Hausdorff Gaps and Limits (Studies in Logic and the Foundations of Mathematics, 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

This Page Intentionally Left Blank

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

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} 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.

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
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 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 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


. . . 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 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 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, 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 .

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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 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 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


w1, in which there is no order embedding of T = (0,l}w* into 7 = (ww, 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


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


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 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 , 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.

,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 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 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