is X

E V[Gla]. Consequently, any D-gap

‘H = ( { a p } ;{ b p } ) contained in K ( X ) cannot be D-separated. Indeed, suppose that a set T D-separates ‘H and let Cc = r ( { t }x w l ) . Then we have the following properties:

and

6#X It follows that

implies

38 < wl(Cp C D b p ) .

x = (5 < wi : cc C D T )

and therefore X E V[Gla], for some a,a contradiction. This proves the claim.

NOW,we define the gap ‘H by induction, as follows. Assume that we have already defined the sets ap and be, for t < a. The family

is at most countable, since at partitioner Se of D such that

a,,. Using a dominating function, we can find a

\a,, C_* Sc

and

dnSc

=. 8

for all

t < r] < a

for every d E D \ Dp.

Indeed, the countable family

is disjoint. Modifying its elements, if necessary, (each by a finite number of points) we may assume that U Re = w . Choose a one-to-one function j : w w x w , onto w , such that the images of the elements of Re are the vertical axes {n} x w , n E w . Then, the sets j [ e ] , for e E D \ Dp are finite on each axis and hence the functions

fe(i) = max[j[e] n ({i} x

w)]

are well defined. All this takes place in some submodel V[GIP] and hence there is a function g E V[G], dominating each function fe, e E D \ Dc. Now, if we set

F

= {(n,i): i

5 g(n)},

$3. PARTITION ALGEBRA UNDER MA

223

then the set Sc = w \ j - ' [ F ] has all required properties. In a similar way we find a set S such that

<
s, E* s

for every

enS=.(i!j

for every e E D

and

\ UE<,D,.

Finally, let us define

D, = {e

E D : 37

< a ( e C, A , n B q ) } .

Arguing as above, there is a partitioner T such that

AanBq GiT

for every 7 < a

enT=,(i!j

for every e E D \

and

D,.

We assume a , = A,\(SUT) and check easily the conditions of the definitions 3.5. The set b, is defined symmetrically by the same means and the proof is complete.

QED Let us note that the set X (determining Ii'(X)) can be always chosen as the Cohen set consisting of Cohen reals added cofinally at earlier stages. Assume that P ( q ) is representable as a partition algebra, say P(R)/[fin, R ] , where R is a mad. Let k = ({z,}; {yp}) be an ordinary gap in P ( w l ) , i.e. the sequences (z, : a < w1) and (yp : p < w1) are strictly increasing and 2, n yp = $3, for all a , p c w1. Using an isomorphism (onto the partition algebra) we can choose a corresponding system of partitioners

Ii' = ( { A , :

<~

1

)

{Bp : P < ~

1 ) )

and, since every gap in P(w1) is filled, a partitioners S (e.g. corresponding to the 2,) such that, for all a,/3 < w1 union

\ S =*

uU ,

Bp n S =*

uWp

A,

for some finite U ,

R

and for some finite Wp G R.

224

cH.9

PARTITIONS O F

ANTICHAINS

3.8. LEMMA.In the notation as above, there exist finite sets U,W finite sets u,, w, G w1, such that a = min u, = min w, and

nEEu,A€ 5. S u IJU

R and

for each a < w1

and ntlEWP

B, nS G 8 U W

for each /3

< w1.

PROOF.Clearly, every intersection

reduces to a finite one p ( a ) = U, n

u,, n.. . u,,,

and we take uq = { a , a l , .. . ,an}.On the other hand, since the function a H p ( a ) is nondecreasing and takes finite sets as its values, p may have only finitely many jumps. Hence, for some am< w l , we have p ( a ) G p(am) for every a < w1

Define U = p ( a m ) . Since the elements of R are almost disjoint, we obtain

Consider now a regular D-gap 3-1 = ({a, : a

<wl} ;

{bp :

/3 < ~

1 ) )

contained in K. In particular, we have a , n B p = * @ andbpnA,=.@

for all a,/3 < w1. Hence, it follows that nCEu,aenuu=*~ for all a,fi

and n,,,,b,nuW=*~

< w1.

$3. PARTITION ALGEBRA UNDER MA

225

Thus, since U and W are disjoint, the partitioner T = S U U subgap

(in

{

€Euo

n b,:

\ W separate the

P < ~ ~ } ) .

VEWP

Further, the regularity of 31 implies that

nEEuo= a , n afl(,)

where h ( a ) = maxu,

nVEwp b, = ba n bf2(@)

where

and

f2(rc)

= maxwb.

If we take f = max{fi,fi}, then the sets a , n a f ( , ) and ba n b f ( p ) are even smaller and hence we obtain the following useful fact.

3.9. COROLLARY. If P(w1) is (isomorphic to) a partition algebra and K,31 are defined &s above, then there exists a function f : w1 + w1, with f(a)2 a,such that the subgap

can be separated. Now, we can finish the proof of the theorem 3.1. In a ground model V , (we assume that V CH), we fix a regular cardinal IE > w1, fulfilling 2<" = n and such that the diamond principle holds for the family H(n) = Ua<" H,, of sets of hereditary power < n, in the form: there exists a sequence (T, : cf(cr) =

w1 A Q

< n) such that

for every

Y C H(n), the set {a < K : c f ( ~ =) w l

and Y n H , = T,}

is stationary in n. For example, as V we can take the constructible model and as n - any regular cardinal > w1 in V .

For every a < K , with c f ( a ) = w1 we choose a cofinal in a sequence h, = ( h , ( [ ) : 6 < w1) E V , such that cf(h,([)) = w , for [ c w1.

cH.9

226

PARTITIONS OF

ANTICHAINS

We define now a finite support iteration (P, : a as follows. P, =

< K ) , with the ccc, by induction

Cs<, P,g (the direct limit) if a is a limit ordinal,

) w Pa+] = P a * C and Pa+2 = Pa+] * D if c ~ ( ( Y = (where C is the Cohen and D - the dominating forcing),

Po+l = P,

* Ta if c f ( a ) = w1 and T,

Pa+] = P,

* E,

P,+1 = P,

* C in all remaining cases.

is a Pa-name of forcing such that P, IF "T, has the ccc",

if c f ( a ) = w1 and T, is a P,-name such that for some D E V('o), P, IF "T, is a D-representation of k ( X , ) " , where X, is the Cohen set C_ a determined by the Cohen reals added at stages h,(E), < w1, and E, is a Pa-name of E * ( H ) for some D-unfilled gap 7-1 contained in K(X,).

P, has the ccc. Let G C P be a It follows from the lemma 3.4. that P = Ca
V [ G ] "MA and c = K " . Suppose that, in the model V [ G ]the , algebra P(w1) is (isomorphic to) the partition algebra determined by a mad R. We shall derive a contradiction. First, let us fix some partitioners Ct, corresponding to the axes {[} x ( < w1. By the lemma 3.2, each family

w1,

for all

contains a subfamily Rc, with card RE= w l , for all [ < w1. Let us denote

and choose an ordinal p < K such that the families D and {Cc : ( < w1) are in V[G)p].For every ordinal y < K , with cf(y) = w1, we have the Cohen set X, C y, X , E V[Gl7]and the corresponding gap

We denote by B the subalgebra of P(w1 x W ] ) , generated by the sets z,(y), y,(y), for all a , /3 < w1 and y < K . Bs is defined similarly, with the additional restriction y 5 6.

$4.MORE O N

227

PARTITION ALGEBRAS

Thus, we have B, = B n V[Gl-y], for y < K . Having an isomorphism h of P(w1 x w1) onto the partition algebra P(R)/[fin,R], we can choose a function r : B + P(R), such that r ( z ) E h ( z ) for all 3 E B and r ( { < }x w1) = Cc,for < w1. Let I: C H ( K ) be a canonical name of r , consisting of some pairs ((z,a)('),q), where z runs over canonical names of elements of B and a - over that of subsets of w. Arguing as in the proof of 2.5, Chapter 8 we check easily that the set

<

Ni = (7 < K : c f ( 7 ) = w1 and V z , a E V('.)(r(t,a)

P,))

is an wl-cub. Denoting ~-ly= E n (V('7) x P,), we infer that r7 = r

n V[G(r]= [rl~][GIy] E V[Glr], for every 7 E Nl.

Similarly, it is easy to see that the set

is an wl-cub. By the diamond principle, the wl-cub N = N 1 n Nz intersects the set {y


: cf(y) = w1

and E n H, = T,},

and hence there exists a y < K such that 7 > p, cf(-y) = w1 and T, is a P,-name of r , E V[Glr], which in turn is a D-representation of k ( X , ) in V[Glr]. By the lemma 3.7, there is a D-unfilled regular gap 'H contained in K(X,). Now, from the definition of the iteration, it follows that we have forced with E*(R),at the stage 7. Therefore each forcing Q('H,), for g E V[Gly] has an uncountable antichain in the model V[G]. On the other hand, the corollary 3.9 shows that for some function f E V[G] the subgap N f can be separated. Since P has the ccc, there is a function g in V majorizing f, i.e. f ( a ) 5 g ( a ) , for all a < w1. Obviously, H ' , is separated as well, (since it is even smaller). Hence, by the lemma 3.3, Q ( H , ) has the ccc in V[G], a contradiction. This finishes the proof of the theorem. QED

$4. MORE ON

In this section we shall consider the following question: is the conclusion of the theorem 1.1 equivalent to the continuum hypothesis? The answer is negative. We prove that, consistently, the continuum can be large and still every Boolean algebra, of cardinality 5 c, is a partition algebra. The construction can be so arranged that the Martin axiom holds, as well. For simplicity, PARTITION ALGEBRAS.

cH.9 PARTITIONS

228

OF

ANTICHAINS

we shall restrict ourselves to a weaker version of MA - for a-linked forcing only, analogously aa in the theorem 3.1 of Chapter 8. Thus, we prove here the following

THEOREM 4.1. (Frankiewicz, Zbierski [FZB]).There is a model of MA (a-linked) plus c > w1, in which every Boolean algebra of cardinality 5 c is (isomorphic to) a partition algebra. The proof of this theorem will take the rest of this section. Let V be a ground model, in which CH holds and fix in V a regular cardinal IE > w1, satisfying n<" = I F . We define a finite support iteration ( P a : a < n), with the ccc. At some stages a we add generic sets X, C w , filling gaps in P(w)/fin generated by some earlier added generic Xp, p < a. Consequently, (see the theorem 3.1, Chapter 8), every algebra of cardinality 5 c will be embeddable into P(w)/fin. More exactly, for any 4 : K + {0,1} let B ( 4 ) be the subalgebra of P(w)/fin, generated by the family { X , : d(a) = 1). For every Boolean algebra 8, with card B 5 C, we find a 4 such that B is isomorphic to B(q5). Simultaneously, at some other stages p, we add generically some sets Ap w, so that (for each 4 as above) there is a mad R(4), consisting of some the A's, such that B is isomorphic to the partition algebra determined by R(q5). In other words, for every B, with card B 5 c, we choose a branch 4 of the binary tree U,<,{O,l)a and force along 4 so that B = B ( 4 ) and B is P(R(~$))/[fin,R(4)].Since n<" = n, holds, all this can be done in IE steps of the iteration. After these intuitive remarks we pass to a precise inductive definition. We use here a well known "booking" technique. At each stage a < IC we choose an enumeration da

= {da(() : ( < K }

with repretitions (i.e. each term d,([) occurs n times) of some objects in V('Q). Let j : IF x IF + K be a pairing function (one-to-one and onto) such that j ( a ,p ) > a for all a,p

< K,

and denote N r ( d , ( ( ) ) = j ( a , ( ) .Hence, every object d (enumerated at any stage) has an arbitrary large number N r ( d ) < K and every ordinal a < K is a number a = N r ( d ) of some object d enumerated at some stage < a. Assume the following notation. For any function 'p : a the subalgebra of P(w)/fin generated by the family { X p : &3)

+

( 0 , l ) let B(p) be

= 1). Thus, if s is a

$4.

MOREON

finite zero-one sequence with dm(s)

229

PARTITION ALGEBRAS

c {p<

(Y

:

cp(p) = 1) and

then B(p) consists of finite sums (of equivalence classes mod fin) of the sets X ( s ) . Hence, gaps in B(cp) can be represented as

L = ( { X ( s ):

s E S} ;

{ X ( t ): t E T}),

where X ( s ) n X ( t ) =* 8, for each s E S and t E T. We may assume always that the conditions 31,.. . , s n E S and X ( s ) C, X ( s 1 ) U . . . U X(sn) imply s E S,and the same for T. Now, we describe the inductive step of the iteration. Assume that P, has been already defined. We enumerate all forcings in V('0) of power < K. and all pairs ( X ,p), (Y, cp), (L, cp) of Pa-names such that P, forces all the following properties: 1. X , Y c w a n d X , Y $ f i n a n d w \ Y @ f i n . 4 B(cp)and L is a gap in B(cp) 3. cp E D,, where D, consists of all $ E V('*) with dm($) 5 (Y and such that if {ye : 6 < 6) is an increasing enumeration of the set (7 : $(7)= l}, then for each 6 < 6 there is a gap Lc in B($lyt 1) satisfying 7~+1= N r ( ( L ~ , $ 1 7 ~ + 1We ) ) . extend the function nr to all just enumerated objects in V('o), as mentioned above. 2. X E B(cp), Y

+

We also assume inductively that there are almost disjoint families R,(cp) C V(po), with the property

Moreover, we assume that where, as in 3 above, the 7~ enumerate the set cp-'(l). each infinite set X E B(cp) is a partitioner of R,(cp) and

p) or Now, the ordinal (Y determines either a forcing Q, or a pair (X, cp), (Y, ( L ,cp) enumerated at some stage ,B < a. Accordingly, we distinguish four cases.

Case 1. a = N r ( Q , ) . Define Pa+, = P,

* Q,,

if P, It- "Q, is a-linked ".

230

cH.9

Case 2.

PARTITIONS OF

ANTICHAINS

= Nr(X,cp). We add a new element A = A,(X,p) under X, which is almost disjoint from the family R,(cp). We may assume that we have a fixed ultrafilter F C B(p) containing X in V(pa) (i.e. P, forces all this). Let S be the almost disjoint associated with the family

Q

S = {w \ 2 : 2 E F } U R,(cp). Thus, the elements of S are pairs p = ( t pwp), , where zp : l ( p ) + {0,1} for some I(p) E w and w p C S is finite. The relation p 5 q holds if and only if zg C zp, w q C w p and for each i, l ( q ) 5 i < Z(p) the condition i E u w g implies z p ( i )= 0. Clearly (Ex.A.1, Chapter 8), we have Po IF "S is o-linked".

By the inductive assumption, the set w \ (Jwp is infinite, for all p. Hence, if G is a generic ultrafilter, then the set

A, = {i E w

:

3 p E G ( t p ( i )= 1))

is infinite, almost disjoint from R,(p) and also A, define P,+1=

S

c* 2,for each 2 E F.

Let us

Pa * S and Ro+l(v) = R a ( 9 ) U { A , } *

Obviously, the inductive assumption is still valid for B(cp) and R,+l(cp). Case 3. a = Nr(Y,cp). Suppose that

i.e. Y is congruent mod [fin,R,(cp)] in V('-) to no element of B(cp). Then, the same is true for the complement w \ Y and it follows that the set

is a proper ideal in B(cp). Let F be any ultrafilter in B(v)extending the dual -J = { w \ t : t E J } . Thus, for every t E F we have I n Y 4 [fin,R,(cp)] and also z \ Y 4 [finlRo(cp)]. Hence, we may apply the almost disjoint forcing S(Y), which adds an infinite set Q under each z n Y, for all z E F

$4. MOREON PARTITION ALGEBRAS

231

and the forcing S(w \ Y),adding an infinite b under each x \ Y ,for x E F, so that both a and b are almost disjoint from R,(cp). Let us define

where A , = a U b. It follows easily that the inductive assumption holds for B ( p ) and R,+l(cp). Since Y intersects nontrivailly the element a U b = A , E R,(cp), we infer that Y is, (from now on), not a partitioner of R,+l(cp). Case 4. a = N r ( L ,cp). Suppose that L has a form

L = ( { X ( s ):

sE

S} ; { X ( t ): t E TI).

We enlarge L by adjoining all the elements of R,(cp). Thus, let

S A = {€ < a : Vt E T ( A , n X ( t ) =* 0)) and

TA = { q < (Y

:

3t E T ( A , E* X ( t ) ) }

and define

Now, we apply the separating forcing Q = Q ( L * ) ,(as in Sec. 5, Chapter 2), i.e. we assume

P,+i = Pa * Q(L*).

If G

Q(L*)is a generic filter, then the set X , = {Z E w : 3 p E G ( z p ( i )= 1))

separates the gap L and is a partitioner of R,(cp). Let cp" be an extension of cp such that

P"(0 =

{

cp(€),

! I ;

if € E dm(cp) if dm(cp) 5 6 < a if € = a.

It follows that p a E D,+l and each element of B(cp") is a partitioner of R,+l(cp) = R,(cp), (we add no new elements to R, in the case).

232

c H . 9 PARTITIONS OF A N T I C H A I N S

We must prove that no infinite element 2 E B(cp") is covered mod fin by a finite union of elements of R,+l(rp"). By the inductive assumption, this is true for

2 E B(cp). Suppose now that a Z E B(cp") has the form 2 = X ( a )n X ,

where X ( a ) E B(cp),

and that 2 G* Ae,u ,...,UAt,, for some sets At ,,..., A€, E R,+l(cp").

Since

R,+l(cpo) = R,(cp) and 2 is a partitioner of R,(p), we may assume Z =* At, U ... U A,cc and therefore 2 is in the model V('a). Let us notice that Q(L') is equivalent to the product QOx 6 1 , where QOand Ql are separating forcing for L* restricted to X ( o ) and w \ X ( a ) , respectively (in obvious sense). Since Qo produces X, n X ( o ) = 2 we infer that Qo must have an atomic element ( u , 2,w ) and (in this case of Z),

z =X , n

n

=.

(Ux ( s )u U A( =* x(a)n U x(s).

~ ( 0 ) ~ ( 0 )

8EU

C'Ea

.EU

It follows that Z belongs to B(p) and hence (by the inductive assumption) is finite. The case of Z = X ( T )n(w \ X,) is treated symmetrically. Thus, the inductive assumption holds for B(cpa) and R,+l(p"). To complete the definition we assume that the direct limits are taken at every limit stage a and that Pa+l = P,

*C

in all the remaining cases (C denotes the simple Cohen forcing).

We shall prove below that P = '&r P, has the ccc. Assuming this for the P be a generic filter. moment let us look how does the forcing P work. Let G Clearly, we have V[G] "MA(a-linked) and c = K".

+

Let us note also that V[Gla] "c < K " , for each a < IC. Let B E V[G]be a Boolean algebra of cardinality c = IC, in V [ G ] .B can be decomposed into a union B = Ua, 1) and for each a < K , B,+1 = B,(b,], for some b,. Assume inductively that we have an embedding f : B, + P(w)/fin such that for some ~ c f ,( b 0 = X , /fin, for each < a. Hence, we have f [B,] = B(cp,)/fin, where Q, is defined on sup (7t : E < a} so that p a is 1 on the 7's and 0, otherwise. Let us denote

<

b(s) =

( l - I e ( € ) d d ) . (I-Ia(q)=l - bq)

1

$4.MOREO N

233

PARTITION ALGEBRAS

where s is a finite zero-one sequence, with dm(s) E a. Every element of B, is a finite h in B, determined by b, sum of elements 4.2. Consider the gap '

K = ( { b ( s ): b(s) 5 b,}; { b ( t ) : b ( t ) * b, = 9)). The image L of K under f is a gap in B(cp,) and L has the from

<

where sf is defined on {yt : E E dm(s)} by the equality s f ( 7 ~ )= s(() for E dm(s). The sequence tf is defined analogously. Obviously, we have f ( b ( s ) ) = X ( s f ) ,for any s. Let p < K be SO large that (P, and L are in V('P). Since the continuum c at the stage /3 is smaller than K , we can find an ordinal yo > /3 such that y, = Nr(L,cp,) and in addition

Let us define an extension o f f to B,+1 by setting

According to the Sikorski theorem (7.4, Chapter 1) we must check only the following equivalences

b(s) 5 b,

I

X ( s f )C, X,,

and

b ( t ) * b, = 0

I

X ( t f ) n X,, =. $3.

The implications to the right are obvious, since X,, fills the gap L. The converse implications follow easily, (see the proof of 3.2, Chapter 8), from the following 4.3. LEMMA. Let q E Q = Q ( L * ) and let X = X , denote the generic set separating L* added by Q. Then we have (for any k E w ) if q It- "X(u) \ k C X", then X ( o ) UsEUq X(s), and if q IF " X ( 7 )n X C k", then X ( T )&* UtEwq X(t).

s

Similarly, q ll- "A( \ k C X" implies A( C, X ( s ) , for some s E u and also q It- "A,,n X k" implies A,, C, X ( t ) , for t E w g .

234

cH.9

PARTITIONS OF

ANTICHAINS

PROOF.From the assumption follows that

X(C)Ec X(S~) U . . . X(sn) U Acl where u s = {sl,.. . ,sn;(l,.

.

U .. U A € , ,

. . tn},see Remark 3.3 Chapter 8.

2 = X(a)\

Since, the set

(X(s1)u . . . u X(Sn))

is in B(cp,), there is an s such that X ( s ) c* 2. Hence,

X(S)G* A,,

U

. . . U A,,,,

and by the inductive assumption, X ( s ) must be finite. Therefore, 2 is finite as well, i.e. X(a)E* X(s1)u . . . u X(sn)= X(s).

u

8EUq

The other assertions are proved in similar way.

QED

Hence, f can be extended to Ba+l with f(ba) = X7-.Since the limit step is obvious, we infer that the algebra B is isomorphic to B(d)/fin, where 4 = 'pa. Now, let as denote

R(4) =

U Ra(Va). a
By the construction, each non-zero element b of B(4) is a nontrivial partitioner of

R(b),(actually, there are n many elements of R(4) under a), and every set Y E P(w) \ B(4) is either a nonpartitioner or is congruent to an element 2 of B ( 4 ) mod [fin,R(4)] i.e.

for some sets At, ,.. . ,A€. E R(4). It follows, as in the proof of 1.1, that the mapping Y H z is a homomorphism from 'P(R(6))onto B(4) with the kernel consisting precisely of trivial partitioners. It follows also that R(4) is a mad and therefore B = B(4) is a partition algebra. If card B < c then we use the theorem 2.5 of [BW]. Thus, it remains to prove that our iteration ( P a : a

< K ) satisfies the ccc.

Let us note that the almost disjoint forcing S is a particular case of Q,(the separating forcing), in which u p = 9, for all p . Hence, all the stages of the iteration

54. MOREO N

235

PARTITION ALGEBRAS

(except the case 1)can be treated uniformly with an obvious modification in the case of a = Nr(Y,cp). Now, the proof will be an easy modification of the theorem 3.1, Chapter 8. Let I' E K be the set of all stages at which we force with a Q,i.e. we apply the case 2 or 3 or 4 of the definition of the iteration. Let E , C Pa consist of all p E Pa such that for each Po-names u$,x i , w$ such that Pll

l

E supp(p) n I' there are

It- "P(0 = b;,z;, w;)"

and dm(s) supp(p), for each s E $Uw$ and also C E supp(p), whenever C E u$Uw$; moreover there is an integer 1 = l ( p ) E w such that each zf is of length I , for all

t E SUPP(P) n r. At any stage p E tc \ I' we force with a Qp, such that Pp It- " Q p is a-linked". Hence we can choose a Pp-name K p of a function, such that Pp It- " K p : Qp

+w

and V n E w ( K i l ( n ) is linked)".

c

Let E: E, denote the set of all p E E,, such that for each 6 E supp(p) \ I' there is an n E w , such that PI( II- " K d P ( t ) )= n".

4.4. LEMMA.For every p E Pa and m E w , there is a q 5 p and I(q) 2 m.

such

that q E E:

PROOF.By induction on a. The only nontrivial case is that of a = /3

p E supp(p).

If

p 4 I',

+ 1 and

then there is an rl 5 pJB,such that

rl

It- " K p ( p ( p ) )= n"

for some n E w .

Applying the inductive hypothesis, we can find an 7-2 5 rl such that 7-2 E E i and I(r2) 2 m. Hence, q = r2 * p ( P ) is as required. If p E I ', then we can find an rl 5 p i p such that r1 IF " p ( B ) = (ti,", z,", w,")" for some Pp-names u!, z!, wt.We may assume that

u{

dm(s) : s E u," U w f } C supp(rl),

cH.9 PARTITIONS

236

and that ( E supp(r1) for each C E

OF

ANTICHAINS

U w!, (modifying T I , if necessary, in an obvious

way). By the inductive assumption there is an r2 5 rl such that 1-2 E E; and I(r2) 2 max{m,I(p(/3))}. Thus, if s or 5 is ui Uw!, then 7-2 determines the segments X ( s ) n I(r2) and A€ n 4.1). Hence, there is an extension 3: of x$ of length I(r2) such that r2

Thus, the element q = rz

1- "(upP,I , wpP) 5 p ( @ ) " .

* (u$,2, w!)

has the required properties.

QED

We can formulate now some sufficient conditions under which two elements p , q E P, are compatible. 4.5. LEMMA.I f p , q E Et, are such that I(p) = I(q) and for each

5 E SUPP(P) n suPP(q) we have if 6 E l?, then zf, = x:, and

if 6 4

r, and PIE It- " I ( c ( p ( [ ) ) = n" ql[

IF lC((q(5)) = m", then n = m;

then p and q are compatible.

PROOF.By induction on a. Only the case of a = 0 + 1 and P E supp(p) nsupp(q) is essential. Assume inductively that there is an rl 5 PIP, qIp such that rl E Eg and 1 ( q ) = 1, where 1 = 1(p) = [ ( q ) . If the assumption of the lemma)

P4

r then we have (by

whence rl

It-

"T

5 p ( p ) and T 5 q(P)"

for some Pg-name

7,

because every counterimage K i l ( n ) is linked. Hence, if we define r = r1 * 7,then r 5 p,q and also r E E, and I(r) = 1, so that the inductive assumption is valid for a.

Now, suppose that

and

E .'I Thus, we have

$4.MOREON

PARTITION ALGEBRAS

237

By an assumption of the lemma, we know that z ! = z{ = I, in V(P,). Applying the inductive hypothesis, we find an rp 5 plP,ql/3 such that rp E Ep and l(rp) = 1. It follows that rp It-

"U n W

g",

=+

where U = X ( s ) or U = A t , for s or [ in u!Uu{, respectively and similarly, W = X ( t ) or W = A , for t or 9 in wi U w;. We show that for some V t 5 rp, P E Ep and l ( r ) = 1 we have ?= It- "U n W c I". The existence of V is proved by induction on P. Suppose e.g. that U = X ( s ) and W = A,. If max dm(s) < 9,then we have

and hence rl1q It- " X ( s )E+ U { X ( t ): t E w&}". By the inductive hypothesis we may define

+ 1 It- " A , n X, =. 9"

(or " A , G+ X,'').

Thus,we have rp17 IF A, f l u =+ g", where u = X ( s ) or u = A t , for any s or t in u:,. By the inductive hypothesis (applied for each u as above), there is an r y 5 rp17, r y E E, and l ( r y ) = 1 such that ry It- "A, nu I", all u. Hence, we define

-

r = ry * (.;,,z;,,w;o

u ( 7 7 1 ) * ( r p ( 0 : 7 < t < P). W = X(t). If s u t is not a sequence, then

Consider the case of U = X ( s ) and Pp It- "X(s)nX(t) = g". Otherwise, X ( s U t ) = X(s)nX(t) and if y = rnax dm(sUt), then X ( s U t ) = X(sUtly)nfX, and hence we have rpIy+l It- " X ( 3 u t l ~ ) n X=+~ 0" (or "X(s U tlr) C, X,"). Thus, ' p l y IF "X(s U tlr)n u g" for each u = X ( a ) or u = At for u or t in u;,. We may apply now the inductive hypothesis. Thus, there is an r , 5 rplr, r , E E, and I(r,) = 1 such that r , It- "X(s U t l y ) n u c I", for u as above. Hence, we may define an element

=.

-

r = ry * (q,, ,:z ,4, u

u t h } ) * ( V ( 0: Y < t < P ) ,

which has the desired properties. All the remaining possibilities can be proved in a similar way.

238

cH.9 PARTITIONS

OF

ANTIC~IAINS

Now, beginning with rB we repeat the above procedure to each pair U,W up to obtain an T 5 rp, satisfying the claim, for all U,W. Let us define r = F * (u," u u t , I,w,"u w;).

Thus, r lies under p and q,

T

E E, and l ( r ) = 1, which proves the lemma.

QED

Now, we can finish the proof of the theorem 4.1.We must show that P,, for any a,has the ccc. Suppose that some P, has an uncountable antichain

C = {qi : i E I}C P a . Since, in view of the lemma 4.4, EL is dense in Pa, we may assume that C C E:. Applying the A-system lemma to the family {supp(qi) : i E I} we find a common root d and an uncountable Co C C , so that the assumptions of the lemma 4.5 are satisfied on d, for all q E Co. This contradicts the lemma and completes the proof of the theorem.

PROBLEMS A. GENERALIZED A-SYSTEMLEMMA 1. Let n > w1 be a regular cardinal such that A" < family of countable subsets C K , card R = n.

K,

for all X

< n. Let R be

a

a) The family R contains n elements, which are order isomorphic each other. Therefore, we may assume R = {A, : a < K } , where A, = {fa(() : ( < 6}, for a fixed 6 < w1, where all functions fa are increasing. b) We have sup ,<%( sup A,) = n. Hence, there is a 7 < 6 such that SUP a < z f a ( 7 ) = c) If 7 is the smallest ordinal, for which sup cr
d) For arbitrary ( <

< K we have

Hence, there is a subfamily S R, card S = n and at most countable set d C p, such that A n B = d , for any distinct A, B E S.

PROBLEMS

239

Thus, a generalization of the A-system lemma holds for some higher cardinalities and countable sets. (Sanin). 2. Assume CH. Every family R, with card R = w2 consisting of countable sets, contains a subfamily S C R, card S = w2 such that for some d, we have

A n B =d

for any distinct A , B E S.

B. Q-SETS The Cantor space C = {O,l}'" is homeomorphic to its w-th power C = C". Thus, we may assume that the elements of C are sequences f = (fn : n < w ) , where fn E (0, I}", for all n E w. 1. Assume that the power-set algebra P ( w l ) is isomorphic to a partition algebra P(R)/[fin, R], for some mad R. a) Let A , be a partitioner of R, corresponding to an x C w1, x are e,(z) E R such that e,(x) A , , for all n < w .

c*

b) If Kn(z) = {f 6 C : fn an Fm6 in the space c. C)

If g n ( z ) = ~ e , ( z ) , for

XA,},

then the set k ( z ) =

# $9.

There

nnEw Ui,,Ki(z)

is

# 8, then

for infinitely many n E w. d) The set F = { g( { a } ) : a < w1 } has power w1 and we have

for arbitrary Y 2 F . Hence, if P ( w 1 ) is a partition algebra then there exist Q-sets (uncountable subsets of the reals R each subset of which is relatively Borel). A. Miller has constructed a model, in which 2'"' = c and there are no Q-sets. Hence, we may consistently assert, that c > w1 and there are Boolean algebras of cardinality 5 c, which are not partition-algebras.

cH.9 PARTITIONS OF ANTICHAINS

240

C. HOMOMORPHIC IMAGES 1. Let (.fa : (Y < Ib) be a limit in ww and {A, : (Y < Ib} - an independent family (of subsets of w ) . Assume the consistent inequality Ib 5 s. a) Every infinite subset X set

T, = ( A ,

w x w contains an infinite

x w ) \ {(n,j)E w x w : j

5 fa(.)}

Y E X, such that no where (Y

< Ib

splits Y. b) There exists a mad R, such that each set T, is a partitioner of R. c) The partition algebra P(R)/[fin,R] contains the field B((0,l}b) as a subalgebra. Hence, if Ib 5 s, then the free algebra of Ib generators is embeddable into a partition algebra. 2. Assume MA and c

> w1.

a) The power-set algebra P(w1) is a homomorphic image of B ( ( 0 , l}b). b) There is a partition algebra A and a homomorphism h : A '3P(w1) (Ex. 1 and the Sikorski theorem). Hence (the theorem 4.1), it is consistent with MA plus c > w1 that a homomorphic image of a partition algebra is not a partition algebra. (see [DF]).

D. THECARDINAL a. Let u denotes the minimum power of a maximal (infinite) antichain in the algebra P(w)/fin. Shelah constructed a model in which s = w1 < a. a) If s < a,then the subalgebra A P(w), generated by fin and a splitting family S of power s, is not a partition algebra. Hence, it is consistent that there is an algebra of power w1, which is not a partition algebra. ([DF]). 2. If s < a,then the power-set algebra P ( w ) is not a partition algebra (and even cannot be embedded into a partition algebra). ([DF])

E. TOPOLOGICAL DYNAMICS

A (topological) dynamicd system (X, G) is a topological space X together with a semigroup G of continuous mappings g : X + X. Any closed subspace A C X

PROBLEMS

241

determines a subsystem ( A , G I A ) , where GIA = { g l A : g E G } , provided A is G-invariant (i.e. g[A] E A , for each g E G). 1. A dynamical system ( X ,G) is minimal (i.e. it has no proper subsystems) if and only if one of the following conditions holds a) Any orbit G ( z ) ( = { g ( z ) : g E G } ) is dense in X . b) For every open U C X , we have X = UPEG g-'[U]. Hence, for a compact X , (X, G) is minimal if and only if, for every open U C_ X there exist finitely many 9 1 , . . . ,g n E G such that

x = g;"u] u . . . u g,"U]. 2. Every compact dynamical system ( X ,G ) contains a minimal subsystem. (Apply the Kuratowski-Zorn principle to the family R = { A X : 2 = A and V g E G ( g [ A ]G A ) } ) . A point z E X is called recurrent for an f : X + X (or for the dynamical system (X,{f" : n E w } ) ) if for each neighbourhood U of I there is an n 2 1, such that f"(z) E U , (i.e. if the orbit of I returns infinitely often to each of its neighbourhoods). 3. Every compact space X has a recurrent point, for any continuous f : X + X. (Every point of a minimal subsystem of ( X ,{f" : 12 E w } ) is recurrent). 4. If ( X ,G) is a compact minimal dynamical system with a commutative group G of homeomorphism, then for every open U C_ X and finitely many 91,. . . ,g k E G there is an n 2 1, such that

u n g;"[U] n . . . n gkn[U]# 8. (Biaszczyk, Plewik, Turek; for metric X - Furstenberg, Weiss). The proof goes by induction on k. Given a U X fix fi, . . . ,f, E G such that X = f;'[U]U . . . U f;' [U], (EX. 1 ) . a) k = 1. There is a sequence {Uj : j E w} of nonempty open sets X such that Uo G U ,Uj C f;,![U] (some m j ) and

(*) Now, if r

g[Uj+l] Uj for all j E W .

< s and m are chosen so that U,.,U, f ; ' [ U ] ,

have fm[Us] C_ U n g-"[U].

then for n = s - r , we

CH.9 PARTITIONS

242

O F ANTICHAINS

+

b) The inductive step k + k 1. There are sequences {Uj : j E w } and {nj : j E w } such that no = 0, the Uj's are as above with the exception that (*) is strengthened now to

Having already the Uj, n j apply the inductive hypothesis to g i i l g i , . . . ,gLilgk to obtain an n,+l such that

U, and

[U], for a suitable r n j + l . Now, if r < s and m are and let Uj+l = W n f;,1+, such that U,,Ua E fG1 [U], then for n = n,+1 . . . n, and V = fm[Ua] we have V U gr[V] U . . . U gi+l [V]C U ,whence the conclusion follows. 5. If ( X ,G ) is a compact dynamical system with a commutative group G (of homeomorphism of X into X ) , then for any open cover X = R and any 91,. . . ,Qk E G there is an n 2 1 that

+ +

u

g;"[U]

n , . . n g;"[U] # (3

for some U E R.

(Apply Ex. 4 to a minimal subsystem). 6. (van der Waerden theorem). If w = A , U .. . U A , is a finite partition, then one part contains arbitrary long arithmetical progressions. (Let g j ( z ) = I j , for j 5 k and apply Ex. 5 to (P[w], G) and the cover P [ w ] = x u . . . U where G

x,,

+

is a group generated by the g j , j 5 k). 7. (Multidimensional case). If w m = A1 U . . . U A , is a finite partition of wm = w x . . . x w (m-times), then one part contains m-dimensional progressions = { a b - d : b E k'"}, for some a E urn,d E w and any k E w . (Furstenberg. For the proof of both versions it suffices to use a theorem as in Ex. 5 for metric spaces X , only).

+

Chapter 10

Small P-set in w* $1. PROPERFORCING. We have mentioned already (Sec. 1, Chapter 6) that, consistently, there can be no P-points in the ultrafilter space w*. In this chapter we prove a stronger theorem: consistently, there can be no small P-sets in w*. The term “small” means here the Suslin property or the countable chain condition (every disjoint family of relatively open sets in countable). Let us note that many such sets exist if the continuum hypothesis is assumed. Indeed, let X be any separable compact space. Then, the Gleason G ( X ) is separable as well (7.6, Chapter 3) and, under CH, it can be embedded into the space w* as a P-set (cf Theorem 4.2, Chapter 4). In the proof we shall apply a different kind of forcing than used previously. Therefore, in this section we introduce the notions connected with so called proper forcing and quote the theorems to be used later. We generally omit the proofs, since they are too long to be included here and we refer the reader to Shelah [S).

For an uncountable set X let

P w ( X )= { A

X

:

card A 5 w }

be the family of all (at most) countable subsets of X . Although, P w ( X )is only just partially ordered by inclusion C, we can still define cub (closed and unbounded) and stationary subsets of P , ( X ) in a similar way as in the case of (tc,c),(cf Sec. 2, Chapter 5). We call a set C P w ( X )to be a cub, if for every a E P w ( X )there is a b E C such that a c b and a, for every increasing chain ( a , : n < w ) of elements of C, the union UnEw is in C.

A set S c P w ( X )is stationary, if S n C # $3, for each cub C c P w ( X ) . 243

cH.10 S M A L L P-SET IN

244

W*

Now, a forcing P, in a ground model V, is called proper, if for every uncountable . that set X in V , each stationary set S P w ( X )remains stationary in V [ G ] Notice X may have new countable subsets in V [ G ]and, even if this is not the case, there can be still new cub sets C in V[G].It is easy to see that any proper forcing P has the following property.

1.1.

If X is uncountable in V , then for every countable set a E V[G], a GX there is a countable b E V , b

cX

and a

b.

Indeed, the set C, = { b C X : a C b } , defined in V [ G ]is , obviously a cub. Of course the set S = Pu(X)in V is stationary and hence S n C, # $9, which proves the claim. It follows immediately that 1.2 the cardinal wlof V remains a cardinal in V [ G ] i.e. , wl(V) = wl(V[G]).

In fact, for any set of ordinals A E V we have

if V

"card A

> w " , then V [ G ]i= "card A > w".

In a similar way, we show that c f ( ~ )> w in V implies c f ( ~ > ) w in V [ G ] . Let us note that any forcing with the ccc is proper. Also, any countably closed forcing is proper.

For any two sets N

M , we write

N 5 M , if for every formula d(z,51,.. . ,z,) and any elements a l , . . . , a , E A4 if the condition

holds for some a E M , then it holds also for some a E N. We shall consider countable sets N such that N 5 H ( K ) for large enough K , (where as usual, H(n) denotes the family of all sets of hereditary power < K ) . Assume that a given forcing P belongs to N and call a q E P N-generic, if for every predense set E 5 P, i f E E N, then the set E n N is predense under q.

245

$1. PROPER FORCING We shall use the following characterization of N-genericity. q is N-generic if and only if for every P-name

1.3.

r E N the condition P Ik ”r is an ordinal” implies q Il- “ r E N”.

Using the above notions we can often decide whether a given P is proper. P is proper if and only if for every countable N

1.4.

(where

K

is regular and

K

5 H(K),

> 2card‘),

if P E N, then for each p E P CI N there is a q 5 p , which is N-generic.

A countable support iteration of arbitrary length of proper forcing is proper. More exactly, if P = CfJAP, is a countable support iteration such that P, Il- ” Q , is proper”

where Q, = P,+l/P,,

(Y

Also, if cf(A) < A.

for each (Y

(i.e. Q, is such that P,+1 = P,

> w and f

: w -+ w and f

< A,

* Q,),

then

P is proper.

E V [ G ] then , f E V[Gla]for some

Actually, we need a slight strengthening of properness. Call a forcing P w-proper, if we have for every continuously increasing chain (N, : i 5 w), where each Ni is countable and N, 5 H ( K ) (for , a regular K > Fard ’), and (N, : i 5 j) E N,+l for all j < w, if P E No then for every p E P n No there is a q 5 p which is Ni-generic, for all i 5 w. Obviously, an u-proper forcing P is proper. We need this more complicated notion because we want our iteration to be bounding, the latter property means that the set of old functions ww fl V dominates: for every f E V [ G ]f, : w

-+

w, there is a g E V such that

f 5.

g.

If the factors of an iteration are both w-proper and bounding, then so is the resulting P (which may fail if w-proper is weakened to proper). Thus, we have (W)

P, is an iteration with counta.ble supports and such that

If P =

1.5.

,
P, IF ”Q, is w-proper and bounding” for all

then P is both proper and bounding.

(Y

< A,

Ct1.10 SMALL P-SET IN w*

246

Finally, let us say that P has the strong PP-property, if for every p E P and a P-name f such that p Ii- ”f : w -iw” and any h : w + w , limi h ( i ) = +m there is a q 5 p and a perfect tree T G w C w , with all levels T n w” finite, and for infinitely many n, T n w” has 5 h ( n ) elements and such that q It-

”

f is a branch of T”.

Recall that a tree T is perfect if every element of T splits, ie.e every s E T has incompatible extensions t l , t 2 E T . Let us note the following fact. If P has the strong PP-property, the P is bounding.

1.6.

Thus, from 1.5 and 1.6 it follows that an iteration with countable supports of w-proper forcings having the strong PP-property is both proper and bounding.

$2. ON P-FILTERS WITH THE CCC. we know that each closed set in the remainder space w* has the form n{A* : A E F } , where F is a filter on the set w . The filters F corresponding to closed P-sets are called P-filters and can be characterized by the following condition (cf Sec. 6, Chapter 4):

if {A, i E w } C F , then there is a set A E F such that

A

s1Ai, for each a E w .

Thus, the dual ideal I = { w \ A : A E F } has the property:

if A, f I, for i E w , then there is a set A E I such that

2.1

A,

c* A,

for each i E w .

Further, the countable chain condition imposed upon the set n{A* : A E F } is equivalent to Every chain (or every antichain) in the factor algebra P ( w ) / F is Countable, and implies, in turn, that the dual I is fat in the following sense:

if Ai E I, for i E w and limminAi = w, i

(2.2)

then there is an infinite set Z G w such that

$2. ON P-FILTERS WITH TIIE ccc

247

Indeed, let us denote ei = A; \ A , where A E I is such as in 2.1. Since min A, is arbitrarily large, we can find an infinite set Y w such that the family { e ; : i E Y} is disjoint. Let {Y, : a < C} be an almost disjoint family of subsets of Y. Clearly, the unions for a < c, S, = U { e i : i E Y,} are almost disjoint each other,

S,

n Sp =. 8

for a

# p < c.

Hence, the family of open-closed sets {S: : cr < c} is disjoint and by the ccc we have

s : n n { B * : B E F )= 9

for some a

< c.

It follows that S, belongs to I and, consequently, the union

U

l€Ya

U ( A ; n A ) u U ( A ;\ A ) :€Yo = U ( A ; n A ) U S, =

=

:€Ye

;€YO

is in I as a subset of A U S, E I . This proves the claim.

For the rest of this section let us fix a given fat P-filter F and its dual I. Now, we shall define a forcing &O = Qo(F).

A partial ordering (T,ST),where T E w , will be called a tree if for every i E T the set of predecessors { jE T : j
for all i,j E T .

st

We introduce a partial ordering for trees: T S t S if and only if (S,5s) is a subordering of (T,S T ) and every branch of T contains cofinally a (unique) branch of S. Let T, denote a fixed tree which is order isomorphic to the Cantor tree (i.e. the binary tree of height w ) and such that T, E I. For any tree T let T ( n )denote the subtree obtain from T by deleting all elements

5 n, i.e. T(n) = T \ (0,. . . ,n}. Obviously we have T ( " )<-t T("')

whenever n 5 m.

cH.10 S M A L L P-SET

248

IN

W*

Let 7 consists of all trees T such that

T


TL")

for some n E w,

and T E I. Clearly, each tree T E 7 has finitely many roots. We define a forcing Q = Q o ( F ) as the set of all pairs p = (Tp,f p ) , where Tp E 7 and f,, : Tp + {0,1}. The ordering on Qo is defined thus p

How does the forcing

5 q if and only if Tp 5 , Tp and f p _> fq. &O

work? Let { b , : a

<

c} be a fixed enumeration of all

branches of To in a ground model V . For a generic filter G P let TG = UpEGTP and f~ = UpEGf p . Clearly, TG is a tree (but TG @ 7, since TG 4 I) and fG : TG + {0,1}. For every branch b, there is a (unique) branch B, of the tree TG such that B , contains b, cofinally. If bp, denotes the branch of Tp extending b,, then we have B, = UpEG&.Let us define

x, = { z

Ew : z E

B, and

fG(i)

= 1)

for a

< c.

Since Tp E I, the set (w \ Tp)n A

is infinite for each p E QOand A E F.

It follows that the set

0:.= { p E P : 3i > n[i E 4 n A and f p ( i )= E ] ) is dense for each A E F , n E w,

Q

< c and E

= 0 , l . Therefore, the intersections

X , n A and ( B , \ X,) n A are infinite for every A E F and a < c. This means that X , is a Gregorieff set lying on the branch B,. Since the branches are almost disjoint, we may say that Qo adds uncountably many almost disjoint Gregorieff sets.

For some technical reasons (Qo is not flexible enough) we shall use a countable product Q ( F ) = Q o ( F ) ordered coordinatewise. Symplifying the notation we may write for a q E &,

niEv

q=(f,":

i<w)

where(dm(f,'),f,') E Q o f o r e a c h i < w .

52. ON P-FILTERS WITH THE

ccc

249

Let wn(q), for q E Q , denotes the sequence (gi : i < w ) , where

(l!f

9i =

f?,

dm(f,P))("),

if i < n if i > n .

Obviously, wn(q) E Q for all n E w .

2.3. LEMMA. For every decreasing chain po 2 pl 2 is a q E Q and an infinite set 2 w such that q

_< wn(p,)

. .. of elements of Q

there

for each n E 2.

(f!"

i < w ) . Since minT,$') 2 n , we may apply 2.2 to define by induction a decreasing sequence Zo 2 Z1 2 . . . of infinite subsets of w such that

PROOF.Let

Tni

= dm(f,Pn), where pn =

u Ti:)

:

belongs to I for each i

< w.

nEZi

There is an infinite 2 5 w such that Z

c*Zi, for each i < w. Let us define

and

Clearly, we have Ti = dm(f,) for all i lemma.

<w

and q = (f,: i

< w ) E Q satisfies the

QED

For any q E Q and n E w let S(q, n) be the set of all sequences s = ( S O , . . . ,sn-1) fulfilling the following conditions. 1. S O , . . . ,sn-1 are finite zero-one functions. 2. The domains t o = dm (SO),.. . ,t , - l = dm(s,-l) maxto

< minT,("), . .. ,maxtn-l < minTfi)l,

whereTo = dm(f,P),...,Tn-l = dm(fi-l). 3. There exist trees Yo,. . . ,Yn-l in 7 such that . . . ,Yn-l (n) - T(n) n-l YJ") = To("), and

are finite trees such that

cH.10 S M A L L P-SET

250

IN W*

Directly from the definition of 7 it follows that the set S ( q , n ) is always finite. Let us denote (for any q, s, n as above)

. . ,Yn-l satisfying the Actually, s * wn(q) depends also on the choice of trees YO,. condition 3 above. We omit this to simplify the notation. From the construction it follows easily that the set {s * wn(q) : s E S ( q , n ) ) is predense under wn(q), i.e. that

C{S* wn(q) :

s

E S ( q , n ) }= wn(q)

(the Boolean sum).

Now, we can prove a lemma from which the properness of Q will follow. 2.4. LEMMA.For any p E Q , n E w and a Q-name T such that Q ll- ” r is an ordinal” there is a q 5 p and ordinals a ( s ) , for s E S ( p , n ) , such that wn(q) IF ”

v

r = a(s)”.

8

PROOF.Let S(p,n) = {so,. . . ,s”-l}. &om the assumption Q II- ” r is an ordinal ” it follows that the set {q

E Q : 3a (a is an ordinal

Aq

ll- ” T = a”)

is dense in Q. Using this we define by induction elements PO,.. .pm E Q such that p o = p , pi+l ~ s ’ * w , ( p , ) f o r i = O , ...,m - l a n d p;+1 ll- ”T = a”

for some ordinal a = a(.’). Let s be such that p = s * w , ( p ) . We may assume that s E S ( p , n ) and define q = s * wn(pm). Obviously, q has the required properties. QED The expected properties of Q are stated in the following 2.5. THEOREM. The forcing Q is w-proper and has the strong PP-property.

52. ON

251

P-FILTERS WITH THE CCC

PROOF.Let K be a regular cardinal, K > 2cardQ, and choose a countable N 5 H ( K )such that Q E N . We want to use 1.4. So let p E Q n N . We have to find a q 5 p which is N-generic. Let { T :~ n E w} be an enumeration of all Q-names such that T~ E N and Q II- ” T ~is an ordinal ”, for all n E w. Using the lemma 2.4 we define by induction a descending sequence po 2 pl 2 . . . and ordinals a ( n ,s) E N such that wn(pn) II-

”

AV

~i

= a(n,s)”

for each n E w.

i
In the n-the step of induction we apply the lemma for all names TO,. . . ,T ~ Notice . that the p‘s and the a’s can be found in N , since N 5 H ( K )and the lemma 2.4 is , Q E H(K)). obviously true in H ( K )(assuming By the lemma 2.3 there exists a q E Q such that for some infinite set 2

w we

have q

Hence, we have

5 wn(pn)

for each n E 2.

also

w,(q)

5 wn(pn)

for arbitrarily large n and all m.

It follows that wm(q) II- ”T,,E N”

for all n,m < w.

Thus, every wm(q) is N-generic, in view of 1.3, and consequently Q is proper. To see that Q is w-proper let (N, : i 5 w) be a continuous increasing chain such that the N’s are all countable and 5 H ( K ) and (Ni : i 5 j ) E Nj+l, for each j < w, and Q E NO. Let p E No n Q. Assume inductively that we have a q 5 p such that q and all the wn(q) are Ni-generic for i = 0 , . . . , j . Since Nj+l 5 H ( K )we can find such a q in Nj+l (because all necessary parameters are already in Nj+l). Arguing as above we construct a qj+l 5 q which is Nj+l-generic and so are all the wn(qj+l),

+

n E w. Hence, qj+l and all the wn(qj+l) are Ni-generic for i = 0,. .., j 1. Thus, we may assert that there is a decreasing sequence p 2 qo 2 q1 2 . . . such that for every j E w, q j and all the wn(qj) are Ni-generic, for all i 5 j and are in Nj. By the lemma 2.3 we find a q E Q such that q 5 wj(qj) holds for infinitely many j . Thus, q 5 p and q is Ni-generic for all i < w and hence also for all i 5 w. This proves that Q is w-proper. Finally, to check that Q has the PP-property, suppose p II- “f : w + w” and let

h : w -+ w be a function diverging to infinity. Set

k, = min{i E w

:

h ( i ) > 2“ . card S(p,n ) }

for n E w.

cH.10 S M A L L P-SET IN w*

252

Using the lemma 2.4 we define by induction a sequence p = po 2 pl 2 w,(p,)

It- ”

A

v

. . . such that

f(i) = a ( s ,4’’for each n E w and some integers a ( s ,i).

i
Let T be the tree built up of the integers Q(S, i) for i E w and s E S(p, i) in an obvious way (the elements a ( s ,i), for s E S ( p , i), form the i-th level of T so that each finite branch ( ( Y ( Q ~ , O ).,.. a ( s i , i ) )is a possible initial segment of f). Of course, all levels of T are finite and the k,-th level has less than h(k,) elements, for all n < w . Now, if q E Q is such that q

5 wn(pn)

for infinitely many n,

then we have q It- ” f is a branch of T”, which finishes the proof.

QED

Now, we can prove the main theorem.

2.6. THEOREM. (Frankiewicz, Shelah, Zbierski [FSZ]). There is a model of set theory in which c = w2 and the remainder space w* has no P-set with the ccc.

PROOF.Let us choose a ground model V in which we have c = w1 and 2w1= w2 and the diamond principle for H(w2): there is a sequence (T, : c f ( a ) = w1 and a < w2) such that for every X E H(w2) the set { a c w2 : cf(a) = w1

and X n H, = T,}

is stationary.

As usual, H ( w 2 ) = U,H, is a decomposition into a continuously increasing chain. Let us define by induction an iteration P = C:;,, P,, with countable supports, as follows: P, = EgoPg P a + l = Pa * Q o

Po+1 = Po

(limits with countable supports if a is a limit ordinal if cf(a)= w1 and T, is a Pa-name of a fat P-filter and Q, is a Pa-name of the associated forcing Q(To) in all remaining cases.

Let G P be a generic ultrafilter. In view of 1.5 and the theorem 2.5 P is proper and bounding. Since card P, 5 w l ” , for all a < w2, we have and

V[Gla] ”c = w1

V [ G ]!= ”C = ~

2 ” .

for all a

< w2,

$2. ON P-FILTERS WITH T H E

CCC

253

We shall prove now that there are no P-filters with the ccc in V [ G ] .Suppose the opposite and let F E V[G]be a P-filter having the ccc. In particular, F is fat. Denote Fa = F n V[GIa].We show that Fa E V[Gla]for cub-many a. Choose a canonical name F H(w2) of F consisting of some pairs ( a , p ) ,where a is a P-name of a subset of w and p E P. The set c 1=

{a < w2 : Va[a E V(Pa)+ F ( a ) c P a ] }

is an w1-cub, which can be proved similarly as in the theorem 2.5, Chapter 8 (notice that card F(a)5 wl). Clearly, F , = F n (V('a) x P a ) is a Pa-name and

and hence Fa E V[Gla],for all a E C1. Moreover, we may assert that the Fa's are fat P-filters, because it is so on an wl-cub subset of C1. Indeed, if R = {A, : i E w } E Fa, then there is an A in F n V [ G ] such , that A c* A,, for all i E w. We know, (cf the remark after 1.4), that A E V[Glp],for some p < w2. Since there is 5 c = w1 countable subfamilies R of Fa, there is an ordinal a* E C1 such that in V[GIa*]we can find some lower bounds A E F for all such R.. Now define a0 = a,ae+1 = a; and Q A = sup ~ < x for a ~limit A. If /3 = sup e<wrae,then p E C1 and Fp E V[GIP]is a P-filter. Obviously, the set { p E C1 : Fp is a P-filter} is wl-closed. In a similar way we show that Fp is fat on an wl-cub subset C1, which proves the claim. A standard reasoning shows that the set

c

is an wl-cub. Hence, C1 n Cz is an wl-cub. By the diamond principle applied for X = F, there is an a < w2, with c f ( a ) = w l , such that T, is a P,-name of the fat P-filter Fa. By the definition, we force with Q(F,) at every such stage a, i.e. Pa+l = Pa

* Q(Fa).

Now, we shall derive a contradiction. Since F, such that p li- "F, E F".

F in V [ G ] ,there is a p E P

Recall that Q(Fa) is the countable product Q t ( F , ) and Q0(Fa) adds uncountably many Gregorieff sets (with respect to Fa).Let XF be the [-th set added by the

ClI.10 S M A L L P-SET IN

254

W*

n-th factor of Q(Fa). Since F has the ccc, the relation X ; E F holds for at most countably many t ‘ s , for any n. Hence, there is a ( such that for all n E w we have w \ X; E F. Since F is a P-filter, there is a set A E F and a function g E V[G] such that A nnEw(w \ X ; ) U [O,g(n)). Thus, for some p1 5 p we have

The same will hold with any function f 2 g replacing g, so we may assume that g E V (because P is bounding). The a-th term q = p l ( a ) of p is a Pa-name of a sequence (f: : n < w ) E Q(Fa). Since, tn = dm(f:) are in I (the dual of F), there is a set B E I and an increasing sequence of integers k~ < kl < . . . such that for all n E w we have

We define an element q‘ 5 q as follows. For a given n E w , we extend tn by adding elements of [k,, k n + l ) \ B on the C-th branch bz and set f i ( i ) = 1, for each

iE

[kn,h + l )

\ B.

Obviously, for every n E w , we have

where r = (plcr) * q’ r It- ”

* ( p ( r ] ): a < r] < wz). Hence

n

(w

\ X T ) u [ O , g ( n ) )\ B n

nEw

u

[kn,k n + l ) = 8”.

nEw

It follows that r IF l1

(W

\ x;)u [o,g(n))c B”.

nEw

which contradits (*), since B E I. This finishes the proof.

QED

It can be proved that, consistently, c = w2 and there is no P-set having C-cc, (every family of relatively open sets has power less than c), in the space w*.

255

PROBLEMS PROBLEMS A. MEASUREON P(w).

A measure means here a finitely additive measure m : P(w) + [0,1], with m ( { i } )= 0 for all i E w and m(w) = 1. 1. Measures on P(w) correspond in a one-to-one way to finitely additive measures on P(w)/fin, with m(1) = 1 and also to a-additive Bore1 measures in the space w*, with m(w*) = 1. Thus, all three notions become the same. 2. Any p E w* determines a zero-one measure

and each zero-one measure has this form. The support supp ( m )is defined (passing to w * ) as the set of all z E w* such that m(U)> 0, for all neighbourhoods U of z. For example supp ( m p )= {p}, for each p E w*.

3. For any rn, supp ( m )is the Stone space of the measure algebra P(w)/A, (where A, = { A w : m(A) = 0)). Hence supp ( m )= n { A * : m(A) = 1). 4. The support of m has always the ccc and hence is a (closed) nowhere dense set.

A measure m is said to have AP (the additive property), if (in terms of P(w)/fin): every countable antichain { u , : i E w} such that m(a) = CiEw m(ui).

P(w)/fin has a.n upper bound u

Note that the u in the above is unique up to the measure null. 5. A measure m has AP if and only if every decreasing chain { bi : i E w} has a lower bound b such that m(b) = inf, m(b;).

P(w)/fin

6. A measure m on P(w) has AP if and only if for every disjoint family { A , : i E w}

there are sets A: =* A,, for all i E w, and such that iEw

iEw

256

cH.10 SMALL P - S E T IN

id*

7. If m has AP, then the support supp (m) is a closed P-set having the ccc. Hence, the corresponding filter (= {A w : m ( A ) = 1)) is a fat P-filter. 8. If p E w* is a P-point, then the zero-one measure inp has AP.

B. LINEARFUNCTIONALS cp:

ON

IF.

By a linear functional on I" = C*(w) we mean here a linear bounded functional I" + R such that cp(f) 2 0, for f 2 0; cp(x.) = 0 for s E fin and cp(1) = 1.

1. If f E CO, then cp( f ) = 0. More generally, if f E 1- is convergent, then cp(f) = limi f(i).

2. The mapping f H f * induces a one-to-one correspondence between linear functionals on "I and linear positive functionals on C(w*). An integral for f E "I

with respect to a measure m on P ( w ) is defined in usual

way:

sw

.s,

3. We have f ( i ) m ( d i )= f*(p)E(dp), where Tst is a a-additive counterpart of m in w * . Hence, the Ftiesz representation theorem holds for functionals and measures on w. We say that a linear functional cp has AP, if for every sequence fn 2 0 having a pointiwise bounded sum fn E Im there axe gn =. fn such that

En

n

n

4. A measure m has AP if and only if the corresponding functional cp, (i.e. p(f) =

swf ( i ) m ( d i ) ,f E I"),

has AP, (Mekler [MI).

c. ASYMTOTIC DENSITY. The r e d d ( A ) = limn $ . card A n [0, n) (whenever exists) is called the (asymptotic) density of an A C w . 1. d is an additive function on its domain. 2. The family D = {A C w : d ( A ) = 1) is a P-filter. (Given

{Ai : i E

W }

ED,

PROBLEMS

257

define k, E w such that

-n1

*

1 card A, n [0, 6") 2 1 - n

and set A = UiewAi n [ki, ki+l)). 3. No ultrafilter p E w* extending D is a P-point. (Otherwise p contains a decreasing chain A0 2 A1 _> . .. with d(An) = 2-". Notice, that d ( A ) = 0 implies d(w \ A) = 1). Let C be the Cesaro matrix, i.e. for f E

I".

4. C : "I -t "I is a linear operator and ~ C ( X A = d(A), ) for any A having density and any linear functional y .

Hence, there are linear functionals density.

'p

such that

' ~ ( x A )=

d(A), if A has the

5 . The density d can be extended to a (translation invariant) measure on P(w).

6. If a linear functional

'p

has AP, then yC has AP, as well.

Hence, if there is a P-point, then there is a continous measure m having AP (A.8 and Ex. 5). (Mekler [MI).

D.MORE ON

P-POINTS

Let X denote the family of all infinite sets z c w . For a E fin and A E X define

(.,A) = {z E X : z c a U A and z n m a x a + 1 = a ) For any F E w * , let X F be the topological space X with the topology generated by the sets (.,A), where a E fin and A E X . 1. The intersection then

( a , A)

n (a, B) is nonempty if and only if (a, A)

n (b, B) = (a U b, A n B).

Hence, the ( a ,A)%constitute a base. 2. The space X F has always the ccc.

a

b or b

c a and

CH.10 S M A L L P-SET IN

258

W*

3. Assume that the Baire theorem holds true in the subspace F 2 X F . a)

Let Ai E F, for i E w . If we set for i E w,

Ri = {(a,A,) : a E fin}

then the sets Ui = U R i n F are open dense in F. b) If A E

niEv

U i , then

A

c* A;, for all i E w. Hence, F is a P-point.

4. Assume now that F is a P-point and let the sets that Ui n F are dense in F, for all i E w .

U,be open in X F and such

a) Let Ri be a maximal family of basic open subsets C Ui. Then, the union U R i is dense in Ui and we may write (Ex. 2), .

.

Ri = {(UL~AL) : k E w}. b) By assumtion, fix an A C, A t , for all i, k E w . We have

($?),A) (aOko,AOko)# Q,

for Some ko.

+

If no = max(A \ A:O) 1 and

then

10

# Q, and A n [0, no) = z0 U Y O , 10 n YO = Q,.

c ) Inductively, if n , , z , , y i are already defined and i is even, then let such that

ki+l

be

( y i , A) n ( a ~ ~ ~ 1Z Q, , ~ ~ ~ ~ 1 )

and let ni+l

= max{min(A \ max(A \

yi+l

= Y i U [A n [nirn i + l )

+ 11,min(A \ n i ) + 1)

n (aiT$l U AiT:1)]

zi+l = A n [ni,n i + l ) \ Y i + l . Then, yi+l is nonempty. If i is odd the definition is symetrical with the positions of zi and yi reversed.

d) If I = UiEwI , and y = UiEwy , , then A = I U y is a partition of A into two U, n F . infinite disjoint sets. Either I or y is in

niEw

Hence, X F satisfies the Baire theorem. From 3 and 4 we obtain the following theorem:

F is a P-point if and only if the Baire theorem is true in the subspace F Of X F .

Chapter 11

Forcing $1. S E T THEORY A N D ITS usually in the following way:

MODELS.

Theorems on consistency are formulated

1.1. The property (expressed by the formula)

4 is consistent with set theory

Equivalently: 1.2. There exists a model of set theory in which

4 holds true.

The intuitive meaning of such theorems is clear. In order to describe the proof technique we define more precisely the logical notions occuring in 1.1 or 1.2. By set theory we mean the axiomatic system of the theory of sets of Zermelofiankel. The primitive notion of the theory is ”set” and the primitive relation is ” E ” (membership). Some fundamental properties of sets are stated and assumed as axioms, from which all the theorems can be derived. The explicite form of the axioms is not important here. It is sufficient to note only that the axioms ensure: 1. Feasibility of set-theoretical operations of pair { A ,B}, union UR (of an arbitrary family R), intersetion R, subtraction A \ B, power-set P ( A ) , product etc. By the axiom of choice every set can be well ordered and we have the maximum principle of Kuratowski-Zorn.

n

2. Definability of subsets: for any set A and any propety expressed by a settheoretical formula #(z) there exists the set { u E A : 4 ( u ) } consisting of all elements a E A with the property 4. 3. Feasibility of inductive processes: for any function-like formula +(z, y) (i.e. 4 is such that Vz3!yd holds) let b = F+(a) denote the unique element b such that 4(u, b) holds. Then, for every ordinal X there is a unique sequence of sets ( A , : (Y < A) such that A , = F+((Ap : ,B < a ) ) ,for all a < A.

259

Ci1.11 FORCING

260

Using this apparatus we develop, in a known way, the algebra of sets, the theory of powers and orderings and we build up all standard number and function systems (beginning with non-negative integers) together with their algebraic and topological structure. In this way, set theory includes all the classical mathematics (i.e. any mathematical theory can be interpreted as a part of set theory). Therefore, the consistency of a given property with respect to set theory has a universal character. How we can imagine the universe of sets? It can be proved (in set theory) that to every set I there corresponds a unique ordinal a = r(z), called the rank of z, so that r(8) = 0 and

1.1.

r(A) = sup {r(a)

+ 1: u E A }

for nonempty sets A.

For example, we have .(a) = a,for every ordinal a. The family R, = {A : r( A) < a} is transitive (i.e. U R , E R,), and the rank hierarchy (R, : a E On), (where On denotes the class of all ordinal numbers), is a continuously increasing sequence satisfying

The universe of sets is equal to the union

U { R a:

E On} = U{R,+I \ Ra

: a

E On},

i.e. every set A occurs at some level of the rank hierarchy:

A E Ra+l \ R,

where a = r(A).

$1. SET

261

THEORY AND ITS MODELS

It follows also that every set A determines a tree TA (describing the structure of A) in which the set A lies on the upper vertex, on the first level below there are elements a of A, below - the elements of elements etc. Every branch in the tree TA has a finite length and terminates with the empty set 0. Now, we shall discuss the other logical notions: formulas, models, and consistency relation. Set-theoretical formulas can be regarded and defined in set theory (as some specific sets) in the following manner. We fix a countable set X(say X = w ) , whose elements will be called variables and a finite set S = { = , E l 1 , +, V}, whose elements are denoted suggestively according to their intended meaning (we may assume e.g. S = { ( i , 8) : i = 0,. .. ,4)). To make the definition shorter, we omit the other connectives and the quantifier 3. Atomic formulas are then defined as triples (I, =, y) or (I, E, y), where I, y E X. More compound formulas are obtained from atomic ones by an application of logical connectives and quantification. Thus, the set F of formulas is the union 7 = U n E w 3 , where , the sequence (3,,: n E w ) is defined by induction as follows: 2-0

= the set of atomic formulas

F n + l = F n u { ( l l d ) : 4 ~ F n I u { ( 4 1 + 1 4 ) 4: 1 4 ~ 3 n I U U { ( V , I , ~:) I E X and 4 E Fn}. Obviously, we simplify the notation and write e.g. I E y instead of (I, E , y , ) , V1q5 Let us note that .F E R,, i.e. any formula 4 E 9 is a instead of ( V , I , ~ etc. ) hereditarily finite set. It is clear that each set-theoretical property can be encoded by some 4 E conversely, any 4 E 3 is (a code of) some property.

F

and,

We can easily define when a variable I occurs in a formula 4 and whether it is bounded or free in q5 in a well known sense. Thus we have a map 4 H v(q5) ) always assigning to any 4 the set of free variables of 4. Of course, the set ~ ( 4 is . to indicate that v(4) = ( 1 1 , . . . ,z,,}. If finite. Customarily we write ~ ( I I , .. ,z,) v ( 4 ) = 0,then 4 is called a sentence. Thus, any formula d(z1 ,.. . ,I , ) expresses some property of 1 1 , . . . ,I , and its truth depends on what specific sets are substituted for q,.. . , z n 1while the property expressed by a sentence does not depended on parameters.

CH. 11 FORCING

262

We denote by ZF the set of sentences corresponding to axioms of Zermelo-Frankel set theory. Let M be any nonempty family. According to the definition of Tarski, a truth relation for M is a relation denoted by M

(s satisfies 4 in M or 4 of s is true in M),

4.91,

between formulas 4 E 3 and substitutions s : u ( 4 ) conditions: 1. For any variables z,y M

I=

(z = Y)[4

M

I=

(I E

M, satisfying the following

+

iff

.(z) = 4 Y )

Y"1

iff

4I) E 4Y).

M

I= (-4)[4

iff

n0nM

I= 4 . 1

M

I= (4

iff

nonM

k d444)I

and

2. For any $,$J E F and +

$)[sI

or

M

t= $[44$>1*

E 3 and a variable I 4 u ( 4 )

3. For any

t=

M I= Vzdsl and if z E u(+), then

iff

M

I= VzdsI

iff

for every a E M, M

M

where s(a/z)(y) =

4Y)l a,

4 4 1

I= +[s(a/z)],

ifY # 2 if y = I .

It is easy to prove that each nonempty set M has (exactly one) truth relation.

4 is a sentence, then the only possible substitution is s = Q, (because 8 ) and we write M 4 (4 is true in M) instead of M +[@I. If (z1,...,zn} and a ] ,... , a , E M, then M k [a1/11 ,. . .a,/z,] or simply If

u(q5) =

M

I= d [ a l , .. . ,a,]

means M

v(4) =

4[s], where s(z,) = ail for i = 1,. . . ,n.

To understand better the truth relation let us denote by q5M the set theoretical property encoded by a formula q5 E 3with a simultaneous limitation of all quantifiers to the set M (i.e. each subformula of the form Vz$ is replaced by Vz[z E M + $1). If u ( 4 ) = {zl, . . . ,I,}, then for arbitrary elements 01,. . . ,a, E M we have M #[a~/z~,. ..a,/z,] if and only if 1.3. 4 M ( ~ l. ,. , . a,), i.e. $ M holds for a1 ,.. . ,a,.

+

$1. SET THEORY A N D ITS

MODELS

263

1.4. EXAMPLE. Let d(x) be "Vy+ E x)", i.e. q5 defines the empty set 8. If, q5[a],then we write a = O M and call a the empty set of for an a E M , we have M M , (since a behaves in M as 8 in the universe). A similar notation is used also for other definable notions.

+

Applying 1.3 we obtain (for any a E M ) :

1.5.

a = (DM

iff M

b 441.

iff V m E M [ m 4 a] iff a n M = 8 .

Hence, if a is an arbitrary set and e.g. M = { a } , then we have a = (B.' On the other hand, if M is transitive, then 1.5 implies immediately that O M = 8 . 1.6. DEFINITION. A family M is said to be a model of a set S of sentences if each q5 E S is true in M :

M

q5

for every q5 E S.

Any model of S is also a model of all theorems which follow from S. A set S is consistent, if for no 4, both q5 and Y#J are theorems of S. Clearly, if S has a model, then S is consistent. We call a sentence q5 to be consistent with S, if the set S U { q5} is consistent. We omit here a precise definition of a theorem, since our consistency proofs go always via construction of a model in the sense of 1.6 for S =ZF (the set of axioms of set theory). This clarifies the meaning of the notions occuring in 1.1 and 1.2. Let us note however, that the existence of models of ZF cannot be proved in set theory itself (this follows from a celebrated theorem of Godel). For the existence proof we need some additional means, for example, it suffices to assume that inaccessible cardinals exist. Indeed, if K is inaccessible, then it is easy to see that the set R, is a model of set theory. It can be proved that if there is any model M of set theory, then there is also a countable and transitive model V . Troughout the book we assume, that "model" means a countable and transitive model of Zermelo-Fkaenkel set theory and assume also that such a model exists. As remarked above the latter assumption is necessary for consistency proofs. Now, let us call a given formula q5(x1,. . . ,x,) absolute, if for every (countable, transitive) model V and arbitrary a l , . . . , a , E V we have

V

b [ a l , .. . ,a,]

iff

q5(al,.

..,a,)

holds.

cH.11 FORCING

264

In Example 1.4 we proved that the formula q5(z) defining the empty set is absolute. It is easy to check that absolute are formulas 4(z,y, z ) defining the following relations

etc., and also the formulas is a function mapping G into y) (z is an ordinal number) (z is a nonnegative integer) (z is the set of nonnegative integers) (y is the family of all finite subsets of z). (z

" 2 : G + y"

04x) "2

E w')

"2

= w"

"y = fin(z)"

In particular if we denote (for a model V) On(V) = { a E V : V

+ On[a]},

then we have On(V) = On n V, i.e. the ordinals in the sense of a model V coincide with the ordinary ordinals belonging to V. By transitivity, the set On(V) is an initial segment of the whole class On. Similarly, we have (for f,A, B E V)

f

1.7.

: A +B

iff V

b " f :A + B",

i.e. functions in the sense of V coincide with ordinary functions in the universe, which belong to V. On the other hand, the power-set operation y = P ( G )or, more precisely, the formula 4(z,y) : Vz[z E y

=

z

s z]

is not absolute. Indeed, we have V

+ 4[a,b]

iff Vw E V[v E b

=

v

s a]

iff b = P ( a ) n V ,

which means that the set P(a)' (the power-set of a in the sense of V) equals to P ( a ) n V. Obviously, P(a)' = P ( a ) n V # P ( a ) for any infinite a E V, since P ( a ) n V is countable (as a subset of V), while P ( a ) is uncountable. Of course, the set b = P(a)' is uncountable in the sense of V, i.e. we have V $[b], where $(z) is a formula " Iis uncountable". Since always w E V (because w v = w ) , no function f : w "3 b belongs to V , by 1.7. This proves that the formula +(I) is not absolute.

$2. FORCING Finally, let Card (z)be the formula " 2 is an infinite cardinal number denote Card (V) = { K E V : V Card [ K ] } .

265

".

Let us

Since every cardinal is some ordinal, we have Card ( V ) C On(V). Finite cardinals n E w and w are absolute: w v = w and nv = n, for each n E w . Uncountable, in the sense of V,cardinals K E V are all some countable ordinals (and hence are not absolute). Let M be another model such that V C M and having the same ordinals V n On = M n On. Clearly, we have Card ( M ) C_ Card (V), but the converse inclusion need not hold. It may happend that e.g. w," (the smallest uncountable cardinal in the sense of V ) is no longer a cardinal in M (however it remains to be an ordinal in M ) , since there can be a function f E M , f : w + wy. If so, then also all sets A E V of cardinality w1 in V will be countable in M .

$2. FORCING. With help of the forcing method, (invented by P. J. Cohen), we can construct from a given model V another model with some prescribed properties.

A partial ordering (P, 5 ,l),in which 1 is the largest element, is called a forcing, if it satisfies

2.1. if non p 5 q, then there is an r 5 p such that r I q. Here r I q means that for no s E P we have s 5 r and s 5 q.

If A is a Boolean algebra, then every dense subset P E A (containing 1) is a forcing (set r = p - q, in 2.1.). Conversely given a forcing P, assume the sets

as basic open. The condition 2.1 implies that each Upis an open domain. Obviously, the map p H Up is then an order embedding of P onto a dense subset of the complete Boolean algebra D(P). Thus, forcings are the same as dense subsets of complete algebras (up to an order isomorphism). Note that each complete algebra is uniquely determined by any of its dense subset (cf Ex.D3, Chapter 1). Hence, we may use Boolean terminology, e.g. p Iq ( p , q are incompatible) means p . q = CD, in Boolean terms.

It follows that compatibility relation pllq

iff 3r E P[r 5 p and r 5 q]

cH.11 FORCING

266 is equivalent to p . q > CD.

If F C D(P) is a filter, then the trace G = F n P is called a filter in P. Thus, a set G C P is a filter in P, if G # (B and the following conditions are satisfied a) if p E G and p 5 q, then q E G

b) if p,q E

G,then there is an r E G such that r 5 p and r 5 q.

The notion of forcing is absolute, ”P is a forcing” for any model V and P E V.

P is a forcing iff V

On the other hand, for any A E V, the condition V

+ ”A

is a complete Boolean algebra”

is equivalent, to the following: A is a Boolean algebra and sup X and inf X exist, for every set X C A, which is in V. Let P E V be a forcing. A filter G P is called generic (or complete) over V, if it satisfies

c

c) G n D # (B, for every dense set D E P, D E V. Generic filters G P are traces of V-complete filters F satisfying if X F and X E V, then infX E F.

c

c D(P), i.e.

filters F

c

The condition c) above can be replaced by c’) G n E # (B, for every maximal antichain E C P, such that E E V. Obviously, by an antichain in P we mean any set of pairwise incompatible elements. Thus, E E P is an antichain in P if and only if E is an antichain in D(P),in a usual sense.

c

Every generic filter G C P is maximal (it is the trace of an ultrafilter F D(P)). Generic (over V) filters always exist, in fact, for each p E P there is a generic G C P such that p E G (this follows from countability of V).

2.2. CONSTRUCTION OF A GENERIC MODEL. Now, we describe the construction of a new model from a given V and a forcing P E V. Usually, V is called the ground model in this case.

267

52. FORCING

First, we define a subclass V(')

V , of P-names, as the union

v(') = U{viP) : a E On n v } , where the sets V;"

E V are defined by induction as follows Vd"

v$ v;"

= {@I, = P(V;') x P) n = U{VjP' :

v

p < a}

for a E On n V , for limit a.

Thus, every P-name T is a relation between some P-names of lower rank (where rank ( T ) = min{a : T E V , " } ) and elements p E P. Let us denote (as for any relation) dm(7) = {t E V(') : 3 p E P[(t,p) E

TI}

and for t E dm(7). ~ ( t=){p E P : ( t , p ) E T } The value T[G]for a filter G C P is defined by induction on the rank of T by the equations 2.3.

Q,[GI=Q, T[G]= { z [ G ]: ~ ( zn)G #

0).

Let V(')[G](or shorter: V [ G ]be ) the family of all values 2.4.

V(')[G]= {T[G]: T E V"'}.

2.5. SOME P-NAMES OF IMPORTANCE. For a E ii by induction on the rank of a by the equations

v define the standard P-name

-

8=8

-a = {(z,1) : z E a } .

Clearly, Z[G]= a (by 2.3) and therefore V C V [ G ]for , any filter G 5 P. Let J? = { ( F , p ) : p E P}. Obviously, we have r[G]= G and therefore G E V[G]. For any z,y E V(') let

CH.11 FORCING

268

Then { ~ , y } (E~V(' ) ) (the Boolean pair) and

Similarly, if we set (for any

Z,

y E WP))

then we have (ZIY)'p"Gl = (4GllY[GI). In particular, V[G] is closed under pairs. Fkom 2.4 it follows that the set V[G] is transitive. For generic ultrafilter G the following

P, V[G] is a model. Its basic properties are stated in

2.6. THEOREM. I f P E V is a forcing in a model V, then for any generic (over

V) filter G E P, the family V[G] is a model such that V C V[G] and G E V[G]. Moreover: 1. The filter G generates V[G] over V, i.e. if a model M is such that V M and G E M, then V[G] C M. 2. On n V = On n V[G], i.e. the models V and V[G] have the same ordinal numbers. It may happen that G E V and then V[G] = V. This is exactly the case of a principal ultrafilter G. A forcing P is called atomless, if the corresponding algebra D(P) is such, i.e. if Vp E P3q E P[q < p]. In atomless algebra each ultrafilter is nonprincipal. Hence

2.7. I f P is atomles, then G E V[G] \ V and hence V[G] # V. In order to prove that some particular sentences are true in V[G] we often use the forcing relation p IF q 5 ( q ,. . . ,.m) defined as follows

,. . . ,zn)is an arbitrary formula and where p E P, ~$(q course, G runs over generic (over V) ultrafilters.

TI

,. . . ,7,

are P-names. Of

$3. COMPLETE EMBEDDINGS

269

The following useful1 equivalence expresses the fundamental property of It-: 2.8.

V[Gl

k TI [GI, . . . ,.r[Gl]

iff 3 p E G P It-

. . . ,T,,).

4(~1,

Forcing relation is definable (in a sense) over V. More exactly: for an arbitrary formula 4(x1,. ..,z,) and any a E On n V the set

K,(4) = { ( p , 7 1 , . . . ,T,,)E P x VLP' x .. . x VLP'

: p It- qqT1,.

. . ,T,)}

is definable over V and hence Ka(4)E V. Using this we can prove that some dense subsets D C P, defined with help of the K a ( 4 ) ,belong to the ground model V. P is said to be dense under p , if Vq 5 p3r 5 q[r E D] holds, i.e. if D is dense in { q E P : q 5 p } . We have the following (for any generic G):

A set D

if p E G, D is dense under p and D E V, then G n D # $3. Below, we list some useful properties of the relation It-.

2.9. GENERAL RULES.

(we write 4 omitting the parameters TI,...,T,,).

a) if p II- 4 and q 5 p , then q It- 4; b) if the set { q : q It- 4} is dense under p , then p It- 4; c) if 4 follows from 41,. . . ,dm and p It- 41 A . . . A d,, then also p It-

4.

SYNTACTICAL RULES.

a) p 1- V X iff~for all T E v(P), p IF ~ ( T / z ) ; we have p It- ~ ( T / z ) ; b) if p It- 324,then for some T E c) plI- $1 A .. . A dm iff p b 41 and.. . and p b ;4, d) pII- -4 iff Vq 5 p non q It- 4; e) if p It- 32[2 E T A 41, then there is a q 5 p and a t E dm(7) such that 9 It- 4(t/.).

Let us note also the following property

2.10. U P It- 324 andp It- ~ ( T / z then ) , thereis a P-name T * such that P It- ~ ( T * / x ) and p It- "T = T*". Clearly, P It- 4 means p It- 4 for all p E P, which is equivalent to 1 II-

4.

$3. COMPLETE EMBEDDINGS. Let P, Q be two forcing and f : P + Q an order isomorphism. Assume that f , P, Q E V so that the isomorphism takes place in V.

CH. 11 FORCING

270

Then the mapping G I+ f [ G j is a one-to-one correspondence between generic (over V ) ultrafilters in P and that in Q. The models V(')[G] and V ( Q ) [ f [ G are ] ] identical. More generally, let f : P +Q where f E V , be an order embedding. We say that f is a complete embedding if, in addition, the images f[E]of maximal antichains E E P, E E V are maximal antichains in Q . Complete embedding are traces of Boolean embeddings F : Z)( P) Z)( Q ) preserving sup's, sup F [ X ]in the subalgebra F [ D ( P ) ]= sup F [ X ] in D ( Q ) , for every X

Z)(P), X E V .

For a complete embedding f : P + Q , the counter image G = f - ' [ H ] of a generic filter H E Q is generic. Moreover, f induces a map f. : V(') + V ( Q ) defined inductively by the equations

f48)= 8, f*(r) = {(f*(.),f(P)) : ( G P )E .I. Then we have

.[GI = f,(.)[H] and hence V(')[G]

where G = f - ' [ H ] ,

V ( Q ) [ H ]In . addition

for every absolute formula q ! ~ ( q , . . . ,zn) and P-names

.. ,sn,

71,.

Let, us above, H Q be a generic over V ultrafilter and let G = f-' [HI.If we denote Q/G = { q E Q : qllf[G]},then H E Q / G and H is a generic over V(')[G] filter in Q / G and

V'Q'[H]= ( V ( P ) [ G ] ) ( Q / G ) [ H ] . If the image j [ P ] of an order embedding f : P + Q is dense in Q , then f is a complete embedding, the correspondence G = f-' [HI is one-to-one, 3.1 holds for all formulas and the both forcings P and Q produce the same models. We say that forcings P and Q are equivalent, if the completions D(P) and D ( Q ) are isomorphic. Of course, equivalent forcings yield the same models. As stated above if P is embeddable (in V ) onto a dense subordering of Q, then P and Q are equivalent. Let us note also that any two countable and atomless forcings are equivalent.

$4. CARDINAL

NUMBERS

271

NUMBERS. When cardinals of a model V remain cardinals of $4. CARDINAL V[G]?We say that a forcing P E V has the ccc (or the Suslin property) in V if every antichain E P, E E V is at most countable in V .

A sufficient condition for absoluteness of cardinals is given in the following 4.1. THEOREM. If P E V has the ccc in V then the models V and V [ G ]have

. C ~ ( I remain E) unchanged, for the same cardinals, Card ( V ) = Card ( V [ G ] ) Also, K E Card ( V ) . A forcing P is said to be (countably) closed, if every descending chain po 2 p l 2 has a lower bound: there is a p E P such that p 5 p,, for all n E w . Of course, the phrase "P is closed in V" equivalent to V

...

+ "P is closed"

means that all decreasing chains from V , ( p , : n E p ) E V , are bounded from below.

4 . 2 . THEOREM. If P E V is countably closed in V , then w," = wy[G1 (i.e. w," remains a cardinal) and

B" n V = B" n V [ G ] In other words, i f f : w

B and f

E

for each set B E V.

V [ G ]then , f EV.

In particular, for countably closed P, we have

v = P ( W )n v [ G ]

~ ( wn)

and (0, l } wn V = (0, I } n~V [ G ] .

Thus, the Cantor space is the same in the both models V and V [ G ] . The theorems 4.1 and 4.2 can be generalized for higher powers aa follows. A Suslin number of P is defined as the smallest cardinal X exceding (strictly) the powers of antichains E P. Thus, the ccc means the Suslin number = w1. Now, if

V

"Suslin number (P) = A",

then every cardinal IE 2 A of V remains a cardinal in V [ G ] . Similarly, a forcing P is A-closed, if any decreasing chain of length < X has a lower bound. Then, if V "P is A-closed",

+

CH. 11 FORCING

272

then every cardinal

K

E Card ( V ) , K

5 X remains a cardinal in V [ G ]and

B A nv = B A n v [ G ] for all A,B E V , V

card A

< A”.

Combining those two results we obtain in particular

4.3. COROLLARY. U P E V is countably closed and its Suslin number is w2 (all this in V ) , then the cardinals of V remain unchanged and also

P(u) n V = P(w)n V[G]. The following lemma of Sanin, called also the A-system lemma, is often used in calculating the Suslin number.

4.4. LEMMA (SANIN).Every uncountable family R of finite sets contains an uncountable subfamily Ro C R such that for some set d (the center) we have A n B = d for all A , B E Ro,A

# B.

The values T[G]can be much different as G runs over generic filters. Consider the case of a function with the domain and range in V .

4.5. LEMMA.Let P E V be an arbitrary forcing and assume that for some sets A , B 6 V , a P-name f and a p E P we have pll-”f:

Z+B”

Then, there axe functions E , F E V such that

E

: A + P(P) n V ,

F :

u

{u) x

E ( a ) -+ B,

aEA

the d u e s E ( a ) are maximal antichains under p and q It-

”

f ( u ) = F(a, q)”

for every a E A and q E E(a).

Notice that, by the assumption, the value f [ G ]is a function and f [ G ]: A + B in , p E G. Then, we have f [ G ] ( a= ) F ( a , q ) ,where q E E(a) is every model V [ G ]with such that q E G, (observe that G n E(u) has exactly one point). Thus, the F (u, q),

55. SELECTED MODELS

273

with q E E(a), are all possible values taken by all f [ G ]at the point a (with p E G). The lemma says that such an F can be found in V .

4.6. CANONICAL NAMES. Any pair of functions El F E V , as in the lemma 4.5, determines a canonical name

Suppose that for some f , A , B E V we have

V

"f : A

-+

B".

From the lemma 4.5, it follows that

f = f(E,F)[G] for some ElF E V. In other words, every function f : A + €3 lying in some generic V[G]has its own canonical name. In the case of B = {0,1} the canonical names can be interpreted as names of subsets of A . Calculating these names in V we obtain easily the following estimation 4.7. If the Suslin number of P in V is A, then for every K E Card ( V ) the power of P ( K )in V[G]is non-grater than (card P")" calculated in V:

2", in V [ G ]is

5 (card P")", in V.

For example, if P has the ccc and card P 5 w1 in V , then 2'( in V equals to 2" in V [ G ]for , all ti E Card ( V )= Card (V[G]).

A forcing P is called 0-linked, if there is a decoinposition P = UnEw K,, such that each part K,, is linked, i.e. any two elements p , q E I<,, are compatible. Obviously, every o-linked forcing P has the ccc.

$5. SELECTED MODELS. For a particular consistency proof it is required sometimes to have a ground model V satisfying some additional properties. 5.1. A MODEL IN WHICH CH HOLDS. Let v be an arbitrary ground model and define P in V as the set of all functions p : dm(p) -+ P ( w ) for which the domain

CII.11 FORCING

274

w r is at most countable. The ordering on P is defined as the inverse dm(p) inclusion (in particular, 1 = $3). p 5 q iff p _> q

Clearly, P is countably closed. Let G P be a generic (over V ) filter and consider g = UG. Obviously, g E V[G](since G E V [ G ] and ) g is a function as a union of a family of pairwise compatible functions. Clearly, the sets

D , = { p € P : ( Y E dm(p)}

fora<w,"

and for A E P ( w ) n V, belong to V and are dense in P. It follows immediately that dm(g) = w," and rg (9) = P(w) n V . But P is closed and hence w," = wr[G1 and P(w)' = P(w)VIGI, by the theorem 4.2. Therefore, we have

E A = { p E P : A E rg ( p ) }

+

whence V[G] CH.

Now,let us define P (within a V )to consist of all p : d m ( p ) + P(w1) such that the domain dm(p) w2 has cardinality at most w1. Then, P is obviously wz-closed and hence the first two uncountable cardinals w1 and w2 are the same in V and V[G]. Also P(wi) n V = P(wi) n V[G]for i = 1,2. Arguing as above we see that g = U G maps w2 onto P(w1) and hence we have

V[G] "2w1 = w Z " . Since we may already assume that CH holds in V , it will hold also in V [ G ]because , P ( w ) is the same in V and V[G].Thus, 2w = w1 and 2w1 = w2 are simultaneously true in V[G](and hence consistent with set theory). Repeating this n times we obtain a model in which a finite part of GCH 2w' = w,+l

for i = 0 , . . . ,n

holds true. Using completely different methods, Godel constructed (as early as in 1938) a model, the constructible model, in which the generalized continuum hypothesis 2" = K+

for all

K

E Card ,

$5. SELECTED MODELS

275

holds true. This model has also other interesting properties.

5.2. THECOHEN FORCING. Let C ( A ) ,for an infinite set A , consist of functions p : dm(p) + ( 0 , l ) with finite domain dm(p) A . The ordering on C ( A ) is the inverse inclusion. If V is any model and A E V , then C ( A ) E V as well (by the absoluteness of fin). The Cohen forcing C ( A ) has always the ccc (in every model to which it belongs). Indeed, suppose that E c C ( A ) is an uncountable subset. In view of the A-system lemma we may assume that for some d d m ( p ) n dm(q) = d

for any p

# q E E.

On d there is only finitely many zero-one functions. Hence, there is an uncountable set EO E such that

pld = qld for all p,q E Eo. But then p u g is a function and obviously pU q 5 p, q, whence p and q are compatible. Thus, E is not an antichain, which proves the claim. Assume now that the sets A , B E V are of the same power in V . If f E V is such that V k "f : A + B" then f lifts to an order isomorphism F : C ( A ) -t C ( B ) , F E V by setting

(5.3.)

F(p)(fa) = p ( a )

for p E C ( A ) and a E dm(p).

In particular, C ( A ) and C ( B ) are equivalent. More generally, assume that

V

k "card A 5 card B".

Then, for some f E V , we have f : A B and 5.3 defines now a complete embedding F : C ( A ) -t C ( B ) . In particular, if A E B, then C ( A ) is a complete C ( B ) is a generic filter (over V ) , then GIA = subforcing of C ( B ) . Hence, if G G n C ( A ) is generic in C ( A ) and

We often make use of the following remark.

CH. 11 FORCING

276 5.4.

H f : w + w is in VC(A)[G], then f is already in a subrnodel V(C(AO)[GIAo], for some countable subset A0 C_ A.

5.5. A MODEL WITH A LARGE CONTINUUM. For any K E Card ( V ) , K and K x w are of the same power in V . Hence, the forcings C ( K )and C(K x w ) are equivalent. Therefore, every generic filter G C ( K )or G C ( K x w ) determines a function g = UG E V[G]mapping K x w onto {0,1}. If we denote

X , = { i ~ w : g(a,i)=o}

foraEOnnV,

then the mapping a H X , is in V[G](since g is) and X , # X g , for a sets D,p = { p E C ( K x w ) : 32 E w[p(a,i) # p ( p , i)]}

# B since the

are easily seen to be dense. It follows that in V[G]there are at least that is V[G]k "2" 2 K " .

subsets of w ,

If, in addition, the cardinal

K

E Card ( V )is such that V

b "K"

K

= K " , then

+

V[G] "2" = K " , in view of 4.7. For K > w r , we obtain the consistency of non CH. In particular for any fixed integer n 2 1, the assertion c = 2" = wn is consistent with set theory (since always w," = wn holds in any model V such that V CH). Generally, we say that the rc-fold Cohen forcing C(n) adds to V K many subsets of w (the sets X , defined above). If K is such that V k " K < ' ( = K " , then from 4.7 it follows V[G] "2' = c" for every X,w 5 X < K .

+

+

5.6. A MODEL WITH CH A N D A LARGE 2"'. Let V be a model of CH (see 5.1) and fix a K E Card ( V ) such that K > w r and V " K < ' ( = K " . Let P be defined in V as the set of functions p : dm(p) + {O,l}, where dm(p) K x w1 is at most countable. Clearly, P is countably closed and its Suslin number (in V ) is w r , (Ex.A.2, Chapter 9). F'rom the corollary 4.3 it follows that V and V[G]have the same cardinals and also V[G] CH. As in 5.5 we can show that 2"' = K. in V[G]. Hence, we have V[G] " C = ~1 A 2"' = K " .

+

+

$5. SELECTED MODELS

5.7. A

DIAMOND PRINCIPLE.

277

The transitive closure of a set A is the set

t(A)= A U U A U U U A U . . .. This is the smallest transitive set containing A . Let H ( K ) denote the family of all sets of hereditary power < K , ( K > w l ) ,

H(K= ) { A : card t ( A ) < K } . Thus, A is in H ( K if) and only if A and all elements a E A and all elements of elements 2 E a , . . . are of power < K . More exactly we have the following characterization

A E H ( K ) iff A C H ( K )and card A

< K.

The family H ( K ) is closed under unions

if R E H ( K ) ,then

uR E H ( K )

and products

if A , B E H ( K ) , then A x B E H ( K ) . Also

K

H ( K )and card H ( K )= 2<".

If, in a given model V , we have card P < K , then we may assume that P E Ifv(.) and then V," H ' ( K ) , i.e. all P-names of rank < K are in H ' ( K ) . In particular, the standard P-names si for a E H ' ( K ) and the canonical P-names of subsets a are all in HV(.). Consequently, canonical names of subsets of H ' ( K ) are also subsets of

s

HV(K).

Assume that K satisfies 2<" = K , in V . There is a function f : K H ( K ) ,onto H ( K ) . If we set H , = f[a], then we obtain a continuously increasing decomposition

H ( K )= u { H p : a < K and c f ( a ) = w 1 ) . The following statement is then referred to as the diamond principle for H ( K ) :

5.8. there exists a sequence (Ta: a < K A c f ( a ) = w 1 ) such that for every set X C H ( K ) the set { a < K : cf(a) = w1 and X n H , = T,} is stationary. The principle 5.8 is true in the constructible model, (for every regular

K

>wl).

If K E Card ( V ) ,where V is any model, is such that K<" = K (and K > w y ) , then V has a generic extension V [ G ]with , the same cardinals, in which 5.8 is true.

CH.11 FORCING

278

To prove this let us denote S = {a E K : cf(a) = w l } , (actually, S can be any stationary set C K ) , and define P E V as the set of all functions p E V defined on S n I(p), where I(p) is an ordinal < K and such that p ( a ) H,, for all cr E S n I(p). The ordering on P is the inverse inclusion. Clearly, P is K-closed and hence the H ( K ) and the H,, a < K are the same (in V and V[G]).The cardinality of P is I E <= ~ IE, whence the Suslin number is IE+. Therefore, Card ( V )= Card (V[G]). We define t = (T,: a E S ) E V[G]by setting

Let, in V [ G ] ,X H ( K )be any set and C C K be a cub. We must show that there is an a E S n C such that X n H, = T,. There is a p E G such that p IF “ X

H ( K ) and C is a cub in

K”.

(For simplicity we identify sets with their P-names). If suffices to prove that for each pa 5 p there is a q 5 po and an a E S n C such that q IF ”Xn H , = q(a) and a E C”

Using an obvious remark for every P E P, there is an s

5 r and a set b E V such that

we define by induction a decreasing sequence (pa : a < K ) and an increasing sequence of ordinals (7, : a < I E ) and a family {b, : a < K } such that (PO is already given) [(PO) < 70 < [(pi) < 71< . . . and 1. p,+l I!- 7, E C and X n HI(,^) = b,”, for all a < K . 2. If a is a limit ordinal and 6 = sup p<,l(pp) belongs to S then

Clearly, the set (7, : a < K } is a cub and belongs to V . Thus, for some limit a < 4 ‘Yo E s and since 7, = sup p<,l(pp), we have p,(y,) = Up<, b p , by 2. From 1 it follows that

$6. ITERATED FORCING

279

which finishes the proof. Note that, applying the above to the model of 5.6, we obtain a model with CH and a large 2w1 and with the diamond for K = 2"'.

$6. ITERATEDFORCING. Let P E V be an arbitrary forcing in a model V and assume that for some P-name Q we have P It- "Q is a forcing".

), Q # @, 5Q is a It is understood that Q here means actually ( Q , < Q ,I Q ) ' ~where partial ordering on Q and IQ is the largest element (i.e. P forces additionally all these properties). In the set

K = { (p, q ) E P x V") : P It- " q E Q"} we shall identify pairs (p,ql) and ( p , q z ) for which p IF "q1 = q2". Then, the set P * Q, of all abstraction classes [p, 41, is a forcing (called the iteration of P, Q) with the ordering

The largest element in P * Q is [ l p , 191. With this definition, the iteration P {q' : p

It-

* Q does not belong to V

since in the set

q = q'} there are usually elements q' of arbitrary high rank, whence

f! V. There are however selectors S E V, i.e. sets S containing one element from each abstraction class [p, q] E P * Q. Identifying P * Q with such an S we shall have P * Q E V. This is proved below together with an estimation of the power. [p,q]

6.1. THEOREM. Assume that P has the ccc in V and let

K

E Card (V) be such

that

V

"card P 5

K

and

K"

= K".

If Q is such that P IF "card Q 5 K " , then also card P * Q 5 K in V. PROOF.We may assert that P It- " Q P IF " q E Q", then the set

K",

by the assumption. If q is such that

X ( q ) = {p E P : 30 < " ( p It- " q = Z")}

CH.11 FORCING

280

belongs to V (by definability) and is easily seen to be dense. Clearly, X ( q ) is the union Ua,.r X a ( q ) , where

X , ( q ) = { p E P : p It- ” q = 2’). Let E,(a) C &(q) be a maximal antichain in X , ( q ) . For a # P we have obviously E,(a) I E,(P), since if some p i E E,(a) and p2 E E,(P) were compatible and r 5 p 1 , m then we would have r It- ” a = p”, a contradition. Thus, UaCrc E q ( a )is a maximal antichain in P, whence E,(a)

# (2)

for at most countably many a.

Hence, we may regarded E, as a function in V , with a countable domain and taking countable subsets of P as its values. We have also 6.2.

K

if E,, = Eqr,then P It- ”91 = q z ” .

Indeed, if p E P is an arbitrary element, then there is a p’ 5 p and an a < K such that p‘ E Xa(q1). Hence, p’ is compatible with some pa E E q l ( a ) . So let r be such that r 5 p’ and r 5 p a . By our assumption p a E E,,(a) and hence

whence r It- ”q1 = q z ” . It follows that the set { T E P : T It- ”q1 = q 2 ” ) is dense which implies P It- ”q1 = q2”. The set 3 of all functions of the form Eq (for some q ) is definable over V and hence F E V . Clearly, card 3 5 tcU . ( K ~ =) K~~ = K in V, by the assumption. For an E E 3 let 7 E = min{a E On(v) : 3q E vip’[E= E,]}. If y = sup {TE : E E F},then y E On n V and the family V,“’ contains representatives of all the abstraction classes, by 6.2. Therefore as S we can take any maximal set S E P x V,“’ of nonequivalent elements, S E V (S exists in V by the maximum principle true in V ) . Also, card S 5 card 3 5 K , which finishes the proof. QED

We note that the theorem 6.1. has an obvious generalization for a P with any higher Suslin number. It is easy to see that the map p H [ p , l ~ is] a complete embedding of P into P * Q. It is often convenient to identify p with [p, l ~and ] regard P as a complete subforcing of P * Q.

$6. ITERATED

281

FORCING

How works the iteration P * Q? This is explained in the following

6.3. THEOREM. If F

P * Q is a generic over V filter in P * Q, then

is a generic over V filter in P and

is a generic over V[G] filter in Q[G] and

PROOF.Let p E G and p 5 p’. Then, [p,l] E F and [p, 11 5 [P’,l], whence [P’, 11 E F and p‘ E G. If pi,pz E G, then [pl, 11,[pz,1 1 E F and hence

Thus, r 5 pi,pz and since [r ,1 1 E F , we have also r E G. Therefore, G filter. Let D E V be a dense subset of P. The set

P is a

E = {[p,q] E P * Q : p E D}

*

is obviously dense in P Q and also E E V. Hence, there is a [p, q] E p E G n D. Thus, G is generic over V.

F nE and then

Now, we shall be dealing with the H. Let q[G] E H and q[G] 5 q’[G]. Hence, p It- ” q

5 q‘”

for some p E G.

Since [p, 11, [I,q1 E F , there is a [PI,QI] E F such that [PI,ql] I[p, 11,[l,q]. It follows that [pl,ql] I [l,q’], whence q’[G] E H. Let ql[G],q2[G]E H . Thus,

Hence, p It- ” q 5 q l , q z ” , which implies q[G] 5 (11 [GI, qz[G], since p E G. This proves that H is a filter. Let E E V[G] be a dense subset of Q[G]. Applying 2.10 we find a P-name e such that e[G] = E and P It- ”e is a dense in Q”. The set

D = {[p,q] E P * Q : p k ” q E e”}

CH.11 FORCING

282

is then dense in P * Q. Indeed, if [pol qo] E P * Q, then P IF "qo E Q and hence

P Il- " q 5 qo A q E e"

for some q.

Therefore, PO,^] I: [Po,qo] and [PO,q] E D. Now, let us take any [p,q] E FnD. Thus, q[G]E H n E, since p Il- " q E e" and p E G. We proved that H is generic over V [ G ] . To prove the last assertion we shall apply the theorem 2.6. We have V V [ F ]and F E V [ F ] .Since G is definable from F , G E V [ F ]and therefore V [ G ] V [ F ] .Also H E V [ F ](as definable in V from F and G ) and hence V [ G ] [ H ] V [ F ] .Similarly, we prove the converse inclusion. We have V C V [ G ] [ Hand ] since

s

F = {[p,q] : p E G and q[G]E H}, we have F E V [ G ] [ H (again ], by definability), whence V [ F ] V [ G ] [ H This ] . finishes the proof.

QED

The above theorem says that to force with an iteration P * Q is the same as to force twice - first, with the P to obtain V [ G ]and then with the Q[G]to obtain V [ G ] [ H ]A. converse of this is also true. 6.4. THEOREM. If G

P is generic over V and H

E Q[G]is generic over V [ G ] ,

then

G * H = { [p, q] E P * Q : p E G and q[G]E H} is a generic over V filter in P * Q and we have

V [ G* H]= V [ G ] [ H ] . PROOF.Denote F = G * H and let [p, q] E F and [p, q] I [p', q']. Then we have p E G and p 5 p', whence p' E G. Also p IF " q 5 q"' and hence q[G]5 q'[G],whence q'[G]E H , since q[G]E H . Thus, [p',q'] E F. If [ p l , q l ] ,[ p z , q z ] E F , then there is a

p'

I PI ,PZ and a q[Gl 5 q1 [GI, qz[Gl. Hence r Il- " q

5 q1 and q 5 qz"

for some r E G.

There is a p I p',r such that p E G. Then [P,q] Thus, F is a filter.

I [PI,q ~ ][Pz,qz] , and

Now, let D E V be a dense subset of P * Q. Define

[P,q] E F.

$6. ITERATED FORCING

283

Clearly, e is a P-name and we have

and also P It- "e is dense in Q ,

6.5.

Indeed, to prove 6.5 let P It- "QO E Q" and let po E P. There is a [p, q] E D such that [p, q] 5 [Po,qo]. Then we have p I PO,p It- "4' E e" and p It- " q I qo". Consequently, the set { p E P : 3 q ( p It- " q 5 qo A q E e")} is dense in P, which implies 6.5. Therefore, there is an element q[G] E H n e[G] and then [p, q ] E F n D,for some p E G. This shows that F is generic over V. The last equality (in the conclusion) is proved in the same way as in the preceding theorem.

QED

If we have two forcings P, Q E V, then the product P x Q ordered coordinatewise

. can also form the iteration is also a forcing with the greatest element ( l p , 1 ~ ) We P* since we have obviously P It- " g is a forcing ". It is easy to see that the mapping

v,

(P,d

b,71

for P E p and q E Q

is a dense embedding of P x Q into P * g. Indeed, if [p, t] E P * is an arbitrary and hence p' IF " t = ij", for some p' 5 p and element, then we have p It- " t E q E Q . It follows that [p', t] = [p',q] and [p',q] 5 [p,t],which proves the claim. In particular, the iteration P *

is equivalent to the product P x &.

For example a Cohen forcing C ( A U B ) , where A n B = 9 is isomorphic to the product C ( A ) x C ( B ) . Indeed, it is immediate to check that the mapping p H (plA,plB), for p E C ( A U B ) is an order isomorphism. Hence, C ( A U B) is equivalent to C ( A ) * C ( B ) .Applying the theorem 6.3 we infer that whenever A , B E V and F C C ( A U B ) is generic over V, then G = FIA (= F n C ( A ) ) is generic over V in C ( A ) and H = FIB is generic over V[G] in C ( B ) and

V [ F ]= V[G][H].

CH .11 FORCING

284

Notice the symmetric position of G and H here. The theorem below has an obvious generalization for arbitrary Suslin numbers

6.6. THEOREM. U P has the ccc in V and P II- “Q has the ccc”, then also P * Q has the ccc.

PROOF.Suppose, for a contradition, that {[po,q,] : a < w,”} E V is an uncountable antichain in P * Q. Define

6.7.

A = {@,pa) :

(Y

<my}.

Obviously, A is a P-name and A[G] = {a < w1 : p a E G} The set {qa[G] :

Q

for G 5 P.

E A[G]} is in V[G] and it is an antichain in &[GI for, if it were

then we would have p ll- ” q 5 go, qp”, for some p E G satisfying p 5 p , , p g . But then

which is impossible (we have only p IF ” q E Q” so we must apply 2.10 to find a q* so that P ll- ”q* E Q” and p Il- ” q * = q ” ) . By the assumption, we obtain P Il- ” { q o : a E A} is at most countable” and therefore P Il-

”

SUP

A < WI”,

since the mapping a ++ qo is one-to-one. Hence, the set

is dense in P and belongs to V , (by definability). Let E E E V . Thus, there are ordinals a p < w y , such that p ll- ” sup A = aP”

D be a maximal antichain,

for every p E E .

36. ITERATED FORCING Now, if 7 = sup {a, : p E E}, then y P ll-

"

285

< w," (since E is countable in V), and sup A

< 7".

On the other hand, we have p a It- '5E A" for each a

< w r , because of 6.7, which is

QED

an expected contradition.

We know already, (cf the theorem 6.3), that an iteration P * Q yields a double extension of V : first V[G] for a G C P and then V[G][H] for an H Q(G].The definition will be now extended to obtain iterations with infinitely many factors producing in effect an infinite chain of consecutive extensions. The definition goes by induction on the length X E OnnV. We say that a forcing P E V, consisting of some sequences (p(a) : a < A), is an iteration of length X (with finite supports), if the following conditions are fulfilled. 1. If X = 0, then P = Po = (8) is a trivial forcing (with l p = 9). Then = {a : a E V} consists of the standard names and V[G] = V. 2. If X 2 1 and: a) X = a 1, then Pa = {pla : p E P} is an iteration of length a and there is a Pa-name Qa such that

+

Pa IF "Qa is a forcing" and for any sequence p = ( p ( 0 : and

P E P P

Iq

i.e. the mapping p

= pla E

E < A)

E Pa

E V we have

and Pa IF " p ( o ) E Qa"

pla I P qIa ~ and p l a [I- " ~ ( a IS. ) q(a)'l,

[pla,p(a)] is an order isomorphism of P onto P * Qa.

b) X is a limit ordinal, then for every a < A, Pa = { p l a : E P} is as1 iteration of length a and we have and

pEP

Iq

G

V a < X [ p I a E Pa

and supp(p) E fin(X)]

va < X [ P l a IQ,,'?la], where supp(p) = {a < X : p ( a ) # 1 ~ - and } Qa is as in 2,a. In 2,b the first equivalence can be replaced by P = Ca<X Pa, i.e. P is the direct P

limit, where

CH .11 FORCING

286

Still in 2,b replacing ”supp(p) E fin(X)” by ”supp(p) is at most countable” we obtain countable support iterations (used in Chapter 10). Also other iterations are in use (differing always in defining the limit stages) but here we consider that with finite supports only. From the definition it follows immediately that an iteration P of length X = 1 has (up to an isomorphism) a form

where QO E V is an arbitrary forcing. For X = n < w we have

Generally, if an iteration P has length X then there exists a sequence (Q,: a < A) E V such that for every a < X we have: 1. Q, is Pa-name and Pa Il-

”Q,is a forcing ”

and P,+1 = P,

* Q,.

2. If p E P, then p(a) is a Pa-name and P, Il- ”p(a) E Q,”.Moreover

Since QOis a standard name, it can be identified with an ordinary forcing Q E V. The form of Qo may follow from a given inductive scheme defining the iteration or Qo can be defined by a separate condition. Let us note some simple properties, (the proofs, by induction on A, are obvious).

~ <~A). 6.8. The largest element in P is l p = ( 1 : CY 6.9. If r E Pa, where a < A, then the extension r

* 1, defined by

belongs to P. 6.10. Lfr E P,, where a

< A, p E P and r 5 p l a , then r * p E P, where

56. ITERATED

287

FORCING

6.11. LEMMA.The mapping r H r * 1,for r E P, is a complete embedding of P, into P. If G C P is a generic ultrafilter, then Gla = { p l a : p E G} is a generic ultrafilter in P, and

V(')[G(a]C V(')[G]. PROOF.Obviously, the mapping r

H

r

* 1 is an order embedding.

Also

Therefore, antichains are mapped into antichains. Let E E P, be a maximal antichain and let p € P. We have (pla)lls, for some s E E, i.e. there is an r € P, such that r 5 pIa,s. Hence, r * p 5 p and r * p 5 s * 1. This shows that E * 1 = { r * 1 : r E E} is a maximal antichain in P. The second part of the lemma follows from remarks of Sec. 3. QED

6.12. THEOREM. If P, IF "Q, has the ccc", for all a < X (where P,+l/P,), then P has the ccc.

PROOF.By induction on A. If X

=a

Q,

=

+ 1, then the conclusion follows from the

theorem 6.6. Suppose now that X is a limit ordinal and let E = {pe : t < w1} be an uncountable subset of P. In view of the A-system lemma we may assume that for some d we have

+

Let a = maxd 1. By the inductive hypothesis, P, has the ccc and therefore there exist (,v and an r E P, such that

Moreover, there is an s E P such that sIa = r and for any

P 2 a,

if P E SUPP(P€) if P E SUPP(P9) for remaining p. Then, we have s 5 pe,p,, i.e. E is not antichain. a

QED

Any set which is small with repect the length X will occur already at some stage (cf 5.4). More exactly

< A,

CH .I1 FORCING

288

6.13. THEOREM. Let P E V be an iteration with the ccc of length K , where E Card (V) is regular and uncountable in V. If A, B E V are of power < K in V, B there is an a < K such that f E V[Gla]. then for each f E V[G], f : A -i

K

PROOF. We choose a canonical name p = {((a,F(a))''),q) : a E A

and q E E(a)}

of the function f , (as in 4.6). Each of the antichains E ( a ) is at most countable. ) bounded from above by some Therefore, the sets supp(q), for q E U c E A E ( a are a < K . We set

The definition is correct if we identify P, with its image {p * 1 : p E Pa} under the standard complete embedding. Then we have

and hence

f

E V[GlQl.

QED For example, for any iteration of regular length ww

n V[G] =

K

2 w," with the ccc we have

U ww n V[Gla]. O
If K 2 w r and s = (fa : a < w r ) , where f, 7 < K.

: w + w , then also s

E V[G)y],for some

No cautions in 6.14 (and similar operations) are necessary, if we change a little the definition of iteration. Let as observe that for any p , q E P,

Replace each sequence p = ( p ( a ) : a < A) E P by plsupp(p), the restriction of p to its support. Now, any element p of so modyfied iteration P is a function with a finite domain dm(p) A. The greatest element l p is now the empty function 9. The sequence (Po : a < A), of consecutive stages of the iteration, is now a continuously increasing sequence of forcings, in which the inclusion P, C Pg, for

$7. THEMARTINAXIOM

289

Q 5 p, means that Pa is in fact a complete subforcing of Pg. For limit Q < X we have P, = Ug<,Pa. Hence, for any filter G C P, the restriction Gla equals to G n P,.

For convenience, we use frequently the above notation in some proofs.

$7. THEMARTINAXIOM.The following statement is known as the Martin axiom.

For every forcing P with the ccc and any family R, of power < c,

(MA)

consisting of dense subsets of P, there is a filter G G n D # (3

P such that

for each D E R.

Many problems can be settled with help of MA. Since MA, as proved below, is consistent with set theory, so is any of its consequences. Thus, a proof in which MA is used, is a kind of consistency proof. It is easy to see that in the formulation of MA we can demand that p E G, for a p E P given in advance. Indeed, to prove this claim, it suffices to apply MA to the subforcing P, = { q E P : q 5 p } and the family R,,= { D n P, : D E R} and then extend the obtained G to the whole of P.

A rather technical formulation of MA has an interesting topological equivalent. 7.1. THEOREM. MA holds if and only if in every compact space X with the Suslin propety (every disjoint family of open sets is at most countable) any union of less than c nowhere dense sets is boundw. Thus, MA is a form of a well known theorem of Baire. PROOF. Assume MA and let S be a family of nowhere dense sets in X , card S < c. The sets

K N = {U E D ( X ) : V n N =(B}

for N E S,

are dense in the algebra D ( X ) , of open domains. To reduce the power, let us fix maximal antichains EN E K N ,for all N E S, and let P be the subalgebra of D ( X ) , generated by U{FN : N E S} minus the zero 9. Hence, card P < C. The sets

DN=(PEP: anN=(B}

forNES,

290

CII.11 FORCING

P intersecting all the D N , N E S. are dense in P. By MA, there is a filter G = n { B : p E G } has a point I 4 U S. Obviously, we may assume, that Clearly, P contains an element p , fulfillingjj U ,for a given open U and that p E G. Hence, S is boundary.

U

Conversely, assume that we are given a forcing P with the ccc and a family R of dense subsets, card R < c. Of course, the Stone space X of the algebra D(P) has the Suslin property and the sets

u~=U{u,:p E D }

for D E R,

(W,, denotes the basic Stone neighbourhood), are open and dense in X , (the density of a D in P means that {Up : p E D } is a .Ir-base). By assumption, there is an ultrafilter z E n { u D : D 6 R}.Hence, for every D E R there is a p E D such that I E U p , i.e. p E I. Clearly, the intersection G = z f~ P is the required filter.

QED In the proof of the implication to the right, the Martin axiom has been applied to a P, with card P ' < c. Therefore, we obtain 7.2. COROLLARY. A weakened version of MA, in which forcings P with card P < c occur only, implies the full MA.

If CH is assumed, then in the theorem 7.1 we have the ordinary Bake theorem, true for all compact spaces. Hence, MA makes sense only if c > w1. Now, we pass to the consistency proof. 7.3. THEOREM. (Martin). There is a model (of set theory) in which the Martin axiom holds and the cardinal c is arbitrary large.

PROOF.Let V be a model and IC E Card ( V )a cardinal of V such that IC > w," and I C < = ~ IC. We may assume that the diamond principle for H(n) holds in V, see 5.7, (for example, V can be the constructible model and IC any regular cardinal > w,"). Let (T,: cf(a) = w1 and LY < I C ) E V be a diamond sequence. We define by induction in V a finite support iteration P, as follows: for limit a, Pa = Cp
$7. THEMARTINAXIOM

291

From the theorem 6.1 and 6.2 we infer that P = C a < s P a has the ccc and card P = n, in V . Therefore, we may assume that P H ( K ) . Let G C P be a generic filter. By 4.7, the cardinal c = 2" in V [ G ]is 5 (card Pw)w= nw = n and also c 2 n, since we add n many new sets C w . Hence,

V [ G ]b " c = K " . Note that, (by the ccc), the cardinals of V and V [ G ]are the same. Now, we prove that MA holds in V [ G ] Let . Q E V [ G ]be a forcing satisfying

where 4(1)is the formula " X has the ccc", and let R E V [ G ]be a family of dense subsets of Q,V [ G ] "card R < c". We have to find a filter (I C Q,K E V [ G ]such that for every D E R. K nD # 9 By the corollary 7.2, we may assume additionally that

V [ G ] "card Q < c". Therefore, (replacing Q by an isomorph), we shall assume further that Q A, for some X < n. According to the theorem 6.13, the both Q and R must occur at some stage y < K , whence Q , R E V[Gla] for all Q 2 y. We have also

V[Gla] $[Q] for any a , y 5 a < n. Indeed, if Q had an uncountable antichain A in V [ G J a A ] , would be still uncountable in V [ G ]since , ur is the first uncountable cardinal in V , V [ G ]and all V [ G l a ]a, < n, (by the CCC). Let 7 be a canonical P,-name of Q . Thus, we have 7.4.

p IF 4 ( ~ )

for some p E Gly.

Even stronger, we may claim 7.5.

P, It- 4 ( t )

for some P,-name t such that p It- "t = 7 " .

CH. 11 FORCING

292

Indeed, define an auxiliary name rP as follows: dm(rp) = {z E d m ( r ) : 3r 5 p ( r ll- ”z E T”)} and rp(z) = a a maximal antichain in { r 5 p : r It- ”z E 7”). Let E 5 P, be a maximal antichain containg p and set

Clearly, t is P,-name and for any filter F 5 P, we have

In particular, t[G]= Q. In any case, t [ F ]has the ccc, since either p E F and 7.4 applies or p 4 F and then t [ F ]has a single element only. Thus, we proved 7.5 for so defined t . Since P, is a complete subforcing of Pa, for y 5 a,we have

t [ F ]= t[Fly]

for any filter F

Pa,

and hence for all a 2 y.

P, It- d ( t )

Since t has the hereditary power less than X a < n. By the diamond principle, we get

t = Ta

< IE, we have t C Ha for large enough

for some a,y 5 a

< IE.

+

By the definition of P we obtain Pa+l = P, * t and hence Gla 1 = (Gla)* K, where K t[GIa]= Q is a generic, over V [ G J a filter ] , in Q, (cf the theorem 6.3). Of course, K is in V [ G ] since , If E V[Gla 11 E V [ G ] .Finally, since every set D E R

+

is also in V[GIa]we obtain K

n D # 9, by genericity .

QED

In the above proof we have applied a diamond principle to enumerate all the Q’s since this method is used in some proofs in the text to enumerate also some ”big” objects. If we want to have MA only, then can apply an ordinary ”booking” technique, as in the proof of the theorem 3.1, Chapter 8 or 4.1, Chapter 9. Let us make, to end with, some remarks on generalizations of forcing. In the definition we can omit the condition 2.1 and even the assymetry of the partial order, Thus, we may assume only that the relation

$7. THEMARTINAXIOM

293

the theorems go through for such a general notion of forcing. Let us introduce a topology on P assuming the sets Up = { q E P : q 5 p } as basic open. This is the same topology as for an ordinary forcing but now, the sets Upneed not be open for p E P maps P onto a dense domains. Nevertheless, the function f(p) = int set Q = f[P] of the algebra D(P), of open domains. We note that f is not one-to-one, in general, but it preserves the ordering and maps antichains into antichains. This implies that P is equivalent to an ordinary forcing Q = f[P]. Hence, generalized forcings yield no new models but axe often convenient in use.

[up],

Of course, generalized forcings are admitted in the formulation of the Martin axiom.

BIBLIOGRAPHY [ BFM ] B. BALCAR,R. FRANKIEWICZ, CH. MILLS,More on nowhere dense closed P-sets, Bull. Acad. Polon. Sci. 28 (5-6) (1980), 295-299. [ BSP ] B. BALCAR,J. PELANT, P . SIMON,The space of ultrafilters on N covered by nowhere dense sets, Fund. Math. 110 (1980), 11-24. [ BSV ] B. BALCAR,P. SIMON,P. VOJTAS,Refinement properties and extensions of filters in Boolean algebras, Trans. Amer. Math. SOC.267 (1981), 265-283. [ BFZ ] J. BAUMGARTNER, R. FRANKIEWICZ, P. Zl3IERSI
[ DF ] A. DOW, R. FRANKIEWICZ, Remarks on partitioner algebras, Proc. Amer.

[ DvM ] [ DW ] [ vDP ] [ vDvM ]

Math. SOC.113 (1991), 1067-1070. A. DOW, J. VAN MILLAn extremally disconnected Dowker space, Proc. Amer. Math. SOC.86 (1982), 664-672. H. G. DALES,W. H. WOODIN, An introduction to independence for analysts, Cambridge University Press, Cambridge 1987. E. VAN DOUWEN, T. PRZYMUSI~~SKI, Separable extensions of first countable spaces, Fund. Math. 95 (1980), 147-158. E.VAN DOUWEN, J. VAN MILL,ParoviEenko’s characterization of /3w \w implies CH,Proc. Amer. Math. SOC.72 (1978), 539-541.

[ D F ] A. Dow, R. FRANKIEWICZ, Remarks on partitioner algebras, Proc. Amer. Math SOC.(to appear).

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[ DFZ ] A. DOW, R. FRANKIEWICZ, P. ZBIERSK1,On closed subspaces of w * , Proc. Amer. Math. SOC.(to appear).

[ El ] B. A. EFIMOV, Extremally disconnected compact spaces and absolutes, Trans. Moscow Math. SOC.23 (1970),243-285. [ E, ] J. R. ESTERLE, Solution d’un probleme d’Erdos, Gillman et Henriksen et application B l’etude des homomorphismes de C(Ii), Acta Math. Acad. Sci. Hungar. 30 (1977), 113-127; Sur l’existence d’un homomorphisme discontinu de C(Z<), Proc. London Math. SOC.36 (3) (1978), 46-58; Injection de semi-groupes divisibles dans des algkbres de convolution et construction d’homomorphismes discontinus de C ( K ) ,Proc. London Math. SOC.36 (3)

[ F, ] [ F, ]

[ FKZ ]

[ FZ, ] [FZ, ]

[ FZ, ] [ FZ, ] [ FZ, ]

[ GH, ] [ GH, ]

[ GJ ] [ HI]

(1978),59-85. G. FODOR,Eine Bemerkung zur Theorie der regrefliven Funktionen, Acta Sci. Math. (Szeged) 17 (1956),139-142. R. FRANKIEWICZ, Some remarks on embeddings of Boolean algebras and topological spaces, 11, Fund. Math. 126 (1985),63-68. R. FRANKIEWICZ, K. KUNEN,P. ZBIERSI
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[ H, ] F. HAUsDORFF,fiberzwei Satze von G. Fichtenholz und L.Kantorovitch, Studia Math. 6 (1936),18-19. [ Hs ] F. HAUSDORFF,Summen von N, Mengen, Fund. Math. 26 (1936),241-255. [ H, ] S.HECHLER,On the existence of certain cofinal subsets of w w , Axiomatic set theory, T. Jech (ed.) Proc. Symp. Pure Math. vol 13, I1 (1974), 155-173. [ J ] T. JECH,Multiple forcing, Cambridge University Press, Cambridge 1986. [ K ] K. KUNEN, Weak P-points in N * , Colloq. Math. SOC. Jho’s Bolyai 23,Topology, Budapest 1978,741-749. [ L ] A. LOUVEAU, Caract6risstion dessous-espaces compact dePN, Bull. Sci. Math. 97 (2)(1973), 259-263. [ P, ] J. J. PAROVICENKO, A universal bicompact of weight N , Soviet Math. Dokl. 4 (1963),592-595. [ P, ] B. POSPISIL, On bicompact spaces, Publ. Fac. Univ. Masaryk 280 (1939),3-16. [ R, ] F. ROTHBERGER, Une remarque concernant l’hypothise du continu, Fund. Math. 31 (1939),224-226. [ R, ] W. RUDIN,Homogeneity problems in the theory of Cech compactification, Duke Math. J. 23 (1956),409-419. [ s, ] s. SHELAH,Proper forcing, Lecture Notes in Mathematics 940,Springer-Verlag, Berlin 1982. [ S, ] R. SIKORSKI, On an analogy between measures and homomorphisms, Ann. SOC. Pol. Math. 23 (1950), 1-20. [ S, ] R. SIKORSKI, Boolean algebras, Springer-Verlag, Berlin 1969. [ S, ] M. H. STONE,The theory of representations for Boolean algebras, Ttans. Amer. Math. SOC. 40 (1936),37-111. [ S, ] M. H. STONE,Applications of the theory of Boolean rings to general topology, Ttans. Arner. Math. SOC. 41 (1937),321-364. [ T ] A. D. TAIMANOV, On the extensions of continuous mappings of topological spaces, Math. Sbor. 31 (1952),459-463. [ W ] R. G. WOODS, Characterisations of some C+-embeddedsubspaces of P N , Pacific J. Math. 65 (1976),573-579.

INDEX Antichain 21, 197 Atom 5 Atomic 5 Atomless 5 Algebra of open domains 6 Almost - containment 31 - disjointness 31 Bake property 14 Bernstein set 115 Binary tree 23 Boolean - algebra 1 - ring 30

Compactification 71 Complete - algebra 6

- embedding 269 Completely separated 67 Consistency relation 263 Cozero-set 68 c-Point 128, 132 cub-set 128, 243 Delta-system lemma 168, 179, 191,

213,238,272 Dense subset 5 Density 83 Density (asymptotic) 246

- ordering 3

D-gap 217

Bounding forcing 246 Cantor

Diamond principle 174,183, 192,

201,225

- space 5, 27

Disjoint elements 21

- tree 23

Dominance ordering 33

ccc (countable chain condition) 46,49,51,

Dominating

53,287

- family 34

C-embedded 93 C*-embedded 93 Centered family 11 Chain 20 Character

- forcing 49,204

Dynamical system 240, 241 Element filling (separating) a gap 18 Embedding 10

- of a point 115

Extremally disconnected 83, 93, 176

- of a space 153

Factor algebra 13

Cohen

- forcing 55,221, 275

Family linked with respect to 144 Fat ideal 176, 246,255

- model 55, 110, 138,212

Field of sets 1

297

298

INDEX

Filter - in algebra 11 - on a set 16 - in Z ( X ) 69 F-convergence 119 F-limit 119 Forcing 275 - with the ccc 271 Mchet filter 16 Free generators 27 F-space 94,95,97,101, 102, 103,114, 117,

188, 122, 151, 153,162 Functional with AP 256,257 Gap 17 - on a set 36 Generic - filter 266 - model 267,268 Gleason space 82, 137, 164, 176,243 Gregorieff set 248 Hechler forcing 51,60,194 Homogeneous space 143 Homomorphism 9 Ideal 11 Idempotent 30,177 Independent family 17, 214 Irreducible map 81 Isomorphism 10 Iteration - with finite supports 60, 156, 173, 183, 186,

192, 195,202,203,205,226,230,285 - with countable supports 245,252

Kaplansky problem 196 Limit - in P(w)/fin 32,53,60,203,204,206 - in ww 35,203, 204, 206

Lower class 17 Maximal filter 11 Martin axiom 47,49, 50, 51,57, 58,

179, 183, 197,215,289 MA(a-linked) 185,228 Meager set 14 Measurable cardinal 139 Measure 21 - algebra 21 - algebra of m 27 - on P(w) 255 - with AP 255,257 Model 263 - with CH 273 - with large continuum 276 - with large 2"' 276 - with diamond principle 277 - with MA 290 Near to 38 Normal measure 91 Novik number 138 Nowhere dense set 14 wl-cub set 131 w-proper forcing 245 OK-point 143 Open domain 6 ParoviEenko - algebra 42, 210 - space 96,105, 155

Partition algebra 209,212, 215 Partitioner 209 P(c)-point 80, 109 P(c)-principle 47, 116 x-base tree 123, 125 x-character 156 P(tc)-set 99,104

299

INDEX P ( n)-point 99 P-point 78, 139,140, 176,258 P-set 99,103, 106, 107, 176,246 P-set with ccc 243, 252 P-space 78, 90, 164 Power-set algebra 2 PP-property 246 Prime filter 15, 69 Principal filter 11 Proper forcing 244 Recurrent point 241 Remainder space 76,208 Rudin-Kiesler ordering 150 Scale 35,56,58,63,104 Sequentially compact space 61 Selective P-point 141 Semigroup 177 Separating forcing 48,50, 51, 166,175, 181,

207,217,233 Set theoretical formula 261 Set theory 259 a-complete algebra 6 u-disconnected 82 Space of ordinals 99,141, 142 Splitting family 61 Stationary set 134,243 Stone-Cech compatification 71, 87,88, 157 Stone space 66 Strongly zero-dimensional 78 Subalgebra 9 Superatomic algebra 86, 142 Symmetric difference 10 Theorem of - Baire 258 - Balcar, Frankiewicz, Mills 103 - Balcar, Pelant, Simon 125,126

- Balcar, Simon, Vojtd 166 - Balcar, Vojt& 136 - Baumgartner, Weese 210,212 - Bell 160 - Blaszczyk, Plewik, Turek 241 - Comfort, Hindman,

Negrepontis 161 - Comfort, Negrepontis 149 -

Cech 77

- cech-Stone 71 - Dales, Esterle 196

Dixmier 92 - van Douwen 164 - van Douwen, van Mill 105 - van Douwen, Przymusiriski 179 - Efimov 85 - Fichtenholz, Kantorovitch 16 - Fodor 130 - Freniche 30 - Furstenberg, Weiss 241 - Galvin 178 - Gelfand, Kolmogoroff 87 - Gillman, Henriksen 80, 91,97 - Glazer 177 - Henriksen 158 - Henriksen, Curtis 157 - Halmos 92 - Hausdo& 33, 39,59 - Hechler 52 - Hindman 178 - Kunen 48,59,116, 192, 194 - Kunen, Bell 156 - Loomis, Sikorski 92 - Louveau 103 - Marczewski 85 - Mekler 256,257 -

300 - Miller 239 - Namakura 177 - ParoviEenko 44, 96 - PospiSil 116 - Riesz 92, 256 - Rothberger 38 - Rudin 79, 86 - Sanin 239,272 - Seever 122 - Shelah 143,240 - Sikorski 18, 112, 233 - Solovay 201 - Stone 63, 87 - Stone, Nakano 121 - Taimanov 73, 167 - Tarski 25

INDEX U l m , Solovay 130 Urysohn 95 - van der Waerden 242 - Woodin 196 - Woods 162 Truth relation 262 Tychonoff space 68 Type of a gap 36 Ultrafilter 11 - in Z(X)70 Uniform ultrafilter 16 Upper class 17 Weak P-point 79 Weakly Lindelof property 161 Zero-set 68 -