Combinatorial Set Theory

COMBINATORIAL SET THEORY

Neil H. WILLIAMS University of Queensland, Australia

1977

NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM. NEW YORK . OXFORD

@NORTH-HOLLAND PUBLISHING COMPANY - 1977 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 permission of the copvrighr owner.

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Library of Congress Cataloging in Publication Data Williams, Neil H. Combinatorial set theory. (Studies in logic and the foundations of mathematics ; v . 91) Bibliography: p. Includes indexes. I. Combinatorial set theory. I. Title. 11. Series. QA248. W56 5 1 1 ’. 3 77-3462 ISBN 0-7204-0722-2

PRINTED IN THE NETHERLANDS

PREFACE

Combinatorial theory is largely the study of properties that a set or family of sets may have by virtue of its cardinality, although this may be widened to consider related properties held by sets carrying a simple structure, such as ordered or well ordered sets. These properties are relevant to either finite or infinite sets, although frequently the questions that pose interesting problems for finite sets are either meaningless or trivial for infinite sets, and vice-versa. There has been a recent great upsurge in the study of finite combinatorial problems, and a significant, though more manageable, increase in interest in the combinatorial properties of infinite sets. This book deals solely with combinatorial questions pertinent to infinite sets. Some results have arisen purely in the context of infinite sets. One of the early results in the subject is the following: given an infinite set S of power K then there is a family of more than K subsets of S any two of which intersect in a set of size less than K . Chapter 1 looks at questions related to this. Problems of a different nature for families of sets are studied in Chapter 4, firstly a decomposition problem and then delta- and weak delta-systems. As a special case is the following: given any family A of Hz denumerable sets there is a subfamily 93 of A of size Hz such that the intersection of two sets from is the same for all pairs from CM. Chapter 3 is devoted to the study of set mappings, that is, functionsf defined on a set S such that given any x in S then f ( x ) is a subset of S for which x $ f ( x ) . Conditions are placed on t h e family { f ( x ) ;x ES } which ensure the existence of a large free set T ( a subset of S such that x 4 f ( y )and y 4 f ( x ) for all x,y from 2‘). Other results stem from problems that have been extensively studied for finite sets, and have been found to yield interesting questions when reformulated to apply to infinite sets. For example, in Chapter 5 we study infinite graphs, and in particular show that for any infinite cardinal K there is a graph with chromatic number K which contains no triangle, or indeed no pentagon; however any graph which contains no quadrilateral has chromatic number at most KO. Chapters 2 , 4 and 7 are devoted to various extensions of Ramsey’s classical theorem, which (in its finite form) states: given integers n, k, r if the

VI

Preface

/r-clcnient subsets o f a linitc set S are divided into r classes then provided that

S is sul’ficiently large there will be some k-eleinent subset of S all the n-element

sulwts ol’which l’all in the one class. Chapters 2 and 4 cover ordinary partition iclations, polarized partition relations and square bracket partition relations 101cardinal numbers, whilst Chapter 7 is concerned with ordinary partition relations f o r ordinal numbers. Familiarity with the standard notions of set theory has been assumed throughout. An Appendix summarizes those properties of cardinal and ordinal numbers and their arithmetic which are basic t o a study of this book. The development of infinitary combinatorial theory over the past twenty ycars or so has been greatly stimulated by associates of the Hungarian school, under the encouragement of Paul Erdos in particular. A glance at the list of references gives some indication of how many of the results in this book show his influence. T o all those people who have created this subject, I here acknowledge my debt and record my gratitude. Brisbane. 1976.

Neil H. Williams

FOREWORD ON NOTATION

The following is a brief summary of standard notation in use throughout this book. More technical notation is introduced throughout the text as the need for it becomes apparent. The Index of notation (pp. 206-208) provides a ready reference to the page on which a symbol is first defined. Set membership is denoted by €, and C is the inclusion relation with C denoting proper inclusion. The set of all subsets of a set is 'Ipx,so 8 x = { y ; y C_ x }. The union of all the sets in x is written Lbc and the intersection n x , so u x = {z;

3E

~ E (~ ) ~}nx ; = {z;

v ~ EE ~ ) .(} ~

Set difference is written x - y , sox - y = { z E x ; z e y } . The set of unordered pairs, one member from A and the other from B is A o B , so A o B = { {x, y }; x E A, y E B and x # y }. Ordered pairs are written ( x ,y ) , and sequences as ( x a ;a! < /3). The length of the sequence ( x a ; a! < /3) is /3, written Pn(x, ; a! < /3> = 8. If x , y are two sequences then x"y is the concatenation o f x a n d y , that is, the sequence obtained by placing the entries fromy in order after those from x. If A is any set, then the domain and the range of A are defined by dom(A)

= {x;

3 y ( ( x ,y ) € A ) ) ;

ran(A) =

b;3x((x,y ) € A ) } .

That f i s a function with domain A and range contained in Bis indicated briefly by writingf: A + B. The set of all such functions is *B, so

* B = {f;f:A+B}. Iff: A -+ B , then the value off at x is f (x). The restriction off to a set X is writtenf r X , so

frx={

( X , ~ ) E ~ ; ~ E X ) ;

and f [ X ]is the range off

r X , so

f[XI = { f ( x ) ; xE X ) . The words set and family are used synonymously. However an indexed family, written (Ai; i E I ) , stands for that function A with domain I and

Foreword on notation

X

A(i) = Ai for each i in I. The Cartesian product of an indexed family is written X(Ai; i E l). A decomposition of a set A is a family A of sets such that A = UA; this is the same as a partition of A. The partition A is disjoint if the sets in A are pairwise disjoint. For elements a, b of A and a partition A of A , the notation

a

b(mod A )

means that there is some Ak in A such that a, b E A k . The cardinality of a set X is written 1x1,and [XIK, [XIcK, ... denote {Y C X;IYI = K }, {Y C_ X ; IYI < K } , ... .The operations of cardinal addition, multiplication and exponentiation are written q i6',q . 6' and q', while the corresponding ordinal operations are written a + 0,@ and.!a The infinite cardinal sum and product of an indexed family (qi;i E 1) of cardinals are written C(Q~;i E Z) and n(qi;i E l ) , whereas the ordinal sum and product of a well ordered sequence (a,,; v < (3) of ordinal numbers are written &(a,,;v < 0) and llo(a,;v < /3)Let X be a set ordered by a relation<. By t p ( X , i ) or tp(X) is meant the B to mean that order type o f X under<. For subsets A , B o f X , writeA i a
[XI*= { Y C X ; t p ( Y , < ) = a } . The ordinal numbers are defined so that if a is an ordinal, then a = { /3; (3 is an ordinal and

< a} .

IfA is a set of ordinals then sup A is the supremum o f A , so sup A = UA. Gzrdinal numbers are identified with the initial ordinals. The sequence of infinite cardinals is NO,HI,H2, ..., Ha,... . The cardinal successor to a cardinal K is denoted by K + ; the iteration of this n times by K("+). The cofinality of a cardinal K is written K ' , so K ' is the least cardinal such that K can be written as a sum of K ' cardinals all less than K . The cardinal K is regular if K ' = K , and otherwise K is singular. A cardinal of the form X+ for some X is a successor cardinal;other cardinals are limit cardinals. The cardinal K is a strong limit cardinal if 2' < K whenever X < K . Regular limit cardinals are weakly inaccessible; regular strong !imit cardinals are strongly inaccessible. The cardinal beths (starting from K ) are defined by induction:

lo@)= K ,

ln+l (K)

=2

U K )

The Generalized Continuum Hypothesis (GCH) is the statement: 2K = K +

Foreword on notation

xi

for all infinite cardinals K . The GCH has been assumed throughout this book whenever it leads to a simplification in the statement or proof of a result. Theorems reached with its aid have the letters GCH appended to their result number (as in Theorem 1.6 (GCH)). In many cases the full strength of the GCH is not required, or the result could have been reformulated in a more involved form so as to avoid GCH all together. It is left to the interested reader to observe when this is so. The Greek letters K , A, L, q,O are used throughout to stand for cardinal numbers, and usually K , A are infinite. Other small Greek letters denote ordinal numbers, as d o k, 1 (except that w is always the least infinite ordinal). The letters m, n always stand for non-negative integers.

CHAPTER 1

ALMOST DISJOINT FAMILIES OF SETS

9 1. Almost disjoint families One of the early results in combinatorial set theory was the following theorem of Sierpinski [84]. He proved that any infinite set of power K can be decomposed into a family of more than K infinite subsets in such a way that IA BI < MI, Bl for any two different subsets A , B from the family. Two setsA and B will be called almost disjoint when IA nBI < MI, Bl. Almost disjoint systems of sets were examined in more detail by Tarski [94]. It seems appropriate to start this book with an investigation of this and related problems. Definition 1.1.1. The degree of disjunction 6(d) 01 a ramily d is the least cardinal 8 such that 14 nA21 < 8 for all pairs A 1, A2 in d. We shall be concerned with finding families d of subsets of a given set of infinite power K with degree of disjunction at most 8 for fixed cardinal 8. We need consider only the case when 8 G IA I for all A in in d (and so certainly 8 Q K ) since otherwise the condition S(d)< 8 imposes no restriction on d. Clearly one can always find such a family of power K , consisting in fact of pairwise disjoint subsets, so the problem is to find when there is a f a m i l y d with I d I > K and 6 (94)Q 8. An upper bound for I SQ I is given by the follow ing theorem. Theorem 1.1.2. If an infinite set of power K can be decomposed into a family d with Idl = h and 6 ( d ) Q 8, then h G K' . Proof. We may suppose that in fact it is the set K which has been decomposed. So suppose K = U d where 1941 = A and 6 ( d ) Q 8. For each A in d with IA12 8, chooseA* in [Ale.The condition 6(d) < 8 implies that the mapping

2

Almost disjoint families of sets

a.1.1

which sends A to A * is one-to-one. Thus { A E d ; I 2 8) has cardinality at most that of [ K I e , and so Id1 < I [ K ] " I = K'. The existence of large almost disjoint decompositions will follow from the next theorem. The method of proof is essentially that of the original Sierpinski paper.

Theorem 1.1.3. Let K be infinite, suppose r] 2 2 and let 8 be the least cardinal for which K < qe. Let h be a cardinal with X < 8 and A' = 8'. Then evely set o f power K can be decomposed into a family of qe almost disjoint sets each of power A. Proof. We show that such a decomposition is possible for a particular set of power K ; it then follows that this can be done for any set of power K . Since 8 is least for which K < qe we have 8 < K and consequently X < K . Thus K A = K . The condition X < 8 and A' = 8' means that there is a sequence (a(v);v < N of ordinals a(v) with a(v)< 6 which is cofinal in 8. Put S = U{a(u)r];v < A}. The choice of 8 ensures that lar(')r]I < K for each v, so IS1 < C { I a ( ' ) ~ I; u < A) < K * X = K . We shall define a decomposition of S. For each function g in ' r ] , put

s,=

{grct(v);v
so IS,l = A. And for differentf, g from '77, let /3 be the least oidinal such that f @ ) # g(/3). Then whenever a(v) > 8, for any h in S with dom(h) 2 a(v) it follows that h 4 Sf f S,l . Thus S I , nS l, < 1/31 < A. This means that if J = {Sf;f E 'q}, then J is a family of ve almost disjoint subsets o f S each of power X. If IS1 = K , we have finished. If IS1 < K we must add to S to build it up to power K . Choose any set S1 of power K disjoint from S, and divide S1 into a family d1 of pairwise disjoint sets each of size X (which is possible since K * A = K ) . Then the family J U J 1 decomposes S U S1 in the required manner.

As a first corollary toAthis theorem, we deducedhe Sierpinski result mentioned above. For given the infinite cardinal K , let 8 be least such that K < 2', so necessarily 8 is also infinite. Then by Theorem 1.1.3. with r] = 2 and A = 8, any set of power K can be decomposed into a family of 2e almost disjoint subsets each of power 8, that is, into a family of more than K almost disjoint infinite sets. If the Generalized Continuum Hypithesis is assumed, the theorem leads to the following result. Given infinite cardinals K and h with X < K and A' = K',

Ch. 1.1

Almost disjoint families

3

then every set of power K can be decomposed into a family of K + almost disjoint subsets each of power h. This follows from Theorem 1.1.3 by putting 1) = 2, and noting that by the GCH, 8 = K is the least value of 8 for which K < 2e. However, this result can be deduced without making use of the CGH, by adopting a different method of approach. When A = K , this was known by Erdos in 1934 (see Erdos, Gillman and Henriksen [34, Lemma 4.11). A different proof is given by Sierpinski [87].

Theorem 1.1.4. Let K and A be infinite cardinals with A < K and A' = K ' . Then evew set of power K can be decomposed into a family of more than K almost disjoint subsets each of power A. Proof. The argument splits into two cases, depending on whether K is regular or not. Case 1 : K regular. Here K = K ' = A' Q A Q K , so K = A. Let K be any set with IKI = K , and let 3c be a decomposition of K into K disjoint sets each of power K . We shall show that no family 93 with I931< K + is maximal (under the subset relation) in the class 3of all families extending CK consisting of almost and so disjoint subsets of K each of power K . Zorn's Lemma applies to 3, there is a maximal family d in 3.But then we must have Id1 2 K', and the theorem holds in this case. So let CM be a member of 8 with ICM I < K + . Since 3( C CM, in fact I931 = K . Let (B,; p < K ) be an enumeration (always without repetitions) of cx3. For each p with p < K , note that IU { B , nB,; u < p)l < K since 6(93) < K and K is regular. Hence we may choose inductively elements b, when p < K so that b,EB,-U{B,;v
Almost disjoint families of sets

4

Ch. 1.1

< K ~ } Since . I u { K , fIB,; p < K u ) l < A, . K , < K = IK,I , this is always possible. Put B = u { X u ;u < K ' } , so IBI = Z@,; u < K ' ) = A. Note that given p with p < K , if 7 is least for which p < K , then Xu f B, l= 8 whenever u > 7, and so B n B, C_ U { X u ; u < 7).Hence IB f BPI l < Z(A,; u < 7) < A.

Xu C_ K ,

- U {B,; p

It follows that the family cx3 U { B } is in (9, and so cx3 is not maximal. If we assume the Generalized Continuum Hypothesis, then we can show that the condition A < K and A' = K ' of Theorem 1.1.4 is also necessary for the existence of such a decomposition. This result goes back to Tarski [94]. Unfortunately there appears no way to avoid the GCH for this proof. We need first the following easy lemma.

Lemma 1.1.5. Let A be a cardinal, let /3 be an ordinal with A not cofinal in /3. For each set A with A E [p]" there is an ordinal a with a < /3 such that I& n~i 2 A.

Proof. Given A with A E [/3]>', take a subset A of A with order type A. Since A is not cofinal in /3, there is a with a < /3 such that A C_ a. Then A C a f Al so that la nAl B [All = A . Theorem 1.1.6 (GCH). Let K and A be cardinals with A' # K ' . m e n any decomposition d of a set of power K into subsets each of power at least A with 6(d)< A, has power at most K .

Proof. We may suppose that d is a decomposition of the set K , so suppose PQ C [K]"' with 6 ( d ) < A, and we need only consider the case A < K . So in fact A < K , because A' # K ' . Since A' # K ' we know A is not cofinal in K . Applying Lemma 1.1.5, for each A . in d there is an ordinal a with a < K and la nA1 B A. For each (Y put 94, = { A E d ; ais least for which la nAl > A}. Then 94 = U { d , ; a < ~ } .Thus I ~ I Z<( I d , l ; a < ~ ) . We seek an estimate of ld,I. Since 6 ( d ) < A, for distinct A 1, A 2 from d,we have I(a nA l )

fl(a"A211

< IAl n A 2 l <

A

9

and so a n.4, # a n A 2 . Thus if cx3, = {a n A ; A E d,}then Icx3,l= I &I. Now 6('33& < A and g,is a decomposition of the set a,so by Theorem 1.1.2 we have Ig,l< Ial'. However A < K and la1 < K so by GCH, Icx3,1< K . Thus ld,I<~.Consequently 1941 < Z ( l d , l ; a < ~ ) < ~* K and the theorem is proved.

Ch. 1.2

Almost disjoint functions

5

Corollary 1.1.7 (GCH). No set ofpower K can be decomposed into a fami& d of more than K sets each of powergreater than A such that 6 ( d ) S A. Proof. Such a decomposition is certainly impossible when A' f K ' , by Theorem 1.1.6. So suppose A' = K ' . Since A' < A and (A')' = A', in fact K ' f A'. But now Theorem 1.1.6, with A replaced by A', yields the result. It is worth remarking that if A is finite (but K still infinite) then Theorem 1.1.6 and Corollary 1.1.7 remain true, with the same proofs, and in fact GCH is not needed. 52. Almost disjoint functions In this section we shall discuss a refinement of the question considered in section 1. Rather than seeking families of almost disjoint subsets of an arbitrary set, we shall look at families of functions from a fixed set L to a fixed set K , where ILI = A and IKI = K . We shall in fact from the start identify L with A and K with K , and so shall consider subfamilies of ' K , t h e set of all functions mapping A into K . Let 9 be a subset of ' K . Thinking of the members of 9 as sets of ordered pairs, we can define 6(9), the degree of disjunction, as in Definition 1.1.1. So in fact 6(?) is the least cardinal 17 such that [ { a
Definition 1.2.1. A set T is said to be a transversal of the family 4 if IT nA I = 1 for each A in 4. Given a family SQ = { A , ; v < A} of A pairwise disjoint sets each of power by identifying each A , with K , any transversal of s4 can be identified with a function in ' K . In this-way, a family 9 with 9 C ' K and 6 ( 9 ) S 8 corresponds to a class of transversals of the family d , any two of which meet in less than 8 points. Thus the original problem is equivalent to finding the maximum size of a class 7 of transversals of s4 with S(9)< 8. Clearly one can always find K pairwise disjoint transversals, so we shall wish to know for which values of A and 8 is it possible that 171> K . K,

6

Ch. 1.2

Almost disjoint families of sets

We shall need to assume the Generalized Continuum Hypothesis almost throughout this section. We start with several easy remakrs.

Lemma 1.2.2 (GCH). Suppose either 0 < A and K + A'. 1.7 ' K and 6 ( 7 ) < 6 then 171< K .

< A or else t9 = A and

K+<

Proof. For a contradiction, suppose that under these conditions,there is 7 with 9C ' K such that 6 ( 9 ) < 6 and 171= K'. For distinct f, g in 9, put E(J g ) = {a! < ~ ; f ( a ! =g(a)}, ) SO IE(Lg)i < 0. Put E = u {aft g); { L g } E [TI2},so IEl < A. If a! is in A - E thenf(a) # g ( a ) for distinctf, g in 7, and so 171< K , contradicting that 191= K + . Lemma 1.2.3 (GCH). Let A' < K . Zf 9C ' K with S(7)

< X then 191< K +

Proof. If AS K clearly 171< K + , so suppose K < X and consequently A must be singular. Take cardinals A, where T < X' such that K + < A0 < A1 < A; < ... < A and X(A,; T < A') = A. For a contradiction, suppose in fact that 1912 K++. Since 6 ( 9 ) S A we have a decomposition [TI2= u ~ (, { A g } ; If ngl < b}.At this stage, we shall appeal to the relation K++ (K'): which follows from Theorem 2.2.4. This symbol means that whenever the class of unordered pairs from a set S of power K++ is decomposed into at most K parts there is H with H E [SIK+such that all the pairs from H fall in the same part of the decomposition. In the situation here, this ensures that there is a family Q with Q E [TIKfsuch that for some fixed T with T < A' always lfngl< X, for allf, g in Q. But now if 81 = { g r &+;g E g }, it follows that 91 is a family of K + functions mapping from into K , with S ( 9 f ) < A,. This is not allowed by Lemma 1.2.2. Consequently 171< K + . --f

Lemma 1.2.4.Let K ' = A'. Then there is FF with 7 C ' K , I91 and 6(7)GA. Proof. There are ordinals yo where u < K ' and 6, where T < A' such that < y1 < ... < K = sup {yo; u < K ' } and 60 < 6 < ...< h = sup {6,; T < A'}. We shall show that no family Q with 9 E [ ' K ] " ~ is maximal in the class of all families of almost disjoint functions in ' K ;as in the proof of Theorem 1.1.4 this will give the result. So suppose Q E [ ' K ] ' ~ , and write Q = {gv;v < K } . Define g with g E ' K by choosing g(a) so that if T is least for which a! < 6, then g(a!)E K - {g,(a!); v < 7,).Given p with p < K , if u is least such that p < yo then g ng,, C {(a,g(a));a < 6,) and so lg ng,lG 16, 11 < A. Thus g can be added to 9 and still gives an almost disjoint family. yo

Ch. 1.2

Almost disjoint functions

I

The following lemma is a theorem of Erdos, Hajnal and Milner [ 3 6 ] . Lemma 1.2.5. Let 6 ( 9 )
K+

= h Then there is

9with 9 C ' K such that 191= and

Proof. The plan for the proof is as follows. By thinking of the ordinals less than K + arranged according to order type, it is clear how to define pairwise disjoint functionsg, for u with u < K + mapping into K', with the domain of g, all ordinals with u < p < K + - simply define g,(a) = a. For each a with a < K + we have la![ < K , and so we may take a well ordering (Em; u < a)of (a subset of) K of order type a,and use the above construction on these well orderings to provide pairwise disjoint tail ends of functions f, : K + + K where u < K'. One has only to arrange the front ends so that any two of the functions meet in a set of size less than K , and this gives a family 9as required. Formally, then, we construct functionsf, : K + + K where u < K' by induction as follows. Let u with u < K + be given, and suppose that the functionsfp when p < u have already been suitably defined. If a > u, putf,(a) = tm. Then when p < u certainly f,(a) #fp(a!). To define f,,(a!) for values of a where a! Q u, let {fp; p < u } = { f v p ; p < min(u, K ) } and choosef,(a!) in K - {f,da); a! < 0 ). Then if p < u, for some fl with p < K it follows that fp =f,p and consequentlyf,(a!) #f,(a) whenever fl Q a Q u. Thus Ifp nf,l < IpI < K . Then if 7 = if,; u < K ' ) , we have 171= K + and 6 ( 7 ) < K as desired. We can now turn to the problem stated at the beginning of this section; to find the maximum cardinality of a family 9 where 9 C ' K with 6 ( 9 ) Q 8. The case 8 < X is much easier than when 8 = X, so we dismiss this case first. Theorem 1.2.6 (GCH). Suppose 8 < h Let 111 be the maximum of the cardinalities of families 9 where 7 C ' K with 6 ( 7 ) Q 8. Then 111 = K unless A = K', in which case m = K + .

Proof. Suppose 7 C ' K with 6 ( 9 ) Q 8. IfK < 8 then K + < A, and Lemma 1.2.2 shows 171Q K . If K > 8, consider 7 I'8' = {f r 8 + ; f € 9)and note that since 6 ( 9 ) < 8 the map which sendsf tof 8' is one-to-one. Now 9 r 8' is a decomposition of 8' X K , a set of power K , into sets of size greater than 8 with 6( 9 r 8') < 8. So by Corollary I Q K I ~ I Q K . I ~ =K e a n d ~ + < X , a g a i n L e m r n a1.2.2 1.1.7, I ~ ~ ~ + andhence gives 191< K . Only the case K = 8 and X < K + remains. Since 8 < X we must have X = K'.

Almost disjointfamilies of sets

8

Ch. 1.2

Considering 7 as a decomposition of h X K , from Theorem 1.1.2 we have 191< ( K + ) ~= K'. However, Lemma 1.2.5 shows that the value 1 9 1= K + can be achieved, and so in this case ti1 = K ' . We are left with the situation when 6' = h, and this is rather more involved. Indeed, the question cannot be fully answered on the basis of the usual axioms for set theory. Theorem 1.2.7 (GCH). Let ti1 be the maximum o f the cardinalities o f almost disjoint families 9 where 9 C_ ' K . (i) If h' # K' and h' # K', then 111 = K . (ii) I f h' = K ' , then in =.'K (iii) If A' = K + , then K + < 111 < K + + . Proof. Let us do (ii) first, so suppose h' = K ' . Then A' < K and Lemma 1.2.3 shows 111 < K'. Since Lemma 1.2.4 gives nt 2 K ' , in fact 111 = K.'. Towards (iii), suppose A' = K'. If h is regular then h = K'. We have the trivial bound 111 < I'K I = K ~ = + K", and Lemma 1.2.5 shows K + < 111. So suppose h is singular, and choose cardinals X,where 7 < h' such that K + < A0 < XI < ... < X and h = C&; 7 < A'). We first construct a family 9 where 9 C ' K with 191= K + and S(9)= K , and so show ni 2 K ' . For T such that let (tW;v < 7 + 1) be a K < T < K , for each ordinal a with X,< a < well ordering of K of order type 7 + 1. For v such that K < v < K + define g, : h + by ~

I

0 if a < h,

g,(a) =

,&,ifa>h,.

Then if E.C # v, we have g,(a) #gv(a) whenever a > max(A,, A), and so g l , ng,l < h,,,axb,v)< A. Then 9 = {g,; K < v < K ' } has the properties claimed. To establish that III G K", suppose for a contradiction that we have a family 9 of almost disjoint functions in ' K with 191>K+++. The argument goes as in the proof of Lemma 1.2.3, taking the decomposition

only using now the relation K+++ +. (K+):+ which is a consequence of Theorem 2.2.4. This ensures that there is a family 9 with 9 E [TIK+ such that for

Ch. 1.2

Almost disjoint functions

9

some fixed 7 with T < A' we have ( f ngl < A, for a l l 5 g in 9. But now if 8( = {g r A:;g€ g}, one finds 1911,=K + and 6(8(< &, contradicting Lemma 1.2.2. This leaves (i), so suppose henceforth that A' # K ' , K + . If K + < A' then 111 = K from Lemma 1.2.2. If A < K , by considering any family 9 where 9C ' K as a decomposition of A X K into sets each of power K , when 6 ( 9 ) < A by Corollory 1.1.7 171< K and consequently III = K . Only the situation when K < A and A' < K + remains, and here A is singular. Choose regular cardinals A, for u with u < A' such that K++ < A-, < A1 < ... < A and h = Z(A,; u < A'). In fact A' < K , for A' = K would mean A' = A'' = K ' since always A' is regular. We consider separately the two possible cases, A' < K ' or A' > K ' . The constructions that we use come from the paper [70] of Milner. Take first the case A' < K ' . We have claimed in the statement of the theorem that nt = K . For a contradiction, suppose we have a family 7 consisting of painvise almost disjoint functions in ' K with 171= K + . When a < A and 0 < K , put ?ao = {f€ ? ; f ( a ) = 0). Then for each a with a < A, the sequence ( 9 B~< ; K ) gives a disjoint decomposition of 9 into K classes. However the number of such decompositions is K " ' (since each element of 9 may be in any class), thus the number of decompositions is K ~ =+ K + + . Since each A, is regular with K++ < A,, for each u there must be a set Mu in [A,] ' 0 such that the decompositions 0 < K ) are the same for all a in Mu.But then for alr az in Mu,for all f i n 7 we have f ( a l ) = f(a2).Thus each f i n 9 induces a map f* : A' + K , where

f*(u) = f ( a ) for any a in Mu . However, the number of maps from A' to K is K " , and this is just K since A' < K ' . We had assumed that 191= K ' , and so there must be distinct fl, fz in 9 with ft = Thus fl (a)= f2 (a)for all a in Mu,for any u with u < A'. This means

fz.

If1

nf2l2 lU{M,;u<

A'll

=A ,

contradicting thatfl and fi are almost disjoint. The contradiction shows that in fact 191< K . The final case, where A' > K ' , is similar but rather longer. Again the claim is that 111 = K , and again we seek a contradiction from a family ?of K + almost disjoint functions in ' K . Since here K' < A' < K , also K is singular and we may, choose regular cardinals K , where 7 < K ' such that K~ < K < ...< K and K = Z(K,; T < K ' ) . When a < A and 7 < K ' , put ?(a,7)= {f€ 9; K,
Almost disjoint families of sets

10

Ch. 1.2

such that for fore for each u with u < A' there must be a set Nu in [A,] some fixed 7,for all (Y inN, we have K, < f ( a ) < K , + ~ for any choice o f f i n 9. Thus eachfin 9induces a mapf* : A' + K', wheref*(u) is that T for which K, < f ( a ) < K , + ~ for all the (Y in Nu. The number of possibilities for the map f * is (K')' = (A')' < K + , and so there must be ' 9 in [TIK+such that the maps f* are the same for all f i n 9#, say equal tof'. Since f # is a ma f # : A' + K ' . constant and K ' < A' with A' regular, there must be L in [A'] ' such thatf $ is on L, say f#(u) = p for all u in L. So wheneverfe 9#, u E L and (Y E N , , we havef*(a)=f#(o)=p so thatKp < f ( a ) < ~ ~ + ~ . P u t U{N,;uEL};then N= Wl = h and we can consider f 1 N as a map from N into K ~ wheneverfE + ~ 9 ' . So N ; f E9 ' ) is a pairwise almost disjoint family of functions from a set of size A into K ~ + Now ~ . A' < K ; + ~ = K ~ < + A ~so by the immediately preceeding case, ICfrN;fE 9#)1G K ~ + Since ~ . 19 1 ' = K + , there must befl, fz in 9 ' for whichfl 1N = f 1~ N . But then Ifl nfzl 2 PI = A, in contradiction to the fact that 9consists of pairwise almost disjoint functions. This completes the proof of Theorem 1.2.7.

{fr

The simplest case under (iii) in Theorem 1.2.7 where the cardinal iii is not explicitly determined is the conjecture that there is an almost disjoint family 9with 9 C No and 191= Nz. Such a family exists if there is a Kurepa tree, and consequently there is such a family in L, the constructible universe. On the other hand, Silver [91] has shown (subject to the existence of a large cardinal) that it is consistent with the axioms of set theory + GCH that there is no such family. We shall continue this section by making a few remarks concerned with the following situation. Given a sequence (K,,; u < h> of cardinals, what is the size of the largest almost disjoint family of functions taken from the Cartesian product X(K,; u < A) of the K,? (Thus families from "K correspond to the special case where K, = K for all u.) This new situation is particularly relevant if K is a singular cardinal, A = K ' and the K, form a sequence of cardinals below K converging to K . There is the following lemma which will enable us (in certain circumstances) to construct large almost disjoint subfamilies of a Cartesian product.

'

Lemma 1.2.8. Let (A,,; v < 77) be any sequence of cardinals and let (A,,; v < 77) be any sequence of sets with always &I = It@,; P G v). lien there is a subfamily 9ofX(A,; v < 77) with 6(9) < 7) and 191= II@,,; v < 77).

Proof. For each v, identify A , with X@,; 1.1< v). Put 'II = II@,,; v < q), and let (fa;(Y < n) be an enumeration of X(A,,; v < 77). Define functionsg, in X(A,,;

Ch. 1.2

Almost disjoint functions

u < 0 ) for a with a < n as follows: if v

11

< 17 then

r (V + 1). a < n} so 171= n. Further 7 is almost disjoint, for take distinct

gOlw= f a

Put 9 = {g,; functionsg,, gp from 9. If u is the least ordinal such that f,(u) # fp(u) then g,(p)#g&)wheneverp>u.Thus Ig, ngpl< Iul
Corollary 1.2.9. Suppose K is singular with K ' = No and let ( K , ; n < W ) be any increasing sequence of cardinals all less than K with Z ( K ;~n < W ) = K . Then there is an almost disjoint subfamily 9 of X ( K , , ; n < W ) with 191> K . Proof. Note that if n < w then II(K,; m < n) = K , . By Lemma 1.2.8 there is then 9 with 9 C X ( K ;~n < W ) such that S(7) < No and 1 91= II(K,; n < a). By Konig's theorem though,

n(K,; n < W ) > x ( K n

; n < W)'

K

.

The result in Corollary 1.2.9 is special to singular cardinals K with K' = No. When K is singular with K ' 2 HI,for some increasing sequences ( K ~ u; < K ' ) of cardinals below K there is an almost disjoint subfamily 9 of X ( K ~ u; < K ' ) with 1 9 1> K ; for other sequences possibly 191< K for all such families 9. Examples are given in Theorems 1.2.15 and 1.2.13 below. However, first we need to examine some properties of closed unbounded subset;. of the cardinal h.

Definition 1.2.10. Let h be an infinite cardinal. The subset C of h is said to be closed if whenever B E [C]"' then U B E C. The subset C of h is said to be unbounded (or cofinal) in h if for all a with a < h there is y in C with y 2 a. If h is regular, it is easily seen that the intersection of any family of fewer than h closed unbounded subsets of h is again closed and unbounded in A.

Definition 1.2.11. A subset S of h is stationary in h if S intersects every closed unbounded subset of A. It is clear that any stationary subset of a regular cardinal h has power h. We shall make use of the following result of Fodor [42].

Lemma 1.2.12. Let S be a stationary subset of the uncountable regular cardinal h and let f : S -+ h be a regressive function (that is, f ( a )< a whenever a # 0). Then there is a stationary subset T of S such that f is constant on T.

Ch. 1.2

Almost disjoint families of sets

12

Proof. Suppose the lemma is false, so for all 0in S the set f-'(fl) is not sta. tionary in A. Hence there is a closed unbounded subset Cp of A such that cpnf-l(o) = 8 . Put D = {a
a<

Pn+l

E n ' l c , ; ~ < P n }*

a,},

is possible since n{C,; y < being an intersection of Such a choice of fewer that A closed unbounded subsets of A, is certainly unbounded. Put =U ; n < a},so surely 0> a.Also /I ED. For take any y with y < 0, and we shall show that 0E C,; from this it will follow that Q ED. Since y < fl there is n such that y < 0, and hence y < whenever m 2 n. The choice of then ensures that EC , whenever m > n. Then U m > n } E C, since C, is closed, and 0= U m > n ) so in fact /3 E C,. This completes the proof.

{a,

a,,

a,,

a,,

{am;

{am;

c

Given a subset C of the infinite caruinal A, one can form the closure of C - the least closed subset D of A with C C D. When C i s infinite, Icl = ICl. This may be seen by considering the mapf : - C - t C where for any a in - C,f ( a )= least in C with a < a. Clearly f is one-to-one and hence -C I S ICI.In particular, if A is singular there is a closed sequence (Ao; u
c

c Ic

of cardinals below A converging to A. (That the sequence is closed means that the set {A;, u < A ' } is closed in A.) We are now set to return to the investigation of families of functions.

Theorem 1.2.13. Let

K

be a singular cardinal with K '

> NO.Suppose 2A< K

Ch. 1.2

13

Almost disjoint functions

for all h with A < K . Let (K,,; u < K ’ ) be a closed strictly increasing sequence of cardinals below K with K = C.(K,; u < K ’ ) and let ? be an almost disjoint subfamily of X(K,,; a< K ’ ) . Then 191 a,; if a! = U {a!,; n < w } then a! E L n C). Hence by Lemma 1.2.12 there is a stationary subset Sy? of for L such t h a t g is constant on SU), say with value 7 ( f ) . Thus f ( u ) < all u in Sy?. For S in [L]“’ and T with 7 < K ’ , put ?(S,

T)

= {f€ 7 ;Scf) = S and

~(f = 7) ).

For any f in 7,since Sy? is stationary in K ’ then ISy?l = K ‘ ; If for somefl, f z in 7 we have SY;) = Scf2) = S and f l and f z agree on S then f 1 and f 2 agree on a set of size K ’ and so f l = f 2 , by the property 6 ( 9 ) < K ‘ . Hence IT(S,

< K’I,

7)

= K:‘

< 2K7 K’ s K . ’

Now the number of pairs (S, 7)is ( K ’ ) ~ ’ . K ‘

< 2K’.

K‘.

K’

d K . Hence

I ~ I = I U { S ( S , T ) ; S E[ L l K ’ a n d ~ < ~ ‘ } l-
.

This completes the proof. If, in the situation of Theorem 1.2.13, we consider almost disjoint subfamilies ? of X(K:; o < K ‘ ) , again we can find a bound for 1 91.We shall need to quote a couple of results which will not be proved until Chapters 2 and 3.

Theorem 1.2.14. Let K be a singular cardinal with K ’ > Ho-Suppose 2’ < K for all h with h < K . Let (K,,; u < K ‘ ) be a closed strictly increasing sequence of cardinals below K with K = C(K,; o < K ‘ ) and let 7 be an almost disjoint subfamily ofX(KA;o < K ‘ ) . Then 171
Proof. Suppose for a contradiction that there is such a family 9 with 171= K++. Define a function H from 7 to y 7 a s follows: for f in ?,

H(f)= { g E 9;bb < K’(g(0)Gf(0))) - { f .

14

Almost disjoint families o f sets

Ch. 1.2

T h e n a l w a y s I H ( f ) l < ~ . F o r H ( f ) C X ( f ( u ) + l ; c s < ~ ’ ) w h e r e If(u)+ llG~,,, and 6(H(f)) < 6 ( 7 ) = K’; so that Theorem 1.2.13 applies t o the family H ( f ) , and shows IH(f’)l < K . Thus, in the terminology of Chapter 3, H i s a set map of order K + defined on 7. Sitice 171= K + + , Theorem 3.1.1 shows that there is a subset J of 7 of power K + + which is free for H. This means that f 6 H ( g ) and g H( f ) for all J g in J . In particular then, there is a sequence ( f,; p < K + + ) of elements of 7 such that f , 4 H( f,) whenever p < v < K + + . Thus there must be a,, such that f,(u,,) >fv(uPv)whenever p < v < K + + . Consider [ K + + ] ~ partitioned into K ‘ classes according to the value of a,,. By virtue of Theorem 2.2.5 of the next chapter, the partition relation (2K’)+-+ (K”):, holds; since (2K’)+< K + + and K ” > No also the relation K + + -P (NO):< holds. This relation ensures that in the present situation there are ordinals pn (for n with n < w ) such that po < p l < ... < K + + and an ordinal u with u < K ’ such that U,,,~ = u whenever m .< n < w . This means that f,,(u) >fpI (u) > fP2 (0) > ..., which is impossible. This contradiction proves the theorem. Let US impose the stronger condition that the GCH is true for all cardinals below K , that is 2’ = h+ whenever h < K . Then we can use Lemma 1.2.8 t o produce an almost disjoint subfamily of X ( K ~ u; < K’) of power 2K.

Theorem 1.2.15. Let K be a shgular cardinal with K ’ > NO.Suppose 2’ = A+ for all h with h < K. Let ( K ~ u; < K’) be any increasing sequence of cardinals below K with K = C(K,; u < K ’ ) Then there is a subfamily 7 Of X ( K ~ u; < K’) with 6 ( 7 ) < K ’ and 171= 2 K . Proof. For u with u < K ‘ , note that

Thus Lemma 1.2.8 gives a family 7 of X ( K ~ u; < K‘) with 6(9) 171= ~ ( K L u; < K ’ ) . However

< K ’ and

If we combine Theorems 1.2.14 and 1.2.15, then immediately we have the following.

Theorem 1.2.16.Let K be a singular cardinal with K ‘ > No Suppose ’2 = A+ for all h with A < K . Then 2K = K’.

Ch. 1 . 3

Trarrsversals of non-disjoint fainilks

15

Theorem 1.2.16 is of particular interest in view of the principal result of Easton [ 121, which implies that no similar theorem is possible for any regular cardinal K . Theorem 1.2.16 (in fact a more general result) was first proved recently by Silver [92], using powerful methods from mathematical logic. A combinatorial proof of Silver's theorem was then found by Baumgartner and Prikry [5], amongst others. Using similar methods, Galvin and Hajnal [45] gave general bounds for 2 K (where K > K ' > Ho) in terms of the values of 2' ' for X with X < K . The proofs of Theorems 1.2.14 and 1.3.15 come from Calvin and Hajnal [45]. Theorem 1.2.13 is from Erdos, Hajnal and Milner [36], where a necessary and sufficient condition is given o n the sequence ( K , ; u < K ' ) for the existence of an almost disjoint family of more than K functions from X(Ku;

(5


53. Transversals of non-disjoint families In the previous section, the discussion could be viewed as a consideration of the number of almost disjoint transversals of a family of disjoint sets. This section will be devoted to transversals of non-disjoint families. One of the first results will show that in this situation there is not even one transversal for certain families. Consequently, we shall extend the notation of a transversal as follows. Definition 1.3.1. Given a family d ,an g-transversal of d is a set T such that 1 < IA n TI < 77 for each A in d. Thus a transversal in the previous sense is a 2-transversal by the meaning of this definition. When considering a family SQ of sets each of power K , we shall be interested in the existence of an g-transversal for values of g only with g < K , since a K+-transversal T would allow T n A = A for some or all of the meinbersA of d.(In particular, ud itself would be a K+-transversal of d.) Given an arbitrary family d of' sets all of power X,with a specified degree of disjunction 6(d) = 0 , we shall seek the smallest g l o r which we can be sure that an g-transversal for d exists. One can then go on to ask, lor this smallest g, what is the maximum size of an almost disjoint faniily of g-transversals of d.The situation is rather similar to that found in the last section, and we sliall not pursue the matter further. The results of this section are to be found in the paper [ 191 of Erdbs and Hajnal, expressed in rathcr different notation. We start by giving examples of families with n o transversals in the sense of Definition 1.3.1.

Almost disjoint families of sets

16

Ch. 1.3

Theorem 1.3.2 (GCH). (i) Suppose A < K . Then there is a family d o f K sets each o f power K with 6 (d) < A: which has no A-transversal. (ii) mere is a family SQ of K painvise almost disjoint sets each o f power K which has no K-transversal. +

Proof. For (i), take A with X < K and suppose for a contradiction that (i) is false, that is that every family d of K sets each of power K with ti(.&) < X has a A-transversal. Take any f a m i l y q = {& v < K } of pairwise disjoint sets each of power K . Construct a sequence T,, where y < K + by transfinite induction, so that T,, is a A-transversal of T3 U { T,; Y < p } with T,, C U g . Consider 7 = { T , , ; ~ < K ’ } .Here = K , IT,,I=K f o r e a c h p 1 , a n d 6 ( 9 ) S h < K . 2 1 = K’, this contradicts Corollary 1.1.7, and (i) is proved. Since 1 Likewise, if we assume (ii) t o be false, this construction (with A = K ) can be continued t o yield a family 9 = { T P ; p< K ” } with Iu5’1 = K < K ’ , lT,,l = K for each p , and S ( 7 ) < K . Since (K’)’ # K ’ , this contradicts Theorem

Iu71

1.1.6.

When K is regular, the value Id I = K’ of (ii) in this last theorem cannot be decreased. Any family of fewer than K + pairwise almost disjoint sets each of power K has a K-transversal. This follows from the next theorem. Theorem 1.3.3. Any family d of pairwise almost disjoint sets each of power with 1941 = K ’ has a K’-transversal.

K

Proof. This is almost trivial. Suppose d = {A,; v < K ’ } satisfies the conditions of the theorem. Then whenever p < K ’ one can choose x,, from A,-U { A , ; v < p } . If T = {x,; p < K ’ } then for any Y it follows that T nA , C_ {x,,:p< v}, and consequently T i s a K’-transversal for d . The result of Theorem 1.3.3 is the best possible, in the sense that the existence of a K’-transversal cannot in general be improved t o the existence of a A-transversal for any A with A < K ’ . When K is regular, this follows already from Theorem 1.3.2, using GCH. However, GCH is not used in the construction below.

Theorem 1.3.4. mere is a family d o f K ’ painvise almost disjoint sets each of power K which has no A-transversalfor any h with h < K ’ . Proof. Write

K

as a disjoint union, K = U I& u, <;K ‘ ) where always IKuI= K .

Ch. 1.3

Transversals of non-disjoint families

17

Choose ordinals a, where u < K ' such that (YO < a1 < ... < K and K = sup {a,; u < K ' } . Put SQ = {a, U K2,; u < K ' } U { K 2 , + l ; u < K ' ) . Then 1941 = K ' , 6 ( d )= K and MI = K for each A in G I . Suppose T meets every member of d ;then IT flK I 2 K ' . Consequently, for any h with h < K ' there must be u with u < K ' such that I T n &I, 2 A. Then I T n (a,U K2,)l 2 h, and so T is not a A-transversal of d . Together with the observation that any family 94 of fewer than K ' pairwise almost disjoint sets of power K has a 2-transversal, Theorems 1.3.2(ii), 1.3.3, 1.3.4 provide a complete discussion of the situation pertaining to families of pairwise almost disjoint sets of power K , in the case that K is regular. When K is singular, there are still some gaps to be filled. Specifically, what is the situation with a family of 8 pairwise almost disjoint sets each of power K , where K ' < b Q K ? There is the trivial fact that one can always find a @-transversal, and this is the best possible result, for there is an example of a family of 0 pairwise almost disjoint sets each of power K which has no 8-transversal. Such an example is given in Theorem 1.3.6 below. The construction is due to Erdos and Hajnal [ 191, and is quite complicated. We shall need the following lemma (see Hajnal [52, Theorem 91).

Lemma 1.3.5 (GCH). Let X be a regular cardinal and let X be a set with 1x1= A". There is a pairwise almost disjoint family % = { X u ;u < A"} of subsets of X with always IX,l = h such that for all subsets Y of X with I YI = A" there is X u in X with X,, C Y. Proof. Let < be a well ordering of X . Let 1 be the family of all subsets of X of order type h2 under <. (Here and for the rest of the proof, by h2 is meant 2I = A", and the ordinal power.) Since 1 G [XI*, it follows from GCH that 1 we may write 2 = {z,,; p < A+}. Inductively define subsets X u of Z, with tp(X,) = A, for u with u < A", as follows. Suppose that Y is given with u < h+ and that the X,, where p < u have already been defined. If u < h then IZ, - U { X,,; I-( < u}l = A; choose for X u any subset of Z,, - { X , ; p < u } with order type h. Now suppose h S u< A+. Let (X,; a < h) give a well ordering of {X,,; p < u}. Note that since tp(Z,) = h2 there are sets Z, such that Z, = U {Z,; a < A} where tp(Z,) = X and Z,p < Z, whenever /3 < a (that is, x < y if x E Z,p and y E Zue). Since always tp(X,p) = A, there is at most one a for which tp(X,O nZ,) = A. Define ordinals ?(a) with y(a) < h and elements x, with x, E Zq(e) for a with a < h by induction as follows. Choose y(a) such that y(a)> ~ ( 0 and ) also IX,p f Zv(e)l l < h for all 0 with /3 < a.This is possible, by the regularity of A.

u

Ch. 1 . 3

Alnzost disjoint functions of’sets

18

Now choose ,x from Zqp) - U { X u @/3: < a). The choice of y(a) and the ; < A}. Since xu@ <x, regularity of h make this possible. Put X u = { x V a a whenever B < a it follows that tp(X,) = h, and surely X u C Z,. This completes the definition of the sets X,. The X u are pairwise almost disjoint, for suppose say p < u. Then X,, = X , for some a, so X,, fl X u C {x,,~;p < a } and thus IX,, n X,I < A. Take any subset Y o f X with IYI = A+. Then there is a subset Z of Y with tp(Z) = A’, and Z = 2, for some v. Since X u C_ Z, in fact X u C Y. Thus if 31 = { X u ;u < A+} then X has the required properties, and the lenirna is proved.

Theorem 1.3.6 (GCH). For each 8 where K ‘ < 8 < K there is afamif‘y o f 8 painvise almost disjoint sets each ojpower K which has no 8-transversal.

Proof. Let K and 8 be cardinals with K ’ < 6’ < K . Thus K is singular, so there is a sequence (K,: u < K ’ ) of cardinals below K with K = Z ( K , : u < K ’ ) . If K ’ < 8‘, the theorem is easily proved. without GCH. Start with a pairwise disjoint family {A,,: u < K ’ } where always lA,l = K . For each u with u < K ‘ , choose a pairwise disjoint family {A,,; u < 8) of subsets ofA, with always IA,,,I = K , . PutB,= U { A , , ; U < K ’ } , S OI B , ~ = Z ( K , ; U < K ’ ) = K . C ~ ~ sider the family

d = {A,;

U<

K’)

u {B,:

U<

6)

The B, are pairwise disjoint and A , nB, = A , , so IA, nB,I = K,, < K . Thus 94 is a family of 8 pairwise almost disjoint sets each of power K . Let T be a transversal of d.Since then always T n B,, f Q and U {B,; u < 8) C U{A,: u < K’}, surely IT n U{A,; u < K ’ } I 2 0 . Since K ’ < 8’ there must be some CJ with ITnA,I Z 8. Hence d has n o 0-transversal. Now consider the case 8’ < K‘. Start this time from a pairwise disjoint family d1= {A,,; u < K ’ + ) with always IA,I = K . Since K ’ is regular, we can apply Lemma 1.3.5 to dl,t o produce a family of families = {%,: u < K ” } where always 31, E [ d , ] ” and I%, n%,,I < K ‘ whenever p # u, with the property that for any subfamily y of A l with lyl = K“ there is 31, in X with

x

xuc y.

Write 35, = { A w :7 < K ‘ } . For each u and 7,choose a pairwise disjoint family { A w y : y < 8 ) o f s u b s e t s o f A , withalways l A , l = ~ , . P u t B ~ = U{A,,: 7 < K ’ } , so IBI, = Z ( K , : 7 < K ’ ) = K . Consider the family

cio = { A , ; u < K ‘ + } u { B ~u < ; K ’ and y < 8 j . We shall show that cx3 is pairwise almost disjoint. Take distinct B 1 and B2 from 9,and consider the various possibilities. If B 1= A , and B2 = A , for

Ch. 1.3.

Transversals of non-disjoint families

19

some u and r, then B 1 r l B2 = 8 . If B 1 = B , and B2 =Bus for some v, y and 6 then again B 1 nB2 = 8 . If say B 1 =A,, and Bz = B , then B1 n B2 = 8 unless A , € X u , in which case A , = A , for some r s o A , nB , = A , and hence lB1 flB21 =IA ,I = K , < K . The final possibility is B 1 = B , and Bz = B,6 where p # v. Since U 9 = U{A,; u < K ” } it follows that B1 nB2 = U{A, n ( B n~B P 6 ) ;u < K ’ + } . However, either A , n B,, = 8 or A , n B,s= 8 unlessA,€%Y, n % @ , s o B 1n B 2 = U{A,,nB,nBP~;AuEXY, nX,}.Now always IA, nB I, < K , and I%, r l &I < K by the choice of X , so again IB1 f l B21 < K . Thus 93 is indeed pairwise almost disjoint. Hence further, 191 = K ‘ + iK ’ e = 8. has n o 0-transversal. Since 0’ < K ’ < 0 we Finally, we shall show that know 0 is singular, so there is a sequence ( 0 s ; { < 0’) of cardinals below 0 with 0 = Z(0.+;(< 0’). Suppose T is a set such that IT nA,I < 0 for each u with u < K ” . So for each u there is ((0) with {(u) < 0’ such that IT nA,I < Or(,,). Since 0’ < K” and K ’ + is regular there is a subset H of K” with IM = K“ such that {(u) is constant for u in H , say with value {. By the choice of the family X there is Xu in X with 31, C_ { A , : u € H I . Thus ITnA,I < 0s. for all A , in Xu.Hence IT n U %,,I < K ‘ . Or < 0 . Now { B w : y < 0} consists of 6 pairwise disjoint sets, each a subset of UX,. Hence there is some y such that T nB , = 8. Thus 9 can have n o 0-transversal. This completes the proof. +

Let us return t o the case of families of K sets each of power K but where the degree of disjunction is smaller than K . We shall assume the Generalized Continuum Hypothesis for the rest of this discussion. Theorem 1.3.2 shows that for such a family d with 6 ( d ) = A, the best that can be hoped for in general is a A+-transversal. In fact this can usually be achieved. There is first a trivial lemma.

Lemma 1.3.7 (GCH). Let i and h be cardinals such that there is no q with i=7 . 1 ’ and A’ = q’.Suppose 93 is a family with I U ‘33 I < i and 6 (9) < h and IBI 2 h j o r each B in ‘33. Then I91< 1. Proof. We may suppose A < 1. Since 191 < I [UCK~I’I, certainly if IUCY~I+
Iu91+

The following theorem is slightly more general than our immediate need dictates, but the full strength will be used in the proof of Theorem 1.3.12. We shall adopt the temporary notation d [ K ] for { A € d ;IA I = K } .

Theorem 1.3.8 (GCH). Suppose

K

and h are such that there is no q with

K

= q+

Almost disjoint families of sets

20

Ch. 1.3

and A' = 77' (unless K = A+). Let d be any family of K sets with 6 ( d )< h and IA l < K for each A in d.Then there is a A+-transversal T o f A [ K ] with lTn Al < h f o r each A in d. Corollary 1.3.9 (GCH). Under the conditions of the Theorem, any family d of K sets each of power K with S(d)< A has a A+-transversal. Proof (of Theorem 1.3.8). We need only consider A with h < K . The case when = A+ may be shown by a construction similar to that in the proof of Theorem 1.3.3 (and in fact GCH is not used). So suppose henceforth that h+ < K . Let d be well ordered, d = { A , ; v < K } . We still try t o choose xu in A , , but d o so only when this is possible without having xu in any A from d which already meets {x,; p < v} in h points. This ensures that the set of all the xu meets any A in at most h points. So formally, use transfinite induction t o define elements x, when u < K as follows. Write K

93,=

{AE & / A n {~,;p<~}l>h}.

Choose xo from A o . When v > 0. ifA, - W, = 8 put xu = xo and otherwise choose x, from A , - ULB,,. By the property 6(d)< A. any A in 93, is uniquely determined by its intersection with { x , ; p < v } . Since I { x , ; p < v } l < K , Lemma 1.3.7 shows that { A f{x,;p l < v } ; A E 93,) has power less than K : hence I93,I < K . Again using that S(d)< A < K , it follows that IA, n U9,I < K provided A , 4 93,. Thus if M,I = K and IA, n {x, :p < v ) l < h then A , - Ug, f 8 and so xu E A , . Put T = { x u ;v < K } , then T meets every set in SQ [ K ] . And for any A in SQ, if for some p it happens that IA n {x,: p < p}I = h then for all u with v 2 p either x, = xo or else x, 4A. Hence always IT n A I < h. This completes the proof. If h is finite, the same proof holds, and in fact GCH is not needed. It is not known if Corollary 1.3.9 holds without restriction on K and A. The simplest case that is not settled is when K = Nu+],A = Ho.Explicitly:

Question 1.3.10 (GCH). Is there a family d of H,+1 sets each of power with 6(d)= Ho which has no H1-transversal? One can always find a A++-transversalhowever.

Ch. 1.3

Transversals of non-disjoint families

21

Theorem 1.3.1 1 (GCH). For all K and X every family d of K sets each of power K with 6 ( d ) < A has a h++-transversal. Proof. Modify the proof of Theorem 1.3.8 by putting

CM,= { A E d;IA n { x , ; ~ < v ) I > A } . Using now Corollary 1.1.7 one concludes that 133,l carries over.

< K ;the rest of the proof

We are left finally with the problem of finding a transversal of a family = A < K . Theorem 1.3.2 of more than K sets each of power K , where 6(d) still shows that a A+-transversal is the best that can be hoped for in general (since any family extending a family without a A-transversal has, of course, no A-transversal). We can show that a A+-transversal is usually possible prois sufficiently small, and the problem is unsolved for larger values vided I d [ of 1941. The proof will be by induction on IdI, and so that the induction will work we need a more general statement, on the lines of that in Theorem 1.3.8.

Theorem 1.3.12 (GCH). Suppose K and hare such that there is no q with = q+ and A' = q' (unless K = A'). Let K~ be the least cardinalgreater than K with K ; = A'. Then for all i with i < K ~ if , d is any family of i sets with 6(d) < A and IA I < K for each A in d,there is a A'-transversal T of d [ K ] such that IT flA1 Q A for each A in 4. K

Corollary 1.3.13 (GCH). Under the conditions of the Theorem, any family d of i sets each of power K with 6 ( d ) < A has a A+-transversal.

Proof of Theorem 1.3.12. Use induction on i. By Theorem 1.3.8, the result is true when 1 < K . So suppose L is such that K < i < K~ and that the theorem holds for all i* with i* < 1. Let d = { A , ; v < 1) be a suitable family. Write 94 as a disjoint union, d = U{ A,,; v < 1') where always I d,I < L. By the inductive assumption, there is a suitable set To for do,and we seek to extend To to a suitable transversal of d.However, it may happen that certain members of d - d ohave an intersection of size greater than A with Xo = u 940; in particular they may meet TOin more than h points and so To could not be extended to d.So first increase doby adding to it the family Co of all sets A in d with IA fl Xol 2 A, and look at the family cDo = { A fl X o ; A E doU eo}.It turns out that IQ0l < i, and so there is a suitable transversal TO of go, with TOC X o . Now treat dl similarly, putting X I = U dl - XO and looking at cD1 = { A n X l ; A E d 1U el}.We find T1

22

Almost disjoint families of sets

Ch. 1.3

with T I C X I , and TOU T I extends TO and is suitable for do U 941. If we were to continue in thismanner, possibly there would be a set A in some d, such that IA fl X,l> A for more than A values of p with p < v, and so perhaps !A n U{T,; p < v}l > A, meaning that there would be no suitable extension of (U { T,; p < u } ) to all of d.To prevent this happening, we should finish with such an A at the first stage v where IA n (U{X,; p < v})l> A. So at any stage v, we treat also the members of 93, where 93, includes all members of d with IA n U{X,;p < v}l > A, and form Xu from d , U 73,. The formal construction proceeds as follows. Define sets Xu where v < L' with Xu E [ U d]" by induction as follows. Suppose for a particular v with v < 1' that the X, where p < v have already been defined. Put X,* = U{X,; p < v}, so IX,*l< 1. Write 9, = { A € d ;IA n X,*l> A}. Since 6(d)< A, any A in 73, is uniquely determined byA nX,*.Since IX,*l A. Let E be least such that IA nX$l >A. If A E d,, certainly A C X,* U Xu = X z l , and so g < L). The definition of Xt shows that A C X$ U Xt. There are two cases to consider, depending on whether or not g is a successor ordinal. Suppose first that l is a successor, say E = v t 1. The choice of shows IA fl X,*l< A. Since A is a disjoint union, A = (A fl X,*) U (A r l Xu) U (A UXg), certainly A n T C (A f X,*) l U (A fl T,) U (A f Tt). l The choice of T, and Tt ensures that IA n T,I , 1A fl Ttl< A, and it follows that IA n TI < A. Further, if IAl = K then since A < K , either IA nX,l = K or IA nXtI = K ; this ensures eitherA n T, f 0 or A n Tg f 0,so A n T f 0. Now suppose 5 to be a limit ordinal. We have IA nX,*l< A whenever v<[.SinceA n X $ = U { A n X , * ; v < l } a n d A n X , * C A nX,*whenever p < v, it follows that here IA n X$l = A. Again A is a disjoint union, A = (A nX;) U (A n Xt), SO that A n T c (A nXc) f (A ln Tt). Since IA n TtI < A , we have IA f 7l l < A. Moreover if IAl = K then IA nXtI = K and so A n Tt f 0.This completes the proof.

(ud, ug,)

u{

Ch. 1.3

Transversalsof non disjoint families

23

If A is finite, one can carry out the same construction to show the following by induction on I dl (and the GCH is not needed).

Theorem 1.3.14. If A is finite, evely family d of sets each of power K with 6 ( d )< A has an Ho-transversalT (that is, T n A is finite for each A in d). The case when A is finite has been analysed in more detail by Erdos and Hajnal in [19 pp. 109-1131. They show that any family d with 1941 = He,, and 6 ( d ) < A has a [(A - l)(n + 1 ) + 21-transversa1, and they construct an example to show that this is the best possible result. It is not known if the result of Corollary 1.3.13 holds without restriction on the cardinals involved. It is easy to see that for K = Ha,the number K O is Ha+hl. So the easiest cases left unanswered are the following. (See Problem 39 of [ 241 .) Question 1.3.15 (GCH). Is there a family SQ of H,+l sets each of power H1 with 6 ( d )= No which has no HI-transversal? Is there a family SQ of with 6 ( d ) = H1 which has n o HzJtransversal? sets each of power Finally, we observe that Corollary 1.3.13 is always true, without restriction on 1, K , A if the demand for a A+-transversal be eased to require only a A++transversal.

Theorem 1.3.16 (GCH). For all 1, K , A any family 94 of i sets each of power K with 6 (d) Q A has a A++-transversal. Proof. Modify the proof of Theorem 1.3.12 by putting

q,={ A E 9 4 ; A n X , * l > A } ,

c,

={AE~;IA~x,I>A}.

Using Corollary 1.1.7, one sees IgvI, Ie,l carries over.

< 1941 and the rest of the proof

CHAPTER 2

ORDINARY PARTITION RELATIONS

8 1. The relations defined The partition relations K + ( q k ; k < 7)" and K + (q); defined below, which are the principal objects of study in this chapter, have now been discussed extensively in the literature. A special case of the symbols first appeared in Erdos and Rado [28], although several results that can be expressed in terms of these symbols had appeared before (see for example Sierpinski [85], Dushnik and Miller [ 111, Erdos [ 131, also Kurepa [64]). Detailed discussion is, amongst others, to be found in Erdos and Rado [29] and the encyclopedic Erdos, Hajnal and Rado [38], in both of which references to early results can be found. The subject is concerned with generalizations of the following theorem due to Ramsey [80]: If the unordered pairs from an infinite set S are partitioned into finitely many classes, then there is an infinite subset H of S such that all the pairs from H fall in the one class of the partition. In the following definition, K is a cardinal, n a positive integer, y an ordinal, and q k (where k < y) are cardinals.

Definition 2.1 . l . The ordinay partition symbol K + ( q k ; k < y)" means: given any set S with IS1 = K , for all partitions A = { A k ; k < y} of [S]"into y parts, there exist k with k < y and a subset H of S with IHI = q k such that [H]"C Ak. The special case in which q k = q for all k with k < y will be written Symbols such as K -+ (qk;k < y, O r ; 1 < 6)" have the obvious meaning. The negation of the relation K + ( q k ; k < y)" is written K ( q k ; k < 7)". Thus, in terms of this partition symbol, Ramsey's theorem quoted above can be expressed by: No + ( K , ) k for any finite m. Given a partition [S]" = u{Ak; k < y} of the n-element subsets of a set S, K + (77).;

+

Ch. 2.1

25

The relations defined

a subset H of S with [H]" _C A k for some k is said to be homogeneous for that partition. Before a more detailed discussion of the various partition relations, a few simple remarks are in order. The partition symbol could equally well be defined in terms of disjoint partitions only; this is easily seen to be equivalent to the definition given. The use of the ordinal y to index the classes of the partition A is not essential; any set with the same power as y will serve equally well and the truth or the falsity of the relation will remain unchanged. In particular, the cardinals V k in the relation K + (??k; k < 7)" may be permuted without effecting the relation. If K has a partition property and h is any cardinal at least as big as K , then A has that same property. If K enjoys a partition property with y classes and 6 is any ordinal smaller than y, then the corresponding property with 6 classes also holds for K , since a partition with a small number of classes can be extended to a partition with a larger number of classes by adding superfluous empty parts. If K has a partition property in which the homogeneous^ sets have cardinality q k , then for any cardinals 6 k with 6 k < r)k the corresponding property with q k replaced by e k also holds for K . The truth or falsity of the relation K +. ( q k ; k < 7)" can be trivially determined unless the following conditions are in force:

n> 1,2
u{

Proof. Given a partition

[s]"= u{ A,; 1 < 6 ) u u { r k ; 1 < k < 7) of a set s of power K , put ro= U { A l ; 1 < 6 ) . By the relation K ( V k ; k < y)", there is k with k < y and a subset H of S with IHI = q k and [H]" C rk. If k > 0, nothing more need be done. If k = 0, then [H]" C u{ A l ; l < 6 ) and -+

26

Ch. 2.2

Ordinary partition relations

because of the relation qo + (81;I < 6)" there must be 1 with I < 6 and a subset I of H with 1 1 1= O r and [I]" C Al. In any event, there is a suitable subset of S homogeneous for the original partition of [S]". The restriction to finite exponent n is justified by the following theorem from [27], which shows that the weakest ossible non trivial partition relation with infinite exponent, K (No,No)L, is false for all K . -+

Theorem 2.1.3. Given any infinite set S, there is a partition [S]"O = A0 U A1 of the denumerable subsets of S into two classes which has no infinite homogeneous set. Proof. Let < be a relation which well orders [SIHo. Define a partition [Sf' = A, u A, as follows: for x in [slHo,

XEAO*~YE[X]~~(Y<X), X E A1 -vY E [XIHo( X S Y) . Take any set H in [SINo; we shall show H is not homogeneous for this artiR tion. Let X be the least infinite subset of H ; clearly X E A, and so [H] A,. On the other hand, write H = { h i ;i < a} and for each finite k , put Hk = {ho, hz, ..., h2k } U { h ~ i ;+i <~ a}.Let Hko be the least of the H k . Then A , , and SO H k o c Hko+l and Hko < H k o + l , SO Hko+l E A,. Thus His not homogeneous.

[aHo

As a final remark in this section, note that the symbol K -+ ( q k ; k < 7)" can also be defined when K and the 7)k are ordinals, or in fact arbitrary order types. Rather than speaking of the cardinalities of the sets involved, their order type is specified, for a suitable ordering. Some of the results for ordinal numbers will be discussed in Chapter 7. However, in this chapter we shall contine ourselves to the problems involving cardinal numbers. 52. The Ramification Lemma The method to be described in this section gives one of the most frequently applied techniques in establishing partition relations. The method was used in several early papers before it was expressed in the general form given below in the paper of Erdos, Hajnal and Rado [38]. The method has since found application outside the partition calculus, see for example the book of Juhasz [59]for several applications in general topology.

Ch. 2.2

The ramification lemma

21

The technique hinges on the construction of a so-called ramification system. In order to describe what is meant by this, we look first at trees of sequences of ordinals.

Definition 2.2.1. A partially ordered set (T, is said to be a tree if for each x in T the set { y E T ;y <x } of predecessors of x is well ordered (by G). The order type of { y E T ; y < x } is called the order of x. The 0th level of T is the set of all elements of T whose order is a. A maximal linearly ordered subset of T is called a branch; the members of Tare called nodes in T.

We shall adopt the following notation. For each ordinal u we shall denote by SEQ, the class of all sequences of length u of ordinals, that is all sequences v = (va;a < u) where each v, is an ordinal. We shall write Qn(v) for the length of such a sequencev. Bold face Greek l e t t e r s p , ~will be used for sequences of ordinals. The restriction of v to 7,written v r 7 is defined as follows: if 7 2 Qn(v) then v t-7 = v, whereas if 7 < Qn(v) then v r7 = p where Qn(p) = 7 and p(a) = v(a) for all a with a < 7. Given ordinals E and p we construct a tree as follows: the nodes are all sequences v of length at most p with alwaysv(a) < E , and the ordering is "is an initial segment of", that is p < v* 3 u(p = v ru). It is easy to see that this gives a tree-referred to as the full tree on 5 of height p . The trees that will concern us will not necessarily be full trees, but rather certain subtrees. In particular, they need not have the same number of branches passing through every node, nor even through every node at a given level. The trees we shall use will be of the following form. Let an ordinal p be given, and for each sequence j~with Qn(p) < p let there be given an ordinal n(p). Form the set N = { v; Qn(v)< p and V a < !2n(v) (v(a) < n( v a))},and order N as before. Then ( N , <)is a tree, and it is on trees of this form that we shall build a ramification system. (The ordinals n(p) indicate the amount of branching that occurs at the nodep - one new branch for each ordinal less than n(p).) For example, suppose n(0) = 2 (here 8 stands for the empty sequence), n((0))= 1, n((1)) = 3, n((0,O))= 2, n((1, 0)) = 1, n((1, 1)) = 2, n(
8 by associating with each node v of N two sets S(v) and F(v) by induction

on the length of v as follows: (i) If v E N nSE& (so v = 8) then S(v) is some given set S.

28

Ordinary partition relations

Ch. 2.2

0 Fig. 1

(ii) If v E N n SEQ, and S(v) is known, choose for F(v) some subset of S(v) and split S(v) - F(v) into (not necessarily disjoint) sets S, where a
r

Ch. 2.2

29

The ramification lemma

supposes S ( v ) already given, and defines n(v), F ( v ) and S(p) for all sequences pwith p ESEQ,+l, pr u = v andp(u) < n(v). (The value of n ( v ) for v in SEQ, - N is immaterial.) To save repetition, in all references to a ramification system % the notation F ( v ) , S(v), n(v), N, p is to be understood to be as in the description just given. The effect of a ramification system % of height p on a set S is to divide S into the sets F(v) for those v in N with Qn(v)< p together with the S(v) for those v in N of length p and moreover the F ( v ) along any given branch are pairwise disjoint. This is the content of the following lemma.

Lemma 2.2.2. Let 3 ! be a ramification system on S of height p. Then ( i ) S = U { F ( v ) ; v ~ N a n d Q n ( v ) < pu } U{S(v);v~NandQn(v)=p}. (ii) If v € N a n d u, 7 < Pn(v) with u # 7,then F( v r a) n F( v r 7 ) = 8. Proof. For (i), take x in S and suppose x 4 F ( v ) for all v in N with Qn(v)< p. Use induction to define ordinals v, where a < p such that (v,; a < 7)E N and x E S(v,; a < 7)for all T with T < p . Suppose u is given with u < p and the v, where a < u have already been suitably defined. Then it follows that

(v,; a < u) E N and x E S(v,; a < u) .

(1)

If u is a successor ordinal this is trivial. If u is a limit ordinal, for any 0with and so (vp;fl< u) EN. Also

0< u since (v,; a < 0)E N we know vp < n(v,; a < S),

s ( ~ , a; < u) = n{qv,; a < 0);p< u} = n ( s ( ~ , a; < 0);p< u} and so x E S(v,; a < a). Thus (1) holds. Since by assumption x 4 F(v,;a < a), from the definition of a ramification system there must be p in N n SEQ,+, with p T o = (v,; a < a) for which x E S((C).Choose such a sequence p, and define v, by v, = (~(u). This completes the inductive definition. It follows as above that (v,; a < p ) E N and that x E S(v,; a < p ) ; and hence (i) holds. To prove (ii), suppose T < u < Qn(v) for some v in N. Then v i-7 E N and F(V

r T )n U { s ( p ) ; pENnSEQ,+, and p r T=

so in particular F(v p F(V

so F( v

T) f S( l v l-7 t

r 0)c_ s ( v r 0)c S(V r

7

1) = 8. However,

+ 1) ,

r 7)n F(v r u) = 8. This completes the proof.

rT}

=0 ,

30

Ch. 2.2

Ordinary partition relations

In the applications that we shall make of ramification systems, we shall be interested in having a branch through to the top of the tree such that some elements of S get pushed right up the branch, that is, they are never put in any of the F(b) at the nodes along the way. In a system of height p, such a situation is reached by having a sequence v in N f SEQp l with S(v) non-empty; the initial segments v rcu for (Y with (Y Q p give the nodes along the branch. The following theorem gives conditions under which there is such a branch.

Theorem 2.2.3 (The Ramification Lemma). Let '% be a ramification system of height p on a set S with IS1 > K , for some infinite cardinal K . Suppose either (i) p < K ' ; IF(v)l< K i f v E N ; there are cardinals A, for u with u < p such that A,'u' < K ' and In(v t r ) I < A, whenever v E N r l SEQ, and r < a, or (ii) K = A'; p < A'; IF(v)l< X and In(v)l Q A whenever v E N n SEQ, for any u with u < p ; 6 < A + 2'
< K ' . Thus since p < K ' I { v E N ; !2n(v) < p)I < K ' .

n SEQ,I Q A,""

and K ' is regular,

Consequently IU{F(v);v € N a n d Qn(v)
= K

K,

and so from Lemma 2.2.2

.

So there must be v in N n SEQp with S(v) f 8. Now suppose the conditions in (ii) hold. This reduces to (i), with A, = A for all u with u < p. Note by the last condition on A, that if u < p we have IuI < A', so that A'"' = A < K = K ' . Finally, suppose (iii) in force. Use induction on u to show that INnSEQ,l
v r r E N n S E Q , andv(r)
we have

IN n SEQ,I

Q Z(ln(P)l; P E N n SEQ,) ,

Ch. 2.2

31

The ramification lemma

so W n SEQ,I < K by the inductive assumption and the regularity of K . And if u is a limit ordinal, since

vENnSEQ,*

Vr
we have

W n SEQ,I

n(ln(r)l;3 r < a@ E N nSEQ,)) ,

so that INf SEQ,l l < K by the inductive hypothesis and the inaccessibility of K . This establishes that !NflSEQ, I < K whenever u < p . Thus I { v E N ; Qn(v ) < p}I < K , and so IU{ F ( v ) ;v E N and Qn(v) < p}l< As in part (i), it now follows that there must be v in N n SEQp for which S(v) f

K.

0.

We shall make immediate use of the Ramification Lemma to establish a couple of partition relations, and illustrate the method used. Theorem 2.2.4 (GCH). Ify

< K'

then K +

-+

(K);.

Proof. Take any set S with IS1 = K ' , and suppose a partition [S]' = U{ Ak; k < y} is given. We have to find a homogeneous set H with IHI = K . Define a ramification system % on S of height K and order y as follows. Take u with u < K and suppose S(v) given for all v in N n SEQ,. Take p from NflSEQ,+l such that p r u = v . If S ( v ) = 0, put F ( v ) = 0 and S(p) = 0. If S(v) # 0, choose x ( v ) in S(v), put F ( v ) = { x ( v ) } and S(C0 = {Y E S ( V ) - F ( v ) ; { ~ ( v )Y, ) E Ap(o)I

.

This defines % . Then IS1 = K + ; p = K < K + = (K+)I; IF(v)l< 1 < K + and In(v)l= IyI for all v E N ;and if u < K then lyl'" < K + by GCH. Thus Theorem 2.2.3 applies, and there is v in N n SEQ, with S(v) # 0. Choose such a v . Since S ( v ) # 0 we must have S( v u) f 0 for each u with u < K , and so F( v ru) f 0; thus F( v r a ) = { x ( v ru)}.For brevity, write x u = x( v Po) for u with u < K , and note xu # x, whenever u # r by Lemma 2.2.2(ii). When a< r < K , we have { x , } = F(V

rT)c S( v rr)c S(V rut I ) ,

and so { x u , x,} E A,(,) by the definition of S ( v r u + 1). Since always v(u) < y < K ' , there must be K in [ K ] , and Y with v < y such that v(u) = v for all u in K. Put H = { x u ; u E K } , so [HI = K and [HI2 C A". Thus H i s a

homogeneous set of power K as required.

Ch. 2.2

Ordinary partition relations

32

If the Generalized Continuum Hypothesis is not assumed, the same method will yield the following. -

Theorem 2.2.5. For all infinite cardinals K , (2K)+-+ (K');. Proof. Given a partition [SI2 = U{Ak; k < K } of a set S with IS1 = (2K)+, define a ramification system 8 on S of height K + and order K in a manner similar to that in the proof of Theorem 2.2.4. This time IS1 = ( 2 K ) + ;p = K + < (2K)+;In(v)l= K for all v in N ; and if u < p = K + then do'< ( 2 K ) K = 2K < (2K)+,so again the Ramification Lemma applies t o % , and the result follows as before. We turn now to another application of the Ramification Lemma, which has some useful corollaries.

Theorem 2.2.6. Suppose hl + ( V k ; k < y12 and hz + (el;I < 6)'. Let K be a regular cardinal with hl < K and A2 < K and suppose (A . 16I)' < K whenever h< hi and I < h2. Then K + ( V k ; k < y,&; I < 6)2. Proof. Take any set S with IS1 = K and suppose that [S]' is partitioned, [SI2 = U { A k ; k < y } U U { r 1 ; 1 < 6 } . It may be assumed for all subsets H of S with IHI 2 hl that [HI2 p U{Ak; k < y}, since otherwise the result follows from the property hl + (vk;k < 7)'. It suffices to find 1 and H with I < 6, H C S,WI = 01 and [HI2C rl. Define a ramification system 8 on S of height hz as follows. Take u with u < h2 and v in N n SEQ, and suppose S(v) already defined. If S ( v ) = $, put F(v) = 8 and n ( v ) = 0. Otherwise, choose as F ( v ) a subset of S ( v ) maximal with the property [F(v)l2 C_ U{Ak; k < y}, so IF(v)l< A l . Then if y E S'(v) - F( v), by the maximality of F ( v ) there is x in F ( v ) such that {x, y } E rl for some I with 1 < 6 . Thus there is a decomposition

S ( v ) - F ( v ) = U (S(P);P E SEQu+l , p I-u = v and p ( u ) < n ( v ) ) , where for p in SEQ,+l with p I@) < 6 ,

S(P)=

b E S(

J;

-

r u = v , for some x(p)

in F ( v ) and I(p) with

~ ( 4{x(P), ; V } E rl(,,)} .

This defines F(v), n ( v ) and S(p) for allp in N n SEQ,+l with p Tu = v , and so completes the definition o f % . In this description, In(v)l< IF(v)l . 16 I. This means that the conditions of

Ch. 2.2

The ramification lemma

33

Theorem 2.2.3 are satisfied b y % , and so there is v in N f l SEQx2 for which S(v) f $4. So F ( v To) # 8 for each u with u < h2.Put x , = x ( v ~ u + I), SO xu # X , when u # r (since x u E F ( v To) and x, E F( v rr)). When u < r < h2, we have

x , E F ( v r 7 ) C _ s ( V r r ) ~ s ( V ur + I ) ,

and so {xu, x,} E F r ( v ro+ 1), by the definition of S( v r u t 1). However, X2=U{Al;Z<6}whereAl= { u < X 2 ; Z ( v r u + l ) = l } . B y therelation h2 ( e l ;1 < 6)', there are lo with ZO < 6 and a subset HO of X2 with IHol =Oro and I(v r u t 1 ) = lo for all u in Ho. This means that if u < r < K and u EHo then {xo, x,} E Fro. Thus if H = { x u ;u E Ho}, then [HI2C_ Fro. Since IHI = Ole, this proves the theorem. +

Corollary 2.2.7 (GCH). ZfS


then K +

-+

(K', ( ~ ' ) 6 ) ~ .

Proof. We shall use Theorem 2.2.6 with K = K ' , with X1 = K + (noting that trivially K + (K'):) and with A2 = K ' (noting that K ' ( K ' ) : ) . By GCH, if A < K + and i < K ' then (A * 16 I)L G K' = K < K ' , so the conditions of Theorem 2.2.6 are satisfied, and K + (K', ( K ' ) s ) ~ follows. -+

-+

-+

Corollary 2.2.8. suppose K is strong& inaccessible, y < K and qk m e n K ( K , (Qk; k < ')'))2.

< K for k < y.

-+

hoof. Since K is regular there is a cardinal 6 with 0 < K , y < 8 and l)k < 0 for all k. From the inaccessibility of K , we have 8+ < K . Since 0' is regular, e+ .+ (eg so in particular e+ (qk; k < y)'. Apply Theorem 2.2.6 with x1 = K (noting that trivially K ( K ) ; ) and h2 = 8'. Because K is strongly inaccessible, certainly (A . Iyl)' < K whenever X < XI and t < h2, so we conclude that -+

-+

K+(K,('?k;k
Theorem 2.2.9. For allfinite m, No -+ (No)&. Proof. Take any set S with IS1 = No, and suppose a partition [SI2= u{Ak; k < m } is given. Define a ramification system % on S of height o as follows. Take u with u < o and suppose S(v) has already been defined for each v in N fl SEQ,,. If S(v) = 8, put F ( v ) = 8 and n ( v ) = 0. Otherwise, put n(v) = m, choose x(v) in S(v) and put F ( v ) = { x ( v ) } . F o r p in N nSEQU+, with

34

Ordinary partition relation

Ch. 2.3

p r o = v, define S(v) = { x E S(v) - F ( v ) ; { x , ~ ( v )E} A,(,)}. This defines the ramification system 3. Choose numbers ui for i with i < w by induction so that always S((uj; j < z?) is infinite, as follows. Supposing uj already defined for all j with j < i so that S((uj; j < z?) is infinite (this is trivial if i = 0), since S((u j ; j < i))is split into only finitely many classes S((u0, ..., ~ i - u)) ~ , one of these must be infinite. Choose ui so that S((u0, ..., uidlr ui)) is one of these. Then for the sequence v = (vi;i < w ) , the element x ( v r,u) is defined for all u with u < w . Write x , = x ( v To). Note that whenever u < 7 < w then x , E S ( v t 7) C S( v ru + I), and so { x 7 , x u } E A,(,,) by the definition of S(v l-u + 1). There must be v with u < m and an infinite subset K of o such thatv(u) = u for all u in K. Put H = { x u ;u E K } ; then IHI = No and [HI2 C A,. So H is an infinite homogeneous subset of S,as required. The original theorem of Ramsey asserts more than Theorem 2.2.9, namely that the relation No -+(No): holds for all finite m and n. This may be proved by induction on n , combining the proof just given for n = 2 with the method of the Stepping-up Lemma from the next section.

9 3. The stepping-uplemma This section is devoted to a theorem which enables the results obtained in the last section for n = 2 to be "stepped-up" to results in the case n 2 3.

Theorem 2.3.1. (The Stepping-up Lemma). Let K and X be infinite cardinals such that A < K ' . Let n 2 1 and suppose X (Vk;k < 7)" where lyll'l < K ' whenever u < h Then K + (Vk i1;k < y)"+l. --f

Proof. Takeany set S with IS1 = K and suppose given a disjoint partition A = { A k ; k < y} of [S]"-+'. Define a ramificationsystem 3 on S of height X as follows. Take u with u < A, and suppose S ( v ) has already been defined for eachv in N nSEQu. If S(v) = 0,put F ( v ) = 0 and n ( v ) = 0. Otherwise, choose X ( V ) in S ( v ) and put F(v) = ( x ( v ) } . Write C ( v ) = U { F ( v r 7);7 < u}, so IC(v)l< Iu + 1I. Define a partition r(v)of S ( v ) - F ( v ) by asserting y

= z(mod T(v)) * a U { y }

a U {z} (mod A ) for all a in [C(v)]" .

Put n ( v ) = lT(v)l, so n(v) < lyl''+llN, and let S(p) for p in N

fl SEQu+I with

Ch. 2.3

The stepping-up lemma

35

pF o = v range over the classes of T(v). This completes the definition of%. A, whenever 7 o. Since A is If o A, put A, = lyl'Ol";then n(v l-7) infinite, the assumption lyl'"l< K ' whenever o A ensures that Ad"' K' for o h. Thus the conditions of the Ramification Lemma are fulfilled by% . Hence there is v in N flSEQh for which S(v) f 0.Choose such a sequencev; then always S(v To) f 0 so F(v I' o) = { x ( v To)}. Choose x k from S ( v ) , and when u A put x u = x( v To). Then x u # x , when 7 u A. If T < o < A then x u € F ( v To) CS(v To) c ( 4 - 7 + l), and x A ES(v) C S(v l-7 + 1) so both xu, x A E S(v r7 + 1). Thus both x , and x A are in the same class o f r ( v r T ) , and from the definition of T(v r7),if r1 72 7"

<

<

<

<

<

<

<

< <

< < ...<

< 7,

{xT1, ..., x T n 9 X U )

Put X

= {x,; 7

< A};

{xT1, ..., x,", X A } (mod A )

.

then there is a partition [XI" = U{ A;; k < A} where

A;= { a E [ X ] " ; a U { X A } E A k } . By the relation A + (Vk; k < y)", there are k with k < y and a subset H of X with IHI = q k such that [HI" C A;. But then [ H U { x A } ] " ' ~ C A k , and IH U { X L } = ~ Vk 1, so the theorem is proved. The results obtained in the last section may now be extended as follows.

Theorem 2.3.2 (GCH). Suppose n 2 1 and y < K ' . 7'hen (i) K(" +) -, , (ii) K ( " + ) + ( K , (~'b)"+' .

(KF+'

Proof. Of course, when K is regular then (i) follows from (ii). Use induction on n. The cases n = 1 are Theorem 2.2.4 and Corollary 2.2.7. To go from n to n + 1, use Theorem 2.3.1, noting that K ( ( " + ~ ) + ) is regular and that when u < K @ + ) by GCH, IU1+ <,("+I <,(("+I)+) lyll0l < 1

Without assuming the Generalized Continuum Hypothesis, Theorem 2.2.5 steps up as follows.

Theorem 2.3.3. Ifn > 1 then (>"(K))+

+ (K+);++'.

Proof. Again by induction on n. The case n = 1 is Theorem 2.2.5. To go from n to n + 1, use the Stepping-up Lemma, noting that ( 2 . + , ( ~ ) ) +is regular and that when u < (&(K))+, Kioi

K>n(K)

2~ . > n ( ~ )= 2 2 n W ) =

&l+l(K)

< (&+l(K>)+

*

36

Ordinary partition relations

Ch. 2.4

4. Results for singular cardinals We turn now to partition relations which hold for singular cardinals K . A special technique, that of canonical partitions, is available which reduces the question to a consideration of partition properties of the regular cardinal K ' . We shall assume that Generalized Continuum Hypothesis throughout. Without this assumption, the methods developed in this section work if we assume that, as well as being singular, the cardinal K in question is a strong limit cardinal, that is, 2' < K whenever X < K .

Definition 2.4.1. Let A be a disjoint partition of [S]" and let C? be a pairwise disjoint family of subsets of S. The partition A is said to be canonical in C if for all a, b in [U C]",if la n CI = Ib n CI for all C in C , then a = b (mod A). The following lemma gives a method of constructing canonical partitions. See Erdds, Hajnal and Rado [38, $91.

Lemma 2.4.2 (GCH). Let K be singular and let ( K ~ u; < K ' ) be an increasing sequence of cardinals below K with K = Z(K,,;u < K ' ) . Let S be any set with IS1 = K and take any pairwise disjoint partition A of [S]" into fewer than K classes. Then there is a pairwise disjoint family C = { C,,; u < K ' ) o f subsets of S with always IC,l = K ,,such that A is canonical in C . Proof. We shall give the proof for n = 3 : an entirely similar proof can be given for general n. So let the disjoint partition A = {A,; k < y} of [SI3be given, where y < K ' . To produce the required family { Cu;u < K ' } we first find a family {A,,; u < K ' } of' subsets of S such that for each i with 1 < i < 3 , if a, b E [A,]' and c E [ U { A , ; T < u}I3-' , then a U c - b U c (mod A ) .

(1)

This family is refined to a family {B,; u < K ' } such that for some chosen sets c7, with Ic7j( = j , for all i, j with 1 < i, j < 2 and i + j < 3, and for all T with r < K ' if a, b E [Bopand d E [U{B,;r < u}I3-'-j, then a U csi U d

b U csj U d (mod A ) .

(2)

Finally {B,,: u < K ' } is refined to { C,,; u < K ' } such that for some chosen sets dJ/ with Id+I = 1, for all r, $ where r , $ < K ' if a, b E [C,,]', thenaUcsl Ud+ = b U c T l U d + (modA).

(3)

Results for singular cardinals

Ch. 2.4

31

Thsn A is shown t o be canonical in {C,;u < K'}. We start by taking an arbitrary well ordering i of S and a pairwise disjoint decomposition S = U{S,; u < K ' } with always IS,[ = K,. Use transfinite induction t o define ordinals t ( ~with ) t ( ~ ) K<' and sets& in [SE(,)]~'for u with u < ~ ' a s f o l l o w s . P u t A ( u ) = ~ { A , ; T < U F} o. r e a c h a i n [S-A(u)I3,say a = {xl,x2, x3} w h e r e x l < x 2 < x 3 , define a functionf, : U{ [A(u)]';i<3}-ry by, for c in [A(u)]', f,(c) = k *

{Xi,

...,X 3 - i }

uC E A k .

Now IA(o)l = Z(K,; T < a) < K, so the number 7) of possible functions is yK", so q < K by GCH. Choose t ( u ) such that t ( u ) > t ( ~for ) all T with T < u, and also KE(,) 2 q+++, KY. It follows from Theorem 2.3.2 that A+++ (A)':, so certainly KC(,) + (K,):. Hence there is A , in [S5(,,)]Ku such thatf, is the same for all a in &,I3. This defines A , so that (1) holds. ] each 7 with T < K ' . InChoose, for j = 1,2, some set c,i from [ A ( T ) ]for ductively define ordinals n ( ~ )with n ( ~ )< K ' and sets B, in [A,(,)]K" where u < K ' as follows. With B(u) = U{B,;T < u } , for each a in [S - B(u)]', say a = {xl.x 2 } where x1 i x 2 , define a functiong, mapping-from {
g,(C,i,

d) = k

* {Xi, X i }

U c,i U d E A k

.

Again, the number of possible such functions is less than K , so n(u) can be chosen so large that n(u) > n ( ~whenever ) T < u and so that there is B , in [A,(,)]Ka such that all a in [B,]' have the same g,. This gives B , such that (2) holds. ) ] ~each 7, and inductively define ordinals p ( u ) Now choose d, in ~ B ( T .for and sets C, in [Bp(,)] as follows. Put C(u) = U{C,; T < u}. For each x in S - C(u) define a function h, where h, : {(cT1, d+);7, < K') + y by

+

hX(L',l,d+)=ke {X}uC,1 U d + E A k . As before, p ( u ) can be chosen so large that p(u) > p(7) whenever T < u and so that there is C, in [B,,(u)]Kusuch that h, is the same for all x in C,. This defines C, such that (3) holds. It remains t o show that A is canonical in C = { C,; u < K ' } . Take a, b in [U ?I3 with la n C,l = Ib n CI, for all u. We must show a = b(mod A). There are at most three ordinals <,u, T with < u < 7 such that Q meets C,, C,, C,. Write a = al U a2 U a3 where a l = a flC,, a2 = a n C,, a3 = a n C, (allowing a1 = 8 o r a l = a2 = 8 when a C C , U C, or a C C,). Similarly write b = b l U b2 U b3. If a = a3 then a, b E [C,I3 so by ( I ) , a b (mod A). Otherwise, write

<

38

Ch. 2.4

Ordinary partition relations

= C a @ ( T ) ) j where la31 = j . Since a3 E C, C_ Bp(T)C_ A,@(,)) both a3, ~3 EA,@(,)) with la31 = Ic3[.And a1 U a2 CA(n(p(r)))since both n and p are increasing functions so by (1)

~3

al

U a 2 U a 3 -a1 U a 2 U c3

(mod A ) .

Similarly,

b l U b2 U b3

f

b l U b2 U c3 (mod A ) .

If a1 = 8, it follows from (2) that a2 U c3 bz U c3 (mod A), and so a b (mod A). The only remaining possibility is that [all = la21 = la31 = 1. In this case, put c2 = dp(,). Here a2 E C, C Bp(,,) ,c2 El?,(,) and al C B(p(u)),so by (2) alUazU~3falUc?Ucj(rnodA).

Similarly,

b l U b2 U c3

b l U c2 U c3 (mod A ) .

Finally, because of (3), Uc2 U c 3

so again a proved.

- b l U c 2 U c 3 (mod A),

b (mod A). Thus is all cases, a

b (mod A), and the lemma is

The following remarks concerning the proof of Lemma 2.4.2 will be relevant to later applications. The pairwise disjoint decomposition S = u{S,; u < K ' } with S l ,I = K , of S could have been specified in advance. In the course of the proof an increasing function f : K ' +. K ' was constructed (where f(a) = &p(o)) such that C, E [Sf(,,)IK". In particular, if with respect to the well ordering < of S,we had specified the S, so that S, < S, whenever u < r then it follows that C, < C, whenever u < 7.

Theorem 2.4.3 (GCH). Let K be singular. Then K +. ( K ,

Q k ;k

< 7)'

if and Only if K '

Proof. For the easier direction, suppose K tition [K']'

=A

+(K',

+(K,

qk;k

< 'y)2 . <

~ kk ; Y ) ~and ,

take any par-

u u{ A k ; k < 'y} .

Take pairwise disjoint sets S, for u with u < K ' such that S I I, S = u{S,; o < K ' } then IS1 = K . For k with k < y define


and if

Ch. 2.4 r k =

and put

39

Results for singular cardinals

{{X,y};3

{U,7}E

[K']'(XES,,yES,and

r = [SI2- U { rk;k < y}. Then

{X.y}EAk)},

[S]' = r U U { r k ; k < y } .

Since MI = K , by the assumption K ( K , q k ;k < y)', either there is H in [SIK with [HI' C r, or else there are k with k < y and H in [SIqk with [H]'C rk. In either event, put Z = { u < K ' ; Hn S, # 8). In the first case, 1 1 1= K ' and [Z]' C A . In the second, for each u we must have W n SI, < 1 and so 1 1 1= r)k, and [r]' C Ak. Thus K ' ( K ' , q k ;k < 7)'. For the other direction, we shall use Lemma 2.4.2. Suppose K ' ( K ' , q k ; k < r)', and for a set S with IS1 = K let any partition -+

-+

-+

[S]' = A U U { A k ; k < y}

be given. Suppose for each k with k < y that there is no subset H of S with IM = q k and [HI2C Ak. We must show that then there i s H i n [S]" with [HI2 C A . Take a family C = { C;, u < K ' } such that the given partition is canonical in e . We may suppose that always IC,l> K ' . Since q k < K ' < lc,l, for each u we have [C,]' p Ak for each k. Thus from the definition of canonicity, always [C,]' L A . Choose C with C C UTC,; u < K ' } such that I C n Cl, = 1 for each u, so ICI = K ' . The partition of [S]' restricts to give a partition of [q2, and when k < y by assumption there is no subset H of C with IHI = 7)k for which [HI2 C Ak. By the relation K' ( K ' , q k ;k < Y ) ~ ,there must then be Ho in [C]"' with [HO]' C A . Put H = U { C , ; C , ~Ho#~},soIHI=~.ThenifxEH~C,anduEH~C,, since [Ho1' C A , by canonicity {x,u } E A . Thus it follows that [HI2 C A , and the proof is complete. -+

Corollary 2.4.4 (GCH). Suppose K is singular with K ' = A+ for some X Then K ( K , (A')&)' f o r any 6 with 6 < A'. -+

Proof. From Corollary 2.2.7 we have A+ from Theorem 2.4.3.

-+

(A',

(A'));,

and the result follows

Corollary 2.4.5 (GCH). Suppose K is singular with K ' = A+ for some X Zf n 2 0 and 6 < A', then K(n+)

-+

(K,

(h')~"'' .

Proof. By induction on n , starting from Corollary 2.4.4 and using Theorem 2.3.1.

Ordinary partition relations

40

Ch. 2.5

$ 5 . The case n = 2

This section starts with some special partitions of the pairs from an infinite set which serve to demonstrate the failure of certain partition relations with n = 2. Using these and the positive results from earlier sections, an almost complete discussion is then given of the truth of the relation K -+ ( q k ; k < T ) ~ .

Theorem 2.5.1. Suppose y 2 2 and Kk k < 7)then K $, ( q k ; k < y)2.

< q k for k with k < 7.If K < n(K k ;

Proof. Take sets S k with lSkl = Kk for k < y. k t S be a subset of the Cartesian product X {Sk; k < y } with IS1 = K . Define a partition A = { A k ; k < y} o f [s12as follows: forJ g in S,

{.f g )

Ak

*f(k)fg(k).

If H is a subset of S homogeneous for A , say with

< lAkl < K k < q k .

[HI2 C A k , then IHI

Corollary 2.5.2. If K is any cardinal, then 2K $, (3);. Corollary 2.5.3 (GCH). Let K be singular and suppose there are cardinals l)k where k < with 2 < q k < K and K < n ( q k ; k < 7) m e n K + p ( q k ; k < y)2. Proof. By virtue of Corollary 2.5.2, we need only consider the case when

y < K . Since K is singular, there is an increasing sequence ( K , ; u < K ’ ) of car-

dinals K , all less than K with K = Z(K,; u < K ’ ) . We can choose inductively distinct ordinals k(u) where u < K ’ with k(u) < y such that q k ( 0 ) > K , , for otherwise there would be 7 with 7 < K ’ such that q k < K , whenever k 6 {k(u); u < 7). But since y < K and 7 < K ’ , by GCH this would lead to the contradiction

< n(T)k; k < 7)< KJ’l . n ( T k ( , ) ; U < 7 ) < KJ7’ . K “ ’ < K . However, GCH implies that n(~,;u < K ’ ) = K + , so by Theorem 2.5.1 we have K + f+ ( K , ; 0 < K ’ ) , and COflSeqUently K + $, ( q k ; k < 7). K

Corollary 2.5.4. IfK is singular then K + f+ ( K I : ~ . The basic method of proof of the next theorem is due to Sierpinski, who used it in what appears t o have been the first proof of the relation 2’O P (N,, N1)2. The more general setting appears in Erdos and Rado [ 2 8 ] .

Ch. 2.5

The case n = 2

Theorem 2.5.5. Let K and X be infinite cardinals. Suppose Z(8'"'; a < A) m e n 8" ( K , x + ) ~ .

+

41

< K.

Proof. Let < be the lexicographic ordering on the functions in *8; that is if

f, g are distinct members of "0 and if a is the least ordinal wheref(a)fg(a), then

f
.

Notice the following facts about well ordered and conversely well ordered subsets (under<) of "8:

, then IXI < X . (2) T o prove ( l ) , suppose X = { f, ;p < (1 where f, < f,,whenever p < v < (, If X C "8 and X is well ordered by<, then IXI

and suppose without loss of generality that [ is a limit ordinal. For each p with p < ( define:f in "8 as follows: if a is least such that &(a) ff,+l(a), then (if < A)

f,*@

f,+l(P) =

( 0

ifPGa

9

ifP>a.

T h u ~ f , 4 f , * < f , + ~ , and in fact i f p < v < ( thenf,*dfp+l< f,, * f,,whenever p < v < A+. We shall show that the functions in X with large p have increasingly greater initial sections in common. More precisely, for each a with a < X there is p(a) with p(a)< A+ such that i f p 2 p(a) and Y G a,then f&)

= f,(,)(~)

.

(3)

We shall define p(a) by transfinite induction. So let a be given (with a < A) and suppose that p(P) has already been suitably defined for all with P < a. Since C(lp(P)I; 0< a)< X . la1 = X, there is v with v < A+ such that p(p) < v whenever < a. Thus if p 2 v and y < a,then f,(r) =f,,(y), and if v < p1< 1-12, then f, (a)2 fp2(a).. There must be p(a) such that v G p(a) andf,(a) =f&,(a) whenever p 2 p(a). For if not, inductively one could find ordinals v(n) with v = v(0) <

42

Ch. 2.5

Ordinary partition relations

v( 1) < 4 2 ) < ... where fu(o)(a)>f,(l)(a)>fv(2)(a)> ..., giving an infinite descending chain of ordinals (all less than O), which is impossible. This defines p(a) so that (3) holds, and completes the definition. Now

< A' = x , and so there is v with v < h+ for which p(a) < v for all a with a < h. But then C(lp(a)I; a < h)

fu(a)= f v + l ( a ) for all a ; that is f , = f,+l, which is ridiculous. Thus (2) is proved. We are now ready to prove the theorem. Take any well ordering < of '0, and use it to define the following partition {Ao, A, } of ['el2:

{.L g) E A0 *f < g and f < g

I

{f,g) E A1 * f e g a n d f > g . If X is a subset of '8 with [XI2 C_ A,, then X is well ordered by < and so [XI < K by (1). And if X is a subset of '8 with [XI' C_ A,, then X i s conversely well ordered by < so 1x1< h by ( 2 ) . Corollary 2.5.6. For all K , if h is the least cardinal for which K 2' p ( K ' , h')'.

< 2',

then

Proof. Apply Theorem 2.5.5 with 8 = 2 and K replaced by K'. By choice of h we have X < K and 2Ia' S K whenever a < A. Thus Z(2'"'; (Y < A) < K . K = K < K + , and the result follows. Corollary 2.5.7 (GCH). I ~is K any cardinal, then K + -f ( K ' , (K')')'

.

Proof. From the GCH, C ( K ' ~a' < ; K ' ) = K . K ' = K C K +SO Theorem 2.5.5 applies to show K ~ f' . ( K ' , K')+)*. Theorem 2.5.8. I f

K

is uncountable and not strongly inaccessible then K f i ( ~ ) Z .

Proof. If K is singular, then there is an increasing sequence ( K , ; (7 < K ' ) of cardinals with always K , < K which is cofinal in K . Define a partition A = { Ao,A,} of [K]' as follows: if (Y < 0 < K then {a,P}EAo * 3

U
((W
{a,P } E A, otherwise.

[a'

If H is homogeneous for A , then [HI< K . For if C A, then H can have atmost oneelement in { a < ~ ; ~ , ~ a < ~ , + ~ ) f o r e a c hthat o s IHI
Ch. 2.5

43

The case n = 2

and if [HI'

C_

A1 then H m u s t be completely contained in some {a < K ;

K,

If K is regular but not strongly inaccessible then there is A with X < K S 2'. If we take the least such h then Z(2'"'; a < A) < K , so by Theorem 2.5.5 we have 2' ( K , A')'. In particular then, K ( K , K ) ' .

+

+

We conclude this list of negative relations with the following theorem.

Theorem 2.5.9 (GCH).

Z ~ K is

singular then K +

P (K+, (3)K')'.

hoof. By the GCH, I [ K + ] ~ I = K + so we may put [ K + ] ~= { A , ; a < K + } . For e a c h a , p u t d , = { A p ; P < a a n d A p C a } , s o ld,I<~and [ K + ] ~ = u { d a ; a < K ' } . Choose an increasing sequence of cardinals ( K , ; u < K ' ) all less than K such that K = Z(K,,; u < K ' ) . Choose subfamilies d,,of d ,with I d a u I S ~ , , such that d, = U{ d a uu;< K ' } , for each a. Let u be fixed with u < K ' . Use induction on a where a < K + to define subsetsf,(a) of a with If,,(a)l< Ku+l as follows. Put fa(&) = (3 if a < K . Suppose K < a < K + and f,(@ has been defined for all with P < a. Since P 4 P certainly P 4 fa@). Thus, in the notation of Chapter 3, 3 1, we can consider f, : a +$a as a set mapping on a,andf,, has order at most K , , + ~ . We appeal to Theorem 3.1.2. Since I d,,I S K,, < K , there is a subset Fa,, of a which is free forf, and for which IF,, nAl = K for eachA in d,,.To say that F,, is free for f, means that Fa, n Uf, [F,,] = 8 . Choose f,{a) from [Fa,]' K g so that A nf,(a) f 0 for each A in do,,,. Thenf,(a) C_ 01 and If,(a)I < K , < K , , + ~ . This completes the definition off,(a). We are now ready to define a partition [K']'

=A

u u{ A,;

0

< K '}

which will prove the theorem. For u with u < K ' , if a,P E K' with fi < a then define {a,PI E A,

*P Efu(a)

Y

and put A = [K']' - U{A,; u < K ' } . Take any set H from [ K + I K + . Take A with A E [HIKso A = Ap for some P. Since 1HI= K + , we can choose a with a E H s u c h that B< a andA C a.ThenA E d,,for some u with u < K ' . The definition off,,(a) ensures that there is y with y € A nf,(a). Then y E H and {a,y} E A,,; hence [H]' A. And if for any u there were a,P, y in K', say with y < 0 < a,such that [ {a,0,y}]' C A, then 0,y Ef,(a) C Fa, and y Ef,,(P) so that y E Fa,, flUf, [Fa,] contrary to the fact that F,, is free forf,. This shows that the above partition has the correct properties to prove the theorem.

Ch. 2.5

Ordinary partition relations

44

We are now ready t o discuss the symbol K + ( q k ; k < y)' (where 2 < y< K and 2 < qk < K ) . We shall assume the Generalized Continuum Hypothesis throughout. We shall restrict the discussion t o the case where at most one of the q k has value K . For it follows from Theorem 2.5.8 that if even K ( K ) : then K is inaccessible (or K = No), and so, excluding K = No, it is consistent with the usual axioms for set theory that there are n o such cardinals. Cardinals with the property K + ( K ) : are known to be weakly compact, and consequently are larger even than the first inaccessible cardinal (the last a result due t o Hanf [57]). Case A . qo = K and q k < K for k with 1 < k < y. If K is inaccessible then K + ( K , q k ; 1 < k < y)' provided y < K , by Corollary 2.2.8. This is the best possible result, because of the trivial relation K +(~,(3),)'. If K is a successor cardinal, say K = A+, then K + ( K , (A'),)2 for any y with y < A', by Corollary 2.2.7. This is the best possible, in view of A+S(A', (A')')' from Corollary 2.5.7 and A+ f. (A', ( 3 ) ~ l )The ~ . last relation holds by Corollary 2.5.2 if A is regular, and Theorem 2.5.9 otherwise. If K is singular, suppose first that K ' is a successor cardinal, say K ' = A+. Then by Corollary 2.4.4 we have K -+ ( K , (A)',' provided y < A', and this is the best possible positive result. This follows since K ( K , (A')')' by virtue of Theorem 2.4.1 applied t o Corollary 2.5.7, and K ( K , (3)kt)' by applying Theorem 2.4.3 to the relation A+ (A, (3)ko)' established above. On the other hand, should K ' be a non-successor cardinal then K ' is inaccessible or countable. By Theorem 2.4.3, the relation K + ( K , q k ; 1 < k < 7)' is equivalent t o K ' + ( K ' , Q ~ ; 1 < k < 7)'. This second relation is false if any of the q k exceed K ' , involves the relation K ' + ( K ' ) : if any q k is K ' , and has been discussed above otherwise. Case B. v k < K whenever k < y. This case is completely settled by the following theorem. +

+

Theorem 2.5.10 (GCH). Suppose q k ifand only if n(qk;k < y) < K .

+

< K whenever k < y. Then K

+

(qk;k
Proof. Suppose H(qk; k < y) < K . Then certainly y < K and q k < K for all k. I ~is K inaccessible, by Corollary 2.2.8 we have K + ( K , q k ; k < 7)' so indeed K + (qk;k < y)'. If K is singular, in fact the q k must be bounded away from K . For suppose some set { qk:k E K } is cofinal in K , where we may suppose IKI = K ' . From Konig's Lemma follows the contradiction

< Z(qk; k €0< H(qk; k E K ) < H(Qk; k < 7) < K . Thus there is 6' with 6' < K such that q k < 6' whenever k < y, and we may supK

Ch. 2.5

The case n = 2

45

pose in addition that y < 8. From Theorem 2.2.4 it follows that 6'' + (@)$, and hence K + ( q k ; k < y12 since e++< K . Thus we may suppose K is a successor cardinal, say K = A+. Since n(qk; k < y) < A, it follows that y < A and always q k < A. If A is regular, by Theorem 2.2.4 we have that A+ -+ (A): provided y < A, and so it follows that K + ( q k ; k < y12. If A is singular, since f l ( q k ; k < y) < A certainly I { k ; q k = A}I < A'. Moreover, there can be no set { q k ; k EK } of those q k with q k < A which is cofinal in A, since otherwise (assuming IKI = K ' ) we have the contradiction EK) < n(?j'k;kE K ) < n ( q k ; k

< y)< h . Thus there is 0 with 0 < A such that Qk < 0 whenever qk < A, and we may suppose also A' =G0 and y < 8. So to show K + (qk;k < y) it would be enough to establish that A' + ((0)7, (A)s)~, where y < A, 6 < A'. However, from TheoA<

x(Qk;k

rem 2.2.4 it follows that

A+ -+ (A, (A)&)* and e+++ (e+)$, so A + (el: , whence A' + ((f9)7, (A)s)~ by Theorem 2.1.2. This completes the proof of the sufficiency of the condition n ( q k ; k < y) < K . Suppose flow n ( l ) k ; k < 7) 2 K (where q k < K ) , and We show that K ( Q k ; k
+

+

K

= x(qk;k

€9< n ( q k ; k EK) < n ( q k ; k E y) .

+

+

By Corollary 2.5.3 then K + ( q k ; k < T ) ~ ,so all the more K (qk;k < 7)'. If no such set of the r)k is cofinal in K , there is K O with K O < K such that alWays q k < K O . Then n(Qk; k < 7) < K $ l , SO 1712 K and K (Qk;k < 'y)2. Finally, suppose K is a successor cardinal, say K = A+. If A is regular, n ( q k ; k < y>< A"' so 1712 A. The result follows from the relation A++ (3):, given by Corollary 2.5.2. So we may further suppose that A is singular. If I { k < y; q k = h}l 2 A', since by Corollary 2.5.4 we have A+ (A):!, it follows that K ( q k ; k < y12. n u s we may assume I { k < y; 7)k = A}I = 8, where 0 < A'. As before, if there is some set { q k ; k E K } with q k < A for k E K which is cofinal in A then A < f l ( q k ; k < y), so K ( q k ; k < y)2 by Corollary 2.5.3. And if there is no such set of the q k cofinal in A, there must be &,with ~0 < A and q k < whenever ?)k # A. Since n(qk;k < y) < A' * and A' = A it follows that IyI 2 A. However, A+ (3): from Corollary 2.5.2, so here also K ( q k ; k < 7)'. This completes the proof of Theorem 2.5.10.

e

+

+

+

x0

+

+

A?'

46

Ch. 2.6

Ordinaty partition relations

$6. Results when n 2 3 The situation with regard to partitions of pairs now being settled, we can turn our attention to partitions of the n-element subsets for n with n 2 3. Here a few unsolved problems remain. The Stepping-up Lemma (Theorem 2.3.1) seems to be essentially the only method of proving positive relations in this case. If GCH is assumed, the Stepping-up Lemma ensures that if K (qk;k
-+

Lemma 2.6.1. If

K

is singular then

K

f . ( K , 413.

Proof. Take any set S of power K , and write S as a disjoint union, S = U(S,; a < K ' } where always IS,[ < K . Define a partition {Ao, A,} of [SI3as follows: fora in pi3,

a €A1 a E A,

0

(3 a, T <

K')

otherwise.

(a< Tand la nS,I

= 2 and la

nS,l = 1 ) ,

[a3

A1. And i f H i s a subset o f S with I f H E [SI4 it is easy to see that [HI3 C A,, whenever H nS, # 8 then IH nSol < 1 for all u with u < T. It follows that IHI < K' or IHI < K ' 4 IS,I for some 7 ; in any event IHI < K .

Theorem 2.6.2. Suppose for some n with n 2 3 that K

-+

(K):.

K

-+

(K,

n t 1)". Then

Ch. 2.6

Results when n 2 3

41

and so it is enough to show hoof. If K + ( K , n t l)n then clearly K + ( K , that if K + ( K , 4)3 then K +(.)if. So suppose K -+ ( K , 4)3. Take any disjoint partition [ K ] =~ A. U A, and suppose that there is no set I in [ K ] ~such that [II2C A,. We must show that there i s I i n [ K ] with ~ [ZI2 C A,. Define a disjoint partition [ K ] =~ roU rl as follows: for a,0,y from K with a < B< y, { a , P , y } E r o + {a,O)EAo or { S 9 ~ } E A 1 ,

{a,0,y} E rl

{a,0)E A l and { P , y} E A.

.

Suppose H = {a,0,y, 6 ) is a subset of K with [HI3C rl.If we assume a < p < y < 6 , s i n c e both {a,/3,y}Erl and {.3,y,6)Erl we findboth y} E A, and (0,y} E A,, which is impossible. So there is no such set H. From the relation K + ( K , 4)3 it follows that there must be then a set H from [ K ] ~with [HI3 C ro.For a from H, put

{a,

Hl(a)= { p E H ; a < p a n d ( a , P } E A , ) . Take 0,y from H l ( a ) . If (0, y} E A. then {a,0, y) E Fl contradicting that [HI3L ro,and so (0,y} E A,. Thus [H1(a)I2 C A, and hence IHl(a)l< K. Inductively choose elements a, for v with v < K SO that a" E H - ({a/i;P < v l

u uw

1 (ap); P <

vI> .

Since I M = K and K is regular (by Lemma 2.6.1) such a choice is always possible. And if I = {a,;v < K } then I E'[ K ] ~and [A2 C Ao, as required. Theorem 2.6.2 is a result due to Hajnal [54, Theorem 21. We shall conclude this part of the discussion by stating the following theorem. Theorem 2.6.3. If n 2 3 and y < K then K

--f

(K); ifand only

if^ + (K);.

This theorem was first stated in Rowbottom [ 8 1 ] .For a detailed proof see, for example, Kleinberg [61]. From Theorems 2.6.2 and 2.6.3 it follows that all non-trivial relations of the form K + ( K , q k ; k < 7)" where n 2 3 are equivalent to the relation K + ( K ) ; . We now turn to relations of the form K -,(qk;k < 7)" where q k < K for all k. There is a simple result first. Theorem 2.6.4. Suppose K K + $(qk t l ; k < y ) " + ' .

+ (qk;k < y)" where all the q1( are finite. m e n

48

Ch. 2.6

Ordinary partition relations

+

Proof. Assume K (r)k; k < yr.Let S be any set of power K + , and take a well ordering < of S of order type K'. For each x in S , put L(x) = { y E S ; y < x}, so always IL(x)l G K . Thus for each x there is a partition A (x) = { A k ( x ) ;k < y} of [L(x)]" which for each k has no set H with H E [L(x)]'~ such that [qn C A k ( x ) . Define a partition A = { Ak; k < y} of [SIn+l as follows: i f x l , ..., x , + ~E S withxl < ... < x , + ~ , then {XI.

..., X n + l } E A k *

{XI,

..., X n )

EAk(Xn+I)

.

C A k . Let x be the For any k, take a finite subset H of S and suppose largest element in H. Then [H - {x}ln C Ak(x), so by the choice ofA (x) it follows that IH - {x}l< v k . Thus IHI < r)k t 1, and the theorem is proved.

+

Starting from Corollary 2.5.2, namely that 2K (3):, repeated applications of Theorem 2.6.4 yield that (n -+ 2)'; (when n 2 1). This shows that (assuming the GCH) the result K @ + ) + (K>.C" whenever y < K ' from Theorem 2.3.2 is the best possible in y (at least for K regular). With a different method (see Corollary 2.6.8 for the case n = 2) the related negative relation 3 n ( ~ ) (n t 2);" can be establishsd. This is the same as the relation above if the GCH is assumed, and is otherwise stronger. It should be compared with of Theorem 2.3.3. the relation (>n(~))+ + (K');'' For the rest of this section, we shall be considering partitions of K2. As before, let < denote the lexicographic ordering of K2. For distinct elements A g i n K2, let S(A g) be the least (Y wheref(a) #g(a). Then i f f i g i h , cerg) # S(g, h). Choose and fix a well ordering Q of K2.The notation tainly { A g, h}< will be used to indicate that in the set { A g, h } the relation f Q g Q h holds. We introduce the following notation:

+

+

Koo = { { A g, h}< C K2 ; f i g and g< h } Kol = {{f,g, h } < C K 2 ; f i g a n d h i g } , K l o = { { A g, h)< C K 2 ; g i f a n d g < h ) , K l l = {{Ag,h}gGK2;g
=KooUKii,

Po = { { A g, h } g C K ; &(A g) < S(g, h ) } , =

{ { A g, h}< C K ; S(ft g) > S(g, h ) } .

There are a number of simple observations to be made. Firstly, if [XI3 C K then either C Koo or [XI3 C K l l , so that X is

[a3

Results when n > 3

Ch. 2.6

49

either well ordered or conversely well ordered by the lexicographic ordering

<. For we may suppose 1x12 4, and take distinct setsa, b from [XI3. Let

a U b = { f ~ l,,f. . . , f p } ~ , ~ o p > 3 . S i n c e[XI3 C K , always {f,,f,+l,f,+~}EK, for q = 0, 1, ...,p - 2. Thus eitherf, < f,+2 orf, >f,+1 > fq+2, and

it follows that eitherfo f l > ...>fp.Thus botha, b are in KO0 or both a, b are in K 1 1 , and the result holds. Suppose X is an infinite subset of K2.say 1x1= A, with either [XI3 nKol= $9 or [XI3 n K l o = $9. Then there is X' in [XIAwith [X'I3 C K. For we may suppose X = {f,; a < A} where f, Qfp whenever a < 0. Assume in fact [XI3 nKol = $9. Suppose there are a,/3 with a < 0 < A for whichf, fp whenever a agree on X, so [XI3 C K11 C K. The argument is similar if [x13 nKlo = $9. The last remark is not quite as easy to prove, so we state it as a lemma. Lemma 2.6.5. Suppose

1x1> h where A is regular, and [XI3 C K. Then there

is X' in [XI' with [X']' C Po.

Proof. Again we may suppose X = {f, ;a < A} where f, Q fp whenever a < P < A. Use transfinite induction to define ordinals u(a) where a < K with always u(a) < A as follows: (i) u(0) = 0. (ii) Let t(a)be the least value of 6(fu(,), f p ) for P with P > u(a),then o(a + 1) is to be the least P with 0 > u(a)for which 6(fu(,), fp) = f(a). (iii) If a is a limit ordinal, u(a)= U{ u(y); y < a};the regularity of A ensures that u(a) < h. We shall prove the following: if y 2 u(a + l), then S(f,c,),f,)

=

t(4

(1)

if a < P, then [(a)< t(P).

(2)

From (1) and ( 2 ) it follows that if a < P < y < A then S(fU(,), fu@))

(3) < 6(fu@)9 fu(,)) for if a < < y then u(P) 2 u(a + 1) and u(y) > u(0 + l), so from (1) it fol3

9

lows that 6(fu(,), f&)) = [(a)and 6(fu@),

fUc7))

= t@).Thus (3) follows from

Ch. 2.6

Ordinary partition relations

50

(2). Hence if

We are now ready to state and prove three theorems which give part-converses to the Stepping-up Lemma of $3. Theorem 2.6.6. suppose qo, q1 are infinite with qo regular. If K then 2K + ( q k ~;k
+

( q k ;k


Proof. From the assumption K f . ( q k ; k < y)*, there iS a partition A = { A k ; k < y} of [ K ] such ~ that for each k with k < y, there is no H in [ K ] ‘ ~ with [HI2 C_ Ak. We shall use this partition to construct a partition of [“213. For each k , put A: = {{.K g, h)< € P o ;

g), S(g, h ) ) E

Results when n > 3

Ch. 2.6

51

Define a partition { r k ; k < y} of ["213 as follows: rk =

A: if 2 < k < 7 ,

rl = K U ~A T ,~ ro = ["213 - u{ r k ; 1 < k < y} .

Suppose there was X in ["2]"l with [XI3 C rl. Then [XI3 n K l o f 8, so there would be X' in [XIq1 with [X'I3 C K. Since K fl Kol = 8 and PO C K we would have [X'I3 C AT. Likewise, suppose there was X in ["2Iq0 with [XI3 5 ro.Then [XI3 nK o l = $, so there would be X ' in [XIq0 with [X'I3 C K. By Lemma 2.6.5 there would be further X " in [X']"O with [X"I3 C Po.But then we would have [X"I3 C A t , since [X"I3 C ro n Po C A,*. So to prove the theorem, it is enough to show: for every k with k < y,if [XI3 C_ A: then

1x1< q k i1 .

(1)

Suppose on the contrary that for some k there is X with X E ["2Ivk and [ x ]c~A:. We may suppose x = {for; Q < q k i1) where f, <,fp whenever Q < /3. Since [XI3 C PO C_ K , we know X is either well ordered or conversely well ordered by <, and S(f,, fp) < 6(fp, f,) whenever Q < /3 < y. From this it is easy to see that if Q < /3< y then &(fa, fp) = &(fa, f,). Consequently, if Q < (3 then {W,, fa+d6 ( f p , f p + d = {W,, f p ) , W p , f p + d ) , and so {s(fa,f,+i),6(fp,fp+i)} E A k , since {fapfppfp+i)EAk*.P u t H = {&(fa,fa+l); a < 7)k),so [HI2 C Ak. HoweverH E [K]", and this contradicts the choice of the partition A . This proves (l), and completes the proof of Theorem 2.6.6.

Theorem 2.6.7. Suppose qo is infinite and regular. If K 2" + ( 4 , q k i1 ; ~ ~ ) 3 .

+

( q k ;k

< y ) 2 then

hoof. Take a partition A = { Ak; k < y} of [ K ] as ~ in the proof of Theorem 2.6.6, and define A: as above. Consider the partition { r}U { r k ; k < y} of ["213 where:

r = K ~ ~ , rk= A : if 1 < k < y , ro= ["213 - (r u u{rk;1 < k < TI) . If[XI3 G r t h e n IXI<4,since { f o , f l , f 2 ) < € K O , and { f 1 t f 2 , f 3 } g E K O 1 would require both f 2 < f l andfl < f 2 . If there is X in ["2]"Owith [XI3C ro

52

Ch. 2.6

Ordinary partition relations

then [XI3 n K O , = 8, so as in the preceeding proof there would be X " in [XIvo with [X"I3 C_ A,*. lust as above, for no k can there be X in ["2Ivk with [XI3 C_ A:. Thus the partition { r}U { rk;k < y} serves to prove the theorem.

Proof. We again start from a partition A = { A k ;k < y} of of Theorem 2.6.5, and define A,* as before. Put also A,** =

[K]'

as in the proof

{{.L g9h } g €Pi; {S(L g), S(g, h ) } E A,) .

This time, consider the partition ["213,where

{ro,rl}U {rik; k < y and i = 0,1}

of

ro= K ~ rl~ =, K , ~ ~ rOk= A,* for k with k < y , rlk= A,** for k with k < y . This is indeed a partition of ["213,since ["213= K o l U K l o U K and K = P o UP,. If [XI3 C roor [XI3 C r,, we know 1 x1< 4. As before, for no k is there X with X E ["2]'" and [XI3 C A,*. An entirely similar argument shows that for no k is there X in [K2]'k with [XI3 C A:*. So for no k is there X in ["2Iqk with [XI3 C_ rOk or [XI3 C_ rlk.This proves the theorem.

Corollary 2.6.9. ~

-h (4);.

Z ( K )

Proof. By Corollary 2.5.2, a l ( ~ ) -jb (3); so by Theorem 2.6.8 we have 2"'") $, (4,4,(4)~,(4)K)3, that is,> 2 ( K ) f , (4);. As the final theorem in the discussion of the case n = 3, we mention without proof the following theorem from [38].

Theorem 2.6.10 (GCH). Let K be singular and let X be the immediate predecessor of K ' if this exists; otherwise let = K ' . m e n K + f , ( K , ( 4 ) ~ ) ~ . By using the Stepping-up Lemma from $ 3 to obtain positive results and the negative relations established in this section, a discussion of the case n = 3 can be given, similar to that for n = 2 of the last section. The only problems

Square bracket relations

Ch. 2.1

53

to which a solution is not available are of a technical nature. Apart from questions involving inaccessible cardinals, the simplest unsolved problem is the following. (This is Problem 1 in the list [ 2 4 ] . )

Problem 2.6.11 (GCH). Is H w w + l + l

f i (Hww+l, ( 4 ) ~true? ~ ) ~

$7. !3quare bracket relations The square bracket relation is in some sense the opposite to the partition relation of the earlier sections. This relation was first introduced and discussed by Erdos, Hajnal and Rado [ 3 8 , 8 181.

Definition 2.7.1. The relation K + [ q k ; k < y]" holds if given any s e t s of power K , for all disjoint partitions A = { A k ;k < y } of [S]" into y parts, there exist k with k < y and a subset H of S with [HI = q k such that [HI" flAk = 8. The special case in which q k = q for all k is written K + [ q ] ; ,and in general the notation conventions of 8 1 will be adopted for the square bracket symbol as well. Clearly, the relations K + (qo, q l ) nand K .+ [qo,ql]" are equivalent. And if K + (Q0,l)l)" then K [ Q k ; k < 7'1" (for 2 2 ) . The simple remarks made in 0 1 about the relation K (qk;k < y)n have analogues for the relation K -+ [ q k ;k < 71".We remark only that given cardinals q k where k < y, if the relation K + [ q k ; k < 61" is true for some 6 with 6 < y, then also the relation K -+ [ q k ; k < y]" holds. The relation K + [ q k ; k < y]" is trivially true when y > K , so we always suppose y < K . We shall concentrate on the case n = 2 , and assume the GCH throughout our discussion. From Theorem 2.5.10 we know that K [ K O , K I ] ' if K o , K l < K . n u s all relations K + [ q k ; k < 71' are true if more than one of the 7)k are less than K . Consider first the case when K is a successor cardinal, say K = A'. Froin Corollary 2.2.7 we have K + ( K , A')', and so K + [A', K ] ' . We know from Corollary 2.5.7 that K -f ( K , (A')+)2, and hence K -f [(A')', K I 2 . This leaves unsettled only relations of the form K -+ [(A')', ( K ) ~ where ] ~ 2 < y < K . In fact even the weakest of these, namely K + [(A')', ( K ) ~ is] false ~ , - see Theorem 2.7.4 below. This we shall prove now. We start with a trivial lemma. -+

--f

+

Lemma 2.7.2. Let d = { A a ;CY < K } be a family of at most

K

sets (possibly

Ch. 2.1

Ordinav partition relations

54

listed with repetitions) each of power K . l l e n there are painvise disjoint sets B, with B, E [&IK.

Proof. Take any one-to-one and onto map p : K X K K . Choose elements x,p for a,0with a,p < K by induction on p(a,13)so that xolp EA, - {xrs ;p ( y , 6)
Lemma 2.7.3 (GCH). Let S be any set with MI = K'. m e n there is a disjoint partition [SI2 = U { I'k ;k < K ' } with the property that whenever A E [SIK, B E [SIK'and k < K + then there are a, b with a € A , b E B for which {a, b ) E rk. Proof. Note that such a partition provides an example to show that K + P [K'] Identify S with K'. Let ( A , ; a < K + > be a well ordering of [ K + ] ~For . each r with r < K + , put ,

2..

A T ={ A , ; a < r a n d A , C r } , so 194,1< K . Suppose for each r with r < K + that we could find a function f, : r -+ r such that if A E d,,k < r thenf,(u) = k for some u in A .

(1)

We could then construct a suitable partition { r k ;k < K ' } as follows: given u, r in K' with u < 7, {a, r } E r k

*f7(u)=

k.

This partition has the required property, for take A from [ K ' ] ~ , B from [ K + ] ~and + k with k < K'. Let TO be the least ordinal for which k < 70 and A E A,,.Since IBI = K + , there is r in B with ro < 7.By (I), there is u in A such that f,(u) = k , that is, { u, r } E rk,as required. Thus to prove the lemma, it suffices to find functionsf, : r r with the property (1). Clearly we may suppose d,# Q, for otherwise any function f,: r + r satisfies (1). Thus 72 K , so 171 = K and we may write d,={ A T , ; a < K } . Applying Lemma 2.7.2, there are pairwise disjoint sets B, where a < K with B, E [ATa(YlK. Take one-to-one and onto maps q7, : B, r, and define f, by the following: if for some (necessarily unique) a we have u E B, thenf,(u) = q7,(u), otherwise the value off,(u) is arbitrary. To see that (1) holds, suppose A E A, and k < 7. Thus A = A , for some a. There is u in B, with q7,(u) = k , that is,f,(u) = a. Since B, C A,, this proves (1) and completes the proof of the lemma. -+

-+

Ch. 2.1

55

Square bracket relations

Theorem 2.7.4 (GCH). Forall

K , K+

f . [(K')',

(K+),+I2

.

Proof. If K is regular, the result follows from Lemma 2.7.3. Se we need consider only the case that K is singular. We shall use the partition from Lemma 2.7.3 to refine the partition used in Corollary 2.5.7 to show that K' (K',

(K')+)2.

For Corollary 2.5.7 we defined the disjoint partition with

[ , ' K ] ~ = A0

U A1

t.6 g } E A0 *f Q g a n d f < g t t.6 g } E A0 * f Qg and f > g , where
[,'K]~+

and k


then [HI2nAlk f

8.

(1)

Thus the partition [ , ' K ] ~ = A0 U U { A l k ; k < K ' } is a partition which serves to prove the theorem. Take any subset H of ,'K with IHI = K ' . For f in H, put L H ( n = { g E H ; g < f } and RH(f) = { g EH;f < g } . We first observe there isfin H with ILH(f)l = IRdfll = K +

.

(2)

For suppose (2) is false, so for every f in H either ILH(f)l< K or IRH('lS K . There must be a subset I of H with III = K + such that either ILH(j)l S K for all f in I or else IRH(f)I < K for all f in I. Suppose in fact the first alternative holds - the argument is similar in the other case. We can choose inductively elements fa from I , for a with a < K + , so that fa E I - U{LH(fp); P < a) . Thenfp )has power less than K' (as follows from (1) and (2) in the proof of Theorem 2.5.5). This establishes (2). We are now ready to prove (1). Given H in [ K ' ~ ] K + , choose f i n H with ILH(f)l= IRH(j)I = K + . Take any A in [RH(j)]".Since IA I = K , ILHWI = K + and
56

Ch. 2.7

Ordinary partition relations

such a pair { a , b } also b< f < a so that { a , b} E A l . Thus {a, b } E A , = A l k . This proves (I), andcompletes the proof of the theorem.

nr k

We shall now consider the square bracket relation for singular cardinals K . Let us suppose further that K ' is a successor cardinal, say K ' = A'. From 55 we know K +. (A', K)' and K f . ((A')', K)' so the two relations K +. [A', K]' and K [(A')', K ] are in force. This means we must discuss relations of the form K +. [(A')', (K)?]' where 2 < y < K . We shall find from Theorems 2.7.5 and 2.7.6 below that K [(A')', (Kh]' when y < K ' and K +. [ K ] ; when y > K ' , which settles all problems in this case.

+

+

Theorem 2.7.5 (GCH). I ~is K singular and n 2 2 then

K +. [ K ] ;

for all y with

Y>K'.

Proof. Take any set S with IS1 = K , and let a disjoint partition A = { Ak; k < y} of [S]" be given, where y > K ' . By Lemma 2.4.2, there is a family C = { C,,; u < K ' } of pairwise 'disjoint subsets of S with I U el = K such that A is canonical in C . Thus for each sequence ( n u ;u < K ' ) where 0 < nu < n and E(no; u < K ' ) = n there is an ordinal k(n,; u < K ' ) in y such that if a E [ U C]" and la n C,,l = n , for each u, then a E Ak(n,,;,, K ' , there must be some class from A having empty intersection with [ U C I n .Since IU CI = K , this proves the theorem. Theorem 2.7.6 (GCH). Let

K

be singular with K ' = A+. f i e n K

+ [(A')+,(K)~I]'.

Proof. Take any set S with IS1 = K and write S as a disjoint union, S = u ( S , , ; u < A'}, where always IS,I < K . By Theorem 2.7.4, we know A+ [(A')', and so there is a disjoint partition

+

[A']'

=

ru

{ r k ; k < A'}

such that if I E [A']('')+ then [I]' n r f 8, and if I E [A']' then [I]'" r k f for each k with k < A'. We use this partition of [A']' to define a partition of [S]'. When k < A', put

Ak = {{x,y ) C_ S; for some (u, T} in [A']',

x ES,,, y EST and (u, T} E rk} , A=[S]*-U{Ak;k
8

Ch. 2.1

51

Square bracket relations

Then {A} U {Ak; h < A'} gives a disjoint partition of [SI2 into A+ = K' classes. This partition provides an example to prove the theorem. Suppose H is a subset of S with [HI2C U{ Ak; k < A'}. Then IH n SI, < 1 for each u, and [ { u ; H nS, f @}I2 n r = 0. Thus IHI = I { o ; H nSo#Q}l < A'. Thus if H C S with IHI = (A')' then [HI2 n A # 8. And if H C S with [HI2n Ak = Q (for any k ) , then [ ( a ; H n S, f @}I2 n r k = 8 , SO I { u ; HnS,#@}I 2"-'.

K

is singular with K '

=

NO. ?'%en K +. [K];

Proof. Let S be any set with IS1 = K ,and take a disjoint partition A = { Ak; k < y} of [S]",where y = 2"-' t 1. From Lemma 2.4.2, there is a pairwise disjoint family C = { C,; u < w } with U C E [SIKand IC,l< IC,l when o < 7, such that A is canonical in C . Thus

if a, b E [UC]" with la n C,l = Ib n C,l for all o: then a = b (mod A ) . (1) The number of sequences (nl, ..., np>where p > 1 and 1 S ni < n and nl t ...+ np =nis2"-';1ist these a s ( n l ( j ) ,..., npo)(j))f&jwithj<2"-'. Define a partition

r= { r ( k i ; j < 2 " - l ) ; k i < y

foreachj}

of [a]" as follows: if x E [a]", say x =

x E r ( k i ; j < 2"-l)

{(TI,

..., a,}<, then

* for each j , whenever a E [U C I",

with

la n Coil = n i ( j ) for i S p ( j ) then a E Aki.

By (l), this is a valid definition. From Ramsey's Theorem, No + (No)! for 6 fiiite; thus there is a set I in [ u ] ' ~ such that for some sequence ( k j ;j < 2"-'),

[I]" c r(ki;j

< 2"- l ) .

[a"

P u t H = u { C , ; u E I } , then [HI = K and C_ U{Aki;j<2"-'}.Since > 2"-' there is some class of A which has empty intersection with [HI". This proves the theorem.

y

58

Ch. 2.1

Ordinary partition relations

In fact the theorem above is best possible in y, by the fdlowing general remark. Theorem 2.7.8. If K is singular and n 2 2, then

K

+ [K]in-

1.

Proof. Let S be any set with IS1 = K , and write S as a disjoint union, S = U{S,; u < K'} where always lS,I < K . For each sequence (nl, ..., n p )wherep 2 1, l < n l ,..., n p < n a n d n l +...+ n , = n , p u t A(nl, ..., n p ) = { a E [S]"; there are and la nS,,I

(51,

...,up with

u1

< ... < up < K '

= n i (1 < i < p ) } .

This gives a partition of [S]" into 2"-' disjoint classes. And for any H in [SIK since IH fl Sol 2 n for at least n values of u, it follows that [HI" meets every class of this partition. We shall now make a couple of further isolated remarks about the case (n t 1, K ) " , and so K -b [n + 1, K]", for K uncountable and not strongly inaccessible. In most cases there is a bestpossible strengthening of this negative relation.

n 2 3. From Theorem 2.6.2 we know K

+

Theorem 2.7.9 (GCH). Let n 2 3 and suppose K' is a successor cardinal. Then [n 1,

K

e

(K),t]".

Proof. Suppose first that K is regular, say K = A'. The construction of an example is somewhat similar to that in Lemma 2.7.3, only rather simpler. Let ( A a ;a < A') be a well ordering of [A']', and for each T with 7 < A', put

d 7 ={ A a ; a < 7 a n d A , C T } . For each t with < A', we seek to pick sets a ( t , u, k ) from [Aa]"-' for A , in d r; and k with k < t , such that a(.$,u, k ) n a((, 7 , 1) = 0 if (u,k ) f (7, I). Since I d , X .$I< A, by working along a well ordering of d E X t of order type A we can clearly do this. Now define subsets Ak of [A']" fork with k < A' as follows: if { a1,..., a,)C A' with a1 < ... < a,,\then

{ a1,..., a,} E Ak * { a l , ..., a n - l } = a(ol,, u, k ) for some u . Put A = [A']" - U { A k ;k < A'}. Then { A } U { A k ; k < A'} is a disjoint partition of [A']". SupposeHE [A']"'',sayH= {ao,...,a,}<,with [HI" L U { A k ; k < A ' } . Then in particular there are u, 7, k, I such that {ao,..., an-2,an}=a(&,, u, k )

Ch. 2.7

59

Square bracket relations

and { a l , ...,a,} =a(&,, T, I ) , and (u, k ) # (7,0.But a(a,, u, k ) n a(a,, T, I ) = (3 and {ao, ..., a,-2, a,} fl {a1, ..., a,} # 8, so this is impossible. Hence if H E [A+]"+' then [HI" n A # 8. Now take H in [A']'. Choose A in so for some u with u < A' we have A = A u . Take any k with k < A'. Since IHI = A+, there is in H with ( > u, k and A E d 6. Then a([, a, k ) E [A]"-' so that a(& u, k ) U { E } E Ak. Since a(& a, k ) U { 4') C H, this means [H]" f Ak l # 8. Thus the theorem is proved in this case. Now suppose that K is singular, with K' = A'. Take any set S with IS1 = K , and write S as a disjoint union S = U{S,; u < A+} where always ISul< K. We shall use the partition of [A']" constructed above to define a partition {r}U { r k ;k < A'} of [S]". For k with k < h+ and (xl, ..., x,} from [S]", define

[ah,

{Xi,

..., X,}

Er k

* there is

{ 01,..., 0,)

AkySuch that

and put

r =[S]" - U { r k ; k < A + } Just as in the proof of Theorem 2.7.6, this partition has the required property. To obtain positive relations when n 2 3, there is a Stepping-up Lemma similar to that which holds for the ordinary partition relation.

Theorem 2.7.10. Let K and A be infinite cardinals such that A < K'. Let n > 1 and suppose X +. [qk;k < y]" where lyl''' < K' whenever u < h. Then K +. [Qk i1; k < 7]n+1. The proof is entirely similar to that of Lemma 2.3.1. As in the case of the ordinary partition relation, one can hope for a converse to Theorem 2.7.10, namely that under suitable conditions, if K [qk; k < y]" then 2K -f [7)k i 1;k < y]"". Very little along these lines is known. See Problem 17 in [24] and the discussion in [25]. Even the conjecture N2 [N1]i is unsolved. Shore [83] has shown that if Godel's Axiom of Constructibility ( V = L ) is true then ,("+) [~+]:fl(where n > 2), which corresponds to the negative relation K + [K+]:+ in Theorem 2.7.4. The results in this section have depended heavily on the GCH, and without GCH very little seems known about square bracket relations. For example, Calvin and Shelah [47], [48] have shown 2" -f [2"0]k, and H i $t [HI]:.

e

+

+

60

Ordinary partition relations

Ch. 2.8

+

Shelah has established Nn [Nn]1+2 if n 2 1. The relatioi2’O -?- [HI] ‘3 is open. For the ordinary partition symbol, all-relations with infinite exponent n are false. Little seems known about square bracket relations with infinite n. Erdos and Hajnal have shown the following.

Theorem 2.7.11. Suppose

K

and hare infinite. Then K -b [A].;’

Proof. This is trivial if K < A, so suppose K 2 A. Take any set S with MI = K ; we have to find a disjoint partition A = { Ap; < 2’) of [S]’such that [HI’ n Ap # (I for every 0 and every H in [S]’. Suppose first that A = K . Then we can well order [S]’with order type 2’, say [S]’= { A , ; a < 2”). Take any one-to-one and onto map p : 2’ X 2’ -+2‘. Choose setsAo9 from [A,]’ for a,/3 with a, < 2’ by induction on p ( a , 0) so that Aorp # A76 whenever p ( y , 6) < p(a, p). Put rp = {Ao9; a < 2’}, so { rp;/3 < 2’) is pairwise disjoint, and take any disjoint partition d = { Ap; p < 2’} of [S]‘with rp C_ Ap for each p. Then if H E [S]”, for some a we have H = A , and so A,p E [HIh fl Ap. Thus A is as required. Now suppose K > A. By virtue of Zorn’s Lemma, there is a maximal almost disjoint family% with%’ C [S]’.For each M in% we have just shown that there is a disjoint partition A(M) = {Ap(M); p < 2’) of [MI‘ such that [H]’nAp(M)#(I wheneverHE [M]‘,,<<’.Put r p =U{Ap(M);ME%}. By the choice ofA (M) and since IM nNI < A for distinct M, N in%, certainly rp n,?I = (I when /3 # y. Take any disjoint partition A = {Ap; p < 2’} of [S]’with rp C Ap for each 0.If H E [S]’,by the maximality of 371 there is M in371 with W n MI = A. Then [ H nMI’ n Ap(M) # 8 for every 0,and so [HI’ flAp f 8. This completes the proof. Theorem 2.7.1 1 leaves unsettled most of the cases of the following (Problem 14 in [24]). Problem 2.7.12. If K , A 2 p 2 No,is it ture that K -f [A]:p ? If the Axiom of Constructibility is true then Problem 2.7.11 is answered positively in Shore [83]. He shows that if V = L then K [A’] for all K and A. 58. Partitions of all the finite subsets

The partition symbol K K

-?-

(QX.

-?-

(Q);~

represents an extension of the symbol

Ch. 2.8

Partitions of all the finite subsets

61

Definition 2.8.1. The partition relation K + (~2y<" means the following: given any set S with IS1 = K , for all partitions A = {Ak; k < 7) of [S]
Lemma 2.8.1. Let 77 be infinite and suppose K ally with y < 2N0.

+

(7);". m e n K

+

(7);"

for

Proof. Let any disjoint partition A of [ K ] < ~ O into 2N0 classes be given, and use the functions in w 2 to index the classes in A , so A = { Af; f E w 2 } . Take any one-to-one and onto map p : w X w + w with the roperty that always 1, m
ri * 3 f € w 2 ( f ( l )= i

and

{a1, ...,a m } E Af)

.

(1)

Take H from [ K ] ~such that H is homogeneous for r. Then H i s also homogeneous for A . For take any sets { a l , ..., a m } < , {PI, ..., Om }< from [HIm. Suppose' { a 1 ,..., a,} E Af and {PI, L", O m } E A,. Take any I with I < w and put n = p(l, m). Choose y m + l , ..., 7, from H with a m , Pm < ym+l < ... < 7,; since H i s infinite this is certainly possible. Then since { a l , ...,am>E A,, we have {al,..., a m , Y m + l , ..., yn> E rf,) from (1). Then {PI, ...yPm. " / m + l ,- . . , y n } E rf([,since H i s homogeneous for r, and so g(l) = f ( l ) .Since this is true for every 1, we have f = g and so H is indeed homogeneous for A . The following two theorems for 77 = No follow from methods credited by Erdos and Hajnal [ 18, Thm 9b] to G . Fodor. We shall follow the proofs in Morley [74].

62

Ch. 2.8

Ordinary partition relations

Theorem 2.8.2. Let q be infinite. If then K is regular.

K

is the least cardinal such that K

-+

(q)$w,

Proof. Let K be least such that K --* (q)$" and suppose in fact K is singular. Take a pairwise disjoint decomposition K = U { K , ; u < K ' ) where always IK,I < K . By the choice of K , there are artitions. To = rul1of and a partition I'= { ro,rl}of [K']' O , none of which have a homogeneous set of size K . Define a partition A = {Aoo, A o ~ A, ~ oA, l l ) of [K]'"' as follows: i f a E [ K ] < ' O and i = 0,1

8

* 3 a < K ' (a L K , n rUi), a E A1i * [la1 = 1 O r V U < K ' (a p K , ) ]

aE

and

Let H be a subset of K homogeneous forA . Either there is u such that H C K , , or else IH n K,I < 1 for every u. In the first case, H is homogeneous for r,; in the second, { u < K ' ; IH n K,I = 1 ) is homogeneous for r. In either event, IHI < q. Thus A has n o homogeneous set of size q. In view of Lemma 2.8.1, this contradicts that K --* (v):~. Hence K must be regular.

Theorem 2.8.3. Let q be infinite. If K # ( T ) $ ~then 2K (v):~. Proof. Suppose K # [q)2w and take a partition r = { ro,rl ) of [K]'" which has no homogeneous set of size q. We shall construct a partition A = {Ak; k < w } of [K2]
( i , j = 1 ,..., n ) .

Let< be the lexicographic ordering of K2,and choose a partition A = { Ak; k < a} of K2 such that if fo, ..., f, E K2with fo< f l < ...< f, then

{fo, ...,f,} E Ak * CS(f0,

fl),

..., S(fn-1, f,))

E r:

where, as in 56, S(L g) is the least a where f(a) #g(a). Let H be a subset of K2homogeneous forA . Suppose L g, h E H with f S(g, h),

Ch. 2.8

Partitions of all the finite subsets

63

and moreover this order is the same for all triples from H . Suppose in fact S(f, g) < S(g, h ) ; the argument is similar in the other case. Thus whenever f,g, h E H w i t h f < g < h we haveS(f,g)=6(f, h).Put

I = (Wg);f,gEH}

9

then 1 1 1= IHI. Further, Z is easily seen to be homogeneous for r, so IZI < q. Thus IH< q, and so A has no homogeneous set of size q.

Corollary 2.8.4. Zf K is the least cardinal such that K finite, then K is strongly inaccessible.

-+

(77);"

where q is in-

We shall conclude this section by showing that increasing the value of q in the relation K +. (q)$" represents a genuine strengthening of the relation. In this respect, to relate these properties to the ordinary partition symbol studied before, we should remark that Silver [90] has shown that if K is the least cardinal with even the property K + (No)$", then there is a cardinal X with X < K such that X -+ (A);. Silver's proof of this result depends on methods from mathematical logic. A combinatorial proof has since been given by Henle and Kleinberg [ 5 8 ] .

Theorem 2.8.5. Let q l , q2 be infinite cardinals with q 1 < 772. Zf ~1 and ~2 are the least cardinals with the properties K -+ (q&" and K (q&" respectively, t h e n K l < K z . -+

,

,

Proof. Clearly K < ~ 2 so, suppose in fact K = ~2 = K say. Then for each cardinal a with a < K , we have la1 < K so there is a partition A (a)= {Ao(a),A~(a)} of [ a ] < Hwhich 0 has no homogeneous subset of power q l . Define a partition A = {Ao, A , } of [ K ] " ~ as follows: if a , , ..., a, E K with a1< ...
...,a,} E Aj * {a,, ..., &,.-I 1E M a n ) .

Let H be a subset of K homogeneous for A. Then for each a in H , it follows that Hn a is homogeneous f o r b (a),so W n aI < 771. Hence WI G q1 < 772, so A has no homogeneous set of size q2, contrary to the relation K +. (7?2)?.

CHAPTER 3

SET MAPPINGS

8 1. Set mappings of small order Let S be an arbitrary set. By a set-mapping on S is meant a function f mapping S into the powerset bS of S such that x 4 f ( x ) for each x in S . The set map f is said to be of order h if h is the least cardinal such that If(x)l < h for each x in S. A subset S’ of S is said to be free for f if S’ n Uf [S’] = @. Equivalently the;, S’ is free for f if for all pairs x, y from S ’ both x 4 f ( y ) and Y 4 f (XI. We shall investigate in this chapter some conditions on the set map f under which there will be a large free subset. Apparently P. Turin (in about 1930) was the first to ask questions of this nature - they arose in connnection with a problem on interpolation. Specifically, he asked if when S is of the power of the continuum and each f ( x ) is finite, must there be an infinite set? Ruziewicz [82] generalized the problem to the following: if S is infinite with MI = K and f is a set map on S of order h where A < K , will there be a free set of power K ? Solutions to special cases of this problem are in L b i r [66], Sierpinski [86], Piccard [76], [77] and Fodor [40]. In [14] Erdos proved, under the assumption of the Generalized Continuum Hypothesis, that there always is such a free set. Finally Hajnal [53] proved this result without appeal to GCH. It is easily seen that if the set map f has order h with A 2 K then there need be no free set even of size 2 . For let f be the set map on K where f ( a ) = (0 E K ; 0< a } . Clearly f has order K , and there is no free pair for$ We shall now give Hajnal’s proof.

Theorem 3.1.1. Let S be a set with IS1 = K and let f be a set map on S of order X where A < K . f i e n there is a free set of size K for$ Proof. Suppose first that K is regular. (The proof in this case is due to D. Gzir.) We use transfinite induction and choose subsets S, of S for a with a < h SO

Set mappings of small order

Ch. 3.1

65

that S, is a maximal free subset of S - U{Sp;0 < a } . Since being a free subset is a property of finite character, such a maximal set exists. If for any a it happens that &I= K then there is nothing more to do. So suppose that IS,l< K whenever a < A. We shall obtain a contradiction. Since always If(x)l< h where X < K , it follows that for any a we have lUf[S,]l< K and consequently, from the regularity of K , i f S * = U { S , U U f [ S , ] ; a < h} then IS*[ < K . Thus S - S* # 8. Choose x from S - S*.Then x 4 S, U U f [ S , ] for each a. Since S, is a maximal free set, there must be x , in S, such that x , E f ( x ) , for otherwise S, U { x } contradicts the maximality of S,. Now x , # xp when a # /3 since by construction S, n S, = 8, and {x,; a < A} C f ( x ) so If(x)l> A. This contradicts thatfhas order A; and this case is proved. Now suppose that K is singular. Choose a regular cardinal 8 such that K ‘ , h < 8 < K . Take a strictly increasing sequence ( K , ; u < K ’ ) of regular cardinals all less than K with K = X(K,,; u < K ’ ) , where 0 < K O . Write S as a disjoint I I, = K,. Thenfinduces set mapsf, : S, + VS,, union, S = U { S , ; u < K ’ } where S where f,(x) = f ( x ) n S,, and f, has order at most A. Since S I I, = K , where K , is regular with X < K , , there is a set SA in [S,]”” free forf,,, and S!, is then also free for& Put

S,*=S~-u{Uf[ST];7
f[~,*] nU{S;;T>

U)

=8

.

We shall seek sets T,* with T,* E [S,*IKusuch that

f [ T , * ] n U { T : ; ~ < u ) = 8,

u{

(1)

for then if T’ = T,*;u < K ‘ } clearly T’ is free forf, and IT’I = K . We start by finding sets T , in [S,*IKusuch that if Z , = T,; 7 < U } then for a particular pairwise disjoint decomposition { Z U a ;a < 0) of Z, there is an ordinal a(u)with a(u)< 8 such that

Z , n Uf[T,,] C U { Z , ; a < a ( u ) } .

u{

(2)

Use induction on u with u < K ’ to define sets T,, with IT,I = K , together with pairwise disjoint decompositions T , = U { T,,; a < 8 ) of T , where always ITJ = K,. Suppose that u is given with u< K ‘ , and that this has already Z ,, U(T,,; been done for all 7 where 7 < u. Put Z, = U { TT;7 < u } and = 7 < u}, so that Z , is a disjoint union, Z , = U { Z , ; a < 8 ) . For any x in S,*,

f ( x ) nz, = U { f ( x ) nz,; a < 8 ) .

66

Set mappings

Ch. 3.1

Since I f(x)l < X < 8 and 8 is regular there is an ordinal a(x, a ) b i t h a(x, u) < 8 such that f ( x ) n Z,, = 0 whenever a 2 a(x, u). For each /3 with p < 8 put

s$=

{ x E S o * ; a ( x ,u ) < p ) ,

soS$CS,*,if/3
T,*= U{T,,,;a*
r < u we have y E TT, 5Z Z,, for some a with a >a* 2 a(u). From (2) it follows that f ( x ) nZ,, = (4 for such an a;hence y I f ( x ) . This establishes (1) and completes the proof. so IT,*I = K , . Then i f y E T,* and x E T,* where

There are a couple of interesting extensions of Theorem 3.1.1. The first one is taken from Erdos and Fodor [ 161.

Theorem 3.1.2. Let S be a set with IS1 = K and suppose a family {R,; y < q} of K-size subsets of S is given, where q is any cardinal with q < K. Let f be a set mapping on S of order X where X < K . m e n there is a subset S’ of S free f o r 6 and moreover IS’ n R,I = K for each y. Proof. By Lemma 2.7.2, we may suppose that the sets R , are pairwise disjoint. First suppose K is regular. Let p : K + q be any onto map with the property that for each y with y < q we have I {a < ~ ; p ( a=)y}l = K . Let S be the set of all sequences s = (s,; v < cp) where cp < K , s, E RP(,) for each u, sP f s, if p # v, and ran (s) = { sv; u < q } is free for f: Inductively, define sequences s, = (s,,; v < cp), in S for a with a < K as follows. Choose s, as a sequence in S all the entries for which are in S - U{ ran( so); /3 < a},and which has maximal length amongst all such sequences from S. If any s, has length K , then ran( s,) is a free set which has the required property, so for a contradiction suppose cp, < K for all a. Consider the values p(cp,) where a < K . Since p maps into q and A, q < K , there must be A in [ K ] ’ and y with y < q such that p(cp,) = y for all a in A. Put

s*=U{ran(s,)uUf[ran(s,)];a~A}.

Ch. 3.1

61

Set mappings of small order

Sincefhas order X and always Iran( s,)l < K , it follows from the regularity of K that IS*/ < K . ThusR, - S* # 8. Choosex from R , - S * . Then since x 4 S * , there must be for each a in A an element s, in ran( s,) such that ,s Ef(x), for otherwise the sequence s, U { (7, x)} obtained by extending s, one place further with the value x would contradict the maximal length of s,. Now by construction,,s # sop if a # P so since { s;, a € A } C_ f(x) it follows that If(x)l> X, contradicting that f has order h. This completes the proof in this case. The proof when K is singular is very similar to that of Theorem 3.1.1, so we don’t give full details. Choose 8 regular so that K ’ , X, q < 8 < K . In the notation of the last proof, ensure that the sets S, are chosen so that IS, nR,l = K , for y < 77 and u < K ‘ , and take a free set SA such that IS; nR,I = K , for all y. Rather than constructing the one set T,*, construct sets TZ, for all 7 with y < q, so that

f [ T h ]fl u{T$ ; T < u and 6 < q} = 8 . The set

U{ T&; u < K ’

and y < q} is the required free set.

It is easy to see that the result in Theorem 3.1.2 is the best possible, in the sense that the family { R , ;y < q} cannot be replaced by a family {R,; y < K } of K pairwise disjoint sets from [SIK.For write K as a disjoint union K = U{R,; y < K } where always IR,I = K and ensure that y 4 R,; consider the set map f : K + V Kwheref(a) = (y} just when a ER,. If S‘ meets all the R , then f [S‘]= K and so S’ is not free f o r 5 The second extension of Theorem 3.1.1 is from Fodor [41]. Theorem 3.1.3 (GCH). Let f be.a set map oforder X on a set S where MI = K with X < K . Then there is a free subset S’ of S with IS’I = K such that

~U{f(x)nf(y);x,Y E S ’ and x f y N


With the set map f on S assumed given, for a subset A of S let us put

IL4 = U{f(x) nf(y);x, y € A and x # y } . If we could findA in [SIKwith lnAl < K , then we could obtain a free set S‘ in [AIK(so necessarily InS’l< K ) as follows. Note that for x in A - n A , for at most one y in A can x € f ( y ) hold - write y(x) for thisy, when it exists. Inductively choose x, from A - IL4 for a with a < K so that

x,

EA -

Since IU{f(xp);

(HA u {xp;P < a} u M x p ) ; P < a} u U{f(xp); P < 4) *

< a}l < X

*

la1

< K , the choice of x,

is always possible.

68

Ch. 3.1

Set mappings

Clearly {x,; a < K } is the required free set. Thus Theorem 3.1.3 will-follow from the following theorem. (The proof requires the GCH only when K is singular.)

Theorem 3.1.4 (GCH). Let f be a set map of order h on a set S where IS1 = K with h < K . Then there is A in [SIKwith IHAl < K . Proof. Suppose first that K is regular, and let the set map f : S + ?&be !j given. We shall show that for any T from [SIKthere isA in [TIKwith lHAl < K . For a contradiction, suppose for some particular T from [SIKthat this is false; thus whenever A C T and IHA I < K then M I < K . for a with a < h as follows. Put Use induction to define sets& in A,* = u{uf[Ap];P < a} so lA,*I < K by the regularity of K . Use Zorn's Lemma to choose for A , a subset of T - U{Ap;P < a} maximal with the property HA, C A,*. Then A , # @ and M,I < K by our assumption. Put A = {A,; a < h} so M I < K . Thus T - A # @, so choose x from T - A . For each a there must be y, in S , such that f(x) n f ( y a ) A,*, for otherwise S , U {x} contradicts the maximality of S,. Choose x, with x, E (f(x) n f ( y , ) ) -A,*. Then x, # xp whenever a # 0, for if say P < a then x, Ef(y,) -f(yp) yet xp E f ( y p ) . Thus since {x,; a < A} C_f(x) we have If(x)l > A, contradicting thatfhas order A. This completes this part of the proof. If in fact A+ < K , by assuming GCH the result above can be improved as follows. Given any subset T of S with IT1 = 8' where 0 is a cardinal with X < 8' < 8 < K , then there is a subset B of T with IBI = 8' and lHBl < h. To see this, note that by the construction above there is a subset A of T with MI = 8' and [ H A / < 8'. Look at A' = {x E A ; f ( x ) n HA # 8). The sets f(x) for x inA - A' are pairwise disjoint, so if M'I < 8' then B = A - A' has the required property. So we may suppose IA'I = 8'. For each x in A', the set f(x) f l HA is a subset of HA of power less than A. The number of such subsets is at most O k , and O h = 0 by GCH, since h < 8'. Thus there must be a subset B o f A ' with IBI = 8' such that for allx in B the setf(x) fl IIA is the same. But then HB = f(x) fHA'for l any x in B, so lHBl C A, and the claim is established. We are now ready to prove the theorem for singular K . Take a strictly increasing sequence ( K , ; u < K ' ) where each K , is the successor of a regular cardinal, with K ' , A+ < K , < K and K = C(K,; a < K ' ) . Write S as a disjoint union, S = U{S,; u < K ' } where always IS,[ = K , . By the remark above, for each u there are subsets B, of S , with IB,I = K , such that IHB,l< X. Put B,* = U{Uf[B,];T < u } , so IB,*l< h . C(K,; T < u ) < K , , noting that K , (being a

u

a

Ch. 3.1

Set mappings of small order

69

successor cardinal) is regular. Since the sets f ( x ) - HB, f o r x in B, are pairwise disjoint, the number ofx in B, for which f(x) - IIB, meets B,* is at most lB,*l. Hence if

A , = {x E B,; ( f ( x ) - HBo) n B,* = (3) thenlA,I=Ka.PutA=U{Au;~<~’},sH o I = ~ . M o r e o v e r , I I AC U{IIB,; K’). For take x, y in A , so suppose x € A , , y € A , . If u = 7 certainly f ( x ) flf(y) C HB,, so suppose without loss of generality that T < u. Then y € A , C B, so f(y) C B,*. By the definition ofA, then f ( x ) nf ( y ) C mu. Thus indeed IIA C_ U{ IIB,; u < K’), and hence InAl Q h . K ’ < K . This constructs A with the desired property, and completes the proof. a<

We conclude this section with a theorem of Fodor [40] concerning the decomposition of a set into a union of free sets. Theorem 3.1.5. Let S be a set of power K and let f be a set mapping on S of order X where Ho Q X < K. Then there is a family BI with 1% I < X of subsets of S each free forJ such that S = UBI. Proof. Given X in [S]’, define X, where n < w by induction on n as follows: Xo=X,X,+, =Uf[X,].PutX, = U { X , ; n < o } . T h e n IX,I=XandX, is closed under f, in the sense that whenever x E X, then f(x) C X,. So clearly S may be written as a union (probably not disjoint) of K sets each of power h and each closed under f, say S = U{S,; a! < K). Using such a decomposition of S, we shall show the following: For any non-empty subset T of S, there is a family 91 (T) = { T,; v < A} of X subsets of T each free for f, such that y ET-

U B I ( 0 * 3 x E UBI(T)(x E f ( y ) ) .

(1)

Start by putting, for a! with a < K ,

R , = T n (S,

-

U{Sp;P
Discard any empty R,. Then always 1 < IR,I Q A, and so we may write R , = {saU; v < A}. Define subsets T, of T by induction on v (where v < X) SO that T, C {s,~; a! < K} as follows. Suppose for a particular value of v that the sets T,, for I.( with I.( < v have already been defined. We determine whether or not s,, is in T, using induction on a, by specifying that ,,s E T, if and only if the following two conditions are met: (i) (19

v l (x 4f(s,)), VP < 4 s p u E Tu * spv 4f(s,,)) . Vx E U{T,,;I.(<

I0

Set mappings

Ch. 3.1

This defines T,. Put 81 (T) = { T,; v < A}; we show (1) holds. Take y with y E T - U 81(T). Since T =.U{R,; a < K } there is a such that y E R,, and SO y =,s for some v. Sincey 4 U81(T), in particulary 4 T,. From the definition of T,, either there is x in u{ T,; p < v} such that x Ef(y) or else there is spv in T, such that spv E f ( y ) ; in any event there is x in U%(T) such that x E f ( y ) . Thus (1) holds. Further, each T, is free f o r 5 For take s, sp, from T, and suppose say /3 < a. Since f(spv) C_ Sp and s, ES, - U{S,; 7 < a},

certainly s, 4 f(spv). And condition (ii) in the definition of T, ensures that 4 f(s,,). Hence T, is indeed free f o r t We are now ready to construct the family 81 of free subsets of S. Inductively define families 81, of subsets of S as follows. Put U, = S - U{U 81,; p < v}, then 81, = 81(U,) if U, f 0 ; 81, = $9 if U, = 0.Now define 0i = U{ 81,; v < A}. Then certainly BI is a family of free subsets of S, with 1811G A. Suppose for some v that U, # 0,and choosey from U,. Then y E U, for each p with p < v soy E U, - U 81, = Up - U %(Up).Thus by the property (1) for % ( U p ) , we can choose x, from U 81, such that x, E f ( y ) . Now. U, fU lBI, = 0 whenever t < p ; since U 81, C U, it follows that U81, n U 81, = 0 whenever 5 f p. Hence x5 # x p whenever $ # p. Since always If(y)l< A and {x,; p < v} C f(y) it follows that Ivl < A and so v < A. Thus if U, f 0 we must have v < A. We are now set to show that S = u 81. For take any y in S. There must be v with v < A for whichy E U g,,for otherwise y ES - U(u81,; v < A} = U,, and we have just remarked that U, = 0.Since 81, C 81, this showsy E U BI, and consequently S = U BI. This completes the proof. spv

There are various ways in which the problems investigated in this section can be extended. The original question of Ruziewicz can be phrased in terms of the order type of the sets involved, rather than their cardinality. Questions of this nature have been investigated by Erdos, Hajnal and Milner in [37], for countable order types. On the other hand, one can consider set mappings defined on a topological space, and replace the cardinality requirements by conditions of a topological nature. As an example, we mention the following result of Bagemlhl [2]: I f f : R + P R is a set mapping on the set R of real numbers such that alwaysf(x) is nowhere dense, then there is an everywhere dense free set forf.

Ch. 3.2

Set mappings of large order

71

$2. Set mappings of large order Let S be a set of power K and let f : S .+ !$S be a set map on S of order K . We shall investigate conditions on f under which there will be a large subset of S which is free f o r 5 An example was given in the last section of a set map on K of order K with no free set of size 2, so clearly some restriction on f is required. The conditions we shall impose are of the following type. Let q, 8 be cardinals; we demand that whenever X is an q-size subset of S then the intersection of all the f ( x ) for x in X has power less than 8. Questions of this type were first raised by Erdos and Fodor [ 161 and discussed in detail in Erdos, Hajnal and Rado [38]. The following notation is convenient to summarize the results. (This corresponds to the symbol K -+ [ [ K , q , 8 , A]] of [38].) Definition 3.2.1. The relation ( K , q , 8 ) .+ h means the following. Whenever f is a set map of order K on a set S with IS1 = K such that

then there is a subset S' o f S free for f with IS'I = A. We start by noting a couple of lemmas that extend some of the results from Chapter 1, 5 1 on almost disjoint families to families with the present intersection property.

Lemma 3.2.2. If a set S with I S1 = K can be decomposed into a family d with ISQI = h such that 1091< 8 whenever 9E [d]q+ then X < 17 K'. Proof. Suppose S = U d , where d is a family with the properties in the statement of the lemma. For each A in PQ with MI 2 8, choose A* in [A]'. Then for any B in [S]', we have I ( A E d ; A * = B l ) < q. Thus ( A E d ; I A l 2 8) has power at most q . I[S]'I, and so Idl
Lemma 3.2.3 (GCH). Suppose K and 8 are infinite cardinals with K ' f 8'. Then there is no decomposition d of a set of power K into K + subsets each of power at least 8 such that I n9l< 8 for eveiy family 7from

12

Ch. 3.2

Set mappings

Corollary 3.2.4 (GCH). Suppose 0' < K . Then there is no decomposition d of a set of power K into K + subsets each of power at least 8' with In57 < 0 whenever 3 is a family from [d1"'. (If 8 is finite, GCH is not needed.) We are now in a position to prove the following positive result. Theorem 3.2.5 (GCH). Suppose K is regular and 8'

< K.

Then ( K ,

K,

8) +. K .

Proof. Suppose the theorem is false, and let f be a set map on a set S where IS1 = K such that X E

piK3 ~ n f [ x i i< e ,

(1) .

and yet there is no free set for f of power K . We start much as in the proof of Theorem 3.1 .l. Choose by transfinite induction sets S, for a! with a! < 8' such that S, is a maximal free subset of S - U{Sp; 0 < a},so always IS,l< K . Since K is regular with '8 < K and always I f(x)l < K it follows that if S * = U{S,UUf[S,];a!<8'}thenIS*l<~.Hence lS-S*l=~.Let x E S - S *. By the maximality of S,, tKere must be x, in S, such that x, E f (x), and x, # xp whenever a! # 0. Put T = U{ S, ;a! < O'}, then it follows that ITnf(x)l> 0'. Suppose IT1 = A. Assume A+ < K . Then since '

I { T n f(X);X ES - S*}l<

lvn = < K ,

by the regularity of K there must be X in [S - S*]" such that T flf(x) is the same for all x in X. But then f [XI1 2 8+,contradicting (1). The only other possibility is that A+ = K . Consider the family d = {Tnflx); xES- S*}. Then l.41 > O + for eachA in d ,and < 8 whenever YE [dIh+ by (1). Since IS - S*I = K = A', this contradicts Corollary 3.2.4. Thus in either case a contradiction is reached. This suffices to prove the theorem.

In

In31

The symbol ( K , q,0 ) +. A is fairly readily discussed in the case that K is singular. The relation is fully covered by the following theorem. Theorem 3.2.6 (GCH). Let K be singular and suppose that q,6 < K . m e n (i) ( K , V , e ) - K , (ii) ( K , K , e ) +. K ' , (iii) ( K , K , 1) ( K ' ) + .

+

Proof. Take an increasing sequence ( u < K ' ) and always g , 0 < K~ < K .

K ~ u;

< K ' ) of cardinals such that K = E(Ku;

Ch. 3.2

13

Set mappings of large order

To establish (iii), take any set S with IS1 = K and write S as a disjoint union, S=U{S,;U
f(x) = S,

- { x } where x E S,

.

Thenfis a set map on S of order K . If X E [SIKthen X must meet more than one of the sets S, and so nf[X] = 8. And if S‘ is a subset o f S free forf, for each u we must have IS’ nS,I < 1, so IS’I < K ‘ . Thusfis an example which proves (iii). To show (ii), l e t f b e any set map of order K on a set S with IS1 = K , such that InF[X]l< 8 whenever X E [SIK. We may suppose S = K , and write S = U{S,; u < K ’ } where S I I, = K , and S, < S, whenever u < 7.Define a disjoint partition A = { A1, A,, A,} of [K]*as follows: if a,P < K with a < 0 then

{a,PI E A i

* P Ef(a)

{a,PI E A2 * a Ef@) and P 4 f ( 4 {a,P}EA3

* {a,PI$A, U A 2 .

By the Canonization Lemma, Lemma 2.4.2, there is a family C = { C,; U < K ? with IC,l = K , such that A is canonical in C . Moreover (see the remarks after the proof of this lemma) we may suppose that C, < C, whenever u < 7.So for all pairs a, b from UC we have vu(la n C,l = Ib n C,l) * a

For each (I with u <

3

b (mod A ) .

(1)

x’choose a, from C, and put

X u = {T
Y, =

{ T < K ’ ; o< 7 and

3 PE C7({a,,,0) E A,)} .

By (1) and the definition of A*, if 7 E Y , then C, C nf[C,], and so C,; T E Y,}]l> IC,l = K , > 8 . Hence lY,l < K ’ , for otherwise the intersection property off is violated. Inductively choose ordinals a(t) where $, < K ‘ so that

!nf[u{

( J ( t ) E K ’- ({o(c);

{< u u{xu(f)u yu,);c<

;

by the regularity O ~ K this ’ is always possible. Put S’ = {a,([);t < K ’ } so IS‘I = K ’ . And if {< t then u ( f ) 4 X,,) U Y,,) so that {a,(s), a,(~)}E A3.

74

Ch. 3.2

Set mappings

Hence a , ~ )$ f(a,($ and a , ~4) f(a,($ so that S' is free for f: This proves (ii). Finally, to settle (i), let f be a set map on S as in the proof of (ii), except now with the intersection property XE

[ S I ~1nf[x11< e.

(2)

Define the ordinals o(C;)for C; with C; < K ' and the set S' as above, so that by (1) whenever a E C,C), 0 E C,(t) where 5 f C; then a 4 f ( 0 ) and /3 4: f ( a ) . Moreover, in this case each C, is free for f . Once this is established, it follows that if S" = U{ C,(O; C; < K ' } then S" is free f o r 5 Since IS"I = K , (i) would hold. To show that C, is free, note that since q, 0 < K , there are subsets A , , B, of C, with Hal= IB,I = q 4 f3 and A , < B,. By (l), C, is homogeneous for A . However, if [C,]' C A, then B, C nf [A,] and if [C,]' C A, then A , C nf [ B , ] . Either of these contradicts (2), and so it must be that [C,]' C A3. Hence C, is free for f , and the proof is complete.

Let us return to problems involving regular cardinals. Here a few unsolved problems remain. Theorem 3.2.5 gives the best possible result for inaccessible K , so only the case when K is a successor cardinal need be considered. If K = A+ where X is regular, the following strong negative result of Hajnal [52] shows that the restriction 8' < K in Theorem 3.2.5 cannot be relaxed.

Theorem 3.2.7 (GCH). Let X be regular and let S be any set with IS1 = A'. Then there is an almost disjoint family 9 * with I 9 *I = X' of the X-size subsets of S such that IS - U 9 I < h for any subfamily 9 of CM * with power A'. Corollary 3.2.8 (GCH). Let X be regular. Then (A', 2, A)

p A'.

4

Proof of Corollary 3.2.8. Let S be any set with 16,31 = A+, and let 9*be a family on S with the properties given by Theorem 3.2.7. Take well orderings (x,;a<X+) and (B,; a< A') of S and of '?3*. Define sets f(x,) where a< X+ by induction on a as follows. Let y be least such that f(xp) # B, for all 0 with < a;then

f(x,) =

B, if x , $ B,

0 otherwise.

It is clear that if S* = {x ES; f ( x )f 0) then IS*l = A'. Definef*:S*-,'1?S*byf*(x)=S*nf(x).Thenf*isasetmaponS* of order at most A', and for distinct x, y from S* we have I f * ( x ) (7 f * ( y ) l < I f ( x ) nf ( y ) l < X since the family 9 * is almost disjoint. Further, f * has

Ch. 3.2

Set mappings of large order

15

no free set of power A+. For take S‘ with S’ C S* and IS’I = A+. Then

f[S‘]C CM* and If[S’]l= A+ so that IS’ - Uf[S’]l< A by the properties of ‘3 *. Since IS’I = A+ this means S‘ n Uf[S’] # 8. Since S’ C S*, we have S’ n Uf[S’] = S ’ n Uf*[S’], and consequently S’ is not free forf*. Thus

f* is a set map which illustrates (A+, 2, A) f . A+.

Proof of Theorem 3.2.7.Take the set S with IS1 = A+. By GCH, I [S]’] = A+, so there is a well ordering (A,; a < A+) of [S] .’ Put A, = { A p ;/3 < a } . Use transfinite induction to define sets B, in [S]‘ for a with a < A+ as follows. Put 73, = {Bp;0 < a } , and let y be the least ordinal with y 2 a for which there isA in d, with L4 - uq I = A for all cw3 from [73,]”. Since Im,l
d,* = { A E d,;v 73E [W,]<’(IA

-

UCM I = A)),

so 1 < Id,*l < A. Write A,* = {Aau;v < A} and CM, = {B,,,; v < A), listed with repetitions if necessary. Choose elementsx, for Y with v < A so that X,

E A,, - (U{BaU ;E.C


{xW

;E.C


since A,, E d,* such a choice is always possible. Put B, = ,{xaU;v < A). This completes the definition of B,. Then always B, E [S]’. And IB, fl Bpl < A if a f 0. For if say /3 < a then Bp E CM, so Bp = B,, for some v with v < A. Then

BpnB, C so IB,

{X~;E.C
nBpi< A. Thus if

CM” = { B , ; a < A + } , = h+ and cM* is an almost disjoint family. In fact q* then g*C [S]’,

has the property asserted in the statement of the theorem. This we show now. Put d *= { A E [S]’;

v CM€

[73*]<’(lA

-

UCM I =A)}

Then we have

A p E d *and p
(1)

For if Ap E d*certainly Ap E d ,* so that Ap =A,, for some v . Then E Ap n B,. Also we have

x,

S’ G S a n d IS’I = A +

* 3 A E d*(A C S ‘ ) .

For let S’ from [S]” be given, and put

g’= {BEg*;IBflS’I=A).

(2)

76

Ch. 3.2

Set mappings

Consider two cases. Case I . l n ' lA.~Since IS'A+, we may choose A from [s'-ucM']? Then 14 n BI < A for every B in CM*. By the regularity of A, then IA =A for every CM from [CM*]<', and soA E d *. Case 2. ICM'I > A. Choose 9"from [CM']', and put A = S' n UCM". Take any cx3 from [cx3*]<'. Then there is B" in 93' " - 93. Since the sets in CM * are almost disjoint, I(B" n S') nBI < A for each B in CM. Again by the regularity of A, it follows that",(I n S') n UCM I < A. Since IB" n S'I = A, this means that I@'' n S') - UCM I = A, and so IA - U73 I = A. ThusA E d * ,and ( 2 ) is established. We can now see that * has the property claimed. For take any family 33 from [CM*Ih+; we show IS - U31S A. For if not, IS - UCMI = A+ and so by ( 2 ) there is A in A * with-A C S - U'M. Now A = Ap for some 0.Since I'M I = A', there is a with < Q! < A' such that B, ECM . Then by (l), A n B, f 8 which is impossible with A C S - U 9 . Thus IS - UCM I S A, and the proof is complete.

u'x~'I=

-uql

In view of the result (A', 2 , A) f A' for X regular from Corollary 3.2.8, the strongest positive result for regular A would be (A', A', A) + A. In fact this result holds, as is shown by the following theorem. Theorem 3.2.9 (GCH). I f h is infinite then (A+, A+, A)

+

A'.

Proof. Take S with IS1 = A+ and suppose f : S + 13s is a set map of order A with I nf[X]I < A whenever X E [S] '. Choose elementsx, from S for a! with a < A+ by induction so that x,ES-({xp;P
U{f(xp);B
Then x, 4 f(xp) whenever /3 < a. Put T = {x,; a partition {Ao, A,} of [TI2 by, if /3 < a then {x,, xp} E Ao

* xp E

{x,, xp) E A1

* xp B: f(x,) .

a!

< A'},

so

In = A'.

Define

f h )

By Corollary 2.2.7, A+ + (A', A');, so either there is H in [qh'with [HI2 C A0 or there is H in [TIh' with [HI2E A , . Suppose there is H in [TIh+ with [HI2C A,. L e t H = {X,(~);Y
Set mappings of large order

Ch. 3.2

I1

The results for X regular can be summarized as follows: (A', A', 0) + A' if 0 < X (Theorem 3.2.5), and (A', A', A) + h (Theorem 3.2.9), yet (A', 2, X) A' (Corollary 3.2.8). There remains the possibility that IS1 = A+ where A is singular. The positive result of Theorem 3.2.9 cannot be greatly strengthened even in this case, as is shown by the following theorem. (If X is regular, the theorem is trivial, but then Corollary 3.2.8 gives a strong result.)

+

Theorem 3.2.10 (GCH). Z ~ isK infinite then (K', K',

K)

+ (K')'.

Proof. We may suppose that K is singular, and choose an increasing sequence (K,; u < K') of cardinals with always K, < K and K = Z(K,,; u < K'). Write [ K + ]=~{ A , ; a < K ' } , put 4,= { A p ;P < a and Ap C_ a} and choose sub, of d, with Id,,l< K, and 4,= U{ dola; u < K'}. As in the families & proof of Theorem 2.5.9, define subsetsf,(a) of a with If,(a)l < K , + ~ such that A n f u ( a )f 8 for each A in ?A and y 4 f,(P) for all P, y with 0, y Ef,(a). Define a function g : K' + ?K+ by g(0l)'

-

U{f,(Or);

U

< K'}

Theng is a set mapping on K + of order at most K ' . I claim that g has the properties required to prove the theorem. Take X i n [K']~'; we show Ing[X]I < K . For suppose on the contrary that Ing[X]I 2 K ,and chooseA from [ng[X]]", soA = A p for some 0. Since 1x1= K + we can choose a from X with /3 < a and A C a. Then A E ?A,, for some u with u < K', and consequently there is y with y € A nf,(a). Thus y 4 g(a), and so y 4 ng[X]. Yet y € A and A C n g [ X ] . This contradiction shows that in fact Ing[X]I < K . Now suppose S is a subset of K + free for g. We need to show that IS1< (K')''. Since S is free, if we take a,P from S, say where P < a,then P4 g(a) so P E U{f,(a); P < K ' } . Thus we may define a partition [SI2= u{A,; u < K'} where, if u < K', Au =

{{a, a)< E [S12 ;P Efu(a)) .

As in the proof of Theorem 2.5.9, this partition has no homogeneous set of size 3. However, from Theorem 2.2.4 the relation (K')' + (3)it holds, so if IS1 2 (K')' there would have to be a homogeneous triple. Hence IS1 < (K')'', and the theorem is proved. Let us review the situation with regard to the relation (A', ~ , 0 +) L when X is singular. By Theorem 3.2.5, if 0 < X then the best possible result (A', X',0)

Ch. 3.3

Set mappings

78

A' holds, so we need only consider the relation (A', q , A) I. When q = A', from Theorems 3.2.9 and 3.2.10 the relation is true for i = A' and false for i = (A')''. The truth of the relation (A', A', A) (A')' is an open problem. A couple of results for the case q < A are announced in Erdos, Hajnal and Rado [38], but a complete discussion is lacking. Among the simplest unsolved probq , H,) -+ true when 2 < q < H1,and lems are the following. Is , H,) i true when HI < i < H ,? is (H,+l, H -+

-+

-+

-+

$3. Set mappings of higher type So far in this chapter we have considered maps f : S YS.The concept of a set mapping can be extended by considering functions f : [SIq-+QS for any cardinal q. Such a mapping will be said to have type q. The maps of type 1 can be identified with the set maps of the earlier sections. ' -+

Definition 3.3.1. A set mapping on S of type q is a function f :. [SIq -+ YS such that X nf(X) = (I for each X in [SIq. A subset S' of S is said to be free forfif S'n U f [ [ ~ ' ] q =] (I. Set mappings of type greater than 1 were first discussed by Erdos and Hajnal in [ 181. The first result shows that if we are seeking a free set, only set maps of finite type need be considered, for if q is infinite then there is a set map of type q of order 2 which has no free set of size p. (Trivially, any set of size less than q is free.) Theorem 3.3.2. Let q be infinite and let S be a set with IS1 2 q. Then there is a set map f on S of type q and order 2 which has no p e e set of power q.

Proof. Consider first the case where I S1 = q. Write [SIq = {&;a < 2q}. Choose by induction sets B, from [&Iq for a with a < 2" so that B, # A , and B, # Bp for any 0with 0< a. Since I [&Iq I = 2q, such a choice of B, is always possible. Choose x, from A, - B,. Now define a function f : [SIq+ ps as follows: {x,} if X = B, ,

8

otherwise.

Ch. 3.3

Set mappings of higher type

I9

Then always X nf(X)= 8, s o f i s a set map of type 77, and order 2 , on S. Let S' be any subset of S of power 77, so S' = A , for some a. Then B, E [S'Iqand x, E S' nf(B,). Thus x, E S' n Uf[ [S'Iq]and so S' is not free f o r 5 Thus f h a s no free set of size q , and the theorem holds in this case. Now suppose I S1 > q. By Zorn's Lemma, there is a maximal almost disjoint family 3fI with 3fI C [SIq.For each M in 'Yfl, we have just shown that there is a set mapfM : [MIq '$M of type 7) and order 2 on M, which has no free set of size q. By the maximality of %, for each X in [SIqthere is M in% such that IX nicn = 77. Choose such an M,and call it M(X). Note that if Y E [Xfl M(X)Iq then necessarily M(Y)= M(X), by the almost disjoint property of the family %. Define a functionf: [SIq+ by +

f(X) = fM(x)(XfWX)) l . Since f M ( ~ ( n X M(X)) C M ( X ) - ( X f M(X)), l certainly X nf(X) = @.Thus f i s a set mapping on S of type 77 and order 2. Suppose there is S' in [SIqwith S' free f o r k thus S' n Uf[[S'Iq] = 8.In particular, if S" = S' n M(S') then S" n UfM(s#)[ [ ~ " ] q = ] 8,since for any Y in [ ~ " ] q ,

f(r)=fM(~(YnM(Y))=fM(sl(YnMM(S'))=fM(s')(Y). Thus S" would be a set of power 77 free forfM(sf),which is impossible. Thus f h a s no free set of size 77, and the proof is complete. In view of Theorem 3.3.2 then, we need only consider the case of set maps of finite type. Just as a set map of type 1 and order K on a set of size K need not have a free pair, also a set map of type n and order K on a set of size K("-')+ need not have a free set of size n + 1.

Theorem 3.3.3. Let n be a positive integer and let S be a set with IS1= K("-')+. Then there is a set mapping on S of type n and order K which has no free set ofpower n + 1. Proof. By induction on n. When n = 1, we noted in 8 1 that this is so - identify S with K , and for a in K putf(a) = C Y = { p ; p < a } . Now suppose the theorem holds for a particular value of n , and take S with MI = K @ + ) . We seek a set map F : [S]"" + [S]
F(a) =f,(u - {a})where a = max(a)

.

80

a.3.3

Set mappings

Then F is a set map on S of type n + 1 and order at most K . And clearly a free set X for F , with IXI = n + 2, would give a free set X - { a } of size n + 1 for fa, where a = max X , so that F can have no free set of size n t 2. This completes the proof. The value of IS1 in Theorem 3.3.3 is the maximum possible if there is to be no non-trivial free set (at least if the GCH is assumed), as is shown by the next theorem. Theorem 3.3.4 (GCH). Let n be a positive integer and let S be a set with IS1 = K("+? Then eveiy set mapping on S of type n and order K has a free set ofpower K + . Proof. If n = 1 this follows from Theorem 3.1.1, so we shall suppose henceforth that n 2 2. Let f : [S]" [S]
fa@)

=f(a U {XI)

*

By Theorem 3.1.5, always S-a can be written as a union of at most K sets each free forfa, say S-a = U { H , ( a ) ; a < K}. Take a well ordering, < say, of S. Define a partition A of [S]" as follows: if a, b E [S]" with a = { xo, ..., x,- }< and b = {yo, ...,y n - 1}< , then a - b ( m o d A ) * ( V i < n ) ( 3 ~ ~ < ~ )[ X j E H a ( a - {Xi}) and yi EHa(a -

bi}))l .

Then clearly A has at most K classes. Since IS1 = K ( " + ) , by the partition relation (K'): from Theorem 2.3.2, there is a set H i n [SIK+which is homogeneous for A. But then H is free for J: For take a from and x in H - a; it suffices to show that x 4 f(a ). Define an element y in a as follows. If x < for some u in a, theny is the least such u ; otherwise y is the largest element of a. Thusy is the i-th element of a if and only if x is the i-element of a U {x} - { y } . Since H i s homogeneous for A , we have a a U {x} - { y } (mod A). The definition of A shows that for some a,b o t h y €&(a - { y } ) and x EHa(a - { y } ) .Since H,(a - { y } ) is free forfa- {,,I, then x 4 f a - {,,)o); thus x 4 f ( a ) . So H is indeed free forf, and the theorem is proved. K("+)

-+

[a"

One can ask if, in the situation of Theorem 3.3.4, the size K + of the free set is the largest possible. For n = 1 trivially this is so, for n = 2 we shall show in Corollary 3.3.6 below that again K + is the best possible, but for n 2 3 the

Ch. 3.3

81

Set mappings of higher type

problem is open. An upper bound for the size of a free set is given by Corollary 3.3.6. (Compare with Problem 34B in Erdos and Hajnal [24].) First we prove the following result, which should anyway be compared with Theorem 3.3.3.

Theorem 3.3.5 (GCH). Let n = 2,3, ... and let S be a set with MI = K ( " - ~ ) + f i e r e is a set mapping on S of type n and order 2 which has no free set o f power K("-')+. Proof. By induction on n. If n = 2, take S with IS1 = K + , and identify S with K'. From Lemma 2.7.3, there is a disjoint partition [SI2= U{ rr;y < K ' ) such that whenever A E [SIK,B E [SIK+and y E S then there are a in A and P in B for which {a,8) E r,. Define f : [SI2 [S]" by: +-

'

Then f is a set mapping on S of type 2 and order 2. Moreover, f can have no free set of size K + . For take any subset S' of S with IS'I = K + . Choose y from S' and A from [S'IK,B from [S'IK+with y 4 A U B. By the choice of the partition, there a r e a , p w i t h a € A , p € B a n d {a,P}Er,.Thusf({a,P})= {y} so y E f ( { a,P)),and hence S' is not free. This completes the case n = 2. For the inductive step, proceed much as in the proof of Theorem 3.3.3. Suppose the theorem holds for a particular value of n , and take S with IS1 = Identify S with {a;K @ - ' ) + < a < K ( " + ) ) and for each a in S let be a set map on a of type n and order 2 with no free set of f a : [a]"-+ power K("-')+. Define F : [S]"" +$S by, if a E [Sin+' then

F(a) = fa(a - { &}) where a = max(a) , so F is a set map on S of type n + 1 and order 2. Suppose S' is a subset of S with IS'I = which is free for F. Take (Y in S' such that (Y has K("-')+ predecessors in S'. Then clearly S' n a is a set of power ,("-'I+ which is free for f a , contrary to the choice of f a . Thus F has no free set of size K @ + ) . This completes the proof. Replacing K by

K+

in Theorem 3.3.5 gives the following.

Corollary 3.3.6 (GCH). Let n = 2 , 3 , ...and let S be a set with IS1 = K @ + ) . fiere is a set mapping on S o f type n and order 2 which has no free set of power ,("+I.

82

Set mappings

Ch. 3.3

The result in Theorem 3.3.5 shows that a set map on a set S of type n (n > 2 ) and order A where h< IS1 need not necessarily have a free set of size ISI, in contrast to the result in Theorem 3.1.1 for set maps of type 1. However, if IS1 is singular, then the exact analogue of Theorem 3.1.1 holds.

Theorem 3.3.7 (GCH). Let S be a set with IS1 = K where K is singular, and let f be a set map on S of type n and order h where A < K . Then there is a subset S'of S with IS'I = K which is free f o r f : Proof. By Theorem 3.1.1, we need only be concerned with n where n 2 2. Identify S with K . Choose a disjoint partition A = { Ai; i < n + 1) of [Sin+' such that for a in [Sin+', say a = {a,-,,...,a, I<, a E Ai * aiEf ( a - { ai}), for i with i < n a E A,+,

* V i < n (ai4 f ( a - {ai})).

From Lemma 4.2.2 (the Canonization Lemma) there is a pairwise disjoint family C!= { C,; u < K ' ) with IU CI = K such that A is canonical in C and, by the remarks after the proof of that lemma, we may suppose that always IC,l> h and C, < C, whenever u < T . Thus, whenever a, b E [UC]"",

V U(la n Cul = Ib n Cul) * a

= b (mod A ) .

(1)

1 claim [U C In+' C A,,,, and consequently U C is a set of power K free forf, as needed. For take any set a from [UC]n+l, and suppose on the contrary that a € Aj for some i with i < n. Let the i-th element of a be in C,. Choose b from [ U C 1" such that b r l C, = a n C, if u # T,and if b = {Po, ..., Pn-l)< then for B = C, n { P ; Pi-1 < P < pi} we have IBI = A. Since IC,l > h such a choice of b is possible. Then for any from B it follows that is the i-th element of b U {PI.Moreover, for such /3 we have la n C,l= I(b U { 0 )) n Cl, for all u, so since a E Ai it follows from (1) that b U { 0) E Ai, that is, E f ( b ) . Hence B C f ( b ) . Since f has order h, this is impossible. This shows that [UC]"" C A,,,, and completes the proof.

CHAPTER 4

POLARIZED PARTITION RELATIONS

0 1. Partitions of K X K The polarized partition relations that are to be studied in this chapter were first defined in Erdos and Rado [ 2 9 ] and the more tractable cases carefully investigated in Erdos, Hajnal and Rado [ 3 8 ] .Most of the results in this chapter first appeared in [ 3 8 ] .The polarized relations are an extension of the relation K + ( ~ kk;< y)" from Chapter 2, an extension obtained by considering partitions of sequences of finite subsets, rather than partitions of just finite subsets.

Definition 4.1 .l.The polarized partition symbol

means: given any sets S1,..., S , with ISi[ = K~ (where i = 1, ...,m), for all partitions A = ( A k ; k < y} of [Sl]"'X ...X [S,]", into y parts, there are k with k < 7 and subsets Hi of Si with lHil = ~ i (where k i = 1, ..., m ) such that [Hl]"' X ... X [HmInm C Ak. We shall adopt without specific mention conventions in the use of the polarized partition symbol similar to those used with the ordinary partition symbol. In particular, the sequence H 1 , ...,H, in the above definition will be called a homogeneous sequence for the partition. The same simple remarks that were made at the time of the introduction of the ordinary partition symbol apply, mutatis mutandis, to the polarized symbol.

Polarized partition relations

84

Ch. 4.1

In the first two sections we shall be concerned with the special case where m = 2, n l = nz = 1, and usually y = 2 as well. Thus we shall be investigating properties of the symbol

that is, seeking homogeneous pairs for partitions of [S1]’ X [S,]’. We shall always identify [S]’ with S, and so we are dealing with partitions of the Cartesian product S1X S2. A few of the results that will be proved are stronger than can be expressed by the polarized partition symbol. To formulate these compactly, it is convenient to introduce the relation with alternatives.

Definition 4.1.2. The polarized partition symbol with alternatives (Ki)

Kz

+.

(qio

hio hzo

V

q20 V

‘711 V A i i ) ’ ”

V21 V A21

= K Z , for all partitions S1 X Sz= means: given sets S1, S2 with lSll = ~1 and A. U A , , there are k with k = 0,l and subsetsH1 o f S l and H2 of S2 with H1 X H2 C Ak such that either IH1I = q 1 k and = qZk, or else jH1I = h l k and l H 2 l = A2k.

The definition above can be extended to other cases of the polarized partition symbol. This relation will not be investigated for its own sake. However, the methods that we shall use to establish results concerning the symbol without alternatives at times, without extra work, will lead to a stronger result that can be expressed using the symbol with alternatives. In this first section, we shall concentrate on the relation with m = 2, nl = n2 = 1 and further K 1 = ~2 = K . We shall start by deducing several results using the theory of set mappings from the previous chapter.

Theorem 4.1.3. Let

K

be infinite with q < K. Then

Proof. Let S be any set with IS1 = K , so we may identify S with K . Let a partition S X S = A0 U A1

Ch. 4.1

Partitions of K

X

85

K

be given, and suppose I (0 E S; (a,0)E A,}] < 7) for each a in S. We need to find H I , H2 in [SIKwith H1 X H.2 C A,. Define a set mappingf on S as follows: for a in S, put

f ( a )= { 0E S; (a,0)E A, and a # 0) , so f has order at most q. By Theorem 3.1.1 there is a set S' in [SIKfree for f: Choose disjoint sets HI, H2 from [S'IK;then H1 X H2 C_ A,. The value of 7) in this last theorem cannot be increased to q lowing strong negative relation shows.

=K,

as the fol-

Theorem 4.1.4.For all K ,

Proof. Consider the partition

K

X

K

= A0 U A1 where

(a,fl)EAo *a B . Likewise, an attempt to improve Theorem 4.1.3 by increasing the "1" may not meet with success, although this is not as easy to show.

Theorem 4.1.5 (GCH). Suppose (K:).

(K:

K

K

1

is regular. m e n

K

Vz)'.'

K'VK

.

Proof. By virtue of Theorem 3.2.7 there is an almost disjoint family d = { A a ; a < K ' } with always A , E [ K ' ] ~such that whenever 73 is a family in [d1"' then IK' - U q I G K . Take such a f a m i l y d ,and define a disjoint partition K'X K+=A,,UA, by (a,p)EA, * P E A a . This partition has the correct properties to prove the theorem. For each a,by definition {O; (a,0)E A, } = A , , and always IAaI < K'. Also IA, fA,I l < K if a # 7 , so that A1 is as required. With respect to A,, take H1 from [K']~' and H2 from YK' with H1 X H2 C A,; we need to show that lHzl< K'. However, if73 = { A , ; a E H l } then

(0;(a,0)E A,

for some a in H , }

=

u73 .

86

Ch. 4.1

Polarized partition relations

Thus H2 C K + - U 3 3 , and so lH21<

K.

This proves the theorem.

At this stage, we note another negative relation which follows easily from an example constructed in Chapter 2.

Theorem 4.1.6 (GCH). For all infinite K ,

(I:)+

( K

VK'

K'VK

K

VK')'''

K ' V K

u{

Proof. By Lemma 2.7.3, there is a partition [ K ' ] ~ = r k ;k < K ' } such that wheneverA E [ K ' ] ~ , B E [ K ' ] ~ ' , k E K' then {a, b}E r k for some {a, b } where a € A , b E B. Using this partition of [K']', define a disjoint partition K'X K + = A ,U A , by:

(a,0) E

i,{a,01 E ro .

Clearly this partition has the required property. In the proof of the next theorem, we shall need a polarized canonization lemma, similar to Lemma 2.4.2. This can be proved quite easily from Lemma 2.4.2. We shall state and prove only the special case we need.

Lemma 4.1.7 (GCH). Let K be singular and let (K,,; u < K ' ) be an increasing sequence @cardinals below K with K = Z(K,,; a < K ' ) . Let a disjoint partition K X K = { A k ; k < y} be given, where y < K . m e n there are pairwise disjoint families {A,,; a < K '} and { B,,; u < K '} and a function h : K ' X K ' + y such that IA,I = IB,l = K,, and

u

0,

T

< K'

* A u X B, C Ah(,,,,) .

Proof. Fork, 1 with k, 1 < y put A(k, 1) = {{a,

E [ K I 2 ; (a, 0) E A k

This gives a disjoint partition A of [ K ] ~=

U('A(k, I ) ; k, 1 < y}

and (0, a)E A , } .

[ K ] ~ ,

.

By Lemma 2.4.2, there is a family C = { C,,;u < K? where C,, < C, whenever u < T,such that A is canonical in C ,and we may assume IC,,l> K,,. Thus there

Ch. 4.1

Partitions of

are functions ho, hl : K'

+

K

81

X K

y such that

u < ~ < ~ ' a n d a ! E C , , / 3 E C , * { a ! , ~ ) E A ( h o ( u , 7 ) , h l ( u , r ) ) . (1)

For u with u < K ' , choose A , from [C2,]"" and B, from [C20+l]Ku, so certainly the families { A , ; u < K ' } and {B,; u < K ' } are both pairwise disjoint. Take u, 7 with u, 7 < K ' , and consider a! from A , and 0 from B,. If u < 7 then 2u< 27 + 1 so that a < 0 and by (l),

{a,0)E A(ho(20, 27 + l ) , h 1 ( 2 ~27 , + 1)) so that (a,0)E Ah0(20,2T+1);whereas if 7 < u then 27 + 1 < 20 so that 0< a! and by (11, { a,p) E A(h0 (27 + 1,20), h 1(27 + 1 , 2 0 ) ) , so that ( d ) E A h 1 ( 2 : + 1 , 2 y ) . Hencedefineh:K X K + y b y

h(u, r ) =

2r + 1) 1 ( 2 r t 1, 2u)

if u < 7 , if

7

< u.

Then always A , X B, C Ah(,,T), and the lemma is proved.

Theorem 4.1.8 (GCH). Let qo,q1 be cardinals with q i , q :


Then

(1)

Proof. Let S be any set with IS1 = K , so we may identify S with K . Let any disjoint partition S X S = A0 U A1 be given. Consider first the case that K is regular. Define set mappingsfo, fi on S as follows: fo(a)= { P < a;(0,

Al}, fi(a) = { p <

0 ;(Q,

0)E A11 .

Then bothf,, f l have order at most K . If there is X in [SIKwith Info[X]I 2 70, since n fo [XI X X C A, there is nothing more to show. Similarly if there is X in [SIKwith [XI1 2 q l , since X X n f l [XI C A l there is nothing more to do. So suppose that Info [X]l< v0 and lnf1 [XI1 < q 1 whenever XE [SIK. By Theorem 3.2.5, there is a set A in [SIKwhich is free forfo. Letf; : A -+%A be the restriction off, t o A , that isf;(a!) = f l ( a )n A. Again by Theorem 3.2.5 there is a set B in [AIKwhich is free forf;. Take two disjoint sets H I , H2 from [BIK.Take a from H1 and /3 from H2. If a < 0,since A is free for fo we

lnf1

88

Polarizaed partition relations

Ch. 4.1

know a 4 fo(P) and so (a,0)4 A,. If P < a,since B is free forf; we know 0 4 fi(a) and so (a,0)4 A,. So in any event (a,0) E A, and hence H I X Hz

C A,. This proves the relation (1) when K is regular. Now suppose that K is singular. Choose an increasing sequence of cardinals ( K , ; u < K ‘ ) with always q,, q1 < K , < K and K = C(K,; u < K ’ ) . By Lemma 4.1.7 there are pairwise disjoint families { A , ; u < K ‘ } and {B,; u < K ’ } with A,, B, E [SIKa,together with a function h : K ’ X K ’ + (0,1} such that A , X B, C Ah(,,,). Define a disjoint partition K ’ X K ’ = A,* U AT, where (u, T)€ A? *h(u,

7)’

i, (i= 0,l) .

Since K ‘ is regular, the theorem applies to K ‘ . In particular

(;J+ (;:

I)’”

K1’ V K

So one of the following three events happens. (1) There are 11,1 2 from [ K ’ ] ~ ‘ such t h a t I i X I2 C A,*. If we put Hl = U(A,; u E I l } , Hz = U ( B , ; T E I ~ } then l H l l = lHzl = K and HI X Hz C A,. ( 2 ) There are T in K ’ and Ii in [ K ’ ] ~ ’ s u c h t h a t I I X (T}CAT.PutH1 =U{A,;uEIl}solH1l=~,andHIXB, C A,. (3) There are u in K ’ and Iz in [K’]~’ such that { u } X Iz C AT. Put HZ = U { B , ; T E Z Z ) so lHzl = K , andA,X H2 C A,. Since IB,I = K , > q i and lA,l = K , > qo, this proves that the relation (1) holds. The proof is complete.

Theorem 4.1.9 (GCH). Let qo, q 1 be cardinals with q:, q1 < K . n e n

Proof. Let any disjoint partition K X K = A0 U A1 be given. Suppose first that K is regular. Define a set mappingfon K as follows:

f(a) = { P < a;(a,0)E A,} , sofhas order at most K . If there is X in [ K ] with ~ Inf[X]I 2 00, since X X nf[X] C A, there is nothing more to show. Otherwise, for all X in [ K ] we have Inf[X]l< -)o, so by Theorem 3.2.5 there must be S in [ K ] ~which is free for$ Choose subsetsH1,Hz o f S w i t h IHiI=K, lHzl=q1 andHz
~

Ch. 4.1

Partitions of K

X

K

a9

We come now. to one of the most difficult theorems involving the symbol

We shall need two lemmas first.

Lemma4.1.10. For eachfinite n, (KK+)

+

K(:

y1 .

Proof. Use induction on n , starting from the trivial relation with n = 0. So make the inductive hypothesis that (1) is true for a particular value of n. Take sets S, T with IS1 = K , IT1 = K + and let a disjoint partition S+x T = A. U A, be given. Suppose there are no setsH1 in [SIK and H2 in [TIK withH, XH2 C A,. By the inductive hypothesis, there must then be sets H1 in [SIK and H2 in [TI"such that H I X H2 C A,. For eachy in T, put Q l b ) = { x ES;( x , y) E Al}. If there i s y in T - H2 for which IH1 n Ql(y)I'= K ,then (H, n Ql(y)) X (H2 U {y}) C A, and the result follows. So suppose W1 n Ql(y)l< K for ally in T - H2. We shall show that a contradiction results. For each cardinal X with h < K ,put T h = { Y E T - H ~ ; I HnQi(y)l=X}. ~ There must be some h for which lThl = K + . For such a A, write H1as a disjoint union, H I = U { H 1 , ; Q! < A+} where always IHIJ = K . If y E Th then IH1 n Ql(y)l = X < A+ and hence there is a ( y ) with a ( y ) < A+ for which Q , ( y ) nH 1 ~ ( = , )8. Since K + is regular, there is Y in [TxlK+such that a ( y ) is constant for y in Y , say with value a. Thus Ql(y) fl H 1 , = 8 for ally in Y. Hence H 1 , X Y C Ao. But W1l, = K and IYI = K+, so this contradicts the assumed property of the partition. This completes the inductive step, and the lemma is proved.

Lemma 4.1.1 1 (GCH). Let K be singular and suppose a partition K X K+=AoUA1 isgiven for which there are no sets H I in [ K ] and ~ H2 in [ K + ] ~ +with H I X H2 C A,. Let A be a regular cardinal with K ' < A < K. Take any set A in [ K I K and subset B of K+. Then there is a family 34 o f at most K Sets from [A]' and a map f : d + g B such that IB - U f [d]I< K and X X f ( X ) C A0 f o r a l l X ind.

Ch. 4.1

Polarized partition relations

90

Proof. Let A and B be given, with A E [ K ] ~and B C K + . Wnte Qo(/3) = { a < K ; (a,0) E Ao}. Put B 1 = B ; !A n Q0(/3)I < A} and B2 = B - B 1 . From lemma 4.1.10, the relation

{a€

holds. Since certainly A X B1 C A0 U A l , if lBll = K + by this relation either there would be /3 in B 1 such that IA n Q(0)I = K (contrary to the definition of B 1 )or else there would be H1 in [AIKand H2 in [B1IK+withH1 X H2 C A1 (contrary to the choice of the partition). So we deduce that lBll < K . Let ( K , ; u < K ’ ) be a sequence of cardinals with K , < K and K = Z(K,; u < K ’ ) , and write A as a disjoint union A = U { A , ; u < K ’ } where always IA,I = K,. h t d = u{[A,]h;U
< !A n Qo(P)I = Z(M, Hence B2 = U f [ d 1. Thus B

-

n Qo(P)I; u < K ’ }


U f [ d ]= B 1 , and the result follows.

We are now set for the major theorem. Theorem 4.1.12 (GCH). Forall

(:+)+ +

(K K

K,

K V K’)”’

K’VK

Proof. Let S be any set with IS1 = K + , and let a partition S X S = A0 U A1 be given. Assume for all HI, H2 with H1 E [SIK,H2 E [SIK+that HI X H2 A, and H2 X H 1 A , , so we seek H1, H2 in [SIKsuch that H 1 X H2 C A,. Identify S with K’. For a, /3 with a,/3 < K + define

po(a) = {/3< Then if A E

[K’]~

K+;

(a,0) E Ao}; &(O) = { a < K + ; (a,0) E A,}

.

and B C K + , since

A X ( B - U{Po(a); € A } ) C A1 it follows that IB - U { P o ( a ) ;a € A } / < K . Similarly, if B E [ K ’ ] ~ and A C K + then IA - U { Q o ( / 3 ) ; 0 E B ) I < ~ .

Ch. 4.1

Partitions of

K

X

91

K

Suppose first that K is regular. We shall define a ramification system 3 on of height K so that reading along any branch of the ramification tree gives a chain of elements ao,Po, a l ,PI, ... such that always (a,,0), E A,,. The idea 'of the construction is roughly as follows (with the notation conventions of Chapter 2, 52). Suppose we have already constructed the ramification system up to level u, so for a sequence v in N nSEQ, we know S(v). Choose A from [S(v)IK.By the remark above (with B replaced by S(v)), at most K elements 0 of S(v) fail to have a companion a in A such that (a,0) E Po. Discard these elements by putting them in F(v), split up the remaining elements of S ( v ) according to their companion a in A and push them up to the next level. Do this alternately on the first component and the second component of the pairs (a,0). Formally, then, take u with u < K and a sequence v from SEQ,. Suppose S(v) has already been defined. If IS(v)l< K , put F ( v ) = S(v) and n ( v ) = 0. If IS(v)l = K+, put n(v) = K . Choose R ( v ) from [S(v)]"and for p in N n SEQu+l with p To = v let ~ ( c l )range over the elements ofR(v). If u is even, put K+

F(v) = R ( , )

u (S(v) - U { P o ( r ) ;Y ER(V)})

and for v in N n SEQ,+I with p To = v put S(c) = (S(\*) nPo(Y(P)>>- F(v)

If u is odd, put F(v) = R ( v ) U (S(V) - u { Q o ( ~ )Y; E R ( v ) I )

and for v in N nSEQ,,, with p r u = v put = (S(V)n Qo (Y(P))> - F(v)

.

Thus in either case it follows that S ( v ) - F ( v ) = U { S ( p ) ; p E N n S E Q , + 1 and p r u = v } . This completes the definition of the ramification system '8 . By definition P ( v ) l = K , so from the remark above it follows that IF(v)I
.

r

r

92

Ch. 4.1

Polarized partition relations

u, r < K ; we want to show (a,,0), E A,. Suppose u < r. Then-

p,

= r(vr 2r

+ 2) E R ~ Vr 27 + I ) c

r 2 r t 1)

c s (r~ 2u t I ) ,

s(Vr20+ I ) c P , ( Y ( v ~ ~ u + I))=P,(o~,), and so (a,,0), E A,. Similarly if r < u we find a, E Qo(&) and again (a,,&> E A,. This proves the theorem if K is regular.

Now suppose that K is singular. Choose an increasing sequence of regular cardinals ( K , ; u < K ' ) such that K ' < K , < K and K = Z(K,; u < K ' ) . We shall define a ramification system 93 on K + of height K ' (so the Ramification Lemma will apply) in such a way that reading along any branch in the tree gives a sequence of sets&, Bo, A 1, B1, ... such that M a [ ,IB,I > K , and alwaysA,X B, C A,. The construction is similar to that used in the case above, except that Lemma 4.1.1 1 is applied to provide the appropriate splitting up of the set S(v)To define 8 formally, suppose S(v) is given for some v in N f SEQ, l where u < K ' . If IS(v)l < K , put F(v) = S(v) and n(v) = 0. If IS(v)l = K + put n(v) = K . Choose R(v) from [S(v)IK. If u is even, apply Lemma 4.1.1 1 with X = K,, A =R(v) and B = S(v) to obtain a subfamily d ( v ) = {A,(v); a < K } of [R(v)IKUand a mapf, such that A,(v) X fv(A,(v)) C A, and IS(v)- UfV[d(v)]l < K . Put F(v) = R(v)

" (S(v)

-

U f V V (v)l)

9

and for p in N n SEQo+l with p To = v define

S(N = fv(Ap(u)(VN - F(v) If u is odd, use the obvious modification to Lemma 4.1.1 1 to obtain a family d ( v ) and a mapfv as above, only that this timefV(A,(v)) X A,(v) C A,. , Define F(v) and S(p) as above. This completes the definition of the Ramification s y s t e m s . Then, as in the regular case, there is a sequence v in N fl SEQKlsuch that S ( v ) f @ . PutA. ( ~ r 2 u ) a n d B , = A ~ ( ~ , + ~ ) ( l), ~ rthen 2~+ H1 = U{A,; u,< K ' } and H2 = U{B,; a < K ' ) . It follows that IH1I = lH2l = K . And further, H I X H2 C A,. For take u, r with u, r S K ' . We show A , X B, C A,. Suppose u < r. Then

B, = A v ( 2 7 + 1 ) (r~27 + 1) c R ( V r 2 r t I )

c S(V r 27 + 1) ,

Ch. 4.1

Partitions of

K

93

X K

C A,. This completes the proof for singular K , and concludes the proof of Theorem 4.1.12. We shall prove one more negative relation, and then we shall survey the results obtained so far.

Theorem 4.1.13 (GCH).Z ~ K > ’ No then

and if K ‘ = No then

Proof. We shall construct a partition K + X K + = A0 U A 1 which demonstrates the negative relation. As before, to ensure that there are not H1 in [ K + ]and ~ H 2 in [ K + ] ~with + H1 X Hz L A,, for each H1 in [ K + ] ~there must be y such that for all p with p > y there is some chosen element x@) in H1 such that (x@), p ) E A l . In order not to make A 1 too large, we want that any particular element in K + be not chosen as x(p) for too many p . Since WII = K , when choosing x@) we can safely avoid only fewer than K of the earlier x(u). The formal construction is as follows. Let [ K + ]=~ {A,; a < K ’ } . For each /3 with 0 < /3 < K’, since 1 < 1/31 < K we may write /3 = {/3(u); u < K } and { A , ; a < 0) = {Bpv;u < K } . Use transfinite induction on /3 to choose elements x(0, u ) (where /3 < K + and u < K ) such that x(p, u) E Bpv - {x(/3(p), u); p < u and u < u }

.

Since I { ( p , u);p < u and u < u}l < K this is certainly possible. Define a partition K + X K + = A. U A, as follows:

* 3 V < K (a=X(/3, (0, /3) E A0 *(a,0) 4 A1 . (a,0) E A1

V))

,

Then if Hl E [K’]~,H 2 E [ K + ] ~ certainly +, HI = A , for some a and there is /3 in H 2 with a < /3. Thus H1 = Bpv for some u with u < K . Then x(p, u) E Bpv = H1 and (x@,u), 13) E A1 so that H1 X Hz P Ao. Also for any 0, we have I {a;(a,0)E A111 = I {x(/3, u); u < K } I < K and so if H1 E [ K + ] ~ ,0 E K + we have H1X (0) C A,.

94

Polarized partition relations

Ch. 4.1

Note that the choice of x@, v) ensures p


K

and a = g@)andx(a, u ) =x@,v) * v < u .

Take any infinite set B from that

and suppose there is p with p

(1)

< K such

P>sup{P p for all 0in B. Then if we take B1, 02 in B with 0,< l j 2 , we have 01 = /3z(j.~) for some p, and p < p =G ~ ( 0 2 )so that by (l), 4 6 2 ) < v(P1). Since B is infinite this leads to an infinite descending chain of ordinals, which is impossible. Consequently v(0) < p for some 0 in B. Hence Ql(B) C {x@, v); v < p and 0E B } so that lQl (B) 1 < K as claimed. Suppose K ' > NO.Take any B in [ K ' ] ~ ' . Certainly there is p with p < K a n d p > s u p { p ; 3 a , P E B ( a = P ( ~ ) ) } , s o t h a tl Q l ( B ) I < ~ . B u t i f A X B C A , then A C Ql(B),'so there are not A in [ K ' ] ~ , B in [KO]'O with A X B C A , . Suppose K ' = No,and let ( K , ; n < w ) be a strictly increasing sequence of cardinals with K = X ( K , ;n < w). Take any set B in [K+]", and form the partition [BI2 = U{ r,; n < w } where, in the notation above, for a,/3 in B with

a<0,

{a,01 E r,

* 3 cc < K , (a= ~(CC)) .

there is an infinite subset B* of B such that By the relation tsz + [B*I2 C r, for some n. Thus K,+1

>K ,

2 SUP

3 a, E B *

(0

= b@))}

3

so that 1% @*)I < K and hence lQl (B)I < K . Thus there are not A in [ K + ] ~ , such that A X B C A , . B i n [K'] Thus the partition { A , , A , } provides an example which proves the theorem. We shall now summarize the results concerning the symbol

where 0 < q0, ..., v3 =G K , assuming GCH throughout. It follows from Theorem 4.1.1 2 that the relation is true if q0, ..., 773 < K. If any three of 770, ...,773 are equal to K , the relation is false (from Theorem 4.1.4).

Partitions of

Ch. 4.1

K

X

95

K

By Theorem 4.1.9 the relation

is true if qi,773 < K . If K is a successor cardinal, say K = h’, a stronger relation would be

(;)+ (“A

K)l.l

X

;

however it follows from Theorem 4.1.6 that this relation is false. If we now turn to the relation

by Theorem 4.1.8 this holds if 77: ,775 < K . If K is a successor cardinal, K = h’, we can ask for stronger relations of the form

This relation holds if 773 = 1 (Theorem (4.1.3) and if h is regular it is false with 773 2 2 (Theorem 4.1 S).When h is singular, it follows from Theorem 4.1.13 that (1) is false with 773 > H2 or 773 2 Ho, depending on whether X has cofinality Ho or greater. The situation when 1 < 773 < H2 or 1 < 773 < Ho (respectively) is an open problem. Finally, we consider the relation

where q l , 772, 773 < K . The relation is true if either 77: < K or 775 < K (Theorem 4.1.9), or if q:, 77: < K (Theorem 4.1.8). This leaves the possibility K = A’ and relations of the form

to consider. If q1 is finite, the relation is true (Lemma 4.1.10). If A‘

> No the

96

Polarized partition relations

Ch. 4.2

relation is false for q1 infinite (Theorem 4.1.13). When A' = No, the relation is true for q1 = No (from Theorem 4.2.4 of the next section) and is false for q 1 2 N2 (Theorem 4.1.13). For A = No and q1 = N1 it is false (from Theorem 4.1.6 with K = No). The case A > A' = No, 771 = N1 is an unsolved problem.

32. Partitions of

K

xK

+

We shall now seek results concerning the relation

As in the last section, we shall soon run against unsolved problems. We start with the following easy lemma. (Compare with Lemma 4.1 .lo.) Lemma 4.2.1. For each finite n,

Proof. By induction on n. The case n = 0 is trivial, so make the inductive assumption that the relation (1) is true for a particular value of n. Take a disjoint partition K

x K+=A,UA,

and suppose that there are no sets H 1 in [ K ] ~and H2 in [ K + ] ~ + such that H1 X H2 C AO. By the inductive hypothesis, there must then be setsH1 in [ K ] " and H2 in [ K + ] ~ + for which H1X H2 C A,. For each a! in K , putPl(a!) = { p < K + ; (a!,/3)E A1}. Then since IK -H1l = K and (K

-H1)X (H2

-

u{Pl(Q);aEK- H 1 ) ) L A o 9

by the choice of the partition W2 - U{Pl(a!);a! E K - H1}l < K + . Thus l H 2 n U{P,(a!); a! E K - H 1 }I = K + and so there must be a0 in K - H1such that !HZnPl(ao)l = K + . But

(Hi U {ao})X

(H2

nPi(ao)) C A, ,

and so since [H1U {a!o}l = n + 1, the induction step is complete and the lemma proved.

Partitions of

Ch. 4.2

K

X

K?

91

Theorem 4.2.2 (GCH). For any infinite K , i f q < K, then (KK+)

+. (K:

K:y

.

Proof. The case K = No is immediate from Lemma 4.2.1, so suppose K > No. Define A as follows: if K is regular, take A = q , otherwise let A be the larger of (K')', 7'. In either case, q < A < K and A' # K ' . Take any partition K X K + = A, U A,. To prove the theorem, we must either find sets H1 in [ K ]and ~ H2 in [K']~' with H1 X Hz C A,, or else find setsH1 in [K]' and H2 in [K']~' such that H1 X H2 C A,. Put T = { P < K ' ; lQ1 @)I2 A}, where e l @ ) = { (Y < K ;(a,P ) E A l 1. Suppose Ifl = K', and consider the family { Ql(P); € T}. By Theorem 3.2.3, theremust be T ' i n [TIK'such that In{Q,@>;PET'}I>X2q.Since n{Ql@);P E T'} X T' C A,, there is nothing more to be shown. So consider the case when Ifl < K'. Then IK' - TI = K'. Now A' < K , so there is a pairwise disjoint decomposition K = u{S,; p < A'} where always S I l, = K . Since lQl (P)l < A for in K' - T, for all such there is p(P) for which S,co,fl Ql(P)=O. There must be T1 in'[TIK' such that p(P) is constant for P in TI, say with value p . Thus S, n e l @ ) = 8 for all P in T1 so that S, X T1 L Ao. Since S Il, = K and IT11 = K ' , this proves the theorem. Theorem 4.2.3 (GCH). Z ~ is K singular and q < K, then

Proof. We may suppose that K ' < q < K and that q is regular. Choose an increasing sequence ( K , ; u < K ' ) of regular cardinals such that q < K , < K and K = Z(K,; u< K ' ) . Take any disjoint partition K

x

K+=A,uA,,

and suppose there are no sets H1 in [ K ]and ~ H2 in [K']' such that H1 X H2 C Al. We must find H1 in [ K ]and ~ H2 in [ K ' ] ~with H1 X H2 C A,. We shall start by defining a ramification system % on K' of height K' in such a way that associated with each branch in the ramification tree there are sets A , in [ K ] ~ and " B, in [K']~, such that A , X B7 C A. whenever u < T. The construction is similar to that used in the proof of Theorem 4.1.12. To define $, suppose S(v) is given for some v in N n SEQ, where (7 < K'. If IS(v)l< K , put F(v) = S(v) and n ( v ) = 0. If IS(v)l = K' put n(v) = K . Choose

98

Polarized partion relations

Ch. 4.2

B(v) from [S(v)]"". Apply Lemma 4.1.1 1 with h = K,, A = K and B = S(v) -B(v) to obtain a family SQ (v) = (A,(v); a < K ) of sets from [K]", and a mapfV such that IS(v) - Ufv[SQ (v)]l G K and A,(v) X fv(A,(v)) C A,. Put F(v) = N v )

u (S(v) - Ufv [SQ(v)l)

9

and for p in N nSEQa+l with p To = v define S(P) 'fv(Ap(u)(V)) - F(v) . The Ramification Lemma (Theorem 2.2.3(ii)) applies t o % , so we may choose a sequence v from N n SEQKlfor which S(v) f 0.Put A,=A,,(,)(vr and B, = B(v To). Then IA,I = IB,I = K,. Take u, T with u < T < K'. Then B, = B( v r7)

c S(v r 7)c S(v r

and hence U < T < K'

*A,,

CJ

+ 1)

0)

c fvru(Av(u)(v r0))= f v r .(A

x B, c A, .

(1)

PutA = U { A , ; o < ~ ' } a n d B = U { B , ; U < K ' } S OL4l = IBI = K . The original partition restricts to give a partition o f A X B. Hence by Lemma 4.1.7 there are sets C, in [A]", and D , in [B]"" and a function h : K ' X K' -+ { 0, 1) such that 0,T

< K' * c, x D, c Ah(,,,) .

For i with i = 0, 1 put AT = { ( 0 , E ~) K' X A,* U A:. By Theorem 4.1.8, the relation

K';

h(u, T ) = i},

SO

that K ' X

K'

=

holds. By applying it to the present partition of K ' X K', there are three cases to consider. Gzse 1. There are H in [K']"' and 7 with T < K ' such that h(u, T ) = 1 for all u i n H . P u t C = U { C , ; u E H } ; then ICI=Ka n d C X D,CAl.Since IDTI = K, > 17, this contradicts the original choice of the partition of K X K ' . Case 2. There are H in [ K ' ] ~ 'and u with u < K ' such that h(u, 7)= 1 for all T in H. Put D = U{D,; T EH}; then ID1 = K and C, X D C_ A1. Choose a in C,. Then there is p with p < K ' such that a € A , . It follows from (1) that we must have D C U(B,; r S p } . However, this yields the contradiction K

= ID1 G Z(1B71; r G p ) = C(K,; 7 G p )


Case 3. This case must prevail. There are K1, K 2 in [K']"' such that

Ch. 4.2

Partitions of

K

x

99

K+

h(o, 7)= 0 whenever o E K 1 , 7 E K 2 . P u t H l = U{C,; o E K l ) a n d H 2 = U{D,; 7 E K 2 ) . Then H1 E [ K ] ~ H2 , E [ K + ] ~and H1 X H2 C Z A,. This completes the proof of the theorem. In the case of cardinals K with K ’ = Ho,there is a special method available which enables us to prove the following two theorems. Theorem 4.2.4 (GCH). Z ~ K=’ No then

Proof. Take a partition K X K + = A0 U A, and suppose there are no sets H1 in [ K ] andHz ~ in K + +’ such that H I X H2 C A,. We must find setsHl in [ K ] andHz in [K’]”’’ sLch thatH1 X H2 C A,. Write K = Z(K,; n < w ) where K , < K , < K when m < n < w . We shall define setsA, from [ K ] ~ ”and elementsy, from K + such that always A , X {y,) C Al. By Theorem 4.2.2 and Lemma 4.1.10, the following two relations hold: (KK+)

qK:y1+);(

+(+ ; ;)”’.

,

Use these relations to choose inductively A,, B,, C,, y , so that A0 E [ K ] ~ ’ , Bo E [ K + ] ~ withAo + X Bo C_ A, and Co E [ K ] ~ yo , EBo with COX {yo) CAI. B, in withA,XB,CA, For n with n > 0, we wantA, in [Cn-l]Kn, and C, in [ C n - l ] Ky, , in B, - {y,; m < n ) with C, X { y n )C A1. Note that if m G n then A , X {y,) C A m X B, LA, X B , C A1

> n then A m X b n ) c Cm-1 X

9

and if m

c Cn X b n ) C A1

{ ~ n )

*

P u t H l = U { A , ; n < o ) a n d & = { y n ; n< a )so IHII = K , lH~l= No and H1 X H2 C Al. Thus the theorem is proved. Theorem 4.2.5 (GCH). I ~ = KNo ’ then

~

Polarized partition relations

100

Ch. 4.2

Proof. The proof is very similar to that of the preceeding theorem, making use of Theorem 4.2.3 rather than Lemma 4.1.10. Take a partition K X K + = A, U A1 and suppose this time that there are no sets H1 in [ K ] and ~ H2 in [ K + ] ~with H 1 X H2 C Ao; we must find such sets with H I X H2 C A l . Again write K = C(K,;n < o)where K , < K , < K if m < n < w . We shall define sets A , from [ K ] ~and " D , from [ K + ] ~ , such that always A,X DnCA,. If we then put H 1 = U{A,; n < w } and H2 = U{D,;n < w } it follows that W1l = lH2l = K and H I X H2 C A , , so this would suffice to prove the theorem. We have the following relations at our disposal (from Theorems 4.2.2 and 4.2.3):

Use these relations to choose inductively sets A,, B,, C,z,D, as follows. Take A. from [ K ] ~ O Bo , from [ K + ] ~ +such that A. X Bo C A1 and COfrom [ K I K , Do from [BolK0such that COX Do C A , . For n with n > 0, choose A, from [Cn-,lKnand B, from [Bn-lIK+with A, X B, C A1 and C, from [Cn-1IK9 D, from [B,IKn with C, X D, C A , . It follows that if m < n then

A, X D, CA,R X B, CA, X B , C A,

>

and i f m > n then A, X D , C C,-

-1

x Dn C

X D, C A I .

Thus A, X D , C A1 for all m, n and the theorem is proved. This completes the list of special results available. We shall now summarize (assuming GCH) the results concerning the relation

By Theorem 4.2.2, the relation is true if either qo < K or q1 < K , so we need consider only the case qo = q1 = K . It follows from Theorem 4.1.6 that the strongest possible relation

(") +( +);

, " .

is always false.

Partitions of K x

Ch. 4.2

101

K+

For the relation

there are two cases to consider. If K ' > No then the relation is true if 773 is finite (Lemma 4.1 .lo) and is otherwise false (from Theorem 4.1.13). If K' = No, this relation holds when q3 < No (Theorem 4.2.4) and fails when 773 2 N2 (from Theorem 4.1.13). The situation when 773 = H1 is an unsolved problem. There are many unsolved problems concerning the relation

<

where q2, q3 K . From Theorem 4.2.5, the best possible such relation with q2 = v3 = K is true if K ' = No. If K is singular with K ' No then (1) is true with q2 = K , q3 < K (Theorem 4.2.3), and it is not known if this can be strengthened to have q3 = K . If K is regular and uncountable, the only positive

>

results known are those which follow trivially from partitions of K X X where A < K . Some of the simplest unsolved problems are (see [38, Problem 121):

Concerning the first two of these relations, Prikry [78] has shown that the relation

is consistent with the axioms of set theory that we have been using, so there is no hope of being able to prove the positive relations. One can now consider the general relation

where X > K'. It turns out that the results we have already obtained are suf-

102

Polarized partition relations

Ch. 4.2

ficient for a full discussion to be given; the only unsolved problems that arise are equivalent to unsolved problems we have already mentioned. We shall give the discussion for the special case

which will have applications in Chapter 6, but otherwise the reader is referred to Erdos, Hajnal and Rado [38, pp 188-1931 for details. There are two lemmas first.

Lemma 4.2.6. Let

K

and A be infinite with 2K < A’. Then

Proof. Take any disjoint partition K X A = A0 U A,. For each subset A Of K , let B(A) = { 0< A; {a < K ;(a,0) E A,} = A } , SO A = U{B(A);A E$ K } : Since 2K < A’, there is A0 in!@K with lB(A0)I ,= A. Now A0 X B(A0) C A0 and ( K - A o ) X B(Ao) C A,; since either Hal= K or IK - Aol = K the relation (1) holds.

Lemma 4.2.7. Let

and

(J+(’;

K

and A be infinite with 2K < X Then the two relations

$”

are equivalent.

Proof. if A is regular, this is trivial, so suppose A is singular. Let (A,; u < A’) be an increasing sequence of regular cardinals with always 2K < A, < A for which A = Z(A& u < A’). Suppose that the relation (1) holds. Take any partition K X A’ = ro U rl. Take any pairwise disjoint decomposition A = U{B,; u < A’} where always IB,I = A,. Consider the partition K X A = A0 U A1 where for (a,0) from K X A and i = 0,l

(a,0) E Ai * 0 EB, and (a,a>E ri.

Partitions of K x

Ch. 4.3

103

K+

By assumption there are HO in [ K ] ~and H1 in [A]' such that Ho X H1 C Ai where e i t h e r i = 0 or i = 1. Put& = { u < A';B,, n H 1 Z 8); then IH;I = A ' and HO X H i C ri.Thus the relation (2) holds. Now suppose that (2) is in force. Let any partition K X A = A, U A, be given. For each P, put Qo(P) = { Q < K ; (a,0) E Ao).For any u with u < A', I(Qo(P);

P < A,II G 2K < Xu .

Since A,, is regular, there is C,, in [A,,]'" such that QoGO) is constant for P in C,,, say Q,@) = Q(u). Define a disjoint partition K X A' = roU rl by, if (a,u) E K X A' then

( a , u ) E r O* ~ E Q ( U ) . By the relation (2), there are HO in [ K ] ~ ,H1 in [A']' and i in {0,1) such that H o X H 1 c r i . P u t H 2 = U { C u ; a E H 1 ) . T h e n H 2 E [Al'anditiseasytosee that Ho X H2 C Ai.Thus (1) holds.

rheorem 4.2.8 (GCH). For an infinite cardinal K , p u t Z ( K )= lhen the relation

{ K , K', K ' ,

(K')'}.

holds ifand only $ Z ( K )f Z@) l = 8.

Proof. We may assume throughout the proof that K G A. For brevity, write R ( K ,A) to indicate that the relation (1) holds for the cardinals K and A. Suppose first that Z ( K )n = 8, so in fact K + < A. IfK+ < A', by Lemma 4.2.6 R ( K ,A) holds. If A' < K' we must have (A')' < K . Since either (A')'< K ' or (K')' < A' and both K ' and A' are regular, by Lemma 4.2.6 again R(K',A') holds. Since (A')' < K , by Lemma 4.2.7 R ( K ,A') holds, and since K' < A again by Lemma 4.2.7 R ( K ,A) is true. Thus in all cases, if Z ( K )n Z@)= 8 then R(K,A) holds. Now suppose Z ( K )n Z(A) # 8. From Theorems 4.1.4 and 4.1.6, R ( K ,A) is false if K = A or K' = A, so we suppose henceforth that K' < A. This means that A must be singular and either A' E Z ( K )or (A')' E Z(K).If A' = K (or equivalently (A')' = K ' ) since R ( K ,K ) is false, by Lemma 4.2.7 also R ( K ,A) is false. If A' = K ' , since R ( K ,K + ) is false, a l s o R ( ~ A) , is false. If(A')' = K , since R((h')', A') is false again Lemma 4.2.7 shows that R ( K ,A) is false. The remaining possibilities are that (A')' = K ' , (A')' = (K')', A' = K' or A' = (K')'. In each case by either Theorem 4.1.4 or Theorem 4.1.6 the relation R ( K ' ,A') is false.

104

Polarized partition relations

Ch. 4.2

If in fact (A')' < K , it follows that R ( K ,A') is false and then that R ( K ,A) is false, by two applications of Lemma 4.2.7. On the other hand, if K < (A')' this means either K < K ' , K < (K')' or K < (K')''; in any event K is regular and from R ( K ' A') , being false it follows by Lemma 4.2.7 that R ( K ,A) is false. Thus in all possible cases, we conclude that R ( K ,A) is false. The theorem is proved. We shall end this section with a few brief comments on the relation

where y > 2. Few results have appeared in the literature, particularly for infinite y. One can obtain, in trivial way, results from the ordinary partition relation, but this surely does not lead to best possible results.

Theorem 4.2.9. suppose that each k. Then

K

+

(??k;k

< y12 and that ??lk + ??2k < q k for

Proof. Given a partition A = { Ak;k < y} of K X K , define a partition r= { r k ;k < y} of [ K ] ~as follows: i f a , 0 E K with a < 0, then

{a,0 ) E r k *(a,fl)E A k . By the relation K + (?& k < T ) ~ ,there are k with k < 7 and H in [ K I Q k with [HI2 C A k . Take H 1 from [ H I Q l k and H2 from such that H1 < H 2 . Then H1X H2 L A k . The positive results in $ 1 are nearly all proved by methods that are special to the case y = 2, and don't generalize. The methods of $2, on the other hand, can be extended to finite -/ (see [95]).In particular if K ' = No then (assuming GCH) for any finite n

This is the best possible, for if K' = No then trivially

Ch. 4.3

Larger polarized relations

105

Hajnal has proved (see [ 2 5 ] )assuming GCH that

This is the strongest relation of this type which is provable, for a generalization of the method of Prikry [78] shows that the relations

are consistent with GCH.

53. Larger polarized relations In this section, we shall consider a few further results concerning the polarized partition symbol. We first note two forms of the Stepping-up Lemma (Theorem 2.3.1) appropriate for the polarized relation.

,,

Theorem 4.3.1. Let m, n ..., nm 2 1. Let A be a cardinal such that

where the v i k are infinite cardinals with r);k > q 2 k , ..., q m k (for each k). SUPpose K is an infinite cardinal with A < K' and lyllOl< K ' whenever u < A. Then

Proof. Suppose all the conditions stated in the theorem are satisfied, and take any disjoint partition A = { A k ; k < y } of [ K I ~ ~ + x' [ K I Q X ... x [ K I ~ De~ . fine a ramification system 3 on K of height A as follows. Take u with u < A, and suppose S(v) has already been defined for each v in N n SEQ,. If S(v) = 0, put F(v)= 0 and n ( v ) = 0. Otherwise, choose x ( v ) in S(v) and put F(v) = {x(v)}.

Ch. 4.3

Polarized partition relations

106

Place C ( v ) = U { F( v r r ) ;7 < a}. Define a partition r ( v ) of S( v ) - F ( v ) by requiring: y

z(mod l-(v)) * V a l E [C(v)]"' ... V a , E [G(v)lnrn ((al U { y } , a2, ..., a m ) = ( a l U {z}, az, ..., a,) (mod A)) .

Put n ( v ) = Ir(v)l and let S(P)f o r p in N n SEQo+l with p To = v range over the classes of I'(v). This defines !)I. If u < A, put A, = lyle where '6 = l ~ l " ~ * * * " ~ . Then In(v rr)l < A, whenever v E N fl SEQ, and r < u. Moreover, if u < X then A,'u' < K'. Hence the Ramification Lemma (Theorem 2.2.3) applies t o # , and we may choose a sequence v in N n SEQh for which S ( v ) f 8. Then always S(v To) f 8 so F( v Tu) = {x(v To)}. Choose x h from S(v) and for u with u < X put xu = x( v r u ) . Then x, # x u when r < u < A. If 7 < O < X then xu € F ( V To) c S(v To) c S(V + l), and x h E S ( v ) C S(v + 1) so both xu, x h E S( v r r + 1). Thus both xu and x h are in the same class of r(v r r ) , and so if ai E [ { x p ;p < 7}] (where i = 1, ...,rn) then

r7

rr

"'

(al U {xu}, a2, ..., a m ) = (al U

{xh},

az, ..., a,) (mod A )

.

(1)

Put X = {x,; r < A}, so IXI = A, and consider the partition

[X]"'X ... X [ X I h = u { A ; ; k < y } where ( a l , ..., a,) E A;

* (al U {xh}, a2, ..., a,)

E Ak

.

By the partition relation for A, there are k with k < y and a sequence H1, ..., H , where Hi E [XIqiksuch that [Hl]"' X ... X [H,]", C_ A ; . Since q ; k > ' 7 2 k , ..., qml, there is H in [H1]'lk such that if x, E H and x, EH2 U ... UH, then r < u. Hence by (I), [Wn1+lx [H21n2X ...x [H,InrnL &. This completes the proof.

Theorem 4.3.2. Let m, n l , ..., n , 2 1. Let X be a cardinal such that

where the

qjk

are infinite cardinals with A'

> q l k , ..., q m k (for each k ) . sup-

Ch. 4.3

Larger polarized relations

qrnk

k
Then using Theorem 4.3.1 gives

($)

+

+

2,2

(", )

7

.

107

108

Ch. 4.3

Polarized partition relations

It is not clear how close these results are to best possible. There is the following rather trivial negative- relation. Theorem 4.3.3. Suppose K is a strong limit cardinal (that is 2' h < K ) . Then


whenever

Proof. We shall prove the second relation; the first is similar. For distinct elementsf, g in K2,as before let S(f, g) be the least a wheref(a)fg(a). Define a disjoint partition ["2]' X ["2]' = A0 U A1 by Take distinctfl, gl from "2, so S(f1, gl) < K . Then iffo, go are such that S ( f o , go) G & ( f i t g l ) thenfo andgo differ no later than at S(f1, gl). Hence if H i s a subset of K 2 with the property X {{fl, gl}} C_ A0 then IHG21ay+11 where a = S(fi, gl), so [HI< K . Similarly, if {{fo, go}} X [HI2C_ A1 then [HI< K , and the proof is complete.

[a'

From Theorem 4.1.12 it follows that

so applying Theorem 4.3.2 shows

In fact it is not known whether the better relation

holds (see [24, Problem 281). However, this particular relation, if true, cannot be improved to apply to partitions into 3 classes, as the following theorem of Sierpinksi [88] .shows.

Ch. 4.3

Larger polarized relations

109

Theorem 4.3.4 (GCH). For all infinite K ,

Proof. (This is the second case of an inductive construction of which the first case is Theorem 4.1.4. See Kuratowski [63].)For each ordinal a where a < K + we have la + 11< K and so we may write (0; 0 < a ) = { a(u);v < K ) . Form a partition K + X K + X K + = A0 U A, U A, as follows. Given a,0, y if a 2 0,y and /3 = a&), y = a(v)

(a,p,y)EAo ifpEA2 i f u < p ;

This partition has the right properties. For suppose there are a subset HI of and members fl, y of K + with H l X (0) X {y) C A,. Then a < P, y for all a in H 1 . Suppose P2 y, and y = P(v). Then for any a in H I , we must have for ) some p with p < v, so that lHll< IvI < K . Likewise, if 0< y and a = /I& /3 = y(v) then for any a in H 1 , we have a = y(p) where p < v, so again IH1I < K . Similarly, if {a}X H z X {y) C A , or {a)X I S ) X H 3 C A2, then IHzl,lH31<~ Thus the theorem is proved. K+

As a final remark, we comment that Hajnal [55]has shown that K is an uncountable two-valued measurable cardinal, then (in the obvious notation)

for each y where y < K . A similar argument shows

for all finite n and y. (This result was first proved by Calvin.)

CHAPTER 5

THEORY OF INFINITE GRAPHS

6 1. The chromatic number Agraph G is a pair (G, E ) where E C_ [GI2. The elements of G are the vertices of G , the elements of E the edges of G.Two vertices x , y are adjacent, or joined by an edge, if { x , y } E E. A subset S of G no two elements of which are joined by an edge is called an independent set. A subgraph of the graph G is a graph (H, D ) where H C_ G and D C E. The subgraph of G spanned by a A circuit in the graph G is a finite subset H of G is the graph (H, E fl [HI2>. set ofverticesxl, ..., x , such that { X I , x z } , ..., {x,-1, x,}, { x n , X I } EE. A graph of the form (G, [GI2) in which every two vertices are joined is called a complete graph, and will be denoted by K,, where 77 = [GI. A graph of the form (G U H, E ) where G and H a r e disjoint and E = { { x , y } ; x E G and y EH} is called a complete bipartite graph, and will be denoted by K,,e where 77 = IGI, e = W I . With every graph G = (G, E ) there is associated a partition of [GI2 into the two classes E and [GI2 - E, so the study of graphs amounts to the investigation of partitions of pairs into two classes. However, the language of graph theory often provides a more natural way of expressing the results. In particular, the partition symbol K -+ (qo,q1)2 asserts that for any graph G on K vertices, if G contains no independent set of power 70, then G must contain a Kql subgraph. In this section, we shall replace the condition “G contains no independent set of power v0” by weaker conditions of the form “G has large colouring number” or “G has large chromatic number”, and ask what effect this has on the subgraphs of G . Before giving the definitions of colouring number and chromatic number, it is convenient to introduce the following notation. Given a graph G = (G, E ) , for a vertex x of G and subset A of G, let G(x, A ) be the set of vertices in A adjacent in G to x , that is G(x,A)= {YEA: { x , y } E E } .

Ch. 5.1

The chromatic number

111

Definition 5.1 .l.An qcolouring of a graph G = (G, E ) is a well ordering < of G such that IC(x, { y € G ; y < x})l< g for each vertex x of C . The colouring number Col(C) of G is the least cardinal g such that G has an g-colouring. Definition 5.1.2. The chromatic number Chr(G) of the graph G is the least cardinal number g such that G can be decomposed into g independent sets. Thus the chromatic number gives the smallest number of colours needed to paint the vertices of G so that no two vertices of the same colour are joined by an edge. The importance of the chromatic number in connection with the four colour theorem for finite graphs is well known. However, we shall be concerned only with infinite graphs. We show in Lemma 5.1.3 that the colouring number and the chromatic number are related by the inequality Chr(G) < Col(G). That there can be no reverse inequality is shown by the example of the complete bipartite graph K,,r7 which has chromatic number 2, yet colouring number g + 1.

Lemma 5.1.3. For every graph G , Chr(G) < Col(G). Proof. Suppose < is ,an Tcolouring of G = (G, E). Define subsets S, of G for v with v < g by induction on the ordering< as follows: x€S, o v i s l e a s t such that V y < x ( { x , y } E E * y ~ S u ) .

Since < is an g-colouring this is a good definition. Then G = U{S,; v < g } and each S, is independent, so Chr(G) < g.

As a final introductory remark, let us note that if the graph G = (G, E ) has colouring number g, then in fact G has an g-colouring of the smallest possible order type, namely IGl.

Theorem 5.1.4. Let G be a graph on K vertices, and suppose Col(C) = g. 7Ren G has an qcolouring< such that tp(G,< ) = K . The theorem is trivial if g 2 K , and when g < K it is a consequence of the following lemma.

Lemma 5.1.5. Let G be a graph on K vertices and suppose G has an qcolouring ) =K. where g < K . Then G has an g-colouring< such that tp(G, i Proof. Let 4 be an g-colouring of G , and suppose tp(G, 4 ) = t (so

It1 = K ) .

112

Theory of infinite graphs

Ch. 5.1

Let (x+ y < [) be the sequence of elements of G, in their +-ordering. We shall in fact construct a colouring < o f G , with tp(G,<) = K , such that for eachy in G G(y, tx EG;x
(1)

so certainly< will be an 9-colouring. For eachy in G, define the set PRb) as follows: inductively put

PRO(Y)={y};PR,+i(y)= U{C(x,

{W;W

+ x});x EPR,b)},

and then

PR(y) = U{ PR,(y); n < o}. Take any mapf : K

+

[ from K onto g. Define sets G, for a with a < K by

G, = PR(Xf(,)) - U{PR(xf@));P < 4.

Then G is a disjoint union, G = U{ G,; a < K } . Define a well ordering < on G as follows: for x, y in G, if x E G, and y E Gp then

x< y

0

a < or (a= and x 4 y ) .

If x 4 y , y E Gp and x is adjacent t o y in G then both x, y are in PR(xf@)) so either x E Go or x E G, where a< 0;either way x < y. Suppose x < y and that x,y are adjacent in G . If both x, y are in the same G,, surely x 4 y. So suppose x E G, y E Gp with a # P ; thus a < 0. Hence x E PR,(xf(,)) for some n. If y 4 x then we would have y E PR,+I (xfc,)) soy E PR(xf(,)), which is incompatible with y € Gp where 0> a;thus x 4 y. This establishes (1). To prove the lemma, we need only show tp(C,< ) = K . With this aim, note that always IPR(y)l < 77 4 No, for a trivial induction shows that IPR,(y)I G 77 iNo for each n. Since Gp C PR(xf(p)),also IGp( < 77 4 No. Take any y in G, say y E Gp. Then {x E G ; x
u{

I {X E G ; X < JJ}l Hence tp(G,<)

< (9 4 No)

*

101 < K .


In about 1949 Tutte [9] and Zykov [96] showed that for every integer n there is a (finite) graph which has chromatic number at least n yet contains no triangle (that is, a circuit of length 3). Erdos [ 151 improved this by showing that for every pair of integers n, k there is a (finite) graph which has chromatic number at least n and contains no circuits of length shorter than k . One can ask if this result will hold for infinite chromatic number n as well. The theorem of Erdos [ 151 shows easily that this is so for n = No. Otherwise, the first

Ch. 5.1

113

The chromatic number

result in this direction was by Erdos and Rado [30], where they proved that for every K and every finite k, there is a graph with chromatic number >K and no odd circuits of length shorter than k. In [21], Erdos and Hajnal constructed such an example with just K vertices. This construction is repeated in Theorem 5.1.9 below. However, also in [21], Erdos and Hajnal show that the general answer to the problem above is in the negative, for they show that a graph which contains no quadrilateral has chromatic number at most Ho.This result follows from Corollary 5.1.1 1. We start with a couple of lemmas, which will enable us to give the ErdosHajnal example of the graph without odd circuits. The construction is an extension of one used by Specker [93]. We shall use the partition symbol a+ (@$ for ordinal numbers a and 0,which by definition means that whenever a set well ordered with order type a is partitioned into y classes, at least one of the classes has order type at least 0.

Lemma 5.1.6. Let a, 0be ordinals such that a + (a);and

+

09 (09);.

Go);.

Then

+

Proof. Let S be a set well ordered with order type 09, so we may suppose that

S = B X a,with the lexicographic ordering. Suppose S is partitioned, S = U{Ak; k < 7). For x in 0,put &(x) = { y < a;(x, y ) E Ak}. Then for each x , we have

u

a = {Ak(x); k < y} and by the relation a+(a); there is k(x) such that tp(Ak(,)(x)) = a.Put rk= (x < p; k(x) = k}, so 0= U {rk; k < y}. By the relation fl+@)!, there is ko such that t P ( r k O ) = P . h t R = { ( x , y ) E S ; x E r k o andy€Ak,(x)), then R C Ako and tp(R) = 09, so tp(Ako) = a@.

Corollary 5.1.7. IfK is a regular cardinal, y < K and rn < w then K"' where K"' is the ordinal power. Proof. By induction on rn, using Lemma 5.1.6 and noting K

-+

+(K"')+,

(K)$.

Lemma 5.1.8. Let K be regular, let rn be finite. Take any subset X of "'K of order type K~ (in the lexicographic ordering of the sequences in m ~ )For . each 1 with 1 < rn and each sequence (ao,..., 1 ) from ' K there is a set ~ that in [ K ]such T(a0, ...,

V1< m(al E T(ao,...,

* (ao,..., a m - l )E X .

(1)

Proof. By induction on rn. Trivially for rn = 1 ,put T(9) = X. So suppose rn 2 2. For each a,put

X(CU)={ ( a l ,

a m - l ) ; ( a , a l ? . - . , c v ~ - ~ ) E 9X }

114

Theory of infinite graphs

Ch. 5.1

We show that IT1 = K . For otherwise, I7'/ < K and so T is not cofinal in K , by the regularity of K . Hence T C (3 for some (3 where < K . Thus for each a with a 2 (3 there is an ordinal 6(a) with 6(a)< K such that tp(X(a)) < K " - ~ ~ ( c I ) . Now for y with y < a,we know 1Co(6(a);(3 < a < r)l
; a! < 0) + & J ( K " - ~ ~ ( C Y ) ;


we would have tp(X) < K"-' (3 + K " - ~ This contradiction establishes that indeed ITI = K . We are now ready to define the sets T(ao,..., ( Y I - ~ ) . Put T(8) = T. For each a in T(@,apply the inductive hypothesis to the set X(a), and so obtain sets T(a,a', ..., ar-1) (for I with 1 d I < m ) o f power K such that if a1 E T(a,a l , ..., ~ r ~ - ~< ) (I < l m ) then (a,a l , ..., a,,-') E X(a) . If a g T(o), put T(a,a1, ..., a[-1 )

=K.

Then (1) holds, and the lemma is proved.

We are now ready to give the example of a graph of large chromatic number but without small odd circuits.

Theorem 5.1.9. For each infinite cardinal K and each finite j , there is a graph on K vertices with chromatic number K , which does not contain circuits of length 2i + 1 for any i with 1 d i d j. Proof. Suppose first that K is regular. Put G = 2 i 2 + 1 ~ ,the set of (2j2 + 1)-place sequences with entries in K . Let < be the lexicographic ordering on G. For each sequence a in G, let ak denote the k-th component o f a . Let E be the subset of [GI2 such that, for a , b in G w i t h a < b , { a , b } E E ~ a i < b o < a j + l < b l <...
+ l(azE ~

( a...,~a r, - l ) )* (ao,..., a y 2 )E X .

115

The chromatic number

Ch.5.1

Since K is infinite, we may choose ordinals ao, a l , ..., a2j2 and bo, b l , ..., by2 from K so that a0 E T(0) 3

al E T(ao)and al >ao ,

ay2 E T(a0,

.-*9

a y 2 - J and a2j2 > b2j2-j4

> a2j2 b2i2-j) and b2j2-j+l > b2j2-j

b2j2-j E m o p ... b2j2-j-1) and b2j2-j f

b2j2-j+l E m o p

..*9

3

9

9

by2 E T(b0, ..., b2j2-1) and b2j2 > b2j2-1 . Put a = (ao. ..., a2pX b = (bo, ...,by2). Then a , b EX, a< b and { a , b } E E, so that X is not independent. We can now show that G has chromatic number K . For suppose, on the contrary, that Chr(C) = h where h < K . Thus G = U{ G,; u < A} where each G, is independent. By Corollary 5.1.7, at least one of the G, must have order type and this is impossible with G, independent. Hence Chr(G) = K . It remains to check that G has no odd circuits of lengths between 3 and 2j t 1. This is a tedious, but trivial, verification. For example, suppose in fact a, b , c gave a triangle in G . Clearly we may suppose a< b< c. Since { a , b } E E we know a21 < bj, from { b, c } E E we have bi < co, so that azj < CO. Yet with { a , c } E E we have co < aj+l < azj, which is impossible. The verification for odd circuits of length 5 or more is similar. This completes the proof when K is regular. If K is singular, take an increasing sequence ( K , ; u < K ’ ) of regular cardinals below K with K = Z(K,,; u < K ‘ ) . For each u, there is a graph G , = ( G , E,) with IG,I = K , and Chr(G,) = K , which has no circuits of lengths 2i + 1 for i with 1 < i <j . We may suppose the G, are pairwise disjoint. Then clearly

116

Theory of infinite graphs

Ch. 5.1

the graph G = (U{G,; u < K ' } , U{E,; u < K ' } ) has no circuits of odd length at most 2j t 1, and lGl = K-With Chr(G) = K . The following theorem will show, as opposed to Theorem 5.1.9, that any graph with uncountable chromatic number contains even circuits of any length. In fact more holds, for the graph must contain large complete bipartite subgraphs. The next several theorems first appeared in Erdos and Hajnal [21].

Theorem 5.1.10 (GCH). Let h be infinite and suppose 8' < h Let C be any graph which has no Kh+,e complete bipartite subgraph. Then Col(G) < h (If 8 is finite, GCH is not needed.) Corollary 5.1.1 1. f f G is any graph with Col(G) > Ho then G contains complete bipartite subgraphs K N,,i for each finite i.

Proof (of Theorem 5.1.10) Take the graph G = (G, E ) which has n o Kh+,e subgraph. Put IGI = K , and use induction on K . If K < A then the theorem is trivial. So suppose that K > h and that the statement of the theorem holds for any graph on fewer than K vertices. We use an idea similar to that used in the proof of Theorem 1.3.12 to construct a decomposition G = U { H , ; y < K } of C such that the inductive hypothesis can be used to show that Col(H,)< A, where Hy is the subgraph of G spanned by H,, and moreover the orderings that give the colourings of the H , can be pieced together to give a A-colouring of G. Let (o(a);a < K ) be the sequence of all limit ordinals less than K , in their natural order. Write G as a disjoint union, G = U{ Gu(,); a < K } , where always ICw(,)I < h. First define inductively subsets X , of G where a < K as follows. Suppose X p already defined for all p with p< a,and put

x,*=

Gu(,) u U { X p ; < a} if a = w(y) + I for some y , U{Xp; 0 < a} otherwise.

Then define X , by

x, = X,* u {x E G; IG(x, X,*>l>0')

.

We show by induction on a that IX,l< h 4 IaI.So suppose that whenever

0 < a then IXpl< h 4 101. Then IX,*l< h 4 Z( IXpl ; p < a)

< h iZ@ i101;p < a) < h i @ +Ial). la1 = A i la1 .

Ch. 5.1

The chromatic number

117

Consider the family d,where A, = {C(x, X,*);x EX, and G(x, X;t*)l2 O'}, so 94, is a family of subsets of X,* all of power at least 0'. Since G has no KA+,* subgraph, certainly if 94

c d,with Id I >

( h 4 lal)', then I n d I

<0 .

By Corollary 3.2.4, it follows that I 94,l S h + la(.Considering again that G has no Kh+,o subgraph, for any A in 34 ,surely I{xEX,;G(x, X , * ) = A } I < h + .

Consequently

I { X E G;IC(X,X:)I

e'}r s A . I SQ,I

G A . (A

+ iai) = A 4

101,

and so IX,l<

IX,*l+ (A

as claimed. Clearly X: C X , this that

ilal) = A i- la1 ,

c X,*+l,and since w ( y ) is a limit ordinal, it follows from

U { X , * ; a < w(y)} = U{X,;a< w(y)} =X&)

.

(1)

We are now ready to define the sets Hr where y < K. Put

H, = Z ( y + 1)

-

.

X&)

It follows that x&)=

U{Hp;P
(2)

*

For if P < y then

HpLX:(p+l)= U { X , * ; a < w ( P + 1))C u{X:;a<w(y))=X:(,)

7

so certainly U{Hp; 0 < y} c X:(,) We show by induction on y that also C_ U { H p ;0 < y}. Note that if a < w ( y ) there is with /? < y such that o(0)S a < w@+ l), so that

X:(,)

x,* c XZCp+l) c Hp u x:(p, c Hp u U{HS ;6 < PI (making use of the inductive hypothesis). Hence

x:(,)=

U{X,*;a<w(y)}c_U{Hp;P
and the claim is established.

Since G = U{Gu(,); y < K } and Gu(,) X;(,)+l, it follows that G = U{H,; y < K } . Furthermore, the H, are pairwise disjoint.

118

Ch. 5.1

Theory of infinite graphs

Take any y in G , and suppose in fact y EH,. Put H ( Y ) = G(Y,

uW

b ;8

< 71)

*

The construction of the setsH, ensures that IH(y)I < A. T o see this, note that by (1) and (2), Since y E H,, by definition of H, we know y 4 X&,, so from (I), y 4 g+l for any a with a! < w(y). Hence y 4 X,, so that IG(y, X,*)l<8. Also G ( y , X,*) C C ( y , Xi)if a! < 0,since then X,*C Xi.Thus (3) expresses H ( y ) as an increasing union of sets all of power at most 8. Hence IH(y)l < 8'. By assumption, 8+ < A, and so indeed always IH(y)l< h. We can now give a A-colouring of G , and so show Col(G) < A. Let H , be the subgraph of G spanned by H,. Since lH,I < lX&,+l~l < h Iw(y+ 1)1< K , and necessarily each H , has no K,+ subgraph, the original inductive assumption provides a h-colouring
y or (0=y andx<,y)

.

This gives a A-colouring of G , since for any y in G, say y E H,, { x E G ( y , G);x
9

and both sets in the union have power less than A. One can weaken the condition imposed on the graph G in Theorem 5.1.10 to allow the possibility 8' = A, and ask if still Col(G) < h. The next theorem gives a positive answer if lGl is sufficiently small. When IC1 is larger than the bound given, the answer is not known (see Problem 50 in [24]).

Theorem 5.1.12 (GCH). Let 0 be infinite and let €J0 be the least cardinal greater than 8 such that 8; = 8'. Suppose A = 8' and let G be any graph on at most 86 vertices which has no K,,,,, subgraph. Then Col(G) < A.

Proof. Proceed as in the proof of Theorem 5.1.10, except define X, as

x, = x,*u {x E G ; iqx, X:)I 2 e l . Lemma 3.2.3 shows that still I &I< A i1011 ; the restriction IGI < 80 ensures (X t lal)' f 8' so that the conditions of the lemma are met. With this modification, it follows that IH(y)I < 8, so still IH(y)I < ?.

Ch. 5.1

The chromatic number

119

Attempting to weaken the condition still further, to the case 8 = A, meets with failure however. There is a relatively simple example of a graph G with Col(G) = A+ which contains no K,+,+ subgraph (in fact, no K h , h subgraph). A rather more complicated example 1s possible with Chr(G) = A+.

Theorem 5.1.13. There is a graph G on h+vertices which has no Kh,, subgraph but yet Col(G) = A+. hoof. By Theorem 1.1.4, there is a pairwise almost disjoint family d of h+ subsets of the set h each of power A. Put G = h U d and E = { { x , y } E [GI2; y E d and x E y } ; consider the graph G = (G, E ) . Thus G is a graph with IGI = A', and the almost disjointedness of94 ensures that G has no KA,Asubgraph. We show Col(C) = A+. Suppose for a contradiction that Col(G) < A. By Lemma 5.1.5, G has a A-colouring < with tp(G,<) = A+. Since X is not cofinal in A+, there is x in G with h C_ { y E G; y < x}, so in fact x E d . Now C(x, { y E G ; y < x}) = x, so IG(x, { y E G; y <x})l = A, which is contrary to
Theorem 5.1.14 (GCH). There is agraph G on A+ vertices which has no KA,A subgraph and for which Chr(G) = A+. Proof. We shall construct the graph on the vertex set G = A+ X A'. To define the edges, we shall use an almost disjoint family 94 of h+ subsets of G each of power A. First we define the family d.Let

X = {XCG;ldomXI=Xand V,EdomA'(I{~;(a,B)EX~I=X)~. Then 31 C [GIx and IXi = A+. Write 5X = {A',;p < A+}. We want to define the members A , (where v < A') of d SO that

X,nA,,#@ i f p < v .

(1)

Inductively suppose that for some v with v < A+ the A , for p with p < v have already been defined so that A , E [GIh,the A , are pairwise almost disjoint and

I { p; (a,0) EA,}I

< 1 for each a .

Write { A , ; p < v } = { A , ; p < h } and { X , ; p < v } = {X,,,;p
(2)

120

Ch. 5.1

Theory of infinite graphs

always ldom X,,.,l = A, inductively we can choose ap (where p

< A)

so that

a,., E d o m X v p - { a T : n < p } .

Now choose P,., such that (ap,Pp)EXvp

-

U{A,;n
;

since 1 { P ; (a,.,,0) E Xv,.,}l = A it follows from (2) that such a choice of Pp is possible. Finally put

A" =

{(ap,

P,.,);p
.

Then A , E [GI*, IA, nA,I < A whenever p < u, I {a; (a,8) € A , } [ < 1 for each a, and (1) holds. This completes the definition. We can now specify the edges of G . Letf : G -+ A' be a one-to-one map. For (a,P), (y,6) from G, put

{(a,P), (y,6)} E E 0 a < y and P > 6 and (y,6) E A A ~ ,.~ )

(31

Let us check that the graph G = (G, E ) has no K A , Asubgraph. Suppose, on the contrary, that there are disjoint sets Ho, H1 in [GI' with {x, y } E E wheny E H I . We may suppose that the smallest first component of ever x E Ho, any pair from HO U H 1 is from a pair in Ho. Inductively define subsets Hon, H I , (where n < w ) as follows: H0, is that subset of Ho - U { H o m ;rn < n} maximal with the property that the first component of each pair in Hon is less than the first component of each pair in H 1 - U{H1,; rn < n } , and H1, is that subset o f H l - U{Hl,;m < n} maximal with the property that the first component of each pair in H1, is less than that of each pair in HO- U { H o m ;rn S n}. Only finitely many of the Hon and H1, can be nonempty. For otherwise, all are non-empty so we can choose ( a ~ , 02,) , from Ho, and ( ~ ~ 2 n +&,+I) l9 from H I , . By (3), the sequence (0,; n < w>gives an infinite descending chain of ordinals, which is impossible. Thus there are k, I such that IHokl = A and lHlll=A. Suppose say k S 1. Again by (3), for each (a, 0) in H o we ~ have HI1 C Afca,p). This contradicts the almost disjoint property of the Finally, we show that G has chromatic number A'. For a contradiction, suppose in fact Chr(G) S A, so there is a decomposition G = U{C,; u < A} where each G, is independent in G . In particular, since A' is regular, there is u (with u < A) for which, if

D, = { a ; I { P ; (a,P ) E Gull = A'} then ID,I = A'. By the choice of D,, there are X in 5Y and a in D , such that X C G, and a < y for all y in dom X. Suppose in fact X = X,. Since I { 0;

Ch. 5.2

121

A colouring problem

(a,p ) E C,}l= A', there is p, with (a,8) E G, and f ( a ,p) > p , which is greater than the second component of every pair in X. By (I), there is (y,6) in G with (y,6) E X , fl Af(&,p).Then by (3), (a,0) and (y, 6) are joined by an edge in C . Since both ( a , P )and (y,6) are in G,, this contradicts the independence of G,. Thus Chr(G) = A+, and the theorem is proved. The example just given to prove Theorem 5.1.14 clearly contains no complete K N subgraph. ~ In fact Hajnal [56]has constructed a more involved example which contains no triangle. 52. A colouring problem In this section we shall discuss questions of the following form. Given a graph G on K vertices without too many edges (specifically, without complete K A subgraphs for large values of A) when is it possible to paint the vertices of G with few colours so that there are no KO subgraphs of G all the vertices of which are the one colour? Problems of this type have been discussed by Erdos and Hajnal [ 2 2 ] .We introduce the following notation. Definition 5.2.1. The symbol [ K , A] + [g, 61 means the following. Whenever a graph G = (G, E ) is given for which IGI = K and G contains no K h subgraph, then there is a pairwise disjoint decomposition G = U(G,; p < g} of the vertex set such that none of the subgraphs of C spanned by G, (where I.( < q) contain a KO subgraph. We shall refer to the decomposition G = U{ G,; p < g} of the vertices of G as a colouring of G by g colours (not to be confused with an g-colouring < of C as in the previous section). A few trivial remarks are in order. If the relation [ K , A] [q,81 holds then if K or A are decreased, or if g or 6 are increased, the relation will remain in force. We shall always assume A, 6 > 1 and g < K . The relation [ K , A] [ K , 81 is always true. The relation [ K , A] [ 1, 81 is true if and only if 8 2 A. The case when A > K is without interest, since no restriction is then placed on the graph G . If X >K , obvious colourings of the complete graph on K vertices show that then [ K , A] [q,61 if q, 6 < K and also [ K , A] f . [q,K ] if q < K ' ; yet clearly [ K , A] [ K ' , K ] does hold. In view of these remarks, we may restrict the discussion of the symbol [ K , A] [q,61 by supposing that always 1 < 8 < AGK and 1 < q < K . The case 8 = 2 is of particular interest, for a subgraph of C withput a K2 -+

-+

-+

-+

-+

122

Theory of infinite graphs

Ch. 5.2

subgraph is an independent set in G . Thus the relation [ K , A] -+ [q, 21 is true if and only if every graph OR K vertices without a K A subgraph has chromatic number at most q. The example provided by Theorem 5.1.9 of a graph on K vertices with chromatic number K and yet containing no triangle shows that [ K , 31 [q,21 for all q with q < K . In the first theorem below, we shall prove the following generalization of Theorem 5.1.9: for all finite n with n > 2, the relation [ K , n + 11 f . [q,n] holds whenever q < K . The example which establishes this negative relation is similar to that given in the proof of Theorem 5.1.9. For infinite values of 8, in Theorem 5.2.4 we shall show by a different method that [ K ~ 8',1 [q,81 whenever q < K .

+

+

Theorem 5.2.2. Let K be infinite and let n be finite with n 2 2. n e n for every q with q < K , the relation [ K , n + 11 [q,n] holds.

+

Proof. Let q be given with q < K . There is a regular cardinal K~ with Q < K O < K . In view of the earlier remarks, it suffices to prove the negative relation with K replaced by K O . So in fact we may suppose that K is regular. ~ < ...< a n } , where for a sequence a from ""K, Put G = { a E " + l ~ ;
bo
< b l for some k with 1 < k < n .

LRt G be the graph (G, E ) ;we show G has no Kn+l subgraph and that in any colouring of G by q colours there is a monochromatic K, subgraph. To see that G has no Kn+ subgraph, take distinct sequences a', a', ..., a" from G, where we may suppose a!
Ch. 5.2

123

A colouring problem

We shall use Lemma 5.1.8 to choose ordinals a; where 0 < i < n and 0 < k < n such that if u' =
(i,k ) 4 (i', k') i + k < i' + k' or (i t k = if + k' and i < i') . We shall choose the ordinals a; according to the 4 ordering of the pairs (i, k). Apply Lemma 5.1.8 w i t h X = G, to obtain sets T(a0, ..., al) in [ K such that V l < n t l ( q E T(a0, ...,

* (ao,..., a,) E G, .

Since each T(ao,..., al) has power a;

> 4 whenever ( j , I)

] ~

K,

(1)

we can choose a$ so that

4 (i,k )

and a; E T(@)if k = 0; a; E T(a;,

...,a;-l ) if k > 0 .

Then if a' = (a&,...,a;), by (1) always a' E G,. Consider u',B where i <j. Note that

(i, j - 11 4 ( j , 0) 4 (i,j so

< di <

vertices uo, ...,

-

i t 1) 4 ( j , j - i + l ) ,

< d1;also 1 <j - i < n , so that {u', 8)E E. Hence the give a K, subgraph in G,, and the proof is complete.

We shall turn now to infinite values of 6 in our discussion of the symbol [ K , h] -+ [q,61. The principal result is again a negative result, namely [ K ~ 6',1 f . [q,61 (where q < K ) . An example of a graph exhibiting this property is based on the vertex set ' K of all functions from 6 into K . As before, let< be the lexicographic ordering of ' K . We need first the following lemma.

Lemma 5.2.3. Suppose * K is decomposed into fewer than K classes, say ' K = U{ X,; p < q} where q < K . Then one of the classes X , has a subset of power 6 which is well ordered by > (the converse of the lexicographic ordering). Proof. Consider a subset X of ' K which has n o &size subset well ordered by>. Then for each f in X there is (Y with (Y < 6 such that i f g E X a n d g r a = f r a , theng(a)< f ( a ) . For if this failed for some f in X, for all a there would beg, in X such that

(1)

I24

Ch. 5.2

Theory of infinite graphs

r

r

g, a = f a and f(a)< g,(a). Then whenever a < fl < 0 it would follow that g, r a = f r a = g p ( ~ a n d g g ( a ) = f ( a ) < g , ( a ) ; henceg,>gp.Thus tg,; a < 0 ) would be a &size subset of X well ordered by>, contrary to the

r

choice of X. Take a pairwise disjoint family {X,; p < q} of subsets of ' K where each X , has no &size subset well ordered by>. Put A = U{X,; p < q); we show A # ' K . This will prove the lemma. For each f i n A, let af be the least a with the property (1) taken with X = X , for that X , such that f € X,. Hence for L g in X,,, i f a f = a g a n d f r a f = g rag thenf(af)=g(ag).

(2)

For each a, put A , = { f E A ; a a } so then A = U{A,; a < 0). Now define a function h in i= K by using transfinite recursion to define the values h(a) for a with a < 0 as follows. Suppose for some a that all the values h(P) where fl < a are known (so h r a is known). By ( 2 ) , for each p with p < q there is an ordinal,,,$ such that for allg in A, n X,, g r a = h wg(a)=t,,,. Choose h(a) so that h(a)EK

-

{&yp;P
This completes the definition of h. Then h 4 A, fl X , whenever a<0 and p
Theorem 5.2.4. Let 0 be infinite, suppose 17 < K . Then

[K',

0'1

+ [q,01.

Roof. Take any well ordering Q of ' K , and put

Let G be the graph PK,E). Thus G is a graph induced by the partition of [ ' K ] ~ used in proving Theorem 2.5.5. As noted in equation (2) of that proof, there is no subset X of ' K with IXI = 0' which is well ordered by>. Thus certainly G can contain no Ke+ subgraph. We shall show that in any colouring of G with 77 colours there is a monochromatic KO subgraph. Thus G will provide the example we need. So take any colouring of G by q colours, G = U{ G,; p < q}. By Lemma 5.2.3, there is some G, which has a subset X well ordered by>, where IXI = 0 . Let X be the subgraph of G spanned by X. We shall apply to X the partition

Ch. 5.3

Chromatic number of subgraphs

125

relation

6

-+

(So, 6)*

.

(1)

This relation holds by Ramsey theorem if 6 = 1 ,,by Theorem 2.2.6 if 6 is regular and by Theorem 2.4.3 if 6 is singular. Note that Theorem 2.4.3 depends on GCH. In fact the relation (1) can be established without appeal to GCH - see Erdos and Rado [29, Theorem 441. Applied to X , the relation (1) implies that either there is H in [XINowhich is independent in X , or else X contains a KO subgraph. The first possibility cannot occur, however. For if ( f n ; n < o>is an enumeration of H in increasing <-order, we would have an infinite descending>-chain, ... >fi >fi >f o , contrary to > being a well ordering on X. Thus X contains a KO subgraph, and this gives a KO subgraph of G in G,. If we assume GCH, then the preceeding two theorems show that all the non-trivial cases of the relation [ K , h] [ r ] , 61 are false. -+

Theorem 5.2.5 (GCH). Suppose that 1 < 6 < h < K and 1 C r] < K . Then [ K , XI [ r ) , 61.

+

+

Proof. It suffices to show that [ K , 6'1 [ r ] , 61. This follows from Theorem 5.2.2 if 6 is finite, and from Theorem 5.2.4 if 6 is infinite and K regular (for then K' = K by GCH). If K is singular, there is a regular cardinal K O with ,'6 r] < K O < K . Then [ K O , 6'1 P [ r ] , 61 and hence [ K , 6'1 -b [ r ) , 01. 53. Chromatic number of subgraphs A well-known theorem of de Bruijn and Erdos [6] is the following: for any finite k , a graph has chromatic number at most k if and only if every finite subgraph of it has chromatic number at most k . In this section we consider generalizations of this theorem. We ask questions of the form: if every subgraph of G on h vertices has chromatic number at most r ) , then does G have chromatic number at most 6? The results are mostly by way of counter-examples, and many problems remain unsolved. We start by proving the theorem of de Bruijn and Erdos mentioned above. In fact the theorem is a standard consequence of the Compactness Theorem of mathematical logic, but we shall give a combinatorial proof.

Theorem 5.3.1. Let G be any graph, let k be finite. Then Chr(G) < k ifand only ifChr(H) < k for evevfinite subgraph H of G.

126

Theory of infinite graphs

Ch. 5.3

Proof. The "only i f ' is trivial, so suppose every finite subgraph of G = (G, E ) has chromatic number at most k. We must find a colouring of C with k colours in which there are no monochromatic edges. Take a well ordering of G, say G = {x,; a < K } . We shall define a functionf : G + { 0 , 1, ...,k - 1 ) by determining the valuesf(x,) using induction on a. The function f will be such that colouring vertex x, with colour i if and onlyf(x,) = i will give a suitable colouring of G with k colours. We shall definef(x ) so that for each a! the following holds: (1.a) For a l l H i n [G]"O there is h : H + ( 0 , 1 , ..., k - 1) such that (La) if y < (11and xy E H then h(x,) = f(x,) , (ii)

if {x, y } E E n [HI2 then h(x) # h ( y ) .

Suppose for some a! with a < K that f ( x p ) has already been defined so that (1 .p) holds, for all /.3 with 0< a. (The inductive assumption for a! = 0 is interpreted to mean that for all H i n [G]<'O there is h : H - + ( 0 , 1 , ...,k - 1) such that (ii) holds. This is the hypothesis of the theorem.) Suppose that none o f the possibilitiesf(x,) = 0, 1, ..., k - 2 are compatible with ( l a ) . Thus for each 1 where I < k - 2 there is a finite subset HI of G such that x, E HI and for every function h : H -+ { 0 , 1, ..., k - 1) with h(x,) = I and h(x,) = f ( x , ) whenever xy E H with y < a, there are vertices x, y in H , adjacent in G , for which h(x) = h ( y ) . I claim that then definingf(x,) by f(x,) = k - 1 will make (1 .a) hold. For take any finite subset H of G. Put H* = H U Ho U ... U H k P 2 ,so H* is also a finite subset of G. There is /.3 with < a such that if xy E H with y < a then y G 8. By (1 $) applied to H*,there is h* : H* { 0 , 1 , ..., k - 1) such that (i.0) and (ii) hold for H* and h*. In fact, it must be that h*(x,) = k - 1. For if not, say h*(x,) = 1 where I < k - 2, since HI C H* it follows that h* r H I gives a colouring of HI such that (ii) holds, any xy with y < a is colouredf(xy), and XI is coloured I, contrary to the choice ofH,. So indeed h*(x,) = k - 1 . Thus h* r H gives a suitable colouring of H under which x, is coloured k - 1, and k - 1 = f(x,). Since H was arbitrary, in fact (1 .a!) holds. This completes the definition off, and clearlyfgives a colouring of G with k colours such that no two adjacent vertices are coloured the same. -+

Corollary 5.3.2. Every graph with chromatic number NO has a countable subgraph with chromatic number No. Proof. Take a graph G with Chr(C) = No. For every finite k then Chr(G) > k so by Theorem 5.3.1 there is a finite subgraph H k of G with Chr(Hk) > k.

Ch. 5.3

127

Chromatic number of subgraphs

Let H be the union of the Hkrso Chr(H) 2 No.Thus H is a countable subgraph of G with C h r Q = No. One can ask if Corollary 5.3.2 remains true when No is replaced by N1. The answer is in the negative. Assuming GCH, as a corollary to the next theorem we shall show that there is a graph on H2 vertices with chromatic number HI,every subgraph of which on N1 vertices has chromatic number at most No.We first prove the following lemma.

Lemma 5.3.3. Let n be finite with n 2 2 . Take any set S with IS1 = 2 ,1(~) for some infinite cardinal K . Then there is a function f : "S + K such that for X@ ..., x, E s,

tli < n(xi # xi+l) *f(xo,

..., x,--l) +f(xl, ..., x,) .

(1)

Proof. Identify S with 2,-1 ( K ) . Let us say that a function f has property P(n, K ) iff has the property in the lemma. We shall prove by induction on n that for all K there is a functionfwith property P(n, K ) . Consider first the case n = 2 . Let K be given, and identify 2 1( K ) with K2. As before, let < be the lexicographic ordering of K2 and for distinct functions x, y in K2let 6(x, y ) be the least a for which x(a) #y(a). Take any one-to-one map p : K X 2 + K , and define f as follows: ifx=y,

f(x, v)=

p(6(x, y), 0)

[ op ( W , y ) ,

1)

i f x y.

Since 6(x, y ) # 6(y, z) whenever x < y < z, it follows that f has property P(2, K ) . Now suppose n > 2, and make the inductive assumption that for all m with 2 < m < n and for every infinite h there is a function with the property P(m, h). Let K be given. There is thus a functiong with property P ( 2 , 1 n - 2 ( K ) ) and a function h with property P(n - 1, K ) . Now dom(g) = 2 ( 2 1 ( 2 n - 2 ( K ) ) ) = 2(2,-l(~)). Thus we may define f : "(2n-1(~))+ K as follows: if (YO, ..., an- 1 <2 n - 1 (K 1, f(a0, an-i)=h(g(xo, xi), ...,g(xr2-2, xn-1)). Then f satisfies (1). For given ao,...,a,from ~ , - I ( K )with a0 # a1,..., #a,, sinceghas propertyP(2,2n-2(~))ifPo=g(a0,ad, 1= g(anPl, a,) then fi0 # pl, ...,Pn-2 # &-l. Since h has property P(n - 1, K ) , further h(&,- ..., &-2) # h@1, ...,& - I ) , that is, f(a0, ..., ( ~ ~ - 1#f(al, ) ..., an). Consequently f has property P(n, K ) , and the induction step is complete.

...,on-

128

Theory of infinite graphs

Ch. 5.3

Theorem 5.3.4.Let n be finite with n 2 2 . For every K there is agraph G on (2n-l (K))' vertices with Chr(G) > K that that every subgraph of G spanned by at most 2,- ( K ) vertices has chromatic number at most K .

Proof. Write A = (2,-1(~))+, and let G = [A]". Define a subset E of [GI2 as define follows: for a, b in G, say a = (00, ..., a,-l}< and b = {DO, ..., {a,b}EE*Vi
1 (ai=&+l).

Then the graph G = (G, E ) has the required properties. To see that Chr(G) > K , take any decomposition G = U{ Gk; k < K} of G into K classes. Since G = [h]",the partition relation (2,-1(~))+ + (K'); from Theorem 2.3.3 applies. There must be H in which is homogeneous for the partition. For (YO, ..., a, from H with a0 < ... < a, if a = {QO, ..., a,-1 } and b = { a 1 ,..., a,} then {a, b } E E. Since a, b E [HI" both a, b are in the same class G k ,which is thus not independent in G . Hence Chr(G) > K . Now take any subset H of G with IHI < > n - - l ( ~ ) and consider the subgraph H of G spanned by H. Since h is regular there is a with a < h such that H C [a]", and we may suppose la1 = 2,*- l ( ~ )A. function f as in Lemma 5.3.3 for the set a induces a colouring of [a]", and so also of H , such that no two vertices joined by an edge get the same colour. Hence Chr(H) < K . The graph used in the proof above first appeared in Erdos and Hajnal [20] as an example of a graph with large chromatic number but without circuits of small odd lengths. In [ 2 3 ] ,the same authors noted that it gave an example with the properties in Theorem 5.3.4. Corollary 5.3.5 (GCH). Let n be finite, n 2 1. For every ordinal a there is a such that every subgraph of G graph G on Ha+, vertices with Chr(G) = vertices has chromatic number at most H ,. spanned by at most

Proof. The case n = 1 is trivial. For n with n 2 2, use the graph constructed note in the proof of Theorem 5.3.4, with K = H,. To see that Chr(G) = that by GCH, Ha+, =2n-1(H,+l) and use Lemma 5.3.3. In particular, Corollary 5.3.5 in the special case a = 0 and n = 2 gives a graph G on ts2 vertices with Chr(G) = H1 so that every subgraph H of G on HI vertices has Chr(H) < No. It is not known if there is such a graph G with Chr(G) = No. It is also not known if Corollary 5.3.5 can be extended to infinite values of n. It is clear that if G is a graph on H , vertices such that every subgraph on fewer than K, vertices has chromatic number at most No, then

Ch. 5.4

Chromatic number of subgraphs

129

Chr(G) < No. The simplest unsolved problem is the following (GCH): is there a graph G on vertices with Chr(G) 2 H I , yet Chr(H) < Ho for every subgraph H of G on at most N, vertices? To conclude this section, we note that one can ask questions similar to those above but replacing “chromatic number” by “colouring number” throughout. The direct analogue of Theorem 5.3.1 does not hold if k > 2. Erdos and Hajnal [21] have shown that if G is a graph such that every finite subgraph of G has colouring number at most k, then G has colouring number at most 2k - 2. They provide an example to show that this is the best possible result. Neither the proof of their theorem nor the construction of the example is particularly simple. In the case of infinite colouring number however, Theorems 5.1.10 and 5.1.12 lead to the following positive results.

Theorem 5.3.6 (GCH). Let G be a graph such that every subgraph of G on at most K + vertices has colouring number at most K . n e n Col(G) < K ++. Proof. Let G be a graph with Col(H) < K for every subgraph H of G on K + vertices. Note that Col(KK+K ) = K + , and so G can contain no KK+,Ksubgraph. Thus by Theorem 5.1.16, Col(G) < K + + . If Theorem 5.1.12 is used in place of Theorem 5.1.10, the following sometimes better result is obtained.

Theorem 5.3.7 (GCH). Let K~ be the least cardinal such that K O > K and = K . Let G be any graph on at most K~ vertices such that every subgraph of G on at most K + vertices has colouring number at most K . Then Col(G) < K+. K;

Many unsolved problems remain. The simplest is the following: is it true that Col(G) < No for every graph G on Nz vertices such that Col(H) < No for every subgraph H of G on at most N, vertices?

CHAPTER 6

DECOMPOSITION AND INTERSECTION PROPERTIES OF FAMILIES OF SETS

0 1. Decomposition

properties

The problems to be considered in this section are of the following kind. Let 94 = { A v ;K < A} be a family of subsets of a set S where S has power K , such that the sets in d cover S closely, in the sense that for some cardinal g every g-size subset of S meets almost all the sets in d . Can S, except for a subset of size less than K , be written as the union of a small number of sets B,, each with the property that for any pair x, y in B, there is some A in d with x, y in A? With the following notation the problem can be expressed more precisely. Definition 6.1.l.The family d = { A v ;v < A} is said to have c g x r number 17 if for every B in [ U d Iq,

Definition 6.1.2. Given the family d , a set B is said to be connected in SQ if for all x, y in B there is A in d such that x, y EA . Definition 6.1.3. The symbol ( K , A, 77) 3 6' means the following. Let S be any set with IS1 = K and let d = { A v ;v < A} be any decomposition of S which has cover number g. Then there is a decomposition S = S* U U{B,; p < 6') where IS*[ < K and each B, is connected in s4 . Results on the prope 'ty ( K , A, g) * 6' (in different notation) appear in Erdos, Hajnal and Milner [35, 0 8 11, 13, 141. When considering the symbol ( K ,A, g) * 0 , to avoid trivial cases we shall always suppose that 0 < g G K and that 6' < A, K (the latter since each member of 94 is connected and also each {x} for x in S). Simple considerations

Ch. 6.1

Decompositions properties

131

show that if for particular values of K , A, Q and 8 the relation ( K , A, Q) =) 8 holds, then also (K , A, vl) * 8 whenever q 1 S Q and 8 > 8. We shall start by investigating countable decompositions d of a countable set S. The results are simply stated: if d has finite cover number then S itself is almost connected, if d has infinite cover number then S is almost the union of two connected sets.

Theorem 6.1.4. Let n be finite, n > 0. 7%en(No,No, n ) 3 1. Proof. Let S be a set with IS1 = No;take any family d = {A,; v < w } with U d = S which has cover number n. Suppose S - S* is not connected for every S* in [S]<'O, and seek a contradiction. Note for any subset X of S if thereisMin [w]'O with l { v E M ; x 4 A U } I < No f o r e v e r y x i n X t h e n X i s connected, since given x, y in X, each miss only finitely many A , with v in M and so both are in infinitely many A , . Define by induction elements xk of S and infinite subsets Mk+l of o such that

xi@A , i f j < k a n d ~ E M k + ~ ,

(1)

as follows. Put M O= w . Suppose for some k that XI and MI+^ have already been suitably defined whenever I < k. If ( v E M k ; x @ A , } is finite for every x in S - {XI;I < k} then by the remark above, S - {XI;I < k} is connected, contrary to assumption. so we can choose xk from s - {XI;1 < k} and Mk+l from [Mk]'' such that x k ($ A , for every v in Mk+l. Then (1) holds. Put B = {xo, ..., xnPl}, so IBI = n. Then B n A , = 8 whenever v EM,, and M l ,l Q: No.This contradicts t h a t d has cover number n, and so the theorem is proved. In proving the result when the cover number is infinite, we shall make use of the following lemma.

Lemma 6.1.5. Let the family d = {A,; v < w } give a countable decomposition of a set S. If for no finite subset S* of S is S - S* the union of two connected sets, then 94 does not have cover number NO. Proof. Let d be a family satisfying the conditions of the lemma. We must show that there is an infinite subset T ofS such that I { v < o ; A , f l T=@)I=K,. For any subset X of S, let G(X) be the graph with vertex set S - X and two vertices x, y joined by an edge just when {x, y } A for every A in A. Thus the independent subsets of G(X) are exactly the subsets of S - X connected

132

Decomposition and intersection properties

Ch. 6.1

in 94 . This means that the assumptions on 94 ensure that for any finite subset S* of S, the graph C(S*) has chromatic number more than 2. Hence by the theorem of de Bruijn and Erdos (Theorem 5.3.1), there is a finite subgraph of G(S*) with chromatic number more than 2. Inductively choose finite subsets S k of S (where k < w ) such that the subgraph G k of G(U{Sj; j < k } ) spanned by S k satisfies Chr(Gk) > 2. For each k, then S k has only finitely many subsets and so we can choose inductively sets M k from [ o ] ’such ~ that A , n S k is constant, say A , n S k = S Z , for each v in M k , and M k c il{Mi;j < k } . Since M k is non-empty, certainly each S,$ is independent for the graph G k . Since Chr(Ck) > 2 it follows that S k - SZ is not independent. Thus we can choose x k , Y k in S k - s,*which are adjacent in Gkr and so { x k , y k } C A , for all v . For all k , choose v(k) from M k - { v ( j ) ; j < k ) . Form the partition [a]’= A, U A , where for j , k in w with j < k, {jjk } E Ao

*Xk

4 AVO.)

By Ramsey’s theorem (Theorem 2.2.9) there is H in such that either [HI’ C A0 or else [HI2 C A,. Suppose [HI2 C A,; the other case is similar. For j , k in H , i f j < k then x k 4 Avo) by definition of A,. If j 2 k then u( j ) E Mi C M k so A,u) n S k = S:; since x k E S k - S$ again x k 4 A,c). Thus if T = { x k ; k E H } then T is infinite and A , n T = 8 whenever u E { v(j); j E H } , and I ( v ( j ) ; j EH}I = KO. This suffices to prove the lemma.

Corollary 6.1.6.For all infinite K , the relation

(K,

KO, No) * 2 holds.

The result in Corollary 6.1.6 cannot be improved to (No,No,No) * 1 , as the following simple example shows.

Theorem 6.1.7. Forall infinite K , Proof. Put S = { 0,1} X each v,

K

(K, K , K )

91

and consider the familyd = { A , ; u < K } where for

A , = { ( o , p ) ; p < V } u {(l,p);U
94 has cover number K . For take M from andputA(M)=U{A,;uEM}. F o r a n y p w i t h p < ~ ,t h e r e i s v i n M such that p < v and (0, p ) E A,; hence { 0) X K C_ A(M). Take any B in [ S I K , and put M = { v < K ; BnA , = 81, soB n A ( M ) = 8. If !MI = K , we would then [KIK

Ch. 6.1

Decompositions properties

133

have to have B C { 1) X K . But then for any v from M there would be p with v such that (1, p ) E B , and thus ( 1, p ) E B fl A , contradicting v EM. Thus for any B in [S]" it follows that I { v < K ; Bn A , = @}I < K , and so d has cover number K . For any S* in [S]'" there are p , v with v < p and (0, p), (1, v> E S - S*. Since then { (0, p), ( 1, v)} A for every A in d , it follows that S - S* is not connected. Thus the family4 gives an example that shows ( K ,K , K ) 9 1. p2

The relation ( K , h, 7) * is equivalent to a special case of the polarized partition symbol from Chapter 4, as is shown i n t h e next theorem.

Theorem 6.1.8. Let K ,h and q be infinite cardinals. The relation ( K ,A, q) =+ 1 holds ifand only if

Proof. Suppose first that ( K , h, 77) + 1. Take any disjoint partition K X h = A0 U A,; we seek Ho in [ K ] ~and H1 in [A]' such that Ho X H1 C_ A, or HO X H1 C A1. AS in the last proof, put S = { 0 , I} X K and consider the family d = { A , ; v < A} where this time, for each Y , A , = { (i, a ) ;( a ,v) 4 Ai} . Take S* from [S]'". There is a such that (0, a),( 1, a) E S - S* and so S - S* is not connected in d.Thus S = U d is not almost connected, and so by the relation ( K ,A, q) =+ 1, the family 4 can not have cover number q. Thus there is B in [S]* with I ( v < h;B nA , = @}I = A. There are HO in [ K ] and ~ i from { 0, 1) such that (i, a)E B for all a in Ho. Put H1 = { v < A; B fA l, = @},so H1 E [A]'. Then if a E HO and v E H1,it follows that (i, a)4 A , so (a,v) E Ai. Hence HO X H1 C Air and the relation (1) holds. Now suppose that (1) holds. Take a family 4 = { A v ;v < h} where IUdl = K , which has cover number q. Suppose that U d - S* is not connected for any S* in [ U d 1'". Then we can define by induction sets B, and elements xa, y a where a < K as follows: B, is a maximal connected subset of U d - ( { x p ; /3 < a} U { yp; /3 < a});we choose x, from B, and yorfrom ud - (B, U {up; 0 < a}).Thus for no A in d is it true that x, y, E A. Form the disjoint partition K X h = A0 U A1 where

~Y,V)EAO*X,EA,. By the relation (l), there are HO in

[ K ] ~and

H I in [h]' such that HO X H1 _C A0

Ch. 6.1

Decomposition and intersection properties

134

or Ho X H1 C_ A,. Both {x,; a! €IfO) and (y,; a! E H o }have power 7,so since SQ has cover number g it follows that I { v < A; V a €Ho(x, $ A,)}I < A and I {v < A; V a € H o ( y a 4 A,))I < A. Since lHll = A, there must be a l ra2 in Ho and v l , v2 in H1 such that xal € A , , andya2 € A , , (so certainly xa2 $ A v 2 ) .Thus (a1,vl) E A, and (az,v 2 ) E A1 contrary to the property of such that U d - S* is Ho. H I . This shows that there must be S* in [ U d connected, and so ( K , A, 17) =) 1.

Corollary 6.1.9 (GCH). For infinite cardinals K , A the relation ( K , A, K ) holds ifand on& if { K , K ' , K', ( K ' ) ' ) 6 {A, A', A', (A')') 0.

=)

1

Proof. By Theorem 6.1.8 and Theorem 4.2.8. We shall now seek to extend the earlier results on countable families to higher cardinalities. We consider first the case when the family has power equal to a successor cardinal. In Theorem 6.1.10 we shall show that ( K , A', A') =) A (for arbitrary K ) , a result not as strong as the relation ( K , No, No) * 2 established in Corollary 6.1.6. However, at least when K = A' this result is the strongest possible, for in Theorem 6.1.12 we prove (h', A', A') 8 for any 8 with 8 < A.

+

Theorem 6.1.10. Let A be infinite. Then ( K , A', A')

* A.

Proof. We may suppose K 2 A'. Let S be a set with IS1 = K and take any family d = { A , ; v < A'} with U d = S which has cover number A'. We shall show that S = U(B,; p < A} where each B, is connected in s4 ; then the relation ( K , A', A') follows. For x in S, put A(x) = { v < A';x EA,}. Then S is the disjoint union S = So U S1 where So = {x E S;IA(x)l G A) and S1 = {X E S;IA(x)l= A'}. =)

F o r e a c h x i n S 1 , p u t D ( x ) = {yES1; 3 A € d (x,yEA)}.Suppose D(x)l = A'. Then I ( v < A+;D(x) f l A , = 0}1< A, because ?A has cover number A'. Since IA(x)l = A' there must be v in A(x) with D(x) f l A , # 0;thus there i s y with x, y € A , yet y ED(x), which is a contradiction. Hence ID(x)[ < A for all x in S1. For x in S1 inductively define Dn(x) for n with nGoby

DO(X) = {x};Dn+l(x) = U { D ( y ) ; y ED"(x));DW(x) = U { D " ( x ) ; n < w ) . Then IDw(x)l < A, and D(z) C_ Dw(x)whenever z EDW(x). Inductively define subsets T, of S1 as follows: if S1 f u{Tp; < a} choose x f r o m S 1 - u{Tp;p
Ch. 6.1

135

Decompositions properties

< (Y then y @ Tp soy 4 D(z); thus there is A in d with y , z E.4. And always IT,I < h so we can write T, = {x,,; p < A}. For p with p < h let C, be the set of all the x,. Then S1 = U{C,; p < A} and each C,, is connected. So to prove the claim above, it is sufficient to show that So is the union of at most h connected sets. By way of contraduction, suppose that this is false. Then for (Y with (Y < A', define inductively elements x, of So and ordinals v(01) with v(a) < h' as follows. Suppose xp and v(P) already defined whenever P < (Y. Put z E Tp where

u, =So - ({xp;P<(Yl u U { A , ( p ) ; P < a } ) . Now la1 Q X and the sets {xp} and A,@) are all connected, so since So is not the union of h connected subsets, certainly U, # 8. Choose x, from U,. Since always xp E So we know IA(xp)l< A. Hence we can choose v(01) from A+ - ( I@); P < a) u U{A(xp); P < 4) . This completes the definition of x, and v(01). It follows that cu,P
V(P)

'

for if B< (Y this is ensured by the choice of x, and if 2 a by the choice of v@). Put B = {x,; (Y < A'} and I = { v(01);(Y A'}. Then IBI = 1 1 1= A+ and Bf Al , = 8 whenever v E I. This contradicts that SQ has cover number A+. Hence SO,and so also S, can be written as a union of h connected sets. This

<

proves the theorem.

Before showing that this last result is the best possible in the case K = A+, we state a lemma. This lemma provides a strengthening of Lemma 2.7.3.

Lemma 6.1.11 (GCH).Let S be any set with IS1 = K'. Then there is a disjoint partition [S]' = U{ Ak; k < K + ) with the property that whenever A E [SIK and B E [SIK+then there is a in A such that for every k with k < K + there is b in B with {a, b } E Ak. Proof. Modify the proof of Lemma 2.7.3. Again identify S with

K+.

Put

d = { ( A ,g ) ; A E [ K + I K a n d g : A + K + } . By GCH,I A (K +)I = K + for each A in [ K + ] ~and so I dI = K'. Take a well ordering ((A,, g,); (Y < K ' ) of d . For each r with r < K ' , put d 7= { ( A , , g,); so ld,I

(Y

< r and A ,

C r and g [ A , ] C 7) ,

< K . Suppose now that for each r with r < K +

we could find a func-

136

Ch. 6.1

Decomposition and intersection properties

tion f, : T

-+

T

such that

if ( A , g) E d,then &(u) = g(u) for some u in A Consider the partition { Ak; k

{ U , T } E Ak

*f;(U)

=k


.

(1)

of [K']' where for u, T in K + with

(7

< 7,

.

This partition has the new property, for take A from [ K ' ] ~and B from [K'IK+. Suppose on the contrary that for all a in A there is k such that {a,0) E Ak for no 0in B , so there is a functiong : A K + such that always {a,0) 4 Ad,) for all 0in B. Let 70 be the least ordinal such that ( A , g) E dT0. Since IBI = K + there is T in B with T 2 r0. Then also ( A , g>E A, so by (1) there is u inA for whichf,(u) =g(u), that is, { u , T} E A,,). But since u € A and T E B, by choice of g we know [ u, T} 4 A,(,,). This contradiction proves the result. In order to prove the lemma then, it suffices to find functionsf, : T T with the property (1). We may suppose 94, # 0 , for otherwise any function f, satisfies (1). Thus T 2 K ,so 171 = K ,and we may write 94, = { (A,,, g7& a < K}. Since always IA,I = K we can choose inductively u,, for a with a < K such that uTO1 € A , - { u,p; 0< a}.Define f, : T -+ T as follows: if u = u, thenf,(u) = g,,(u); otherwise f,(u) is arbitrary. Then clearly (1) holds, since if ( A , g) E 94, then ( A , g ) = (A,,, g,,) for some a,and then u7& is an element o f A on which bothf, and g agree, as required by (1). This completes the proof of the lemma. -+

-+

Theorem 6.1.12 (GCH). Let h be an infinite cardinal with 6 < A. Then (A+, A+, A+) p 8. Proof. Using Lemma 6.1.1 1, take a disjoint partition [A']' such that A E [A+]* and B E [A+]*+* 3 a € A V k < A+ 3

=

U{Ak; k

< A+}

0E B((aJ3)E A,).

Put S = h+ X h so IS1 = A+, and for /3 with 0< A+ put Ap=

((ak ) E S ; {(U,fl}EAk}

We consider the decomposition of S given by the family 94 = ( A D 0 ; < A+}. First observe that d has cover number A+. For take any X in [SIh+,and for a contradiction suppose that if

I=

{p<x+;xn~~=0}

then 111 = A'. Put X ' = {a< A+; 3 k < h ((a,k ) EX)}; since

1x1= h+ also

Decompositions properties

Ch. 6.1

137

IX'I = A+. By the choice of the partition, there is a in X ' such that for all k with k < X+ there is 0in I with { a ,0) E Ak. Since a Ex' there is ko < h with (a,k o ) E X and then there is 00 in I with {a,00) E Aka, that is, (a, k o ) E A p o .Hence (a,k o ) E X nApo,contradicting that 00 EZ. Thus indeed, whenever X E [ S ] h + then I {/3 < A+; X nAp = @}I < A+, so d has cover number A+ as claimed. However, S is not almost the union of 8 connected sets. Note that since the partition is disjoint, whenever {a. 0) E Ak then it follows that ( a , 0) 4 A, if I # k. Thus if (a,k ) EAp then (a,I) 4 Ap whenever I # k, so no connected sets includes pairs (a, k ) , (a,I) with I # k. Now take any S* from [S]<". There must be a such that (a,k ) E S - S* for all k with k < h. Suppose S - S* = U{B,;p < 8 ) . There must be some p such that for distinct k, I both (a,k ) E B, and (a,I ) E B,, so this B, is not connected. Hence S is not almost the union of 8 connected sets. Thus the family d provides an example to show that (A+, A+, A') f 8 . In almost the same way, we can prove the following result where the covering number is reduced to A.

Theorem 6.1.13 (GCH). Let h be an infinite cardinal with 8 + < A. Then (A+, A+, h) f 8 . Proof. Almost as for Theorem 6.1.12, only using the set S = h+ X 8' If the cardinals h and r ) in the relation ( K , A, q) =$8 are far apart, then positive results can be proved. If A' > q+ then by Theorem 4.2.8

and hence whenever

K

2 q,

If we now apply Theorem 6.1.8, the following result is immediate.

Theorem 6.1.14 (GCH). Let Then ( K , h, q) =$ 1.

K,

X, r ) be cardinals such that K 2 r ) and A' >.'77

If r ) is replaced by r)+ in the above theorem, then GCH is not needed in the proof. In fact we have the following.

138

Decomposition and intersection properties

Theorem 6.1.15. Let

K,

A, q be cardinals such that

K

> r ) and A' > r)'.

Ch. 6.1

Then

(K,h,V')=*l.

Proof. Suppose A' > r)', let S be a set with IS1 = K and take a family 4 = { A , ; v < A} with U 4 = S which has cover number 77'. For x in S, put A ( x ) = { v < A; x E A , } . Then S is the disjoint union, S = SO U S1 where So = { x E S ; IA(x)l < A} and S1 = {x ES;IA(x)l= A}. If lSol = K , we could take S' from [So]"', and noting that then IU{A(x);x € S'}l < A, we would have I {v < A; S' nA , = @}I = A, contradicting that 94 has cover number 77'. Hence (Sol< K . Thus to prove the theorem, it suffices t o show that S1,but for fewer than K elements, is connected. Let B be a maximal connected subset of S1. Then I claim that IS1 - BI < r). For if not, take B' from [Sl - B]"'. S i n c e 4 has cover number Q', then I { v < A; B' nA , = @}I < A. Take any x in B. By the maximality of B, for any v if x € A , then A , C B, so if x € A , then B ' n A , = @ . T h u s I { v < A ; x E A , } I < A , thatis, IA(x)l
Theorem 6.1.16. Let A be a singular cardinal with A' = H,,. Then ( K , A, A) for all K .

=$

A,

Proof. We may suppose K > A. Let (A,, n < w ) be an increasing sequence of regular cardinals with always Ho < A, < A and A = C(A, n < w). Let S be any set with I S1 = K , and take a family 4 = { A , ; v < A} where Ud' = S and has cover number A. Take any set S' from [S]"'. We start by observing that if there is I in [A]' for which B E [S' Ian * I { v € / ; A , nB = @}I < A ,

(1)

then S' can be written as the union of A, connected sets. For suppose (1) is true. For x in S ' , put A ( x ) = {v E /;x € A , } and write S' as the disjoint union S'=SA U S ; where$ = { x € S ' ; M ( x ) l < A } a n d S ; = {x€S';!.A(x)l=A}. For x in S ; , put D ( x ) = { y E S; ; 3 v E /(x, y € A , ) } . It follows that if x € A , then A , n D ( x ) = 8 , so A ( x ) C { v E I ; A , nD ( x ) = 0). Hence ID(x)l< A,

Ch. 6.1

139

Decompositions properties

for if ID(x)l 2 h, by (1) then I { v E I ; A , fDl( x ) = 811 < h contradicting that M(x)l = A. For x in S ; inductively define D"(x) for n with n < w by

.

DO(X) = {X};Dn+l(X)= U { D ( y ) ; y€ D " ( x ) } : D W ( x )= U { D , ( x ) ; n < O} . Then IDW(x)l< A., Inductively define subsets T, of S; as follows: if U { Tp;0< a } , choose x in S ; - U { Tp;0< a } and put T, = DW(x)u{Tp;0< a}. Then always IT,[ < A., It now follows, just as in the proof of Theorem 6.1.10, that S; can be written as a union of A, connected sets. To complete the proof of the observation above, we must show that the same is true of S;. If IS;/ < A, this is trivial, so suppose IS;/ 2 h,. For any x in S; Since hh = A, > NO,there there is an integer m ( x ) such that IA(x)l < is a subset Si of S; with S I [; = h, such that m ( x ) is constant for x in S:, say m ( x ) = m. Thus I U { A ( X ) ; XES;}I < hn . Am < h, SO I { Y E l ; S : nA , # @)I < A. Since ISil = A, this contradicts ( I ) , and so it is impossible that IS;[ 2 .A, Thus the observation is established. We shall show that S can be decomposed into h connected sets; from this the theorem follows. For a contradiction, suppose this to be false. Use induction on the integer n to define sets I, in [A]', S,* in [S]', and I,* in [A]', as follows: Put 10 = h and choose 1; from [Io]'O. Write S1 #

S, = S --

U { A , ; v E 1; for some m with m < n } .

Now I U { I ~ ; m < n } l < C ( h , ; m < n ) < h . s o s i n c e S i s n o t theunionofh connected sets, neither is S,. In particular IS,I > h so ( I ) , applied in particular to S, and I,, must be false. Thus there are S,* in [Sn]', and In+l in [l,]'such This deChoose I,*,, from [I,+l]'n+l. that A , fl S,* = 0 whenever v E finesS,*, In+l and li+], and completes the definition. Put S* = U { S , * ; n< c ~ }and I* = U { l , * ; n< w } so IS*l= h and 11*1= h. However, S* n A , = 8 i f v E I * ,

(2)

for suppose u €1;. If m < n by definition ofS, then S,* n A , = 0. If m > n, then since 1; C I,,, C I n + l ,by choice of S,* and In+l we know A , n S,* = 0. In both cases then, S,* nA , = 0 and so indeed S* fl A , = 8. Since IS*l = A, (2) contradicts that SQ has cover number h. This contradiction shows that S can be decomposed into h connected sets, as required. For small values of K , the result in Theorem 6.1.16 can be improved sub. stantially. From Theorem 4.2.5, when A' = Ho then

140

Decomposition and intersection properties

Ch. 6.1

so by Theorem 6.1.8, the relation (A', h, A) * 1 holds. When K = A, we shall show that ( K , h, A) * 2. In view of Theorem 6.1.7, this is the strongest positive result possible. Theorem 6.1.17 (GCH). IfX ' = No then (A+, A, A)

* 1 and (x. A, A) * 2.

Proof. Only the second relation remains t o be proved, and in view of Corollary 6.1.6 we need only consider the case where X > NO.Let (A,; n < w ) be an increasing sequence of regular cardinals with always H l < h, < h and A = C(h,,; n < 0).Let S be any set with IS1 = A, and consider any decomposition PQ = { A v ;v < h} of S which has cover number h. For a contradiction, suppose that whenever S * E [Sl" then S - S* is not the union of two connected sets. The argument that follows is similar to that in the proof of Lemma 6.1.5. For a subset X of S, let C ( X ) be the graph on S - X with vertices x, y joined by an edge in G ( X )just when {k,y } A for every A in d . Then when S* E [S]
Since H l

< h,

= Ah

< hi < h, from Theorem 4.2.8 it follows that the relation

holds, where y = 2k(n).Thus there are 1; in [I,!JAn,Ml in [AIh and i(n) < y such that a E 1 ; andv€M;*S,

"A,,=X,i(,).

Further, we can arrange inductively that M; C_ fl {Mk;m < n}. Choose inductively I,, from [Z; - U{Im; m < n}]& and M,,from [M;- U { M m ; m< n}]'" a n d l e t I a n d M b e the disjoint u n i o n s I = U { I , , ; n < w } , M = U{M,,;n<w}.

Ch. 6.2

Delta-systems

141

Fix a in I. There is a unique n such that a E In. The set X,i(n), being contained in some A , , is independent in the graph G,. Since Chr(G,) > 2 it follows that S, - X,i(n) is not independent, so we can choose x,, y , from A for all A in 94. Take S, - X,i(n) which are adjacent in G,; thus {x,, y,} a disjoint partition Z X M = A, U A , such that

(a,V) E A0 *x,

4 A , ; (a,V) E A,

*yol 4 A , .

From the proof of the polarized canonization lemma, Lemma 4.1.7, there are sets I,* in [@, M,* in [MIhn and a function h : w X w -+ { 0,1} such that if m, n < w , then

I;

x

C Ah(m,n) ,

and moreover there are increasing functionsf, g : w -+ w with alwaysf(rn) < g(m) such that I,* C_ If(n) and M,* C Mgcn). By Kamsey’s theorem (Theorem 2.2.9) there are H in [ w ] , and i in { 0,l) such that whenever m, n E H with m > n then h(m, n ) = i. Suppose without loss of generality that i = 0. For m, n in H with m > n then I ; X M,* C Ao; thus if a E I; and v EM,* then x, 4 A,. On the other hand, if rn < n then 1; C Z&) C Z i m ) and M,* C Mg(n)C_ Mgcm) C M f ( m )so for a in I; and v in M,* it follows that S, nA , = X,i(f(m));since x, ES, - X,icf(m)) thus x, 4 A,. Put T = {x,; a E U{Z,*; n < w } } and N = U{M,*;n < w } . Then IT1 = X, N I 1= X and T nA , = @ whenever v € N. Thus [ { v < A; T nA , = @}I = A, which contradicts that d has cover number A. This contradiction proves the theorem. For more results and unsolved problems connected with the relation (K, A, r ] ) * 0 the reader is referred to Erdos, Hajnal and Milner [ 3 5 ] .

52. Delta-systems Let d = ( A i ;i € I ) be an indexed family of sets. The family d is said to contain the family 9 = ( B j ; jEJ)if there is a one-to-one m a p f f r o m J into I such that always Bj = AfO.).For the purposes of this chapter, if d contains 9 and a l s o 9 contains d then we may consider the indexed families d and

93 as being the same. In particular, we may always suppose that an indexed family d is indexed by the ordinals less than some cardinal, so d = ( A , ; v < A). A family d = (Ai; i E I ) is said t o be a (A, K)jizmily if [ I [= X and Mil = K for each i. Expressions such as (X,
Ch. 6.2

Decomposition and intersection properties

142

94 = (Ai; i E I ) is said to be a A(X)-farnilyif 1 1 1= A and A ifl Ai = A nA l whenever { i ,j } , { k , I } E [I]'. Definition 6.2.1. Let 0 , K , A be cardinal numbers. The symbol (A, means the following: every (A, K)-family contains a A(O)-family.

K)

-+

A(0)

We shall ask for which values of 0 , K , A the relation (A, K ) A(0) is true. The first result of this kind in the literature seems t o be due to Mazur [67], where the relation (Hl,6) A(H1) is proved. The general problem was first discussed by Erdos and Rado [31]. Michael [68] independently discovered some o f their results. Erdos and Rado returned t o the problem in [33], and there completely answered the question of finding for given K and 0 (not both finite) the least A such that (A, < K ) A(0). In Theorem 6.2.5 below we shall solve the easier problem of deciding the truth of the relation (A, K ) A(0), under the assumption of the GCH. (This relation is easily seen to be equivalent to (A, < K + ) + A@).) For a discussion not depending on the GCH, the reader is referred to Erdos and Rado [33] where the GCH is not assumed. A couple of simple remarks are in order. Every (A, K)-family where A 2 2 contains a A(2)-family, so in considering the relation (A, K ) -+ A(0) only values of 0 with 3 < 0 < A need be considered, and (A, K ) A(A) is the strongest positive result while (A, K ) f . A(3) is the strongest negative result. Clearly if the relation (A, K ) A(0) is true, then if A1 2 A and 0 G 8 also the relation (Al, K ) A(O 1) holds; and moreover if K 1 < K also the relation (A, K 1) A(0) holds. To verify the last claim, suppose given a (A, K &family d . Choose pairwise disjoint sets C,, one for each A in d , such that U CAI = K and consider the (A, k)-family C = { A U C , ;A € d}.Since for A , B from d we have ( A U C A )fl ( B U CB)= A fl B, any A(0)-family contained in C gives a A(0)family contained in d . The discussion is based on the following two lemmas. The proof for the first follows Davies [8]. -+

-+

-+

-+

-+

-+

-+

-+

Lemma 6.2.2.Suppose that A is an infinite regular cardinal, that K that qK < X whenever 77 < h Then (A, K ) A@).

> 1 and

+

Proof. Take any indexed family d = (Au;u < h) where always MuI = K . If A of the A , are the same trivially d contains a A@)-family, so we may assume this does not happen and in fact since A is regular we may assume all the A , are different. Put A' = U ( A u ;u < A), so IS1 < K . A = A. In fact IS1 = A, for if MI = 77 < A

Ch. 6.2

143

Delta-systems

then S has only qK different subsets of power K , and '7 < h. Take any well ordering< o f S with order type h. For each v, let ( x ( v ,a);a < E,) enumerate A , in increasing<-order, so I E , ~ = K . For a with a < K ' , put S, = { x ( v , a); v < A}. S i n c e S = u { S , ; a < ~ ' }and IS1 = h with h regular and K + < 2 K < h , there must be some S, with IS,] = h. In fact, let /3 be the least ordinal with ISpl = A. Put T = u{S,; a
Lemma 6.2.3. Let 8, for v with v < K be cardinals such that 8, i 1 < 8. Put h = Z(n(8,; v < a);a < K ) . Z%en (A, < K ) f. A(8).

Proof. For each a, let P, be the Cartesian product of the 8, with v < a,and put d = U{P,; a < K } . Consider the functions in d as sets of ordered pairs. Certainly Id I = h and I f I < K for each f i n d . Consider 4 as an indexed family of distinct sets, indexed by h; then d is a (A,
K

is infinite and h S

K'.

Then (A,

K)

+ A(3).

Proof. The relation ( K ' , K ) A(3) is an immediate corollary of Lemma 6.2.3. It can also be seen directly by considering the family (a;K < a < K'). Then certainly if h < K ' , also (A, K ) f . A(3).

Decomposition and intersectioriproperties

144

Ch. 6.2

By Lemma 6.2.4, in considering the relation (A, K ) +. A(O) we may henceforth suppose that K' < A. The next theorem covers all non-trivial cases in which A + K 2 No. (Unsolved problems remain if both A and K are finite. See Erdos and Rado [3 11.) Theorem 6.2.5 (GCH). Suppose that A is infinite and K 2 1. Then (i) ij'h is inaccessible and K < A, then (A, K ) +. A@); (ii) if A is singular and K , O < h, then (A, K ) +. A(O); (iii) i f A is singular and K < A, then (A, K ) A@); (iv) if K < A', then (A', K ) +. A@'); (v) ij'h is singular and K , O < A, then (A', K ) + A(O); (vi) i f A is singular and h' < K < A, then (A', K ) f , A@).

+

Proof. Part (i) is immediate from Lemma 6.2.2. For (ii), take K and O with O < X. By Lemma 6.2.2, ((OK)', K ) +. A(@")'). Since (by GCH) (OK)'< A, and O < (OK)', certainly (A, K ) +. A(O). From (ii), (v) follows trivially. Lemma 6.2.2 implies (iv), noting that A+ is regular and that by GCH, hK= h < A' when K < A'. For (vi), take K with A' < K < A. Take a sequence (A,; v < K ' ) of cardinals such that always A, < A and (A,; v < A') is a sequence cofinal in A. For a with a < K ' , always n(h,; v < a )< hla'< A' by GCH, and n(A,; v < A') = A' by Kiinig's Theorem. Hence K,

A+

< C(IA,;

v < a);a < K+)

< A+ . K + = A+

,

+

so by Lemma 6.2.3 it follows that (A', < K + ) f , A@); hence (A', K ) A@). Only (iii) remains, so suppose A is singular. Write A as a disjoint union, h = u{S,; u < A'} where always IS,[ < A. Take pairwise disjoint sets A , for u with u
Lemma 6.2.4 shows that a (K', K)-family may fail t o contain any non-trivial A-family. However, if some restriction is placed on the ( K ' , K)-family then positive results are possible. We shall investigate ( K ' , K)-families d such that every two sets from d have an intersection of power less that 0 , for some 0 with O < K . Since the sets from such an indexed family d = ( A [ ;i E I ) are pairwise distinct, we may identify 4 with the non-indexed family { A t ;i € I } and refer t o the degree of disjunction 6 ( d )as defined in Chapter 4. The following theorem is from Erdos, Milner and Rado [39].

Delta-systems

Ch. 6.2

145

Theorem 6.2.6 (GCH). Let d be a ( K ' , K)-famiZy with 6(d) = 6 where 8 < K . (i) r f e < K ' then d contains a A(K')-familY. (ii) Lf K ' < 8 < K and 9 < K then d contains a A(q)fumily. Proof. Let the family d = { A , ; u < K + } be given, with 6(d)= 8 and IA,I = K for each u. Use transfinite induction to choose subsets fa! Of K + for a with a < K" so that I, is a subset of K' - U{Zp; p < a } maximal with the property that { A , - U{A,; p E f p for some 0with /3 < a};u €1,) is a pairwise disjoint family. For brevity, put [(a)= U{I,;/.?
A(a)= U{A,:p€Z(a)} ;

thus 1, is a subset of K' - [(a)maximal such that { A , - A(a);u €1,) is pairwise disjoint. Thus in particular { p , u ) E [I,]*

since by choice of[, U EK+

-

*A ,

A,

c A(a)

(1)

9

it follows that (A, - A ( a ) )n (A, - A ( a ) )= 8. Further,

I(&) *

n A(&)[2

.

(2)

For take any /3 with 0 < a. Since u 4 f p by the maximality off, there is p(p) in f p such that (A, - A @ ) ) fl(A,@) - A @ ) ) # 8. Hence we can choose xp from ( A , nA,@))) - A @ ) . And if 7 < P thenx, Zxp since x, € A P ( ? )C_ A ( @ . Thus {xp; p < a} C A , n A(a),so indeed IA, n A ( a ) l 2 IaI. Now we can prove (i). So suppose 8 < K ' , and we wish to show that d contains a A(K+)-family. Assume first that I1,l < K for all a with a < 8. Then If(0)l < 8 . K = K , so IA(O)l< K . K = K and IK+ - f(8)l = K + . By (2), IA,nA(O)I 2 8 for each u in K + - 1(8), so we can choose B, from [ A , nA(8)l'. However, [A(O)]' < K' = K and so there must be distinct 1.1, u in K + - Z(8) for which B,, = B,. But then IA, nA,I 2 IB,I = 8, contradicting that 6 ( d )= 8. Thus there must be a with (Y < 0 for which I1,l= K', and in fact let Q be the least such ordinal. Then IA(a)l < IaI . K . K = K . Put

P = { u E f,; IA, n A ( ~ )2I 8); Q =

{U E I,;

IA, n A@)I < el .

Consider what would happen if IPl = K'. Then for each u in P, choose B, from [ A , nA(a)le.Since I[A(a)]'l < K' = K , there must be distinct p , v in P for which B, = B,. But then IA, nA,I 2 B I I, = 0 , contrary to F(d)= 8. Thus in fact IPI < K ' , and so IQl = K + . Since I [A(c~)]<~l < K' = K there must be A in [A(a)]<' and Q' in [Q]"' such that A , nA(a)= A whenever u E Q'. Thus (A, nA(a);v E Q') is a &+)-family and hence by (l), { A , ; v E Q'} is a A(K')-family, and is contained in d . Thus (i) is proved. Now consider (ii). Thus K is singular, K ' ,< 8 < K and 9 < K. By increasing

Decomposition and intersection properties

146

8 if necessary, we may assume that 8' f

K'.

Ch. 6.2

Suppose II(K>I< K . Then IA(K)I <

= K ; . By (2)) [ A , nA(K)I 2 K whenever V E K + - I ( K ) . Thus there is a decomposition 9 = { A , n A(K); v E K + - I ( K )of} a set of

K

'

K

= K and

IK+ - I ( K ) I

power K into K + sets each of power at least K , with 6(30) < 8. By Theorem 1.1.6 this is impossible. Hence II(K)) = K'. Since I ( K=)u{Z,; a < K } , there must be a with a < K such that II,I = K ' , and we may suppose that a is the least such ordinal. Then w(a)l< K . K . K = K . Again by Theorem 1.1.6 there cannot be I in [I,]K(+ such that IA, n A(a)l= K for each v in I. Thus there must be I in [I,]"' such that 14, A(a)l < K for each v in I. There is a cardinal K with K < K and a set I' in [IIK' such that IA, nA(a)l=K 1 for all u in Z'. By Theorem 6.2.5(v), ( K ' , K + A(q). Thus there is I" in [Z'Iq such that ( A , n A(&);u E I " ) is a A(q)-family, and so by (1) also { A , ; v E I " ) is a A(q)-family. This proves (ii).

The result in (ii) in the theorem above is the best possible in the sense that if K is singular then there is a ( K ' , K)-familyd with 6 ( d )= K ' which contains no A(K)-family. For take a sequence ( K ~ u; < K ' ) of cardinals below K with K = Z(K,; u < K ' ) , and let P be the'cartesian product of the sets K,. For eachf in P,put Bf= { f r o ;u < K ' } . Take pairwise disjoint sets C'with ICfl = K , disjoint from all the B f , and put A f = Bf U 'C T h e n d . = { A f ; f E P} is a (K', K ) family with 6 ( d ) = K ' . Suppose { A f ; f €Q} where Q C P is a A(lQl)-family. For distinct f, g from Q then A f n A , = Bf n Bg,and there is u with u < K ' such that f r u 4 Bf flB, (for otherwise Bf = Bg and s o f = g). Clearly the least such u is a successor ordinal, u = T + 1, andf(7) # g ( ~ )Further, . u is the same for all pairsf, g from Q. Hence lQl < K , < K . The restriction that 8 < K in Theorem 6.2.6 is justified by the following example of a (2K,K)-family d with 6Cd)= K which contains no A(3)-family. F o r e a c h f i n K 2 p u t A f = { f r a ; a < ~ } . T h e n the family A= { A f ; f E K 2 } is a (2", K)-family with 6(d)= K . Suppose { A f ; f E I }where I C K 2is a A(lIl)-family. For distinct f, g from I there is a with a < K such that f r a 4 Af n A,. The least such a is a successor ordinal, a = + 1, and moreover is constant for all pairsf, g from I. Thusf(0) #g(@ for distinctf, g from I , SO that 1 1 1< 3. Hence d contains no A(3)-family. Lemma 6.2.4 gives also the result (A, K ) A(3) if X < K , and one may ask if a restriction to the degree of disjunction of the (A, K)-family will lead to a positive relation, as happened in Theorem 6.2.6 when X = K + . However, no such positive relation is forthcoming; the following gives an example of a ( K , K)-family d with 6 ( d )= 2 which contains no non-trivial A-family. For a with a < K , put A , = {(a,0); 0 < K } u { ( p , a);p < a}.Then if y < a < K , clearly A , fl A , = {(y,a)}. Hence (A,: a < K ) has the required properties.

+

Ch. 6.3

Weak delta-systems

147

83. Weak delta-systems An indexed family d = (Ai;i E I) is said to be a weak A@)-family if M i nA,l = IAk nAIl whenever { i , j } , { k , I } E [I]*.Clearly every A@)-family is also a weak A@)-family, but not conversely. We shall consider those cases when the relation (A, K ) A((?) from the last section is false, and ask about the truth of the weaker requirement that every (A, K)-family contain only a weak A((?)-family. = A and

-+

Definition 6.3.1. Let 8, K , A be cardinal numbers. The symbol (A, K ) wkA(0) means the following: every (A, K)-family contains a weak A((?)-family. +

The relation (A, K ) + wk A((?) was introduced by Erdos, Milner and Rado [39], and for infinite A and K they gave a complete discussion (assuming GCH). The same simple remarks that were made in the last section shortly after the definition of the relation ( K , A) A((?) apply also to the relation ( K , A) wk A((?). Lemma 6.2.4 shows that (A, K ) f . A(3) whenever A < K+,,SO a discussion of the relation (A, K ) wk A((?) is pertinent. Many special cases need to be considered, and there is no simple result analogous to (A, K ) A(3). We first establish several lemmas. For the rest of this section, write + ( K ) for the number of cardinals less than or equal to K , so -+

-+

-+

+(K)

= I{q; 77 is a cardinal and 0

Thus if K = H, then

+(K)

< 77 < K)I

.

= IaI 4 No.

Lemma 6.3.2. Suppose A + ((?)$,(K). m e n (A,

K)

-+

wk A((?).

Proof. Let d = (Av;u < A) be any (A, K)-family. For each cardinal 77 with 1)
r,

=

{{P,u } E [A]*; I A nA,i ~

= 77)

,

so [A]' is the union of the rq.By the partition relation A ( ( ? ) $ ( K ) , there is H i n [Ale such that [HI2 C rV for some 77. Then (Av;u EH) is a weak A((?)family contained in d . -+

Lemma 6.3.3. For all infinite cardinals K , the relation (2J'(K),K ) f. wk A(3) holds.

148

Decomposition and intersection properties

Ch. 6.3

Proof. Let (AD; P < $ ( K ) ) enumerate in order all cardinals (finite and infinite) strictly less than K . Start with a pairwise disjoint family (X(g);g E @2for some with P < $ ( K ) } where i f g E p2 then IX(g)l = A;. F o r f i n *(,)2 put

so always Mfl = Zl(Ai; < $ ( K ) ) = K . Consider the (2$('), K)-familyd = ( A f ; f E *(,)2). For distinctf, g from $(')2, as before write S(J g) for the least ordinalasuch thatf(ff)#g(cr), and note t h a t A f n A , = U{X(fTP); P < S ( J g)} so

Thus IAfnA,I=by;,) ifS(f,g)isinfinite,and MgnA I=qS(f,g)(Gcf;g)+l) if S(f, g) is finite. Take distinct functionsf, g, h from $($2. We may suppose S ( J g) # 6(g, h ) , and hence IAf nA,I # IAg flAhl. Thus the family d contains no weak A(3)-family, and the lemma is proved.

Lemma 6.3.4. Let A' = K ;then (A', H,) f . wk A@) ff A' = No then (A+, No) f. wk A(A)-

Proof. Take a sequence (Ao: u < K ) of cardinals such that always 1 < A, < A and A = Z(A,; u < K ) . Choose a pairwise disjoint family {X(g);g E X(A,;o< P) for some 0 with < K } , where i f g E X(A,; u < P) then IX(g)l= Hp+l.Write P = X(A,; u < K ) and for f i n P put

so always Mfl = Z(Hp+,; B < K ) = H,. And from Konig's Lemma, IPI > A, so the family d = (Af;fE P) is a P A ' , H,)-family. For distinct .f g from P we h a v e A f n A g = U{X(frP);P<S(J g))so that

Let cx3 = (Af;fEI) be a weak A-family contained in d , where we may suppose f C F! Then S ( f , g) must be constant for all pairs { J g} from [I]', say with value u (so u < K ) . Then If1 < A,, < A. Hence d contains no weak A(A)family. If A' = No,modify the above construction by starting with IX(g)l = + 1 wheng E X(A,; u < P), and noting that 0is finite.

Ch. 6.3

Weak delta-systems

149

Lemma 6.3.5. Let K be regular, and let f be a function, f : [ K ' ] ~ -+ ( 0 , I}. There is a (K','K)-family (A,; v < K ' ) such that whenever p < v < K' then H,nA,I


i f f ( { / & V})= 0 ; /ApnA,I

=K

i f f ( ( p , V})=

1.

Proof. Start with a pairwise disjoint family {Sap;a < K + and < K } where S has power K. Since K is regular, by Theorem 1.3.3 any family ?'3 of K each , pairwise almost disjoint sets each of power K has a K-transversal, that is, there is a set T with 1 Q IB n TI < K for each B i n g . Thus we can inductively choose sets B, of power K for p with p < K' such that B, is a transversal of {Sap; a < p and 0< K } U {B,; CJ < p }

.

Put Sa'= U{Saj;0 < K } , for a with a < K'. Then the S, are pairwise disjoint. An easy induction shows that always B, C u{S,; a < p } . Put A , = S, nB,, so B, = U{A,; a < p } . Also MWl = K whenever a < p , for then S ,I, n BPI 2 1 for each 0(with 0< K ) and S , n B, C S, n Bp =A,. For v with v < K ' , put

A , = S , U U{A,,; p < v and f ( {u,v})

= 1}

.

Then& C A , C_ u {S ,;a
c U{&,;

so IA, nA,[ < K , since IB,

< p } n U{A,,; CJ < V) = B, nB, nB,I < K by choice of B,. This proves the lemma.

CJ

Theorem 6.3.6 (GCH). Suppose K is an infinite successor cardinal. Then ( K ' , K ) f. wk A(K') and ( K + , K ) wk A(K). -+

Proof. To see that ( K ' , K ) -+ wk A ( K ) ,note that if K = q+ then the number y of cardinals < K is the same as the number of cardinals Qq, so y < 7) < K . By Theorem 2.2.4 the partition relation K + + (K): holds, so (K', K ) + wk A(K) follows from Lemma 6.3.2. To establish that ( K ' , K ) f. wk A(K'), note that K + f . (K')$ by Theorem 2.5.7. Thus there is a function f : [K']' -+ { 0, 1) such that if H C K' and f i s constant on [HI2, then IHJ< K'. Take the (K', K)-familysd constructed in Lemma 6.3.5 for this particular function f . Then LA contains no wk A(K')family.

150

Ch. 6.3

Decomposition and intersection properties

Theorem 6.3.7 (GCH). Suppose K is not a successor cardinal. Then 6 ) ( K ' , K ) f. wk A(K), (ii) if Ho < K < H, and 8 < K then (K', K ) +. wk A@), (iii) $K = Ho or $ K = H, then (K', K ) wk A(3).

+

r

Proof. For (i), given any function f i n ,2, put A f = {f a;a < K } and consider the family d = (Af;fE ,2). Then d is a (K', K)-family of pairwise different sets. For distinct L g in we have A,- flA , = {f r a;a < S(J g ) } so IAfnA,I = IS(Jg)l. Let (Af;fE/) w h e r e I C be a weak A-family contained in d , say IAf nA,I = q for distinct J g in I. Then q = IS(ft g)l < K. Thus for all { J g} from [112we have IS(L g)l = q so S(L g) < 17'. Thus 1 1 1 Q

I {f ra;fE K2and a < q+)l = P+


= q++

Hence d contains no weak A(K)-family. To show (ii), suppose No < K < H, and take 0 with 0 < K . The restriction on K ensures that $ ( K ) < K . Because K is a limit cardinal with $ ( K ) , 8 < K then K + ( 8 ) $ ( K )(from Theorem 2.5.10). Hence ( K , K ) -+ wk A(0) by Lemma 6.3.2, so surely ( K ' , K ) +. wk A(0). Finally for (iii), suppose K = Ho or K = H,. Thus K = $(K), SO Lemma 6.3.3 ensures that (K', K ) wk A(3). This completes the proof.

+

Together, Theorems 6.3.6 and 6.3.7 provide a complete discussion of the relation (K', K ) +. wk A(0). We shall now consider the relation (A, K ) wk A(0) in the case No Q h Q K. The discussion breaks into several cases. (The GCH is assumed throughout.) If h < $(K)', it follows from Lemma 6.3.3 that (A, K ) wk A(3), so we may suppose that $(K)' < h Q K . Suppose first that h is not a successor cardinal. By Theorem 6.3.7(i), (A', A) wk A@), and hence ( h , ~ ) wk A@). ) . wk A(0) If 0 < A then h +. (0)$(,) (from Theorem 2.5.10) and hence ( h , ~ + by Lemma 6.3.2. Now take the case that X is a successor cardinal, say h =.'77 If q is not a successor cardinal, by Theorem 6.3.7(i), (q', q ) wk A(q) and hence (A, K ) wk A(q). However, if 0 < q then q' +. (0)$(,, since $ ( K ) < q and so (A, K ) +. wk A(8) by Lemma 6.3.2. The final case to consider is that h = q' where q is a successor cardinal. Here (q',,q) wk A(q') from Theorem 6.3.6, so (A, K ) f wk A(h). And if 0 < h, since $ ( K ) < q = q' it follows from Theorem 2.2.4 that 'q +. (8)$(,) so (A, K ) +. wk A(0) by Lemma 6.3.2. All cases have now been covered. The remaining situations where the strong relation (A, K ) A(8) fails are given by Theorem 6.2.S(iii), (vi). Theorem 6.2.5(iii) states that ( h , ~ ) A(h) if X is singular (and K < A). In fact the example given to establish this result is an example of a (A, K)-family which contains no weak.A(X)-family, so the -+

+

+

+

+

+

+

-+

+

Ch. 6.3

Weak delta-systems

151

stronger negative relation (A, K ) p wk A(A) holds. There remains the result in Theorem 6.2.5(vi): if A is singular with A' < K < A then (A', K ) f . A@). In this situation one can ask if either of the relations (A', K ) + wk A@') or (A', K ) + wk A(A) hold. The outcome is summarized in the following theorem. Theorem 6.3.8 (GCH). Let A be singular with A' < K < A. If A' is countable or hi< K then (A', K ) f . wk A@);othewise (A', K ) + wk A@'). Proof. By Lemma 6.3.4, (A', Nag) wk A(A) so if < K then certainly (A', K ) f . wk A@). If A' = No then even (A', No) f . wk A(A) so (A', K ) wk A@) for all K . So suppose A' < K < Nhl with A' # No. We shall show that in this situation, (A', K ) -+ wk A@'). Take a (A', K)-family d = (Av;u < A') and for a contradiction we shall suppose that d contains no weak A@')-family. There is a least cardinal ? such t h a t d contains a (A', K)-family 39 with 6(c1B) = 17 (so 0 < r] < K ' ) , and we may in fact suppose that d =39 . . Suppose for the moment that q is a successor cardinal, say 77 = 8'. Take any A in d ,and put

e

Since I [ A l e [= K' < K~ < A, if V(A)I = A' then there would be B in [Ale and J in [J(A)Ih' such that A , f Al = B for all v in J. Then for distinct p, u from J since B C A , fA l, it follows that 8 = IBI < lA, nA,I < 6 ( d ) = 8'; hence IA, nAvl = 8. Thus (A,; u € 4would be a weak A@')-family contained i n d , contrary to the choice of d . Hence IJ(A)I < A for each A from PQ . This means that we can choose inductively ordinals u(a) for a with a < A' so that

v(a)E A+

-

U{J(Au(p));P < a>

Then IAv(,) nAvml<8 whenever p< a,so that93 = (A,(,); a < A') is a (A', K)-family contained in 94 with F(g)< 8. This contradicts the minimum property of 7.Consequently q cannot be a successor cardinal. So further O
It follows that 7'# A'. If 77 = No this is trivial since A' # No. Otherwise since 7 < K < NAlrthen q = N, for some a where 0 < a < A'. Since q is not a successor, 77 = Z(Hp; P < a) and hence 77' < la1 < a < A'. Thus indeed q' # A'. Use transfinite induction to choose non-empty subsets I, of A' with always II,I < A, for a with a < K', as follows. Suppose for some a that Zp has al-

152

Decomposition and intersection properties

ready been defined for all 0 with

Ch. 6.3

< a. Put

/(a)= u{1p;(3
K

. X = h. Define [*(a)by

Pya)= { v < h + ; ~ ~ , n ~ ( a ) i & . & Put d *(a)= { A , nA ( a ) ;v E 1*(a)};then d *(a)is a family of the at least g size subsets o f A ( a )with S(A*(a))< g. Since IA(a)l< h and A' # g' it follows from Theorem 1.1.6 that [1*(a)l I { x , ; a < K+}I = K'; yet [A,/ = K . This is the required contradiction. Thus d must contain a weak A@+)-family, and the relation (A', K ) -+ wk A@+) is established.

CHAPTER 6

DECOMPOSITION AND INTERSECTION PROPERTIES OF FAMILIES OF SETS

0 1. Decomposition

properties

The problems to be considered in this section are of the following kind. Let 94 = { A v ;K < A} be a family of subsets of a set S where S has power K , such that the sets in d cover S closely, in the sense that for some cardinal g every g-size subset of S meets almost all the sets in d . Can S, except for a subset of size less than K , be written as the union of a small number of sets B,, each with the property that for any pair x, y in B, there is some A in d with x, y in A? With the following notation the problem can be expressed more precisely. Definition 6.1.l.The family d = { A v ;v < A} is said to have c g x r number 17 if for every B in [ U d Iq,

Definition 6.1.2. Given the family d , a set B is said to be connected in SQ if for all x, y in B there is A in d such that x, y EA . Definition 6.1.3. The symbol ( K , A, 77) 3 6' means the following. Let S be any set with IS1 = K and let d = { A v ;v < A} be any decomposition of S which has cover number g. Then there is a decomposition S = S* U U{B,; p < 6') where IS*[ < K and each B, is connected in s4 . Results on the prope 'ty ( K , A, g) * 6' (in different notation) appear in Erdos, Hajnal and Milner [35, 0 8 11, 13, 141. When considering the symbol ( K ,A, g) * 0 , to avoid trivial cases we shall always suppose that 0 < g G K and that 6' < A, K (the latter since each member of 94 is connected and also each {x} for x in S). Simple considerations

Ch. 6.1

Decompositions properties

131

show that if for particular values of K , A, Q and 8 the relation ( K , A, Q) =) 8 holds, then also (K , A, vl) * 8 whenever q 1 S Q and 8 > 8. We shall start by investigating countable decompositions d of a countable set S. The results are simply stated: if d has finite cover number then S itself is almost connected, if d has infinite cover number then S is almost the union of two connected sets.

Theorem 6.1.4. Let n be finite, n > 0. 7%en(No,No, n ) 3 1. Proof. Let S be a set with IS1 = No;take any family d = {A,; v < w } with U d = S which has cover number n. Suppose S - S* is not connected for every S* in [S]<'O, and seek a contradiction. Note for any subset X of S if thereisMin [w]'O with l { v E M ; x 4 A U } I < No f o r e v e r y x i n X t h e n X i s connected, since given x, y in X, each miss only finitely many A , with v in M and so both are in infinitely many A , . Define by induction elements xk of S and infinite subsets Mk+l of o such that

xi@A , i f j < k a n d ~ E M k + ~ ,

(1)

as follows. Put M O= w . Suppose for some k that XI and MI+^ have already been suitably defined whenever I < k. If ( v E M k ; x @ A , } is finite for every x in S - {XI;I < k} then by the remark above, S - {XI;I < k} is connected, contrary to assumption. so we can choose xk from s - {XI;1 < k} and Mk+l from [Mk]'' such that x k ($ A , for every v in Mk+l. Then (1) holds. Put B = {xo, ..., xnPl}, so IBI = n. Then B n A , = 8 whenever v EM,, and M l ,l Q: No.This contradicts t h a t d has cover number n, and so the theorem is proved. In proving the result when the cover number is infinite, we shall make use of the following lemma.

Lemma 6.1.5. Let the family d = {A,; v < w } give a countable decomposition of a set S. If for no finite subset S* of S is S - S* the union of two connected sets, then 94 does not have cover number NO. Proof. Let d be a family satisfying the conditions of the lemma. We must show that there is an infinite subset T ofS such that I { v < o ; A , f l T=@)I=K,. For any subset X of S, let G(X) be the graph with vertex set S - X and two vertices x, y joined by an edge just when {x, y } A for every A in A. Thus the independent subsets of G(X) are exactly the subsets of S - X connected

132

Decomposition and intersection properties

Ch. 6.1

in 94 . This means that the assumptions on 94 ensure that for any finite subset S* of S, the graph C(S*) has chromatic number more than 2. Hence by the theorem of de Bruijn and Erdos (Theorem 5.3.1), there is a finite subgraph of G(S*) with chromatic number more than 2. Inductively choose finite subsets S k of S (where k < w ) such that the subgraph G k of G(U{Sj; j < k } ) spanned by S k satisfies Chr(Gk) > 2. For each k, then S k has only finitely many subsets and so we can choose inductively sets M k from [ o ] ’such ~ that A , n S k is constant, say A , n S k = S Z , for each v in M k , and M k c il{Mi;j < k } . Since M k is non-empty, certainly each S,$ is independent for the graph G k . Since Chr(Ck) > 2 it follows that S k - SZ is not independent. Thus we can choose x k , Y k in S k - s,*which are adjacent in Gkr and so { x k , y k } C A , for all v . For all k , choose v(k) from M k - { v ( j ) ; j < k ) . Form the partition [a]’= A, U A , where for j , k in w with j < k, {jjk } E Ao

*Xk

4 AVO.)

By Ramsey’s theorem (Theorem 2.2.9) there is H in such that either [HI’ C A0 or else [HI2 C A,. Suppose [HI2 C A,; the other case is similar. For j , k in H , i f j < k then x k 4 Avo) by definition of A,. If j 2 k then u( j ) E Mi C M k so A,u) n S k = S:; since x k E S k - S$ again x k 4 A,c). Thus if T = { x k ; k E H } then T is infinite and A , n T = 8 whenever u E { v(j); j E H } , and I ( v ( j ) ; j EH}I = KO. This suffices to prove the lemma.

Corollary 6.1.6.For all infinite K , the relation

(K,

KO, No) * 2 holds.

The result in Corollary 6.1.6 cannot be improved to (No,No,No) * 1 , as the following simple example shows.

Theorem 6.1.7. Forall infinite K , Proof. Put S = { 0,1} X each v,

K

(K, K , K )

91

and consider the familyd = { A , ; u < K } where for

A , = { ( o , p ) ; p < V } u {(l,p);U
94 has cover number K . For take M from andputA(M)=U{A,;uEM}. F o r a n y p w i t h p < ~ ,t h e r e i s v i n M such that p < v and (0, p ) E A,; hence { 0) X K C_ A(M). Take any B in [ S I K , and put M = { v < K ; BnA , = 81, soB n A ( M ) = 8. If !MI = K , we would then [KIK

Ch. 6.1

Decompositions properties

133

have to have B C { 1) X K . But then for any v from M there would be p with v such that (1, p ) E B , and thus ( 1, p ) E B fl A , contradicting v EM. Thus for any B in [S]" it follows that I { v < K ; Bn A , = @}I < K , and so d has cover number K . For any S* in [S]'" there are p , v with v < p and (0, p), (1, v> E S - S*. Since then { (0, p), ( 1, v)} A for every A in d , it follows that S - S* is not connected. Thus the family4 gives an example that shows ( K ,K , K ) 9 1. p2

The relation ( K , h, 7) * is equivalent to a special case of the polarized partition symbol from Chapter 4, as is shown i n t h e next theorem.

Theorem 6.1.8. Let K ,h and q be infinite cardinals. The relation ( K ,A, q) =+ 1 holds ifand only if

Proof. Suppose first that ( K , h, 77) + 1. Take any disjoint partition K X h = A0 U A,; we seek Ho in [ K ] ~and H1 in [A]' such that Ho X H1 C_ A, or HO X H1 C A1. AS in the last proof, put S = { 0 , I} X K and consider the family d = { A , ; v < A} where this time, for each Y , A , = { (i, a ) ;( a ,v) 4 Ai} . Take S* from [S]'". There is a such that (0, a),( 1, a) E S - S* and so S - S* is not connected in d.Thus S = U d is not almost connected, and so by the relation ( K ,A, q) =+ 1, the family 4 can not have cover number q. Thus there is B in [S]* with I ( v < h;B nA , = @}I = A. There are HO in [ K ] and ~ i from { 0, 1) such that (i, a)E B for all a in Ho. Put H1 = { v < A; B fA l, = @},so H1 E [A]'. Then if a E HO and v E H1,it follows that (i, a)4 A , so (a,v) E Ai. Hence HO X H1 C Air and the relation (1) holds. Now suppose that (1) holds. Take a family 4 = { A v ;v < h} where IUdl = K , which has cover number q. Suppose that U d - S* is not connected for any S* in [ U d 1'". Then we can define by induction sets B, and elements xa, y a where a < K as follows: B, is a maximal connected subset of U d - ( { x p ; /3 < a} U { yp; /3 < a});we choose x, from B, and yorfrom ud - (B, U {up; 0 < a}).Thus for no A in d is it true that x, y, E A. Form the disjoint partition K X h = A0 U A1 where

~Y,V)EAO*X,EA,. By the relation (l), there are HO in

[ K ] ~and

H I in [h]' such that HO X H1 _C A0

Ch. 6.1

Decomposition and intersection properties

134

or Ho X H1 C_ A,. Both {x,; a! €IfO) and (y,; a! E H o }have power 7,so since SQ has cover number g it follows that I { v < A; V a €Ho(x, $ A,)}I < A and I {v < A; V a € H o ( y a 4 A,))I < A. Since lHll = A, there must be a l ra2 in Ho and v l , v2 in H1 such that xal € A , , andya2 € A , , (so certainly xa2 $ A v 2 ) .Thus (a1,vl) E A, and (az,v 2 ) E A1 contrary to the property of such that U d - S* is Ho. H I . This shows that there must be S* in [ U d connected, and so ( K , A, 17) =) 1.

Corollary 6.1.9 (GCH). For infinite cardinals K , A the relation ( K , A, K ) holds ifand on& if { K , K ' , K', ( K ' ) ' ) 6 {A, A', A', (A')') 0.

=)

1

Proof. By Theorem 6.1.8 and Theorem 4.2.8. We shall now seek to extend the earlier results on countable families to higher cardinalities. We consider first the case when the family has power equal to a successor cardinal. In Theorem 6.1.10 we shall show that ( K , A', A') =) A (for arbitrary K ) , a result not as strong as the relation ( K , No, No) * 2 established in Corollary 6.1.6. However, at least when K = A' this result is the strongest possible, for in Theorem 6.1.12 we prove (h', A', A') 8 for any 8 with 8 < A.

+

Theorem 6.1.10. Let A be infinite. Then ( K , A', A')

* A.

Proof. We may suppose K 2 A'. Let S be a set with IS1 = K and take any family d = { A , ; v < A'} with U d = S which has cover number A'. We shall show that S = U(B,; p < A} where each B, is connected in s4 ; then the relation ( K , A', A') follows. For x in S, put A(x) = { v < A';x EA,}. Then S is the disjoint union S = So U S1 where So = {x E S;IA(x)l G A) and S1 = {X E S;IA(x)l= A'}. =)

F o r e a c h x i n S 1 , p u t D ( x ) = {yES1; 3 A € d (x,yEA)}.Suppose D(x)l = A'. Then I ( v < A+;D(x) f l A , = 0}1< A, because ?A has cover number A'. Since IA(x)l = A' there must be v in A(x) with D(x) f l A , # 0;thus there i s y with x, y € A , yet y ED(x), which is a contradiction. Hence ID(x)[ < A for all x in S1. For x in S1 inductively define Dn(x) for n with nGoby

DO(X) = {x};Dn+l(x) = U { D ( y ) ; y ED"(x));DW(x) = U { D " ( x ) ; n < w ) . Then IDw(x)l < A, and D(z) C_ Dw(x)whenever z EDW(x). Inductively define subsets T, of S1 as follows: if S1 f u{Tp; < a} choose x f r o m S 1 - u{Tp;p
Ch. 6.1

135

Decompositions properties

< (Y then y @ Tp soy 4 D(z); thus there is A in d with y , z E.4. And always IT,I < h so we can write T, = {x,,; p < A}. For p with p < h let C, be the set of all the x,. Then S1 = U{C,; p < A} and each C,, is connected. So to prove the claim above, it is sufficient to show that So is the union of at most h connected sets. By way of contraduction, suppose that this is false. Then for (Y with (Y < A', define inductively elements x, of So and ordinals v(01) with v(a) < h' as follows. Suppose xp and v(P) already defined whenever P < (Y. Put z E Tp where

u, =So - ({xp;P<(Yl u U { A , ( p ) ; P < a } ) . Now la1 Q X and the sets {xp} and A,@) are all connected, so since So is not the union of h connected subsets, certainly U, # 8. Choose x, from U,. Since always xp E So we know IA(xp)l< A. Hence we can choose v(01) from A+ - ( I@); P < a) u U{A(xp); P < 4) . This completes the definition of x, and v(01). It follows that cu,P
V(P)

'

for if B< (Y this is ensured by the choice of x, and if 2 a by the choice of v@). Put B = {x,; (Y < A'} and I = { v(01);(Y A'}. Then IBI = 1 1 1= A+ and Bf Al , = 8 whenever v E I. This contradicts that SQ has cover number A+. Hence SO,and so also S, can be written as a union of h connected sets. This

<

proves the theorem.

Before showing that this last result is the best possible in the case K = A+, we state a lemma. This lemma provides a strengthening of Lemma 2.7.3.

Lemma 6.1.11 (GCH).Let S be any set with IS1 = K'. Then there is a disjoint partition [S]' = U{ Ak; k < K + ) with the property that whenever A E [SIK and B E [SIK+then there is a in A such that for every k with k < K + there is b in B with {a, b } E Ak. Proof. Modify the proof of Lemma 2.7.3. Again identify S with

K+.

Put

d = { ( A ,g ) ; A E [ K + I K a n d g : A + K + } . By GCH,I A (K +)I = K + for each A in [ K + ] ~and so I dI = K'. Take a well ordering ((A,, g,); (Y < K ' ) of d . For each r with r < K ' , put d 7= { ( A , , g,); so ld,I

(Y

< r and A ,

C r and g [ A , ] C 7) ,

< K . Suppose now that for each r with r < K +

we could find a func-

136

Ch. 6.1

Decomposition and intersection properties

tion f, : T

-+

T

such that

if ( A , g) E d,then &(u) = g(u) for some u in A Consider the partition { Ak; k

{ U , T } E Ak

*f;(U)

=k


.

(1)

of [K']' where for u, T in K + with

(7

< 7,

.

This partition has the new property, for take A from [ K ' ] ~and B from [K'IK+. Suppose on the contrary that for all a in A there is k such that {a,0) E Ak for no 0in B , so there is a functiong : A K + such that always {a,0) 4 Ad,) for all 0in B. Let 70 be the least ordinal such that ( A , g) E dT0. Since IBI = K + there is T in B with T 2 r0. Then also ( A , g>E A, so by (1) there is u inA for whichf,(u) =g(u), that is, { u , T} E A,,). But since u € A and T E B, by choice of g we know [ u, T} 4 A,(,,). This contradiction proves the result. In order to prove the lemma then, it suffices to find functionsf, : T T with the property (1). We may suppose 94, # 0 , for otherwise any function f, satisfies (1). Thus T 2 K ,so 171 = K ,and we may write 94, = { (A,,, g7& a < K}. Since always IA,I = K we can choose inductively u,, for a with a < K such that uTO1 € A , - { u,p; 0< a}.Define f, : T -+ T as follows: if u = u, thenf,(u) = g,,(u); otherwise f,(u) is arbitrary. Then clearly (1) holds, since if ( A , g) E 94, then ( A , g ) = (A,,, g,,) for some a,and then u7& is an element o f A on which bothf, and g agree, as required by (1). This completes the proof of the lemma. -+

-+

Theorem 6.1.12 (GCH). Let h be an infinite cardinal with 6 < A. Then (A+, A+, A+) p 8. Proof. Using Lemma 6.1.1 1, take a disjoint partition [A']' such that A E [A+]* and B E [A+]*+* 3 a € A V k < A+ 3

=

U{Ak; k

< A+}

0E B((aJ3)E A,).

Put S = h+ X h so IS1 = A+, and for /3 with 0< A+ put Ap=

((ak ) E S ; {(U,fl}EAk}

We consider the decomposition of S given by the family 94 = ( A D 0 ; < A+}. First observe that d has cover number A+. For take any X in [SIh+,and for a contradiction suppose that if

I=

{p<x+;xn~~=0}

then 111 = A'. Put X ' = {a< A+; 3 k < h ((a,k ) EX)}; since

1x1= h+ also

Decompositions properties

Ch. 6.1

137

IX'I = A+. By the choice of the partition, there is a in X ' such that for all k with k < X+ there is 0in I with { a ,0) E Ak. Since a Ex' there is ko < h with (a,k o ) E X and then there is 00 in I with {a,00) E Aka, that is, (a, k o ) E A p o .Hence (a,k o ) E X nApo,contradicting that 00 EZ. Thus indeed, whenever X E [ S ] h + then I {/3 < A+; X nAp = @}I < A+, so d has cover number A+ as claimed. However, S is not almost the union of 8 connected sets. Note that since the partition is disjoint, whenever {a. 0) E Ak then it follows that ( a , 0) 4 A, if I # k. Thus if (a,k ) EAp then (a,I) 4 Ap whenever I # k, so no connected sets includes pairs (a, k ) , (a,I) with I # k. Now take any S* from [S]<". There must be a such that (a,k ) E S - S* for all k with k < h. Suppose S - S* = U{B,;p < 8 ) . There must be some p such that for distinct k, I both (a,k ) E B, and (a,I ) E B,, so this B, is not connected. Hence S is not almost the union of 8 connected sets. Thus the family d provides an example to show that (A+, A+, A') f 8 . In almost the same way, we can prove the following result where the covering number is reduced to A.

Theorem 6.1.13 (GCH). Let h be an infinite cardinal with 8 + < A. Then (A+, A+, h) f 8 . Proof. Almost as for Theorem 6.1.12, only using the set S = h+ X 8' If the cardinals h and r ) in the relation ( K , A, q) =$8 are far apart, then positive results can be proved. If A' > q+ then by Theorem 4.2.8

and hence whenever

K

2 q,

If we now apply Theorem 6.1.8, the following result is immediate.

Theorem 6.1.14 (GCH). Let Then ( K , h, q) =$ 1.

K,

X, r ) be cardinals such that K 2 r ) and A' >.'77

If r ) is replaced by r)+ in the above theorem, then GCH is not needed in the proof. In fact we have the following.

138

Decomposition and intersection properties

Theorem 6.1.15. Let

K,

A, q be cardinals such that

K

> r ) and A' > r)'.

Ch. 6.1

Then

(K,h,V')=*l.

Proof. Suppose A' > r)', let S be a set with IS1 = K and take a family 4 = { A , ; v < A} with U 4 = S which has cover number 77'. For x in S, put A ( x ) = { v < A; x E A , } . Then S is the disjoint union, S = SO U S1 where So = { x E S ; IA(x)l < A} and S1 = {x ES;IA(x)l= A}. If lSol = K , we could take S' from [So]"', and noting that then IU{A(x);x € S'}l < A, we would have I {v < A; S' nA , = @}I = A, contradicting that 94 has cover number 77'. Hence (Sol< K . Thus to prove the theorem, it suffices t o show that S1,but for fewer than K elements, is connected. Let B be a maximal connected subset of S1. Then I claim that IS1 - BI < r). For if not, take B' from [Sl - B]"'. S i n c e 4 has cover number Q', then I { v < A; B' nA , = @}I < A. Take any x in B. By the maximality of B, for any v if x € A , then A , C B, so if x € A , then B ' n A , = @ . T h u s I { v < A ; x E A , } I < A , thatis, IA(x)l
Theorem 6.1.16. Let A be a singular cardinal with A' = H,,. Then ( K , A, A) for all K .

=$

A,

Proof. We may suppose K > A. Let (A,, n < w ) be an increasing sequence of regular cardinals with always Ho < A, < A and A = C(A, n < w). Let S be any set with I S1 = K , and take a family 4 = { A , ; v < A} where Ud' = S and has cover number A. Take any set S' from [S]"'. We start by observing that if there is I in [A]' for which B E [S' Ian * I { v € / ; A , nB = @}I < A ,

(1)

then S' can be written as the union of A, connected sets. For suppose (1) is true. For x in S ' , put A ( x ) = {v E /;x € A , } and write S' as the disjoint union S'=SA U S ; where$ = { x € S ' ; M ( x ) l < A } a n d S ; = {x€S';!.A(x)l=A}. For x in S ; , put D ( x ) = { y E S; ; 3 v E /(x, y € A , ) } . It follows that if x € A , then A , n D ( x ) = 8 , so A ( x ) C { v E I ; A , nD ( x ) = 0). Hence ID(x)l< A,

Ch. 6.1

139

Decompositions properties

for if ID(x)l 2 h, by (1) then I { v E I ; A , fDl( x ) = 811 < h contradicting that M(x)l = A. For x in S ; inductively define D"(x) for n with n < w by

.

DO(X) = {X};Dn+l(X)= U { D ( y ) ; y€ D " ( x ) } : D W ( x )= U { D , ( x ) ; n < O} . Then IDW(x)l< A., Inductively define subsets T, of S; as follows: if U { Tp;0< a } , choose x in S ; - U { Tp;0< a } and put T, = DW(x)u{Tp;0< a}. Then always IT,[ < A., It now follows, just as in the proof of Theorem 6.1.10, that S; can be written as a union of A, connected sets. To complete the proof of the observation above, we must show that the same is true of S;. If IS;/ < A, this is trivial, so suppose IS;/ 2 h,. For any x in S; Since hh = A, > NO,there there is an integer m ( x ) such that IA(x)l < is a subset Si of S; with S I [; = h, such that m ( x ) is constant for x in S:, say m ( x ) = m. Thus I U { A ( X ) ; XES;}I < hn . Am < h, SO I { Y E l ; S : nA , # @)I < A. Since ISil = A, this contradicts ( I ) , and so it is impossible that IS;[ 2 .A, Thus the observation is established. We shall show that S can be decomposed into h connected sets; from this the theorem follows. For a contradiction, suppose this to be false. Use induction on the integer n to define sets I, in [A]', S,* in [S]', and I,* in [A]', as follows: Put 10 = h and choose 1; from [Io]'O. Write S1 #

S, = S --

U { A , ; v E 1; for some m with m < n } .

Now I U { I ~ ; m < n } l < C ( h , ; m < n ) < h . s o s i n c e S i s n o t theunionofh connected sets, neither is S,. In particular IS,I > h so ( I ) , applied in particular to S, and I,, must be false. Thus there are S,* in [Sn]', and In+l in [l,]'such This deChoose I,*,, from [I,+l]'n+l. that A , fl S,* = 0 whenever v E finesS,*, In+l and li+], and completes the definition. Put S* = U { S , * ; n< c ~ }and I* = U { l , * ; n< w } so IS*l= h and 11*1= h. However, S* n A , = 8 i f v E I * ,

(2)

for suppose u €1;. If m < n by definition ofS, then S,* n A , = 0. If m > n, then since 1; C I,,, C I n + l ,by choice of S,* and In+l we know A , n S,* = 0. In both cases then, S,* nA , = 0 and so indeed S* fl A , = 8. Since IS*l = A, (2) contradicts that SQ has cover number h. This contradiction shows that S can be decomposed into h connected sets, as required. For small values of K , the result in Theorem 6.1.16 can be improved sub. stantially. From Theorem 4.2.5, when A' = Ho then

140

Decomposition and intersection properties

Ch. 6.1

so by Theorem 6.1.8, the relation (A', h, A) * 1 holds. When K = A, we shall show that ( K , h, A) * 2. In view of Theorem 6.1.7, this is the strongest positive result possible. Theorem 6.1.17 (GCH). IfX ' = No then (A+, A, A)

* 1 and (x. A, A) * 2.

Proof. Only the second relation remains t o be proved, and in view of Corollary 6.1.6 we need only consider the case where X > NO.Let (A,; n < w ) be an increasing sequence of regular cardinals with always H l < h, < h and A = C(h,,; n < 0).Let S be any set with IS1 = A, and consider any decomposition PQ = { A v ;v < h} of S which has cover number h. For a contradiction, suppose that whenever S * E [Sl" then S - S* is not the union of two connected sets. The argument that follows is similar to that in the proof of Lemma 6.1.5. For a subset X of S, let C ( X ) be the graph on S - X with vertices x, y joined by an edge in G ( X )just when {k,y } A for every A in d . Then when S* E [S]
Since H l

< h,

= Ah

< hi < h, from Theorem 4.2.8 it follows that the relation

holds, where y = 2k(n).Thus there are 1; in [I,!JAn,Ml in [AIh and i(n) < y such that a E 1 ; andv€M;*S,

"A,,=X,i(,).

Further, we can arrange inductively that M; C_ fl {Mk;m < n}. Choose inductively I,, from [Z; - U{Im; m < n}]& and M,,from [M;- U { M m ; m< n}]'" a n d l e t I a n d M b e the disjoint u n i o n s I = U { I , , ; n < w } , M = U{M,,;n<w}.

Ch. 6.2

Delta-systems

141

Fix a in I. There is a unique n such that a E In. The set X,i(n), being contained in some A , , is independent in the graph G,. Since Chr(G,) > 2 it follows that S, - X,i(n) is not independent, so we can choose x,, y , from A for all A in 94. Take S, - X,i(n) which are adjacent in G,; thus {x,, y,} a disjoint partition Z X M = A, U A , such that

(a,V) E A0 *x,

4 A , ; (a,V) E A,

*yol 4 A , .

From the proof of the polarized canonization lemma, Lemma 4.1.7, there are sets I,* in [@, M,* in [MIhn and a function h : w X w -+ { 0,1} such that if m, n < w , then

I;

x

C Ah(m,n) ,

and moreover there are increasing functionsf, g : w -+ w with alwaysf(rn) < g(m) such that I,* C_ If(n) and M,* C Mgcn). By Kamsey’s theorem (Theorem 2.2.9) there are H in [ w ] , and i in { 0,l) such that whenever m, n E H with m > n then h(m, n ) = i. Suppose without loss of generality that i = 0. For m, n in H with m > n then I ; X M,* C Ao; thus if a E I; and v EM,* then x, 4 A,. On the other hand, if rn < n then 1; C Z&) C Z i m ) and M,* C Mg(n)C_ Mgcm) C M f ( m )so for a in I; and v in M,* it follows that S, nA , = X,i(f(m));since x, ES, - X,icf(m)) thus x, 4 A,. Put T = {x,; a E U{Z,*; n < w } } and N = U{M,*;n < w } . Then IT1 = X, N I 1= X and T nA , = @ whenever v € N. Thus [ { v < A; T nA , = @}I = A, which contradicts that d has cover number A. This contradiction proves the theorem. For more results and unsolved problems connected with the relation (K, A, r ] ) * 0 the reader is referred to Erdos, Hajnal and Milner [ 3 5 ] .

52. Delta-systems Let d = ( A i ;i € I ) be an indexed family of sets. The family d is said to contain the family 9 = ( B j ; jEJ)if there is a one-to-one m a p f f r o m J into I such that always Bj = AfO.).For the purposes of this chapter, if d contains 9 and a l s o 9 contains d then we may consider the indexed families d and

93 as being the same. In particular, we may always suppose that an indexed family d is indexed by the ordinals less than some cardinal, so d = ( A , ; v < A). A family d = (Ai; i E I ) is said t o be a (A, K)jizmily if [ I [= X and Mil = K for each i. Expressions such as (X,
Ch. 6.2

Decomposition and intersection properties

142

94 = (Ai; i E I ) is said to be a A(X)-farnilyif 1 1 1= A and A ifl Ai = A nA l whenever { i ,j } , { k , I } E [I]'. Definition 6.2.1. Let 0 , K , A be cardinal numbers. The symbol (A, means the following: every (A, K)-family contains a A(O)-family.

K)

-+

A(0)

We shall ask for which values of 0 , K , A the relation (A, K ) A(0) is true. The first result of this kind in the literature seems t o be due to Mazur [67], where the relation (Hl,6) A(H1) is proved. The general problem was first discussed by Erdos and Rado [31]. Michael [68] independently discovered some o f their results. Erdos and Rado returned t o the problem in [33], and there completely answered the question of finding for given K and 0 (not both finite) the least A such that (A, < K ) A(0). In Theorem 6.2.5 below we shall solve the easier problem of deciding the truth of the relation (A, K ) A(0), under the assumption of the GCH. (This relation is easily seen to be equivalent to (A, < K + ) + A@).) For a discussion not depending on the GCH, the reader is referred to Erdos and Rado [33] where the GCH is not assumed. A couple of simple remarks are in order. Every (A, K)-family where A 2 2 contains a A(2)-family, so in considering the relation (A, K ) -+ A(0) only values of 0 with 3 < 0 < A need be considered, and (A, K ) A(A) is the strongest positive result while (A, K ) f . A(3) is the strongest negative result. Clearly if the relation (A, K ) A(0) is true, then if A1 2 A and 0 G 8 also the relation (Al, K ) A(O 1) holds; and moreover if K 1 < K also the relation (A, K 1) A(0) holds. To verify the last claim, suppose given a (A, K &family d . Choose pairwise disjoint sets C,, one for each A in d , such that U CAI = K and consider the (A, k)-family C = { A U C , ;A € d}.Since for A , B from d we have ( A U C A )fl ( B U CB)= A fl B, any A(0)-family contained in C gives a A(0)family contained in d . The discussion is based on the following two lemmas. The proof for the first follows Davies [8]. -+

-+

-+

-+

-+

-+

-+

-+

Lemma 6.2.2.Suppose that A is an infinite regular cardinal, that K that qK < X whenever 77 < h Then (A, K ) A@).

> 1 and

+

Proof. Take any indexed family d = (Au;u < h) where always MuI = K . If A of the A , are the same trivially d contains a A@)-family, so we may assume this does not happen and in fact since A is regular we may assume all the A , are different. Put A' = U ( A u ;u < A), so IS1 < K . A = A. In fact IS1 = A, for if MI = 77 < A

Ch. 6.2

143

Delta-systems

then S has only qK different subsets of power K , and '7 < h. Take any well ordering< o f S with order type h. For each v, let ( x ( v ,a);a < E,) enumerate A , in increasing<-order, so I E , ~ = K . For a with a < K ' , put S, = { x ( v , a); v < A}. S i n c e S = u { S , ; a < ~ ' }and IS1 = h with h regular and K + < 2 K < h , there must be some S, with IS,] = h. In fact, let /3 be the least ordinal with ISpl = A. Put T = u{S,; a
Lemma 6.2.3. Let 8, for v with v < K be cardinals such that 8, i 1 < 8. Put h = Z(n(8,; v < a);a < K ) . Z%en (A, < K ) f. A(8).

Proof. For each a, let P, be the Cartesian product of the 8, with v < a,and put d = U{P,; a < K } . Consider the functions in d as sets of ordered pairs. Certainly Id I = h and I f I < K for each f i n d . Consider 4 as an indexed family of distinct sets, indexed by h; then d is a (A,
K

is infinite and h S

K'.

Then (A,

K)

+ A(3).

Proof. The relation ( K ' , K ) A(3) is an immediate corollary of Lemma 6.2.3. It can also be seen directly by considering the family (a;K < a < K'). Then certainly if h < K ' , also (A, K ) f . A(3).

Decomposition and intersectioriproperties

144

Ch. 6.2

By Lemma 6.2.4, in considering the relation (A, K ) +. A(O) we may henceforth suppose that K' < A. The next theorem covers all non-trivial cases in which A + K 2 No. (Unsolved problems remain if both A and K are finite. See Erdos and Rado [3 11.) Theorem 6.2.5 (GCH). Suppose that A is infinite and K 2 1. Then (i) ij'h is inaccessible and K < A, then (A, K ) +. A@); (ii) if A is singular and K , O < h, then (A, K ) +. A(O); (iii) i f A is singular and K < A, then (A, K ) A@); (iv) if K < A', then (A', K ) +. A@'); (v) ij'h is singular and K , O < A, then (A', K ) + A(O); (vi) i f A is singular and h' < K < A, then (A', K ) f , A@).

+

Proof. Part (i) is immediate from Lemma 6.2.2. For (ii), take K and O with O < X. By Lemma 6.2.2, ((OK)', K ) +. A(@")'). Since (by GCH) (OK)'< A, and O < (OK)', certainly (A, K ) +. A(O). From (ii), (v) follows trivially. Lemma 6.2.2 implies (iv), noting that A+ is regular and that by GCH, hK= h < A' when K < A'. For (vi), take K with A' < K < A. Take a sequence (A,; v < K ' ) of cardinals such that always A, < A and (A,; v < A') is a sequence cofinal in A. For a with a < K ' , always n(h,; v < a )< hla'< A' by GCH, and n(A,; v < A') = A' by Kiinig's Theorem. Hence K,

A+

< C(IA,;

v < a);a < K+)

< A+ . K + = A+

,

+

so by Lemma 6.2.3 it follows that (A', < K + ) f , A@); hence (A', K ) A@). Only (iii) remains, so suppose A is singular. Write A as a disjoint union, h = u{S,; u < A'} where always IS,[ < A. Take pairwise disjoint sets A , for u with u
Lemma 6.2.4 shows that a (K', K)-family may fail t o contain any non-trivial A-family. However, if some restriction is placed on the ( K ' , K)-family then positive results are possible. We shall investigate ( K ' , K)-families d such that every two sets from d have an intersection of power less that 0 , for some 0 with O < K . Since the sets from such an indexed family d = ( A [ ;i E I ) are pairwise distinct, we may identify 4 with the non-indexed family { A t ;i € I } and refer t o the degree of disjunction 6 ( d )as defined in Chapter 4. The following theorem is from Erdos, Milner and Rado [39].

Delta-systems

Ch. 6.2

145

Theorem 6.2.6 (GCH). Let d be a ( K ' , K)-famiZy with 6(d) = 6 where 8 < K . (i) r f e < K ' then d contains a A(K')-familY. (ii) Lf K ' < 8 < K and 9 < K then d contains a A(q)fumily. Proof. Let the family d = { A , ; u < K + } be given, with 6(d)= 8 and IA,I = K for each u. Use transfinite induction to choose subsets fa! Of K + for a with a < K" so that I, is a subset of K' - U{Zp; p < a } maximal with the property that { A , - U{A,; p E f p for some 0with /3 < a};u €1,) is a pairwise disjoint family. For brevity, put [(a)= U{I,;/.?
A(a)= U{A,:p€Z(a)} ;

thus 1, is a subset of K' - [(a)maximal such that { A , - A(a);u €1,) is pairwise disjoint. Thus in particular { p , u ) E [I,]*

since by choice of[, U EK+

-

*A ,

A,

c A(a)

(1)

9

it follows that (A, - A ( a ) )n (A, - A ( a ) )= 8. Further,

I(&) *

n A(&)[2

.

(2)

For take any /3 with 0 < a. Since u 4 f p by the maximality off, there is p(p) in f p such that (A, - A @ ) ) fl(A,@) - A @ ) ) # 8. Hence we can choose xp from ( A , nA,@))) - A @ ) . And if 7 < P thenx, Zxp since x, € A P ( ? )C_ A ( @ . Thus {xp; p < a} C A , n A(a),so indeed IA, n A ( a ) l 2 IaI. Now we can prove (i). So suppose 8 < K ' , and we wish to show that d contains a A(K+)-family. Assume first that I1,l < K for all a with a < 8. Then If(0)l < 8 . K = K , so IA(O)l< K . K = K and IK+ - f(8)l = K + . By (2), IA,nA(O)I 2 8 for each u in K + - 1(8), so we can choose B, from [ A , nA(8)l'. However, [A(O)]' < K' = K and so there must be distinct 1.1, u in K + - Z(8) for which B,, = B,. But then IA, nA,I 2 IB,I = 8, contradicting that 6 ( d )= 8. Thus there must be a with (Y < 0 for which I1,l= K', and in fact let Q be the least such ordinal. Then IA(a)l < IaI . K . K = K . Put

P = { u E f,; IA, n A ( ~ )2I 8); Q =

{U E I,;

IA, n A@)I < el .

Consider what would happen if IPl = K'. Then for each u in P, choose B, from [ A , nA(a)le.Since I[A(a)]'l < K' = K , there must be distinct p , v in P for which B, = B,. But then IA, nA,I 2 B I I, = 0 , contrary to F(d)= 8. Thus in fact IPI < K ' , and so IQl = K + . Since I [A(c~)]<~l < K' = K there must be A in [A(a)]<' and Q' in [Q]"' such that A , nA(a)= A whenever u E Q'. Thus (A, nA(a);v E Q') is a &+)-family and hence by (l), { A , ; v E Q'} is a A(K')-family, and is contained in d . Thus (i) is proved. Now consider (ii). Thus K is singular, K ' ,< 8 < K and 9 < K. By increasing

Decomposition and intersection properties

146

8 if necessary, we may assume that 8' f

K'.

Ch. 6.2

Suppose II(K>I< K . Then IA(K)I <

= K ; . By (2)) [ A , nA(K)I 2 K whenever V E K + - I ( K ) . Thus there is a decomposition 9 = { A , n A(K); v E K + - I ( K )of} a set of

K

'

K

= K and

IK+ - I ( K ) I

power K into K + sets each of power at least K , with 6(30) < 8. By Theorem 1.1.6 this is impossible. Hence II(K)) = K'. Since I ( K=)u{Z,; a < K } , there must be a with a < K such that II,I = K ' , and we may suppose that a is the least such ordinal. Then w(a)l< K . K . K = K . Again by Theorem 1.1.6 there cannot be I in [I,]K(+ such that IA, n A(a)l= K for each v in I. Thus there must be I in [I,]"' such that 14, A(a)l < K for each v in I. There is a cardinal K with K < K and a set I' in [IIK' such that IA, nA(a)l=K 1 for all u in Z'. By Theorem 6.2.5(v), ( K ' , K + A(q). Thus there is I" in [Z'Iq such that ( A , n A(&);u E I " ) is a A(q)-family, and so by (1) also { A , ; v E I " ) is a A(q)-family. This proves (ii).

The result in (ii) in the theorem above is the best possible in the sense that if K is singular then there is a ( K ' , K)-familyd with 6 ( d )= K ' which contains no A(K)-family. For take a sequence ( K ~ u; < K ' ) of cardinals below K with K = Z(K,; u < K ' ) , and let P be the'cartesian product of the sets K,. For eachf in P,put Bf= { f r o ;u < K ' } . Take pairwise disjoint sets C'with ICfl = K , disjoint from all the B f , and put A f = Bf U 'C T h e n d . = { A f ; f E P} is a (K', K ) family with 6 ( d ) = K ' . Suppose { A f ; f €Q} where Q C P is a A(lQl)-family. For distinct f, g from Q then A f n A , = Bf n Bg,and there is u with u < K ' such that f r u 4 Bf flB, (for otherwise Bf = Bg and s o f = g). Clearly the least such u is a successor ordinal, u = T + 1, andf(7) # g ( ~ )Further, . u is the same for all pairsf, g from Q. Hence lQl < K , < K . The restriction that 8 < K in Theorem 6.2.6 is justified by the following example of a (2K,K)-family d with 6Cd)= K which contains no A(3)-family. F o r e a c h f i n K 2 p u t A f = { f r a ; a < ~ } . T h e n the family A= { A f ; f E K 2 } is a (2", K)-family with 6(d)= K . Suppose { A f ; f E I }where I C K 2is a A(lIl)-family. For distinct f, g from I there is a with a < K such that f r a 4 Af n A,. The least such a is a successor ordinal, a = + 1, and moreover is constant for all pairsf, g from I. Thusf(0) #g(@ for distinctf, g from I , SO that 1 1 1< 3. Hence d contains no A(3)-family. Lemma 6.2.4 gives also the result (A, K ) A(3) if X < K , and one may ask if a restriction to the degree of disjunction of the (A, K)-family will lead to a positive relation, as happened in Theorem 6.2.6 when X = K + . However, no such positive relation is forthcoming; the following gives an example of a ( K , K)-family d with 6 ( d )= 2 which contains no non-trivial A-family. For a with a < K , put A , = {(a,0); 0 < K } u { ( p , a);p < a}.Then if y < a < K , clearly A , fl A , = {(y,a)}. Hence (A,: a < K ) has the required properties.

+

Ch. 6.3

Weak delta-systems

147

83. Weak delta-systems An indexed family d = (Ai;i E I) is said to be a weak A@)-family if M i nA,l = IAk nAIl whenever { i , j } , { k , I } E [I]*.Clearly every A@)-family is also a weak A@)-family, but not conversely. We shall consider those cases when the relation (A, K ) A((?) from the last section is false, and ask about the truth of the weaker requirement that every (A, K)-family contain only a weak A((?)-family. = A and

-+

Definition 6.3.1. Let 8, K , A be cardinal numbers. The symbol (A, K ) wkA(0) means the following: every (A, K)-family contains a weak A((?)-family. +

The relation (A, K ) + wk A((?) was introduced by Erdos, Milner and Rado [39], and for infinite A and K they gave a complete discussion (assuming GCH). The same simple remarks that were made in the last section shortly after the definition of the relation ( K , A) A((?) apply also to the relation ( K , A) wk A((?). Lemma 6.2.4 shows that (A, K ) f . A(3) whenever A < K+,,SO a discussion of the relation (A, K ) wk A((?) is pertinent. Many special cases need to be considered, and there is no simple result analogous to (A, K ) A(3). We first establish several lemmas. For the rest of this section, write + ( K ) for the number of cardinals less than or equal to K , so -+

-+

-+

+(K)

= I{q; 77 is a cardinal and 0

Thus if K = H, then

+(K)

< 77 < K)I

.

= IaI 4 No.

Lemma 6.3.2. Suppose A + ((?)$,(K). m e n (A,

K)

-+

wk A((?).

Proof. Let d = (Av;u < A) be any (A, K)-family. For each cardinal 77 with 1)
r,

=

{{P,u } E [A]*; I A nA,i ~

= 77)

,

so [A]' is the union of the rq.By the partition relation A ( ( ? ) $ ( K ) , there is H i n [Ale such that [HI2 C rV for some 77. Then (Av;u EH) is a weak A((?)family contained in d . -+

Lemma 6.3.3. For all infinite cardinals K , the relation (2J'(K),K ) f. wk A(3) holds.

148

Decomposition and intersection properties

Ch. 6.3

Proof. Let (AD; P < $ ( K ) ) enumerate in order all cardinals (finite and infinite) strictly less than K . Start with a pairwise disjoint family (X(g);g E @2for some with P < $ ( K ) } where i f g E p2 then IX(g)l = A;. F o r f i n *(,)2 put

so always Mfl = Zl(Ai; < $ ( K ) ) = K . Consider the (2$('), K)-familyd = ( A f ; f E *(,)2). For distinctf, g from $(')2, as before write S(J g) for the least ordinalasuch thatf(ff)#g(cr), and note t h a t A f n A , = U{X(fTP); P < S ( J g)} so

Thus IAfnA,I=by;,) ifS(f,g)isinfinite,and MgnA I=qS(f,g)(Gcf;g)+l) if S(f, g) is finite. Take distinct functionsf, g, h from $($2. We may suppose S ( J g) # 6(g, h ) , and hence IAf nA,I # IAg flAhl. Thus the family d contains no weak A(3)-family, and the lemma is proved.

Lemma 6.3.4. Let A' = K ;then (A', H,) f . wk A@) ff A' = No then (A+, No) f. wk A(A)-

Proof. Take a sequence (Ao: u < K ) of cardinals such that always 1 < A, < A and A = Z(A,; u < K ) . Choose a pairwise disjoint family {X(g);g E X(A,;o< P) for some 0 with < K } , where i f g E X(A,; u < P) then IX(g)l= Hp+l.Write P = X(A,; u < K ) and for f i n P put

so always Mfl = Z(Hp+,; B < K ) = H,. And from Konig's Lemma, IPI > A, so the family d = (Af;fE P) is a P A ' , H,)-family. For distinct .f g from P we h a v e A f n A g = U{X(frP);P<S(J g))so that

Let cx3 = (Af;fEI) be a weak A-family contained in d , where we may suppose f C F! Then S ( f , g) must be constant for all pairs { J g} from [I]', say with value u (so u < K ) . Then If1 < A,, < A. Hence d contains no weak A(A)family. If A' = No,modify the above construction by starting with IX(g)l = + 1 wheng E X(A,; u < P), and noting that 0is finite.

Ch. 6.3

Weak delta-systems

149

Lemma 6.3.5. Let K be regular, and let f be a function, f : [ K ' ] ~ -+ ( 0 , I}. There is a (K','K)-family (A,; v < K ' ) such that whenever p < v < K' then H,nA,I


i f f ( { / & V})= 0 ; /ApnA,I

=K

i f f ( ( p , V})=

1.

Proof. Start with a pairwise disjoint family {Sap;a < K + and < K } where S has power K. Since K is regular, by Theorem 1.3.3 any family ?'3 of K each , pairwise almost disjoint sets each of power K has a K-transversal, that is, there is a set T with 1 Q IB n TI < K for each B i n g . Thus we can inductively choose sets B, of power K for p with p < K' such that B, is a transversal of {Sap; a < p and 0< K } U {B,; CJ < p }

.

Put Sa'= U{Saj;0 < K } , for a with a < K'. Then the S, are pairwise disjoint. An easy induction shows that always B, C u{S,; a < p } . Put A , = S, nB,, so B, = U{A,; a < p } . Also MWl = K whenever a < p , for then S ,I, n BPI 2 1 for each 0(with 0< K ) and S , n B, C S, n Bp =A,. For v with v < K ' , put

A , = S , U U{A,,; p < v and f ( {u,v})

= 1}

.

Then& C A , C_ u {S ,;a
c U{&,;

so IA, nA,[ < K , since IB,

< p } n U{A,,; CJ < V) = B, nB, nB,I < K by choice of B,. This proves the lemma.

CJ

Theorem 6.3.6 (GCH). Suppose K is an infinite successor cardinal. Then ( K ' , K ) f. wk A(K') and ( K + , K ) wk A(K). -+

Proof. To see that ( K ' , K ) -+ wk A ( K ) ,note that if K = q+ then the number y of cardinals < K is the same as the number of cardinals Qq, so y < 7) < K . By Theorem 2.2.4 the partition relation K + + (K): holds, so (K', K ) + wk A(K) follows from Lemma 6.3.2. To establish that ( K ' , K ) f. wk A(K'), note that K + f . (K')$ by Theorem 2.5.7. Thus there is a function f : [K']' -+ { 0, 1) such that if H C K' and f i s constant on [HI2, then IHJ< K'. Take the (K', K)-familysd constructed in Lemma 6.3.5 for this particular function f . Then LA contains no wk A(K')family.

150

Ch. 6.3

Decomposition and intersection properties

Theorem 6.3.7 (GCH). Suppose K is not a successor cardinal. Then 6 ) ( K ' , K ) f. wk A(K), (ii) if Ho < K < H, and 8 < K then (K', K ) +. wk A@), (iii) $K = Ho or $ K = H, then (K', K ) wk A(3).

+

r

Proof. For (i), given any function f i n ,2, put A f = {f a;a < K } and consider the family d = (Af;fE ,2). Then d is a (K', K)-family of pairwise different sets. For distinct L g in we have A,- flA , = {f r a;a < S(J g ) } so IAfnA,I = IS(Jg)l. Let (Af;fE/) w h e r e I C be a weak A-family contained in d , say IAf nA,I = q for distinct J g in I. Then q = IS(ft g)l < K. Thus for all { J g} from [112we have IS(L g)l = q so S(L g) < 17'. Thus 1 1 1 Q

I {f ra;fE K2and a < q+)l = P+


= q++

Hence d contains no weak A(K)-family. To show (ii), suppose No < K < H, and take 0 with 0 < K . The restriction on K ensures that $ ( K ) < K . Because K is a limit cardinal with $ ( K ) , 8 < K then K + ( 8 ) $ ( K )(from Theorem 2.5.10). Hence ( K , K ) -+ wk A(0) by Lemma 6.3.2, so surely ( K ' , K ) +. wk A(0). Finally for (iii), suppose K = Ho or K = H,. Thus K = $(K), SO Lemma 6.3.3 ensures that (K', K ) wk A(3). This completes the proof.

+

Together, Theorems 6.3.6 and 6.3.7 provide a complete discussion of the relation (K', K ) +. wk A(0). We shall now consider the relation (A, K ) wk A(0) in the case No Q h Q K. The discussion breaks into several cases. (The GCH is assumed throughout.) If h < $(K)', it follows from Lemma 6.3.3 that (A, K ) wk A(3), so we may suppose that $(K)' < h Q K . Suppose first that h is not a successor cardinal. By Theorem 6.3.7(i), (A', A) wk A@), and hence ( h , ~ ) wk A@). ) . wk A(0) If 0 < A then h +. (0)$(,) (from Theorem 2.5.10) and hence ( h , ~ + by Lemma 6.3.2. Now take the case that X is a successor cardinal, say h =.'77 If q is not a successor cardinal, by Theorem 6.3.7(i), (q', q ) wk A(q) and hence (A, K ) wk A(q). However, if 0 < q then q' +. (0)$(,, since $ ( K ) < q and so (A, K ) +. wk A(8) by Lemma 6.3.2. The final case to consider is that h = q' where q is a successor cardinal. Here (q',,q) wk A(q') from Theorem 6.3.6, so (A, K ) f wk A(h). And if 0 < h, since $ ( K ) < q = q' it follows from Theorem 2.2.4 that 'q +. (8)$(,) so (A, K ) +. wk A(0) by Lemma 6.3.2. All cases have now been covered. The remaining situations where the strong relation (A, K ) A(8) fails are given by Theorem 6.2.S(iii), (vi). Theorem 6.2.5(iii) states that ( h , ~ ) A(h) if X is singular (and K < A). In fact the example given to establish this result is an example of a (A, K)-family which contains no weak.A(X)-family, so the -+

+

+

+

+

+

+

-+

+

Ch. 6.3

Weak delta-systems

151

stronger negative relation (A, K ) p wk A(A) holds. There remains the result in Theorem 6.2.5(vi): if A is singular with A' < K < A then (A', K ) f . A@). In this situation one can ask if either of the relations (A', K ) + wk A@') or (A', K ) + wk A(A) hold. The outcome is summarized in the following theorem. Theorem 6.3.8 (GCH). Let A be singular with A' < K < A. If A' is countable or hi< K then (A', K ) f . wk A@);othewise (A', K ) + wk A@'). Proof. By Lemma 6.3.4, (A', Nag) wk A(A) so if < K then certainly (A', K ) f . wk A@). If A' = No then even (A', No) f . wk A(A) so (A', K ) wk A@) for all K . So suppose A' < K < Nhl with A' # No. We shall show that in this situation, (A', K ) -+ wk A@'). Take a (A', K)-family d = (Av;u < A') and for a contradiction we shall suppose that d contains no weak A@')-family. There is a least cardinal ? such t h a t d contains a (A', K)-family 39 with 6(c1B) = 17 (so 0 < r] < K ' ) , and we may in fact suppose that d =39 . . Suppose for the moment that q is a successor cardinal, say 77 = 8'. Take any A in d ,and put

e

Since I [ A l e [= K' < K~ < A, if V(A)I = A' then there would be B in [Ale and J in [J(A)Ih' such that A , f Al = B for all v in J. Then for distinct p, u from J since B C A , fA l, it follows that 8 = IBI < lA, nA,I < 6 ( d ) = 8'; hence IA, nAvl = 8. Thus (A,; u € 4would be a weak A@')-family contained i n d , contrary to the choice of d . Hence IJ(A)I < A for each A from PQ . This means that we can choose inductively ordinals u(a) for a with a < A' so that

v(a)E A+

-

U{J(Au(p));P < a>

Then IAv(,) nAvml<8 whenever p< a,so that93 = (A,(,); a < A') is a (A', K)-family contained in 94 with F(g)< 8. This contradicts the minimum property of 7.Consequently q cannot be a successor cardinal. So further O
It follows that 7'# A'. If 77 = No this is trivial since A' # No. Otherwise since 7 < K < NAlrthen q = N, for some a where 0 < a < A'. Since q is not a successor, 77 = Z(Hp; P < a) and hence 77' < la1 < a < A'. Thus indeed q' # A'. Use transfinite induction to choose non-empty subsets I, of A' with always II,I < A, for a with a < K', as follows. Suppose for some a that Zp has al-

152

Decomposition and intersection properties

ready been defined for all 0 with

Ch. 6.3

< a. Put

/(a)= u{1p;(3
K

. X = h. Define [*(a)by

Pya)= { v < h + ; ~ ~ , n ~ ( a ) i & . & Put d *(a)= { A , nA ( a ) ;v E 1*(a)};then d *(a)is a family of the at least g size subsets o f A ( a )with S(A*(a))< g. Since IA(a)l< h and A' # g' it follows from Theorem 1.1.6 that [1*(a)l I { x , ; a < K+}I = K'; yet [A,/ = K . This is the required contradiction. Thus d must contain a weak A@+)-family, and the relation (A', K ) -+ wk A@+) is established.

CHAPTER 7

ORDINARY PARTITION RELATIONS FOR ORDINAL NUMBERS $ 1. Introductory remarks As noted in Chapter 2 , the definition of the ordinary partition relation for cardinal numbers can be extended to order types. In this chapter we shall consider the partition symbol for well ordered types, that is, for ordinal numbers. The definition is as follows.

Definition 7.1.1. Let a,y,ak (where k < y) be ordinal numbers and let n be a positive integer. The ordinary partition symbol a + (ak;k < 7)" means the following. Let S be a set ordered with order type a. For all partitions A = ( A k ; k < y} of [S]" into y parts, there exist k with k < y and a subset H of S having order type ak such that [HI" C A k . The various conventions concerning the use of the partition symbol adopted for cardinal numbers in Chapter 2 will be followed without further comment for ordinal numbers. The problems concerning the symbol for ordinal numbers are considerably more ramified than those for cardinal numbers. We shall confine most of the discussion to the case n = 2 , and frequently y = 2 as well. Even so there are many unsolved problems, and we shall not attempt to cover even all the cases where progress has been made. We shall limit our treatment to a few special Cases where a reasonably complete discussion is possible. Thus 8 2 is devoted to partitions of [wN12into two classes, mainly for finite a. In $3, we prove Chang's theorem for w w .And in $4 we shall consider relations of the form w1 + (a1,...,ak)', and prove that w 1 + (a): for countable a and finite k. In this chapter we shall need to distinguish clearly between the order type of a well ordered set and its cardinality. The problem is particularly acute with the initial ordinals. When order type is to be emphasized, we shall write

154

Ch. 7.1

Ordinary partition relations

w , w l , w 2 , ..., a,, ... for the sequence of infinite initial ordinals, although the sequence Ho,Hi, HZ, ..- H,, ... of infinite cardinals denotes the same se-

quence (as a sequence of sets). It will be left to the context to distinguish between the symbols $ for ordinal exponentiation and K~ for cardinal exponentiation. In particular, s mbols such as a : , &, ... stand for the ordinal . . operation, whereas Hi,H, , ... indicate cardinal exponentiation. The symbols Co, II, are used to indicate the ordinal sum and the ordinal product of a well ordered sequence of ordinals. It is easy to see that a partition relation between cardinal numbers is equivalent to the same relation between the corresponding initial ordinals. Unlike the situation with cardinal numbers, for ordinal numbers results with n = 1 are not trivial. In [73], Milner and Rado consider this situation. They give an algorithm to determine in finitely many steps for any sequence ( a k ; k < y) of ordinals the least a such that a .+ (ak;k < y)'. We mention a couple of results from [73], but otherwise refer the reader to the original. There is the following lemma.

x,

Lemma 7.1.2. (i) Suppose a + ( a k ; k < y)' and /3 ( P k ; k < y)'. Then 09 ( a k P k ; k < 7)'. (ii) Let y be finite. Suppose a,, (a,,k; k < y)' for p with p < p. Put a@)= ~o(cu,;P < P ) and a&) = no(a,,k; p < P). Then a ( p ) (a&); k
-+

-+

+

Proof. A special case of (i) was stated as Lemma 5.1.6, and tiic proof of the general case hardly differs from the proof of Lemma 5.1.6. Let S be a set well ordered with order type 09, so we may suppose S = X a under the lexicographic ordering. Take any partition S = U{ Ak; k < y}. For x with x < p put Ak(x) = { y < a;(x, y ) E Ak}. For each x , we have a = U{ Ak(x); k < y}, and so there is k(x) such that tp(Ak(x)(x)) 2 Put r k= {x < 0;k(x) = k}, so p = U( r k ;k < y). Then there is ko such that tp(rk0) = pko. Put R = {(x, v)E S;X E r k , and y E Ako(x)}. Then R L Ak, and tp(R) 2 @k,pk,, so tp(Ako) 2 a k O P k O . This proves (i). To prove (ii), for each u with u < p , put a(v) = n,(a,,; p < v ) and a k ( v ) = IIo(a,k; p < u). Use transfinite induction on p . The case p = 1 is trivial. Suppose p is a successor ordinal, say p = u t 1. Then a(u) -+ ( ~ ( u ) ; k < y)' by the inductive hypothesis, and a, -+ (a,k; k < y)'. So by (i), a@) (a&); k < 7 ) ' .Now suppose that p is a limit ordinal. Let S be a set well ordered by a relation < with order type a@), and suppose S is partitioned, S = u{Ak; k < y}. For each p with p < p there is a subset T,, of S with tp(T,,) = a@).By the inductive hypothesis there are always k(p) and a subset H,, of T,, with tp(H,,) 2 akcr~)(p)and H,, C Aku). Put r k= { p < p ; k(p) = k}, +

Ch. 7.1

155

Introductory remarks

so p = u{ r k ; k < y}. Since y is finite and p is a limit ordinal there is some ko such that r k o is cofinal in p. Then for each p in r k o we have tP(Ak0) = tP(Ak@)) Hence tp(&,)

ak@)(p)= ako(p> .

tp(H&)

2 SUp(ako(/l); /l E r k o ) = olko@). This COmpkteS the proof.

Theorem 7.1.3. If m is finite then w"

+

(w")h.

Proof. From Lemma 7.1.2(ii), noting that w + (a);. The second, somewhat surprising, result from [73] concerns partitions into infinitely many classes. It has been referred to as the Milner-Rado paradox.

Theorem 7.1 A. For all 0, $a

< up+'

+ (up"; k < a)'

then a

Proof. (Note that trivially up+' + (wp+l):, so certainly up+'+ (up"; k < a)'.) It suffices to show that if a < up+' then w; (up"; k < w)' . This we prove by induction on a.The case a = 1 is trivial. Suppose a is a successor ordinal, a = y + 1, and uJ (up"; k < w)'. Take any set S ordered by a relation < with order type a;, so S = U{S,,; 1.1 < wp} where tp(S,) = wJ and S,, < S, whenever p < v < u p . By the inductive hypothesis, for each p there is a decomposition S,, = A,,k; k < w ) where tp(A,k) < up". h t A0 = A1 = f$ and for k with k 2 0,

+

+

u{

Ak+2 = Then S =

u

AMk ;P < u p 1

u { Ak; k < w } and tp(A0) = tp(A1) = 0 ,

tp(Ak+,)

< &(Opk; p < U p ) =

Ur' <

W r 2

.

+

This gives a partition of S which demonstrates w$ (up"; k < a)'. Now suppose a is a limit ordinal and that wJ (wpk;k < a)' whenever y < a. There is a sequence (a@);p < up>of ordinals below CY with a(0)< a(1) < a(2) ... such that a = sup {a@); p < u p } , Let S be a set ordered with order type ZO(w;@);p < up), so t p ( 9 2 w;, and write S = U{S,,;p < up}where tp(S,,) = w j @ )and S,, < S, whenever p < v < wp. By the inductive hypothesis, for each p there is a decomposition S,, = U{A,,k; k < w } where tp(A,,k) < wpk. As before, put A0 = A1 = 8 and for k with k 2 0,

+

Ak+2 = U{Apk;P<wp) ; then this decomposition of S shows w; f . (up"; k < w)'. This completes the induction, and proves Theorem 7.1.4.

156

Ch. 7.1

Ordinary partition relations

We shall conclude this section by noting a couple of negative relations.

Theorem 7.1.5. For all P, $ a


then a

+

(up + 1

ul2

Proof.If Icy1 < Np this is clear, so suppose Jal= Np. Let < b e a well ordering of a of order type u p . Define a partition [aJ2= A, U A, by: if u, T < a with 0<7,

{u,T)EA,*u
{u,T}EApu>T.

Take a subset H of a.If [HI2 C A, then both < and Q agree on H, so tp(H, <) = tp(H, q;hence tp(H, <) < up. If [HIz C_ A, then H enumerated in increasing Corder gives a descending <-chain of ordinals; hence H is finite. Thus this partition of [.I2 suffices to prove the theorem. Results of Kruse [ 6 2 ] extend Theorem 7.1.5 as follows for values of n with n 2 3.

Theorem 7.1.6. Suppose n 2 3 . For all 0, ifa

< up+'

then a f . (a0t 1, n + 1)".

+

Proof. By Theorem 7.1.5, a ( u p + 1, a)"-'. Thus there is a disjoint partition A = {A,, A,} of [a]"-' such that there is no subset A of a of order type upt 1 with [A]"-' C A,, nor an infinite subset B with [B]"-' C A'. Define a disjoint partition r = { ro,PI) of [a]" as follows: if u1 < (12 < ... < u,, < then {al,..., u , ) E r o * {ul,..., u,-l)EA, or (02 ,..., o , } E A o , {u, ,..., u , ) E r l *

{u1,..., ~ , , - ~ ) E A , a n d {u2 ,..., u , ) E A 1 .

Take H from [a]"", say H = { ul,..., u,+'}, listed in increasing order. If both { u,, ..., un} E rl and { u2, ..., a,+,} E rl we would have the contradiction { u2, ..., un} E A, n A,, so [HI2C rl. Suppose there is a subset H of a with tp(H) = u p + 1 such that [HI2C ro. Put

B = {u E H ; for all {al, ..., u n n - l from } [HIn-', if 01,

..., un-.l G o then {ul, ..., un-,) € A l } ,

so [B]"-' C A,. Put A = H - B ; then B < A . And in fact [A]"-' C A,. For take T ~ ..., , T ~ from - ~ A where < 7 2 < ... < ~ " - 1 . There are u1, ..., unVl in B with u1 < ... < un-l G and { ul, ..., unPl} 4 A'. Let u1, ..., u, list { ul, ..., un-,} U { 71, ..., T"-,} in order (so m = 2n - 2 or 2n - 3 depending

157

Countable ordinals

Ch. 1.2

on whether u n - ] < r 1 or un-l = rl). Always { ~ f + ..., ~ ui+"} , E ro, and { ul, ..., un-l} 4 A1. Hence an easy induction on i shows that always (ui+2, ..., ~ i +E~A,. } In particular (71, ..., T n - i } E A,, so indeed [A]"-'C A0 as claimed. Now tp(B) + tp(A) = tp(H) = u p + 1, so either tp(B) 2 w or tp(A) 2 u p + 1. This contradicts the choice of the partition A. Hence if H C a with tp(H) = w p + 1 then [HI2 ro.Thus the partition r of [&In suffices to prove the theorem. $ 2 . Countable ordinals

In this section we shall discuss partition relations of the form a-+(a~,al)" where a is a denumerable ordinal. From Ramsey's theorem, a (w, w)". By Theorems 7.1.5 and 7.1.6 (with /3 = 0), a f . (w + 1, and a f . (w+l, n+l)" if n 2 3. Hence the only relations of interest are those of the form a+(ao,m)2 where m is finite. Some of the first such relations to be established are in Erdos and Rado [ 2 9 ] ,where it is shown that wm (w + I, m)2 and wm f . (o+ 1, m + Moreover, for each k and m it is shown that there is a least integer Zo(m,k ) such that wlo(m, k ) (wk, m)2,and that a f . (wk, m)2 if a < wlo(m, k). In [ 3 2 ] ,this result is generalized to arbitrary /3 by showing that there is a least integer Zp(m, k ) such that oplp(m, k ) -+ (wok, m)2, and that a f . (opk, m)2 if a < oplp(m, k). Erdos and Rado conjecture in [ 3 2 ]that in fact Ip(m, k ) = lo(m, k ) for all /3. This was later proved correct by Baumgartner [ 3 ] .The reader is referred to the papers mentioned for the proofs of these results. We shall concentrate on partition relations for off, the ordinal powers of w, with countable a.The first results will be concerned with w" for finite n. For each positive integer n, put -+

-+

-+

W(n)= { ( a o , ...,

" w ; a O< a l

< ...
,

and let < be the lexicographic ordering of W(n), so (ao, . . . , a n - l ) < (bo, ..., b,-l)*ao
or (a0 = bo and a1 < b l ) or

... .

Under this ordering, W(n) has order type a".For a sequence a from W(n), the i-th component of a will be denoted either by ai or a(i), and similarly with b, c, ... .

Definition 7.2.1. A subset S of W(n) is said to be free in the i-th component

Ordinary partition relations

158

a.1.2

if for all (ao, ...,an-,) in S there is an infinite subset A of o such that for all a in A there are ..., a;-1 in w such that (ao, ..., q - 1 , a, a:+l. ..., aL-1) E S. Lemma 7.2.2. Let S be a subset of W(n)and take m with m < n. Then tp(S) > urnifand only if there is a subset Tof S which is free in m different components.

Proof. Suppose there is T with T C S such that T is free in m components. We shall show tp(T) 2 urn(and so also tp(S) 2 urn) by induction on m. This is trivial form = 1. So take T, free in m + 1 components. Let i be least such that T i s free in the i-th component, and letA be the infinite set of i-th components given by the definition. For each a in W(i t 1) which is an initial segment of a sequence in T, the set T(U) = { b E T; b r (i t 1) = a } is free in m components and so by inductive assumption tp(T(a)) 2 w". Since tp(T) 2 &(tp(T(a));a(i) € A ) thus tp(T) 2 wrnw = urn+'. Now suppose tp(S) 2 urn, and use induction to show that there is a subset of S free in m components. We may suppose tp(S) = urn. If m = 1, let i be maximal such that there is some sequence a of length i for which the set ~ ( a ) ={ E S ; ~r i = a } is infinite (where i = 0 and a = 8 is allowed). For such an a, put A = { bi; b E T(a)}; then the choice of i ensures that A is infinite. Thus T(u) is a subset o f S free in the i-th component. Now suppose the statement is true f o r m , and take a subset S of W(n) with tp(S) = urn+'. Let i be the largest integer such that there is a sequence a in W(i) for which the set

T(u)= { b ~ ~r i ;= ab} has order type a"". (Again i = 0 is allowed). So by the choice of i, for such an a , if

T,(u) = { b E T(u);b j = X} then tp(T,(a))

< urnfor each x in a.Put

X = { x E o ; tp(Tx(a))=wm}, then X is infinite (since urn+' = tp(T(a)) = &(tp(T,(a));x E a)). By the inductive hypothesis, each T,(a) for x in X is free in m components. Hence there is an infinite subset A of X such that all the T,(u) for x in A are free in

Ch. 1.2

Countable ordinals

159

the same m components. But then if T = U{ T,(u);x € A } , it follows that T is free in the i-th component and m later components; thus T is a subset of S free in m + 1 components. This completes the proof. The first paper in which partition relations for anare proved is Specker [93], where it is shown that w2 (a2, m), for every finite m, and un (an,3 ) , if n 2 3. The following simple proof of the first of these relations is from Haddad and Sabbagh [50]. -+

Theorem 7.2.3. For all finite m, the relation w 2 + (u2, m ) , holds.

Proof. Since tp(W(2)) = w 2 , it suffices to consider partitions of [W(2)I2. SO let any partition [W(2)I2 = A, U A 1 be given. Define a partition of [wI4 into 16 classes, [wI4 = u{A(io, ..., i3);io, ..., i3 = 0 , l } , by: for a, b, c, d in w with a < b < c < d,

{a, b, c, d } E A(io, ..., i3) 0 { ( a ,b),(c, d)} E Aio and { ( a ,c), (b,d)} E Ail and

{(a, d), (b, c)} E Ai, and {(a, b),(a, c ) }E Ais . By Ramsey's theorem there is H in [w]"O such that H is homogeneous for C A(io, ..., i3). Let ( h k ; k < w ) enumerate H this partition of [wI4, say [HJ4 in increasing order. If ij = 1 for any j , the following gives a subset I of W(2) of power m with [I], C A , : { (h2k, h 2 k + l ); k < m} if io = 1 ,

{ ( h k ,hrn+,J;k < m} if il

= 1,

{ ( h k ,h 2 r n - k ) ; k < m} if i 2 = 1 , {(ho,hl+k);k<m}ifi3= 1 . On the other hand, if ij = 0 for all j , write H as a disjoint union,H= U{Hk; k < w } where each H k is infinite, and put

I = { ( h , h ' ) ; h E H oa n d h ' E H h + l a n d h < h ' } . Then I C W(2), tp(0 = w 2 and [ I ] , C A,. Thus the theorem is proved. Lemma 7.2.4. The relation w 3

-+

(a3, 3), is false.

160

Ch. 1.2

Ordinary partition relations

Proof. Consider the following disjoint partition [W(3)I3 = ro U rl where {a,b}EI'l * u ~ < u ; < b o < a z < b l


rl

rl,

Suppose [ { a , b , c}]' C where a b ic. Since { a , b } E certainly a2 < b l , f r o n i { b , c } E r l followsbl
<

>

>

> rl

[a2ro.

ro.

vides an example which proves the lemma. Theorem 7.2.5. If 3 < m < w then 0"

+ (am,3)2.

Proof. The case m = 3 is Lemma 7.2.4. Form with m > 3, it is enough t o prove the following: (1) there is a one-to-one m a p f : a"' -+ w 3 such that if X is a subset of urn with tp(X) = w"' then tp(f[X]) = w 3 . (In this situation, one says thatfpins wm to w3.) For suppose (1) established. Take a partition A = {A,, A,} of [a3]' such that there is no H in [w3I3 with [HI' C_ A,, nor H with tp(W = w 3 and [HI2 C A,. Define a partition A* = {A:, AT} of w"' by, for x, y in am,

f[a

Then H in [wmI3 with [HI2 C_ AT would give in [w3I3 with If[H]]2CAl, whereas if H C urnand tp(W = amwith [HI2 C A: thenf[H] C w 3 , tpCf[H]) = w 3 and [f[H]]' C_ Ao. Thus A* shows urn (urn, 3)2. We prove (1) by observing that for any integers k, I : (2) if wk can be pinned to w', then wk+l can be pinned t o w ' + l . Since any one-to-one map f : urn-'-+ w pins urn-*to w , two applications of (2) establish (l).To prove (2),suppose indeed that wk can be pinned t o w! Write wk+las a disjoint union, wk+l= U {Si;i< w } where tp(Si) = u kand Si<Sj whenever i<j, and similarlyw"'= U{Ti;i
+

Ch. 1.2

Countable ordinals

161

Calvin and Larson [46] investigate just which countable ordinals can be pinned to w 3 (the use of “pin” in this context is due to them), and as a consequence they show: if a is a countable ordinal and a + (a,3)2 then a = 0, 1, u2 or a = ww’ for some 0. It was noted by Erdos and Hajnal [24, see Problem 61 that for each n and k (k 2 3) there is a least integerf(k, n ) such that unP (ak, f ( k , n))2.(So from Lemma 7.2.4,f(3,3) = 3.) The exact value off@, n ) is not known in general. However, Nosal [75] has s h o w f ( 3 , n ) = 2n-2 + 1. This result appears as Theorem 7.2.9 below. The proof depends on the existence of cartain canonical partitions of W(n).

Definition 7.2.6. Two pairs { a , b } , {a’, b ‘ } from [W(n)I2are said to be similar if for all i, j with i, j < n,

ai < bj * a: < b; and ai > bj * a;.> b; . A partition A of [W(n)]’ is said t o be canonical if for all { a , b } , {a’, b‘} from [W(n)12whenever {a, b } is similar t o {a’, b ’ } then {a, 6 ) 5 {a’, b’} (mod A). The existence of canonical partitions was first proved by Hajnal, and independently later by Calvin.

Theorem 7.2.7. Given any disjoint partition A of [ W(n)12 into finitely many classes there is an infinite subset H of w srtch that the induced partition on [Pn w(n)12 is canonical. Proof. L e t S be the set of all strictly increasing functions f : n -+ 2n. Forfin 7 and x from [uI2“, say x = {XO, XI, ..., ~ 2 1 }< ~ put - xf = (Xf(0) ..., X f ( n - l ) ) . Suppose A = {A,,, ..., A,_ l } . For each x in define a function Fx : [SI2+rn by

Fx({.6 g}) = i

{xfixg} E Ai

.

Since there are only finitely many functions mapping from [TI2t o m , by Ramsey’s theorem there is an infinite subset H of w such that all the sets x from have the same F,, say F, = F. Take any similar pairs { a , b } , {a’,b’} from [H“fW(n)I2. l There are x, y in [HI2” andf, g in -7 such that a = xf, b = xg,a’ = y f and b‘ = y r Since Fx = F,, thus { a , b } = {xf, xg} E A F ( { f , g }and ) {a’, b‘} = {yf.Yg} E AF({f,g}), so { a , b } { a ’ , b ‘ }(mod A ) as required. Let n be fixed, with n 2 3. We define the following subsets of W(n):for i,

162

Ch. 7.2

Ordinaiy partition relations

j , k with 0 Q i< j < k

< n, put

... and aiPl = bi-' and ai< ... < a k _ l < b i < ... < b j - l < a k < ... < a , _ l < b j <

ECk = { { a , b } E W(n);a.

= bo

and

Consider the disjoint partitionr" = {I':,

ry = u{E$k;0 < i < j <

... < b , _ l } .

G}of W(n), where

k < n}

This partition has several useful properties, which we collect in the following lemma.

Lemma 7.2.8. For this partition r"of W(n), (i) there is no subset H o f W(n)of order type u 3such that [HI2L I$, (ii) there is no subset H o f W(n)with IHI = 2"-2 t 1 such that [HI2C ry, (iii) there is a subset H o f W(n) with IHI = 2n-2 such that C G.

[a2

Proof. For (i), take any subset H of W(n) with t p Q = u3. By Lemma 7.2.2 we may suppose that H is free in three components, say i, j , k where 0 < i < j < k. Thus given a in H , there is b in H such that b T i = a t i and b(i) >a(k- 1). There is c in H such that c r k = a r k a n d c(k) > b( j - 1). There is d in H such that d r j = b r j and d( j ) > c(n - 1). Then { c , d } E EZk, and so [HI2 n ry # 0.This proves (i). To prove (ii), use induction on n. When n = 3, the partition r3is the Specker partition used to prove Lemma 7.2.4, so (ii) is true when n = 3. Suppose (ii) is true for a particular value of n. Take a set A from [W(n + l)]q+' such that [AI2 C rl" ; we wish to show that q < 2,-'. Let A = {ao, ..., listed in lexicographic order. Note that for a, b with a < b and {a, b } E rl , always a(z) < b(i). Further the definition of any EZ:' ensures that b(l)fa(n), for if j = 1 then a(n) < b( j ) = b( 1) and if j > 1 then b( 1) Q b( j - 1) < a(k) Qa(n). Hence there is I, with 1 Q 4 , such that

'

ao( 1) Q a'( 1) Q

... Q a[( 1)
1) Q ... < as( 1)

.

(1)

We show first that Z < 2"-2. Observe for a , b with a b and {a, b } E I'"' that if a(0) < b(0) < b( 1) < a(n) then also { a , b'} E I'y'l where b' = (a(O), 6(1), b(2), ...,b(n)). For if a(0) = b(0) this is trivial, and if a(0) < b(0) then {a, b } EE:$ for some j , k , where j > 1. But then { a , b'} EEfh', so indeed {a, b'} E q + l . n u s if

A' = { (ao(O), am(l), ... a,(n),; )

then [A'I2 C I'"'.

Consequently, if

A" = { ( a m ( l ) ,..., a,(n)>;m Q Z }

112

< I}

Ch. 7.2

163

Countable ordinals

then [A"I2 C I'l. So by the inductive hypothesis, I t 1 < 2"-'. Return now to (1). Thus if 1 = 4 , nothing more is necessary. So suppose 1< q. Observe first that a1+1(0) = a,2(0) =

...= a,(O) .

(2)

For suppose (2) is false. Then there is rn with I + 1 < rn <(I such that
A* = {(a,&), ..., a,(n)>; I t 1 < rn < 4 ) ,

E.

as above [A*I2 C_ So again by the inductive hypothesis M*I < 2"-' , that is, 4 - I < 2"-2. Since I < 2n-2, thus 4 < 2"-'. This completes the inductive step, and proves (ii). Finally, we establish (iii). Again use induction; the case n = 3 is trivial. So suppose for some n that (iii) holds. The partition of { a E W(n t 1); a(0) = 0) induced by rn+'is isomorphic to the partition I-" of W(n), so.by hypothesis there is a subset H of W(n t 1) with IHI = 2"-' and a(0) = 0 for ail a in H such that [HI2 C I'y+'. Let H = {ao, ...,u p } , listed in lexicographic order (so p = 2"-2 - 1). Then


h t B = { b l ; O < l < p } a n d C = { c l ; O < l < p } . It iseasy tosee that {al,a,} is similar to both { b l , b,} and {cl, c,}. The definition ofr"" ensures that rn+lis a canonical partition of [ ~ ( +n1)12, so since c ry'l also [B]' C I'y" and [CI2 C I'"'. Moreover, for any I, rn with 1, rn < p there is k such that

[a2

bl(0) < ...< bl(k - 1) < ao(n) = ~ m ( 0 < ) bl(k) < ... < bl(n)


... < c m ( n ) , (1)

Ch. 7.2

Ordinary partition relations

164

and hence { bl, c, 1 E E:;;, so { bl, c,} € r'"'. Hence [B U C]' C r':", and since IB U C I = 2 n - 2 -+ ?"T2 = 2"-', the induction step is complete. This proves (iii), and comgletes the proof of the lemma.

Theorem 7.2.9. Zfn < 3 then w n f. (a3, 2"-2

+

and w n -+ ( a 32"-2)2. ,

Proof. From (i) and (ii) of Lemma 7.2.8, the negative relation is established by the partition r" of W(n). For the positive relation, take any disjoint partition A = {Ao, A,} of [W(n)I2such that there is no subset H of W(n) of order type w 3 with [HI2CAo and no 2"-2-element subset H of W(n) with [HI2C A , , and seek a contradiction. By Theorem 7.2.7 there is an infinite subset of w on which the induced partition is canonical, and we may suppose that A itself is canonical. For i, j , k with 0 < i <j < k < n and integers ao, a,, a2 put

f.. (ao, a,, a 2 ) = (0, 11 k

1,

..., i-

1, i -+ ao, ..., j- 1 + ao, j + a,,

k - 1 + a l , k + a 2,..., n - 1

...,

+a2),

m f i j k ( a 0 , a l , a2) < . f i j k ( b o , b l , b 2 )9 (ao, a l , a 2 ) < (bo,b l , b2>.Define a disjoint partition A * = (A;, AT} of [W(3)I2by ( a l p 02, a 3 ) E

A;

* f i j k ( a l , a29 a 3 ) E A0

.

Clearly A * is canonical since A is. A subset of W(3) homogeneous for A* gives a subset of W(n)homogeneous for A in the obvious way. Hence A * is a canonical partition of [W(3)I2having no H of order type w 3with [HI2 C A,* and no H of size 2"-2 with [HI2 C_ AT. A check of the possible canonical partitions of [W(3)I2(see Milner [72]) shows that apart from the Specker partition r3 all have either H of order type w 3 with [HI2 in class 0 or else H of arbitrary large finite size with [HI2 in class 1 . Hence A* must be the Specker partition r3.Thus if (ao, a l , a 2 ) < (bo, b l , b 2 ) then {fijk(ao,al,a2),fijk(bo,

b1,bz)}€A,~ao
An easy check shows ( f i j k ( 0 , 1, n + l ) , f i j k ( n , 2 4 2n + 1 ) ) E E t k , and hence f 8. Since A is canonical, it follows that f$)k C A,. Since i, j , k were arbitrary, thus y'I C_ A,. Hence by Lemma 7.2.8(iii) there is a 2"-*-element subset H of W(n) with [HI2 C_ A , , contrary to the choice of A . This is the required contradiction, and the proof is complete.

E t k flA,

In particular, if we put n = 4 in Theorem 7.2.9, then we find u4-+ (0~,4)~ and u4 f . (a3, 5)2. These particular results were also obtained by Calvin, by

Countable ordinals

Ch.7.2

165

Hajnal (see [24]) and by Haddad and Sabbagh [Sl]. Earlier results for w4 were obtained by Milner [71], where it was shown that w4 + (a3, 3)2, w4 (w3 + 1, 3)2 and a! (a3, 3)2 if a! < w4. We conclude this section by proving the following theorem of Erdos and Milner [26]: wl+pm+ 2 9 ' where rn < w and p < wl. This result dates back to 1959; a proof also occurs in Milner [69]. The theorem does not give best possible results -- for example with p = 2, it gives w 2 m + i + ( w 3 , 2 ~ 2 whereas by Theorem 7.2.9 in fact + (a3, 2m)2 - but it seems to be the best general result of this type known.

+

+

Theorem 7.2.10. Let a!, /3 be countable, let k be finite. If a"+ (w"O, k)2 then we'@-+ 2k)2. Corollary7.2.11.1frn<wandp<wl then w1+Ctm+(w1+p,2m)2. Proof. By induction on rn using Theorem 7.2.10, noting that trivially wl+p (w1+pCc, 2)2.

+

k)2. Take any set S ordered Proof of Theorem 7.2.10. Suppose w" + and let a partition [SI2 = A0 U A1 be by a relation < with order type given. Assume there is no H in [S]wl+pwith [HI2 C A. and no H in [S]2k with [Ill2 C Al, and seek a contradiction. For any x in S, put Aj(x) = {Y ES; {x, U) E Aj}

*

We shall use the following observation. Suppose given a family { A , ; u < 6) where each A , is a subset of S of order type wQ.For x in S, put M(x) = { u < 6 ; tp(Ao(x) nA , ) = ma} . Then AE[S]We~tp{x€A;tp(M(x))=6}=w". Consider first the case when 6 is an ordinal power of w , say 6 = w y . Suppose (1) is false for a particular set A from [S]W~. Put A' = { x E A ; tp(M(x)) < w 7 } , so tp(A') = w". By the relation w" + (wl'p, k)2 and the choice of the partition, there is H in [A']ksuch that [q2 C_ A1. Since w7 + (a'>: (by Theorem 7.1.3) and tp(M(x)) < orif i E H, there is u with v < orsuch that u 4 U{M(x);x EH } . Thus for all x in H , tp(Ao(x) nA,)
Ch. 1.2

Ordinary partition relations

166

and [H U r]* C Al, contrary to the choice of the partition. Thus (1) holds when 6 = w7. For arbitrary 6 , write 6 as a finite sum of powers of w and successively use the above result finitely often to prove (1). From (1) we shall prove the following. Suppose given a family { A , ; v < u p } where each A , is a subset o f S of order type w" and moreover A,, < A , whenever fi < u < up.Suppose also finitely many values vo, ..., u, less than upare given. Then for any A in [S]"" there is x in A and an order preserving function g : up+. upsuch that g(uo) = vo, ...,g(u,) = v, and tp(Ao(x) fl Ag(")= w a whenever v < wp. (2)

For we may suppose vo < ul < ... < v,. Let A in [SIw" be given. Define subsetsBo, ..., B,+l o f S a s follows: Bo = A , B m + 1 = { X E B,;

tp(Ao(x)

fl A,,)

= a"}, (m = 0, 1,

...,n ) .

Successive applications of (1) with 6 = 1 show that always tp(B,+l) = wQ.In particular, if x E B,+l then tp(Ao(x) n A,,) = w" for each m. Define subsets CO,..., Cn+l of S as follows. Put Mo(x) = { v ; v < vo and tp(Ao(x) nA , ) = a"},

co =

{X E B n + 1 ; tp(Mo(x))

=~

09 )

and for m with m = 0, 1, ..., n put

M,+l(x) = { v ; u, cm+l

< u < v,+,

and tp(Ao(x) f A,) l = ma) ,

= C x E C m ; v m + 1 +tp(Mm+l(x))=vm+l)

9

(with the convention u,+~ = up). Successive applications of (1) show that always tp(C,) = war.In particular, tp(C,+,) = a" so Cn+lf 8. Choose x from for each m. Cn+l.Then tp(Mo(x)) = uo and v, + 1 + tp(M,+l(x)) = Hence if

M(x).= { v < up;tp(Ao(x) nA,) = w"} , then M(x) is the ordered union M(x)=Mo(x)u { v O } ~ M l ( x ) u{vllu ...U { ~ n } U M n + l ( x ) 9

and so tp(M(x)) = up.Let g : up+M(x) be the enumeration of M(x) in order. Theng(uo) = uB ...,g(un) = v,, and ( 2 ) is proved. Since lwpl = No, there is a sequence (7,;n < w>of ordinals below upin which every v with u < upis repeated infinitely often. We shall define by induction on the integer n elements x, of S and subsets A(n, v ) of S for v with

Ch. 7.2

Countable ordinals

167

v < u p . Write

s = U{A(O, v); v < UP} where always tp(A(0, v)) = oOrand A(0, p )
A*(% v) = { x E A ( n , v ) ; x > x , for all m with m < n } , then tp(A*(n, v)) = a".Use ( 2 ) on the family {A(n, v); v < up}with vo = yo, ...,v, = y, and A = A*(n, 7,) to find x , in A*(n, y), and an order preserving map g, : up-+ upwith g,(y,) = y m if m < n such that tp(Ao(x,) n A(n, g,(v))) = ooL whenever v < up.Put A(n + 1, v) = Ao(x,) nA(n, g,(v)). Then tp(A(n + 1, v)) = oOrand since g, is order preserving also A(n + 1, p ) < A(n + 1, v) whenever p < v < up.This completes the construction. Put H = { x n ;n < a}.The construction ensures that if n > m then A(n, v) C Ao(x,) for all v ; in particular since x , E A(n, yn) thus x , E Ao(x,). Hence [HI2 C A,. We shall show that tp(H) = u"~,contrary-to the choice of the original partition: This will provide the contradiction required to prove the theorem. Take integers m, n ; say m < n. Since

A(n, v) C A ( n - I,gn-l(v)) C A ( n - 2,g,-2g,-l(v))

c A ( m , gm ... gn-zgn-l(v)) in particular

9

c ...

168

Ch. 7.3

Ordinary partition relations

Now given m, the set { n < w ; m < n and y, = y m } is infinite. Further the definition of x, ensures that if p < n and yn = y p then x, > x p . Thus tp(H n A ( m , ' y m ) ) = w . Since { y m ;m < w } = UP, it now follows from (5) as claimed. This completes the proof. that tp(H) = wwp=

53. Chang's Theorem for ow One of the long-standing problems in the partition calculus for countable ordinals has been to decide whether or not the relation ww (ww, 3)2 holds (see Specker [93]). A positive answer was finally given by Chang [7] in a lengthy tour de force. His result was subsequently extended by Milner who showed that ww +. (aw, rn)' for all finite m. In [ 6 5 ] ,Larson gave a new short proof of this. We shall follow Larson's method. We shall make use of the following lemma, a consequence of Ramsey's theorem. -f

Lemma 7.3.1. Let S E [wIK0.Suppose for each a from there are given an integer m(a)and a partition A (a) = { Ao(a), Al(a)}l of [SIm('). Then there is an infinite subset H of S such thatfor each a from the set { h E H ; h > max(a)} is homogeneous for A (a). Proof. Inductively define setsHk from [sINo and elements hk offfk, for finite k , as follows. Put Ho = S, and let ho be the least element of Ho. If Hk and ho, ..., hk have already been defined, list all subsets a of {ho, ..., hk} with max(a) = hk, say a0, a l , ..., a[. Put Hko = H k and use Ramsey's theorem to choose successively infinite subsets H k i + l from H k i so that Hki+l is homogeneous forA(ai). h t H k + l = { h E H k [ + l ;h > h k } ; then H k + l is homogeneous for each A (ai). Let hk+l be the least element ofHk+l. Put H = {hk; k < w } . Take a from and suppose max(a) = hk. Then { h EH; h > h k } is a subset ofHk+l, and hence is homogeneous forA(a), as required.

Theorem 7.3.2. Let m be finite. Then ww +. (ow, m)2. As in the previous section, for positive integer n put

< ...< a n p l } , and order W(n) lexicographically. Put W = U{ W(n);0 < n < w } and order W ( n ) = {(a0,..., a , - l ) ~ " w ; a o < a l

the elements of W first by length of sequence, and then lexicographically in

Ch. 7 . 3

169

Change theorem for ww

each W(nj. Under this ordering, W has order type ww. So to prove Theorem 7.3.2, it suffices to take any partition [WI2 = A, U A , and show that, given m, either there is a subset H of W with tp(H) = ww such that [HI2 C A0 or else there is a subset H of W with IHI = m such that [HI2C A , . Let m be fixed. Take any partition A = { A o ,A , } of [WI2 for which there i s n o s e t H i n [W]" with [HI2 C A I . We must f i n d H i n [WIw" with

[HI2 L A,. By Corollary 7.2.1 1, certainly the relation w"" -+ (w", 2m)2 holds. By the relation applied to the partition of W(mn) induced byA , there must be a subset W'(n) of W(mn) with tp(W'(n)) = w" such that [W'(n)12C A,. Put W' = u{W'(n); 0 C n < a}, so tp(W')= ww and for { a , b } from [W'I2 with ln(a) = In@) we know { a , b } E A,. It will suffice to find a subset H of W' with tpQ = owsuch that for { a , b } from [HI2with In@) # In@) still { a , b } E Ao. By redefining the partition A , we may suppose that W' = W. For the rest of this section we shall use the convention that if a is a finite increasing sequence of integers then Q is the set of entries ofa and conversely, if a is a finite set of integers then a is the sequence of the elements of Q arranged in increasing order. The concatenation of sequences a a n d b is a b. c\

Definition 7.3.3. For a, 6 from Wwith In (a) < ln(b)and for integer k, say { a , b } has form 2k [or form 2k + I ] if there are non-empty finite sets of integers ao, al, ..., Q k and bo, bl, ..., bk-1 [bk] such that (i)

a O < b O < a l < b l < ...
(ii) a =ao "a1 "al

... ak ; A

b = b o " b l h...hbk-l for form 2 k ; b = b o A b l ^...^bk for form 2k + 1; (iii) if c = (laol, lao U all, ..., lao U ... U akl) and

d = (Ibol,Ibo U b,l,

..., Ibo U ... U bk-l

[U bk]l)then

c
Lemma 7.3.4. Let S E [wIK0.There is H in [SIN0,H = { h l ,h2, h3, ...I listed in increasing order, such that for all positive integers I there is i(1) = 0,l so that if { a , b } is any pair of sequences of form I and shape { c , d } with a, 6, c, d C { h E H ; h > h l } then ( a , b } E A,(,,.

Ch. 1.3

Ordinary partition relations

170

Proof. For each possible shape { c , d } with c, d C S and each a. in [S]<'O with c < a. < d, choose a partition A (c, ao, d ) of suitable size subsets of S into two classes so that: whenever x = a. U bo U a l U b l U ... where if a = a o " a l " ...,b = b o " b l "... then {a,b}hasshape {c,d}and { a , b } E A i , then x E Ai(c, ao, d ) . By Lemma 7.3.1. there is HO in [SIKosuch that for c, ao, d from Ho there is i(c, ao, d ) so that if a U b C { h E H o ; h > max(d)} and {ao "a, b } has shape { c, d } then (a0

"a, b ) E Ai(c,,o,d,

.

For each finite subset c ofHo, choose two partitionsA (c), A*(c) into two classes of the c(0) + IcI, c(0) + IcI - 1 size subsets of Ho so that: if x = a. U d where la01 = c(O), Id1 = IcI, c < a.

< d and

i(c, 110, d ) = i, then x E Ai(c) ; if x =ao U d where

11101 = c(O),

i(c, ao, d ) = i, then x

Id1 = IcI - 1, c
E AT(c)

< d and

.

Ap ly Lemma 7.3.1 twice to find H1 in [HolNO such that for c from [H,IR0 there are i(c),i*(c) such that for appropriate sized subsets a@d of { h E H1;h > max(c)}, if Id1 = IcI then i(c, ao, d ) = i(c) ; if Id1 = IcI - 1 then i(c, ao, d ) = i*(c) . Finally, define partitions A (2k), A(2k t 1) of [Hllk+' by c E Ai(2k) * i*(c) = i; c E Ai(2k

+ 1) * i(c)= i .

Use Ramsey's theorem repeatedly to find infinite subsets Hl+1 of HI minus its least element such that HI+^ is homogeneous for A (1) (where I2 1). Let hl be the least element of and put H = { h,; 0 < I < o}. Then for all positive 1 there is i(1) such that if c E [ ( h E H ; h > hl}lk+' then i*(c)= i(l)or i(c) = i(l),depending on whether 1 equals 2k or 2k t 1. It follows that H has the property required. For take a pair { a , b } of form I and shape { c , d } where a, b, c, d C { h E H ; h > hl). Since H C H1 C Ho, then { a , b } E Ai(c,ao,d) (where a. is the sequen,:e of the first c(0) elements of a), and

i(c, ao, d ) = i*(c) or i(c)(depending on whether I is even or odd) = i(l)

.

This completes the proof.

Ch. 1.3

Change theorem for ww

171

Lemma 7.3.5. Given any infinite subset H of w, for all positive integers 1, m there is M in [ W]" such that whenever a, b E M then a, b C H, { a , b} has form I and if { c , d } is the shape of {a, b} then c, d C H.

Proof. Suppose 1 = 2k or 1 = 2k t 1. Choose increasing finite sequences ci, U i j from H for i, j with 1 < i < m, 0 <j < k t 1 in the order that follows, so that always the last element of one sequence is less than the least element of the next: cl,alo, C2,a20, -., crn,arno,all~a21,-..,am1

3

a12, a22, ..., U m 2 , ..., alk, ...,amk, arnk+l,.-)a2k+19alk+l

9

such that ln(ci) = k + 1, In(aio) = ci(0) and ln(Uij) = C i ( j ) - ci(j - 1) (for i, j with 1 < i < m , 1 <j < k ) , 1n(aikzl) = cj(k) - Ci(k - 1). If1 = 2k t 1, put ai = aio ail ..."aik whereas if 1 = 2k, put ai = " aio ail ... fzik-1 aik+l and put A

A

"

A

M = { a i ;1 < i < m } . If 1 = 2k + 1 then { a i , aj} has form 1 and shape { ci, cj}, whereas if 1 = 2k then { a i , aj} has form 1 and shape { c i , dj} where dj = (cj(O), ..., cj(k - 2), cj(k t 1)). Hence M has the property required in the lemma. Lemma 7.3.6. Given any infinite subset H of w, say H = { h l ,h2, h3, ...} listed in increasing order, there is a subset X of W with tp(X) = owsuch that whenever a, b € X with In (a)
Proof. Choose increasing finite sequences ci, ai, a(i, j, k ) from H for i, j , k with 1 < i < w and 1 <j < i < k < w as follows. Suppose for given k that ci, ui, a(i, j , 1) have already been chosen whenever 1 <j < i < 1 < k. We choose ck,a k , a(i, j , k ) where 1 <j < i < k in the order that follows so that always the last element of one sequence is less than the least element of the next. and greater than Ensure also that the least element of c k is greater that every element in the sequences already defined. The order is: Ckr akr a(1,

1, k ) , 4 2 , 1, k ) , a(2,2, k ) , 4 3 , 1, k ) , ..., a(k, k, k )

(that is, the a(i, j , k ) are chosen according to the lexicographic ordering of (i,j ) , where 1 <j < i < k ) . The sequences have lengths as follows: ln(ck) = k t l,ln(ak) = ck(O), ln(a(i, j , k ) ) = ci( j ) - ci(j - 1).

112

Ordinary partition relations

Ch. 1.3

Put Xi =

{ai"a(i, 1, kl)=a(i, 2 , k2)" ..."a(i, i, ki); 1 < i
< k2 < ... < ki < o},

X = U{Xi; 1 < i < o } , S O XC W. For fixed i, all the sequences in Xi have the same length, namely

ci(0)t (ci(1) - ci(0)) t ... + (ci(i)- ci(i - 1)) = ci(i) . Further, the choice of a(i, j , k ) ensures that if k l

< k2 < ... < ki then

a(i, 1, k , ) < a(i, 2 , k2) < ... < a(i, i, ki) ,

and hence the order type of X i , in the lexicographic ordering on W,is wi. Moreover, if i <j then ci < c, so in particular ci(i) < ci( j ) ; thus all sequences in X i precede those in Xi. Hence X has order type ow. Take a , b from X with In@) #In@). Since all sequences in any X i have the same length, we may suppose that a E Xi and b E Xi where i <j. No sequence from X i has any value in common with a sequence from Xi. Thus it follows that { a , b } has form 1 for some positive integer 1. And if { c , d } is the shape of { a , b } then c is a subsequence of ci and d is a subsequence of c j ; thus c, d C H. Finally, ci(0) > h2*1; this ensures that c(0) > hl. Thus X has the properties required, and the lemma is proved. We are now set to prove Theorem 7.3.2, namely that ow+ (ow, m)' for all finite m.

Proof of Theorem 7.3.2. Take a partition [ WI2 = A, U A, for which there is no set H in [W]" such that [HI2 C A,. As noted before, we may suppose that whenever a, b E W with ln(a) = In@) then { a , b } E Ao. Take an infinite subset Hof w with the property given by Lemma 7.3.4. Thus i f H = {Al, h2, h3, ...}, listed in increasing order, then for every positive I there is i(l) = 0, 1 so that for all pairs {a, b } of form I , if { a , b } has shape {c, d } and a, b, c, d C { h E H ; h > hl} then { a , b } E A,(,). Given I, apply Lemma 7.3.5 to { h E H ; h > hl} to find M in [W]" such that all { a , b } from [MI2 have form I, and if the shape is { c , d } then a, b, c, d C { h E H ; h > hl}. Thus i(1)= 0, and this is true for all 1. Now use Lemma 7.3.6 to find a subset X of W with tp(X) = ww such that all { a , b } from X with In@) # In@) have form I for some I , and moreover if {c, d } is the shape 0 1 {a, b } then a, b, c, d C_ { h E H ; h > h,}. Thus { a , b ) E Ai(,), that is {a, 6) E A,, for all such {a, b } . Hence [XI2 C Ao, and the proof is complete.

Ch. 7.4

173

Partitions of / w 1 / 2

94. Partitions of [ a l l 2 A complete discussion is possible for the relation w1 + ( a k ; k < T ) ~ Some . results are immediate from Chapter 2. From Corollary 2.5.2 we know w1 (3):,, so only partitions of [ 0 1 1 2 into finitely many classes are relevant. From Theorem 2.5.8 comes the further negative relation w 1 ( ~ 1 ) ; .

+

+

Theorem 7.4.1. For sfinite,

w1

-+ (w1,

(w + I),)

2

.

Proof. Applying Theorem 2.2.6 to the two relations H1+ (Hl); and Ho+(Ho), 1 gives the relation N1 + (N1, Only slight changes to the proof of Theorem 2.2.6 yields the marginally stronger result w 1 + (wl, (w + 1),)2. Referring to that proof, with the notation in use there, the changes are as follows. Start by taking a well ordering < of the set S of order type K . When one comes to define the set F(v) for v in N n SEQ,,, put C(v)= { x ~ S ( v ) ; x < fyo r s o m e y i n U ( F ( V ~ T ) ; T < U ) } ,

and take for F(v) the union of G(v) and the original F(v). Since K is regular, IC(v)l < K and so still IF(v)l < K . The elements xu are defined as before, using a sequence v in N n SEQk2 for which S(v) # 0.Choose x in S(v). Then for each u it follws that x 4 F ( v r u + l), so in particularx 4 G(vru t 1). Sincex ES(v To t 1) and xu E F( v r u) it follows that x # xu;hence xu < x. From x E S(v u + 1) we know {xu, x} E r r ( v r o+l).It follows that HA = HO U {x} is a subset of K of order type at least qfo+ 1 with [HA]2 rro.

r

Theorem 7.4.1 is the strongest positive result of its type. The relation (w1, w t 2)2 was proved by Hajnal [52] assuming the continuum hypothesis. Whether the relation can be proved without this assumption is an open problem. See Erdos and Hajnal [24], Problem 8, and their discussion of this problem in [ 2 5 ] . w1

+

Theorem 7.4.2(CH).

w1

+

(017

w + 2)2.

Proof. Take any set S ordered with order type w1 by a relation<. By Corollary 3.2.8 there is a set mappingf: S-+CpS of order tE1 with If(x)nflv)l< Ho for any pair x, y from S, such that f h a s no free set of power H1.Since always If(x)l < Ho, we can choose inductively elements x, of S for a with a < w1 so that

a
andf(x,)<

{xp}.

174

Ordinary partition relations

Ch. 7.4

Put T = {x,; a < wl) so tp(T) = wl. Define a partition [TI2= A, U A1 by, if a < 0 then {x, xp) E A1

* x,

Ef(xp)

.

Any subset HO of T with [&I2 C A, is free forf, so tp(H0) < 01. Suppose there isH1 in [T]w+2with lH112 C A l . I f x a n d y are the o - t h and (a+ 1)-st elements o f H l thenH1 - {x, y ) C f ( x ) n f ( y )so If(x) n f ( y ) l = No,contrary to the choice off. So there is no such set H1,and the theorem is proved. We have left to consider relations of the form 01 -+ k < s ) ~where s is finite and the (Yk are all countable. Trivially Ramsey's theorem gives w1 (a),'. Erdos and Rado [29] showed that w 1 + (w + n); for any finite n. In [60], Hajnal strengthened this to w1 (w2, for any finite n. The next result was due to Calvin (1970, unpublished) that w 1 -+ (03);. Later still Prikry [79] obtained w1 + (a2+ 1, a)2 for every countable a. Finally Baumgartner and Hajnal [4] settled the problem by proving the best possible result, namely w 1 +(a),' for every cduntable a and finite s. In fact they proved somewhat more. They showed that if cp is an order type such that cp + (w)k then cp + (a),"(for all countable a and finite s). Their proof uses deep methods from mathematical logic. They show first that the relation cp (a),"holds in a particular model of set theory, and then use absoluteness criteria to conclude that the relation is true in the real world. Calvin [44] has since given a combinatorial proof of the Baumgartner-Hajnal theorem. We shall devote the remainder of this section to Calvin's proof for the relation -+

-+

-+

a 1

+

(4,".

Theorem 7.4.3. Let a be a countable ordinal and let s be finite. Then w 1 +(a)%. For the rest of this section, we adopt the convention that the letters a,0, y, ... (with maybe subscripts, superscripts or the like) range over only the ordinals less than w 1. Let the partition A of [all2 be given, [ a l l 2 = A0 U ... U As-l . For any a,we shall show that there is H in [ollw" such that H i s homogeneous for A ; from this Theorem 7.4.3 follows. We introduce the following property, somewhat weaker than being a homogeneous set for A.

Ch. 1.4

115

Partitions of / w 112

Definition 7.4.4. A subset X o f w l is almost homogeneous (for A ) if wheneverA € [XIw" (for any a) then for all p with < a there are B in [A]"', A' in [Alwa and i with i < s such that B < A ' and B e A' C_ Ai, where B e A' = {{x, y } ; x €B, y € A ' and x # y } . Definition 7.4.5. The symbol 0 +.AH(a(O), ..., Q(S - 1)) indicates that whenever X is an almost homogeneous subset of o1with tp(X) = up,then there are i with i < s and A in [X]""(') such that [AI2 C_ Ai. The next lemma reduces the problem of proving Theorem 7.4.3 to the problem of constructing almost homogeneous subsets of arbitrarily large (countable) order type. Lemma 7.4.6. For all a there is p such that 0 +.AH(a,

..., a).

Proof. We shall show the following: given ordinals a(O), ..., a(s - 1) such that for all i with i < s and all y with y < a(i) there is p(i, y) such that p(i, y) +. AH(a(O), ..., a(i - l), y, a(i -t l), ..., a(s - l)), then there is such that 0 +.AH(a(O), ...,a(s - 1)). It follows by induction on ao, ..., that for all ao,..., as-l there is /3 such that 0 +AH(ao, ...,&+I); in particular the lemma is true. If any a(i)is zero, the claim above is trivially true (with arbitrary), so suppose that always a(i)2 1. Choose sequences (a(&n); n < w>such that = &,(w"("~); ri < a). Put a(i, 0) G a(i, 1) G ... < a(i) and

p(n) = max @(i,a(i, n)); i < s} and 0 = Z0(fl(n);n < a). We shall show that

0 +.AH(a(O), ...,a(s - 1)) .

(1)

So take X from [wllW' such that X is almost homogeneous. Then there are X o in [X]w'(o), A. in [XIwp and i(0) with i(0)< s such that X o < A0 and X o c A. G Ai(o . Inductively we can contjnue and find X , in [A,-1 ]WP(,) A, in [A,-l]wB and i(n) such that X ,
if n < m, then X ,

<X ,

and X , 8 X ,

C_

A,(,)

.

(2)

There i s N in such that i(n) is constant for n in N, say with value j. For each n, the choice of p(n) ensures that P(n) +.AH(a(O), ..., a(j - l), a( j , n), a(j + I), ...,a(s - 1)). Thus for each n, either there are i with i # j and C, in

Ordinav partition relations

176

Ch. 1.4

[Xn]wff(i) such that [C,]' C Aj, or there is B, in [Xn] w"(i'") such that [B,,]' C A , . If for any n the first of these alternatives is true this gives i and C in [XIwdd with [CI2 C Ai. On the other hand, if for all n the second alternative holds, consider B where B = U{B,;n EN}. By ( 2 ) , tp(B) = Z;o(wff(i~"); n EN) = a") . From ( 2 ) , B,e B, C_ Aj if { n, m ) E [N]', and also [Bn]' C A . by the choice of B,. Thus [B]' C A,-, so also in this case there is B in [X]udn with [B]' C A,. Thus the relation (1) holds. This completes the proof. We shall later make use of the following lemma to construct large almost homogeneous sets from smaller ones. Lemma 7.4.7. For each n with n < w let X,, be an almost homogeneous subset of wl. Suppose X , < X , whenever m < n, and suppose there are integers i(m)(with i(m) < s ) such that

m < n *X ,

Q

X, C

Then U{X,; n < w } is almost homogeneous. Proof. Take any subset A of U{X,; n < w } with tp(A) = 0". Take any (3 with (3 < a.If A C X,, U ... U X , for some m, then by Theorem 7.1.3 there must be k with k < m such that tp(A n X k ) = wQ.Since X k is almost homogeneous

this gives B in [A fl Xklw' and A' in [ A n X,]"" such that B < A ' and B s A' C A,- for some i. On the other hand, ifA is cofinal in U{X,; n < 0) then tp(A nX,) 2 w p for some m. Then if B E [A n X,]"' and A' = A - ( X , U ... U X,) (so tp(A') = a")it follows that B s A' C A,(,). Thus U{ X,,; n < w } is indeed almost homogeneous.

The construction of almost homogeneous sets will depend on refining pairs ( A , X ) of subsets of w1 (with X uncountable) such that A @I X i s "almost contained" in Ai for some i. We shall write G(i) for the set of pairs ( A , x> where A is in [wllW" for some 01, X is in [wlIHl , A < X and A @ X is "almost contained" in Ai. "Almost contained" is to have the sense that n o matter what subset A' of A of the same order type as A or what uncountable subset X' of X are chosen, for every (3 smaller than a there is a subset A" in [A']"', not cofinal in A', and an uncountable subset X " of X ' such that A"@ X " is "almost contained" in Ai. This leads to the following inductive definition of G(i). The relation ( A , X ) E C(i)where i < s, A E [w1IWLy and X E [allH1 is

Ch. 7.4

177

Partitions of / w 1 / 2

HI = 1) then (A, x> E C(i) if and only ifA < X a n d A 8 X C Ai .

defined by induction on a.If a = 0 (so

If a > 0, supposing the relation (A', X') E G(i) has already been defined whenever tp(A') = upwhere 0< a,then

U,X ) E C(i) if and only ifA < X and whenever A' E [A]"" and X ' E [ X I N 1for , every 0 with 0 < a there are A" in [A']"'

not cofinal in A' and X " in [ X ' I N 1such that (A", X " ) E C(i) .

It is clear from the definition that if ( A , x> E C(i) and X ' E [XI"l , A ' C A with tp(A') = tp(A), then U', X ' ) E G(i).

Lemma 7.4.8. Given A in [wllw"and X in [ul]"'then there are A' in and X' in [XI" such that ( A ' , X ' ) E C(i),forsome i with i < s. Proof. Since H I < No and = N,, we may suppose A < X. If (Y = 0, for i w i t h i < s p u t X i = { x E X ; A s ( x } C A i ) . Forsomeiwehave lXil=H1; then ( A , X i ) E C(i). Now suppose a > 0, and suppose inductively that the lemma is true whenever tp(A) = upwith 0< a. Suppose the lemma is false for particular A0 in [u1IWa and X o in [ u l l Hthus 1;

1x1

vi< s VA' E [

A ~ VX' ~ wE ~[ X ~ I " ( ( A ' ,x')4

,

c(i)).

(1)

Inductively define chains A . 2 A 2 ... 2 A , from [Aolw" and X o 2 X1 2 ... I X , from [ X O ] " so ~ , by (1) ( A , X i ) 4 C ( j )wheneverj< s, as follows. Since in FAj, Xi) 4 C ( j ) ,by the definition of C( j ) there are Aj+, in [Ailwa, [Xi]" and P ( j ) < a such that VA" E [~i+,lw") VX" E [ X ~ + ~(A" I " not ~ cofinal in

Ai+l =$(A",X " ) 4 G ( j ) ) .

(2)

Let 0 = max { b ( j ) ; j < s}. Choose B from [A,lwP not cofinal in A,. By the inductive hypothesis applied to B and X,, there are B' in [B]"' and X' in [X,]" such that (B', x'>E C ( j ) for some j. If 0(j) = 0,since X' C X , C and B' C B C A , C Aj+, with B' not cofinal in A . + l , this contradicts (2). If /3(1) < 0 by definition of C ( j ) there are B" in [Bl]w'o.) not cofinal in B' and X " in [X']" such that (B",X " ) E G ( j ) . Now X " C Xi+,, B" C Aj+l and B" is not cofinal in Aj+l so again ( 2 ) is violated. In either event a contradiction has been reached, and thus the lemma cannot be false for A0 and X O .This completes the proof.

178

Ordinary partition relations

Ch. 1.4

Lemma 7.4.9. Let ( A , X ) E C(i). Then there are x in Xand a subset A' of A with tp(A') = tp(A) such that A ' s { x } C Ai. Proof. Let A and X be given with ( A , x> E C(i), and suppose tp(A) = wa. If a = 0, the result in trivial, so suppose a > 0. Make the inductive assumption that the lemma is true for all pairs (Ao,Xo>where tp(A0) < tp(A). For a in A and p with p < a define ~ ( a p), = {x EX;

3 B E [ AIW'

(a < B, B is not cofinal in A ,

and B o {x}L A,)} . P u t X * = X - U{X(a,/3);aEA a n d p < a } . We shall show that X* # C. Suppose on the contrary that X* = @.Then IX(a,/3)1= H1 f o r s o m e a a n d p . L e t A ' = { a f E A ; a < a ' } , t h e n i t follows from Theorem 7.1.3 that tp(A = wa. Since ( A , X ) E G(i), there must be X" and A" in [A']'' not cofinal in A' such that A"@ X " E G(i). in [X(a, Thus by the inductive hypothesis applied to A" and X", there are x in X" and B in [A"]"' such that B s {x} C Ai. However then a < B and B is not cofinal in A , so x 4 X(u, p). This contradicts x EX" C_ X(u, 0). Hence X* # Now choose x EX*. Take ordinals a(n) where n < w such that always a(n) < a and wa = Co(w(u(n); n < w). Then x 4 X(a, a n ) for all a in A and all n. Thus we can choose inductively sets B, in [A]WJn1but not cofinal in A , for n with n < w , so that sup B,-l < B, and B, s {x}C Ai. Put A' = U{B,; n < a};then A' E [A]Waand A ' s {x} C Ai. This proves the lemma.

f$

The next lemma will serve as the starting point for an inductive construction of an almost homogeneous set.

Lemma 7.4.10. For n with n < w let (A,, x> E G(in).Then there are x in X and subsets A ; o f A , with tp(AL) = tp(A,) such that A ; s (x} C Ain for each n. Proof. Suppose given X and the A , such that (A,,,X) E G(i,). Put X, = {x EX; 3 -4; C An(tp(A L) = tp(An) and A ; 8 { X} C Ai,)}. If X = U{ X,;n < O} then there is some X n with IX,l= N l . Thus (A,, X,) E G(i,), s o b y Lemma 7.4.9 there is x in X, and A' contained in A , with tp(A') = tp(A,) such that A ; s {XIC Ai,; this contradictsx EX,. T h u s X # U{X,; n < a}, and so the lemma holds..

For the remainder of this section, for each non-zero fix upon a sequence of ordinals (a(n);n < w ) such that always a(n) < a(n + 1) < a and wQ=

Ch. 1.4

Partitions o f l w l l 2

179

Co(w"("); n < b).We shall eventually construct almost homogeneous sets of order type upfor arbitrarily large p by induction on p. We shall do this by finding for each n almost homogeneous sets D, with tp(D,) = up(')such that there are subsets B, in [D,]wp(n)which fit together to give an almost homogeneous set of type up.In order to d o this, we shall need a way of choosing, w times, elements in each D, in such a way that the set of those elements which are chosen at every step constitutes a subset of D, of type

,P(n).

We now look at a method of making such choices. In general, let a set A in [w1lw" be given. Let also a finite family d of subsets of w1 each of order type an ordinal power of w be given. For each integer k, define operations Sk on such sets A and such finite families d as follows: if a = 0, then SAA) = { { A } } ,if a > 0, then S&)=

w"(k)

{{B,C};BE [A]

, C E [AIWaandB
if d = {Ao, ...A,} then S k ( d ) = {{Bo, CO,..., B,, C,}; {Bi, Ci} ESk(Ai) for i = 0, ...,rn} with the understanding that if tp(Ai) = w0 then Bi= Ci= Ai.Thus { A } ) = S&). If d is a finite pairwise disjoint family then so is any member of S k ( d ) . GivenA from [w1Iwa, suppose we take a sequence of families ( d , ; n < w ) such that do = { A } and there is some integer k so that always

Sk(

E S k + n ( d n)

when n 2 0 .

Thus d provides two disjoint subsets of A , one small, one large; d z provides similarly two subsets of each of'these; and so on. If we put A , = ud,, then A n is to be the set of elements o f A chosen at step n. Thus'n{A,;n<w} is the set of elements chosen at every step.

Lemma 7.4.1 1. In the above situation,

n { A , ; n < o}E [AIw".

Proof. By induction on a,where tp(A) = w". If a = 0 then for every k and n we have Sk+,(A) = { { A } } so n{A,; n < w } = A E [AIWO. SO now suppose that a > 0 and make the inductive assumption that for any p with < a,given n < w ) where CM 0 = { B } and for all n, gn+l E Sl+,(?3,) a sequence (qn; for some finite I , where tp(B) = op,then tp( n{ UCM n;n < w } ) = up, Now take the sequence (A,; n < w>described above, w i t h d = { A } where tp(A) = wa. Inductively choose members B , and C, o f d n+l with tp(C,)=w" as follows. Since d , E S k ( A ) there are Bo in [A]Wdk) and COin [A]w"such

180

Ordinary partition relations

Ch. 7.4

since d , + 2 E that Bo < CO and {Bo, CO)G d1.Supposing C, E there are subsets B,+l of C, with tp(Bn+l) = oa(k+n+l) and tp(Cn+l) = we such that B,+1 < Cn+l and {Bn+l, Cn+l} C d n + 2 - Thus always Bn < Bn+l. Consider now the set Bo. There is one subset of Bo in d1,namely Bo itself. In 942 there are two subsets of Bo and together they give a member, say g g, of Sk+l((X 0). In J Q j , these get divided t o give a member of Sk+2(Ca g), and so on. In general, if we put cx3 on = { X E d n + 1 ; X C Bo} then Sk+n+l(dn+l)

~ O O ={ B o ) ;i f n

2 0 then%,+l

ES(k+l)+n(qOn).

by the inductive hypothesis, if

Since tp(B0) =

then Bb E [Bo]Wa(k). We can repeat this with B1, B2, ... . For any m , put ' 3 m , = { X E s Q n + m + l ; X C B m ) . Then g m n =

{Bm 1;Bmn+1

ES(k+m+l)+n(cIomn)

.

So by the inductive hypothesis, if

BL = n{Uq,,;n < 01

then BL E [Bm]Wa(k+m). Now for any n we have

BL

C_

U g m , C Udm+n+1 -Am+n+i ,

and if 1 < m then

BL CB, CCm-l C-Ci-1 C U d , = A , . Hence always BL C fl{ A , ;1 < a}.Thus if B = U { B L ;m < w } then B C { A , ; 1 < o}and tp(B) = Co(oa(k+m):m < o)= we. This proves Lemma 7.4.1 1.

Lemma 7.4.12. Take any finite painvise disjoint f a m i l y d and any uncountable set X such that for all A in 94 there is iA with iA < s such that ( A , X ) E G(iA). Then for any integer k there are a f a m i l y 9 in S k ( d ) and a set X ' in [XI* 1 such that for all B inc1s , B C A E d =*(B,X')EG(iA). Proof. Write d = { A o , ..., A m } . Suppose tp(A,) = oaiand ( A j , X ) E G(ij). Take subsets A; and AT of A , with A; < A,? such that tp(A;) = oe,(k) and tp(A *) = we, (but if a, = 0, put A; =A,? = A , ) . Choose sets X i from [XIK

Ch. 7.4

Partitions of / w 1 J 2

181

for j with j < m + 1 so that X o = X and Xj+l is in [X.]' such that there is A for which ( A j , Xj+l)E G(ij).This a subset A; ofA; with tp(A;) = uaj(k) choice is ossible since ( A j , X ) E G(ij). Put X ' = fl{ X i ;j < m + 1) so X ' E [XIR1and (A;, X ' ) E C(ij) for each j. Also ( A T , X ' ) E C(i.) since i' tp(Af) = tp(Aj). Now {A;, A T } E S k ( A j ) so if we put CM = { A d , A ; , ..., A;, A : } then CZJ E S k ( d ) and 'I3 with X' has the required property. In the next lemma we shall finally construct the almost homogeneous set of order type upfor arbitrary 0. The more complicated statement in the lemma is needed for the inductive construction.

Lemma 7.4.13. Let be given. Take any set X in [ul]' and any finite pairwise disjoint family d = ( A o , ..., A k } with ( A j , x ) € C(ij)(wherej = 0 , ...,k). Then there are B in [XIwpand subsets A; of A j with tp(Ai) = tp(Aj) such that B is almost homogeneous and A; @ B C A,, (wherej = 0 , ...,k). Proof. The plan of the proof is this. We shall construct the almost homogeneous set B by finding almost homogeneous sets B, with tp(B,) = up(,)and B, < B , whenever m < n such that there is i(m) with i(m) < s for which Bm @ B, C A,(,) whenever m < n ; by Lemma 7.4.7 this will suffice. At stage n in the contruction we shall introduce a set D , of order type up(n) from which the elements of B, will be drawn. When finding D,, we shall choose elements of all the earlier D, and of the Aj, say in sets D& and AT, such that D& @ Dn C A,(,) and A T @ D , C Aij. Those elements that are chosen at every stage will form the B , and A;. The proof is by induction on p. The case p = 0 is given by Lemma 7.4.10. So suppose that p > 0, and make the inductive assumption that the statement of the lemma is true for any y with y < 0. Put d o= d and X o = X. By Lemma 7.4.12 we have a familye 0 in S o ( d 0 ) and a set XA in [XOlN such that for all C in C if C C A j then (C,X;) E C(ij).Since p(0) < p by the inductive hypothesis applied to eo and XA there are an almost homogeneous set Db in [XA]wp(0)and subsets C'of C for C in C? with tp(C') = tp(C) such that C'@ Db C Ai. for that i j for which (C,XA) E Gi.. Note that also { C';C E- C,} E S o ( d o ) .by Lemma 7.4.8 there are Do in [ D i ] w p ( oand ) XI in [X,]' such that (Do, X1)E G(i(0))for some i(0)with i(0)< s. Then Do is also almost homogeneous. Put d l = {C'; CE'ZJ o } U { D o } . Since Do C Xo and necessarily A , < XO for each j , it follows that d is a finite pairwise disjoint family. And further, for each A in d l there is i(A) such that ( A , X1)E C(i(A)).Thus we can repeat the process. This leads to the following inductive construction of finite pairwise disjoint

182

Ordinav partition relations

Ch. 7.4

familiesd , uncountable sets X,, and almost homogeneous sets D, in [X,]"'p(n)! Suppose that 4,and X , have already been constructed:'such that for all A i n d ,there is i(A) with i(A) < s for which ( A , X,) E C(i A ) . By Lemma 7.4.12 there are a family C, in S,(A,) and a set XA in [X,] L l ) such that for each C in C, if C C A with A E d ,then (C, X i ) € G(i(A)). Noting that p(n) < 0, by the inductive hypothesis a plied to en and XA , we can find an almost homogeneous set DL in [Xi]Wp and subsets C' of each C in C ,with tp(C') = tp(C) and C'e DL C A i c ~ for ) that A in d ,where C' C C C A . We mention that { C';C E en}E S n ( 4 , ) . Using Lemma 7.4.8 gives Dn in [DL]"p(n' and Xn+, in [XnlN such that (D,,X,+,)€ C(i(n)) for some i(n). Then D,, being a subset of D;,is almost homogeneous. Also C' < D, for every C', since C' C C, D, C X , and C< X , (for each C in en). Put -d,+, = { C ' ;C E en}U {D,}, so d,+,is a finite pairwise disjoint family. And for each A in d , + l ,there is i(A)such that ( A , X,+,) E G(i(A)). Now take any one of the original sets Aj (so j = 0, ..., k). In each of the families 9'2, consider those members which are subsets ofAj. As in Lemma 7.4.1 1, this gives'a sequence of subfamilies ( d i nn;< w ) where

8)

Ajo =

{Aj};Ajn+l E S n ( d j n ) .

Hence by Lemma 7.4.1 1, if A; = n{U d j , ; n < w } then A; C_ A j and tp(AJ) = tP(Aj). d ,consider the subsets of D,. Take any D, (where m < a),and in each s There are n o n e i n d o , ..., dm; in d,+1 there is D, alone; in .dm+2 there are two giving a member of S,+2(D,); in Am+3 these are further divided, and so on. Thus there is a sequence (Q;, n < w ) with (D , ,Cd such that am0 =

{Dm};g r n n + l

ES(rn+l)+n(Gmn).

Thus by Lemma 7.4.1 1, ifB, = n{UCbmn;n < w } then B, C D, and tp(B,) = tp(D,) = Then each B, is an almost homogeneous subset of X. And if m < n then B, < B, since B, C D,, B, C D, C_ X , and D, < X,. Further, if m < n then B, 8 B, C A,(,) since D, 8 X,+, C_ A,(,,,) by choice of D, and Xm+l, and B, C D,, Xn C X,+l. Put B = U{B,; m < a}.By Lemma 7.4.7, B is (~ <)w; ) = up. almost homogeneous. And tp(B) = Z O ( ~ ~ m Finally; we show I ; @ B C Ati if j = 0, ..., k. Given A j , for any integer n let dj,= { A E d , ; A C A j } so U -4, gives the set of members of A j chosen at the n-th step. An easy induction on n shows that ( A , XA) E C(ij) for every A in dj,. Then the choice of DL and d +,, ensures that A t e Dh C A,, for every A' in djn+,, so ( U d j , + l ) e DL C Aij. Since A; C_ UoQj,+, and

Ch.1.4

Partitions of [ w 1 ] 2

183

Bn C Dh,thus A;@ B, C Aii; hence A;@ B C Aii. This completes the proof of the lemma. Lemma 7.4.13 contains the last step necessary for the proof of Theorem 7.4.3.

Proof of Theorem 7.4.3. Let the countable ordinal a be given, and take the partition [w1I2 = A, U ... U As. By Lemma 7.4.6 there is P such that P+AH(a, ..., a). From Lemma 7.4.13 there i s B in [ w l ] w psuch that B is almost homogeneous for this partition. So by Definition 7.4.5 of the symbol P+AH(a, ..., a),there isA in [BIw" such thatA is homogeneous for the partition. Hence there is surely a subset of w1 of order type a homogeneous for the partition. Thus indeed w1 + (a),".

APPENDIX

CARDINAL AND ORDINAL NUMBERS $1. Set Theory

In this appendix we shall list a set of axioms for set theory and develop briefly the properties of ordinal and cardinal numbers that have been used in this book. For a more detailed treatment and the proofs that are not given here, see for example Bachmann [ 11 or Sierpinski [89]. Any one of the standard systems of set theory provides an adequate foundation for the material in this book (for example, the systems of ZermeloFraenkel (Fraenkel [43]), Bernays-Godel (Godel [491) or Morse-Kelly (Kelly [60])). We shall list the axioms for the Zermelo-Fraenkel system (with the Axiom of Choice), ZFC. The system is formalized in first order predicate calculus with identity = and one other binary predicate symbol E for set membership. The axioms are as follows. Axiom I (Extensionality). VaVb[a=b*Vx(xEa*xEb)].

Informally, a = b if and only a and b have the same members. Axiom I1 (Pair set). VaVb 3 c V x ( x E c * x = a o r x = b ) .

Intuitively, for any two setsa, b the pair set { a , b } exists. Axiom 111 (Union set). Va 3 c V x ( x E c * 3y(x E y a n d y € a ) ) .

If a is a set, so is Ua, the union of the members of a.

Ap. 2

Ordinal numbers

185

Axiom IV (Power set).

.

V a 3 c vx(x E c " V y ( y E x * y Ec))

For any set a, the power set p a of a exists. Axiom V (Replacement). For each formula q(x, y ) (may be with further free variables) in which z, a, b do not occur: Vx Vy Vz(q(x, y ) and q(x, z ) * y = z ) Va 3 b V y ( y E b

0

3 x ( x E a andcp(x,y)))

.

Informally, if q(x, y ) is a function-like formula then for any set a there exists theset { y ; 3 x ( x E a a n d q ( x , y ) ) } . Axiom VI (Infinity). 3 b( 3 x(x E b ) and V y ( y E b

* 3 x(x

E b a n d y Ex)))

Ensures the existence of an infinite set. Axiom VII (Foundation (or Regularity)). "a( 3 x(x E a) * 3 x(x € a and V y ( y € a * y

4 x))) .

Intuitively, every non-empty set a has afoundarion member, that is, a member disjoint from a. Although Axiom VII is one of the standard axioms of ZFC, the results in the text do not require its use. Axiom VIII (Choice). Va 3 f ( f i s a function and dom(f) = a) and

b'x(xEaand 3 y ( y E x ) * f ( x ) E x ) . Every set has a choice function.

52. Ordinal numbers A variety of definitions of the ordinal numbers can be given, all producing the same family of sets. (Bachmann [ 1, p24] proves the equivalence of five such definitions.) A convenient definition is the following. A set x is said to

Append&. Cardinal and ordinal numbers

186

Ap. 2

be transitive if members of members of x are members of x,?hat is, if y Ex * y

c x.

A set x is said to be well-founded if every non-empty subset of x has a foundation member. Then we pose the following.

Definition A.2.1. A set x in an ordinal i f x is a transitive, well-founded set all the members of which are transitive. (If the Axiom of Foundation is assumed, it is unnecessary to require specifically that an ordinal be well-founded, since the axiom asserts that every set is well-founded.) It is easily seen that every member of an ordinal is again an ordinal. Define the ordering < between ordinals by, for ordinals a,/3 p
Then each ordinal a is the set of all smaller ordinals,

a = { /3; /3 is an ordinal and 0 < a} . Each ordinal a is well ordered by < (that is, by < restricted to the members of a). Any descending sequence of ordinals must be finite, for if ...< 012 < a1< a. is an infinite descending sequence, by transitivity {an;1 < n < w } is a subset of ao,but yet it has no foundation member, contrary to the wellfoundedness of ao.IfA is any non-empty set of ordinals then both nA and U A are ordinals; nA is the least member of A and U A is the supremum of A (the least ordinal a with 0 < a for every 0 in A ) . If a is a non-zero ordinal such that a = Ua then a is said to be a limit ordinal. If a (with a # $) is not a limit ordinal then there is an ordinal 0 such that a = 0 U { 0). In this case a is said to be the successur of 0, and one writes a = / 3 t 1. Thefinite ordinals are defined to be those ordinals which are well ordered by>, and one putsO= $, 1 = 0 t 1 , 2 = 1 t 1,3 = 2 t 1,etc. Let w be the set of all finite ordinals; then w is an ordinal and in fact w is the smallest limit ordinal. There are the following methods for giving a proof by transfinite induction over the ordinals. Let q(a) be a formula (maybe with other free variables).

Theorem A.2.2 (Transfinite induction, first form). Suppose that for all ordinals 0, if V y < B(q(y)) then ~(0).m e n q(a)holds for all ordinals a.

Ap. 2

Ordinal numbers

187

-

Theorem A.2.3 (Transfinite induction, second form). Suppose that

(9 d o )

9

(ii) c p ( 4 cp(a + 1) , (iii) i f a is a limit ordinal then VP < a(cp(lj)) * q(a) . Then cp(a) holds for all ordinals a. Both theorems are proved by considering the least counter-example to

cp(a),should there be one.

Corresponding to the two forms of proof by induction, there are the following methods of giving a definition by transfinite recursion over the ordinals. (There are more general forms of definition by transfinite recursion; however the two given below suffice to justify the uses of transfinite recursion made in the text.) Let cp(x, y ) be a function-like formula (maybe with further free variables).

Theorem A.2.4 (Transfinite recursion, first form). Given the ordinal a: there is a unique function f with domain Q such that for all with /3 < a,

f(P)

= y f o r thaty such that cp({f(y);Y < PI, v) .

Theorem A.2.5 (Transfinite recursion, second form). Given the ordinal a and a set u, there is a unique function f with domain a such that for all /3 with /3 t 1 < a and all limit ordinals y with y < a, (i)f(O) = u , (ii) f ( 0 + 1) = y for that y such that cp( f(/3),y ) , (iii)f(Y) = U { f ( P ) ;P < 7). The operations of ordinal addition, multiplication and exponentiation are defined by recursion as follows: a + o = a , a + ( p + l ) = ( a t p ) +1 ,

a t y = U { a + 0;P < y} if y is a limit ordinal; a0 = 0 ,a@

+ 1) = (4)t a ,

ary = U{ OQ; 0 < y } if y is a limit ordinal; (Yo=

1 ,cyP+l=(cyP)a,

aT = U ( $ ;

< y} if y is a limit ordinal.

Sets with the order types a + 0, Orp, 2 can be realized as follows. Take sets
a, b ordered by
188

AQ. 2

Appendix. Cardinal and ordinal numbers

pose a flb = (b and order a U b as follows:

x

<1

y

0

(x,y E a and x

or (x € a a n d y Eb). Then tp(a U b, <1) = ar + p. For multiplication, order a X b by the reverse lexicographic ordering, that is (X
(U,X)<2 (U,.V)

and U


.

Then tp(a X b, < 2 ) = 4?. For exponentiation, let Z(a, b ) be the set of all functions f : b -+a such that f(x) is non-zero for only finitely many values of x. Order Z(a, b ) by last differences, that is, forf, g from Z(a, b ) i f y is the greatest element in b such thatfly) # g ( y ) then

f <3 g "f(Y1
g(y) *

Then tp(z(a, b), <3) = (up. The basic properties of the ordinal operations will not be developed here. We mention however the following result.

Theorem A.2.6. Evely ordinal g has a representation in the form where p1 > 0 2 > ... > Pm and m,n 1, n2, ..., n , are finite. Proof. Let $, be given. There is a least ordinal ar such that successor ordinal, say ar = p + 1. Thus

5 < w";

U{ wPn;n < w } , and so there is n with n < w for which wPn< 5' < wp(nt UP


then ar is a

= wPw =

1). Thus

wPn
g = CjPn t y . Now re eat the process on y; there are 0 2 , n2 with n2 < w and 7 2 with for which

y2 < w

f:

P2

y = o n2t-12,

and since y < wP and wP2< y necessarily 0 > /32. Repeat with 72, and continue in this way. For some finite m we must have y m = 0, for otherwise

Cardinal arithmetic

Ap. 3

189

...< P3 < 8 2 < P would be infinite descending sequence of ordinals. Hence t = won + w P2 n2 + ... + w’mn,,,

.

In fact the representation of the ordinal t given by Theorem A.2.6. is unique; this is called the Cantor normal form for the ordinal t . The operations of addition and multiplication can be extended without difficulty to infinitary operations acting on a well ordered sequence of ordinals. The infinite sum Zo(tcY;a < b) and infinite product no(&; a < 0) of the sequence (la;a < 0) are defined by recursion on the length 0 of the sequence as follows:

$3. Cardinal arithmetic Two sets a and b are said to be similar, written a % b, if there is a one-toone and onto function f : a -+ b. Define a b if a is similar to some subset of b, and a < b if a ib and a # b. There are a number of classical results concerning the relhtion of similarity, provable without using the Axiom of Choice.

Theorem A.3.1 (i) (Schrbder-Bernstein). If a 4 b and b 4 a then a % b. (ii) (Cantor) I f a is any set then a < pa. (iii) (Hartogs) For every set a there is an ordinal a such that CU K a. However, the Axiom of Choice is needed if a reasonable concept of “size” is to be based on the relationd. With Choice, it follows that iis a total ordering, and that there are ordinals of every possible size.

Theorem A.3.2. The Axiom of Choice is equivalent to each of the following: (i) the relation < is total, that is, Va V b (a < b orb < a). (ii) every set is similar to some ordinal, that is, Va 3 a (a % a).

Appendix. Cardinal and ordinal numbers

190

Ap. 3

An ordinal a is said to be an initial ordinal if a if not similar to any smaller ordinal, that is < a * a p. The sequence w o ,wl, w2, ..., w,, ... of the infinite initial ordinals is defined by recursion: wo = w , and if a > 0 then

+

o,= least initial ordinal greater than u p for all p with


For this to be a valid definition, one must check that given the u p for all p with 0 < a,there is a larger initial ordinal. Put a = u{wp; p < a};by Hartog’s Theorem there is an ordinal y such that y $ a. Whenever 0 < a then u p C a so o p 4 a ; thus 7& wp and hence wp< y. Then 7 is an ordinal greater than all the wp and similar to none of them. The least such ordinal y will be an initial ordinal. The similarity relation has the properties of an equivalence relation (although the equivalence classes are “proper classes”; they d o not exist as sets in the system ZFC). One seeks a representative from each equivalence class, to serve as a “number” representing that size of set - the representatives SO found will be called cardinal numbers. Since the Axiom of Choice is assumed, by Theorem A.3.2(ii) there are ordinals in each similarity class and hence exactly one initial ordinal, so this initial ordinal is a good choice. Thus we make the definition: K is a cardinal number if and only if K is an initial ordinal. When the infinite initial ordinal w, is used as the representative of the similarity class and ignoring its structure as an ordinal, it is usual to denote it by H, rather than w,, so we define: Nor= w,. Then the sequence of cardinal numbers is 0, 1, 2,

..., K O , HI,Hz,..., K,, ... .

Each set is similar to exactly one cardinal number; thus we define the cardinality, Ix I, of a set x by : 1x1 = that cardinal number K such that x

=K .

The operations of cardinal addition, multiplication and exponentiation are defined as follows. Given cardinal numbers K , h take any sets a, b with la1 = K , Ibl = h and a n b = 8 for (i), and define:

where ba = {f;f : b + a } , the set of all functions mapping from b into a. These definitions do not depend on the choice of the sets a and b.

Ap. 3

191

Cardinal arithmetic

Theorem A.3.3. If

K

is an infinite cardinal then

K’

=K.

Proof. Suppose K = H,, and use transfinite induction on a. The case (Y = 0 is easy, so we assume (Y > 0. Make the inductive assumption that H$ = Hp whenever P < (Y.We shall show that then H, X H, = H,; from this the result follows. Since clearly H, 4 H, X H,, we have only to show that HX , Ha< N,. Note that for ordinals u, 7 with u, 7 < N, also u + 7 < N, (the ordinal sum of u and 7). For since u < H, either IuI < No or Iul = N, for some y with y < a,and likewise for 7.Hence there is /3 with 0 < (Y such that lul, 171< Hp. But Iu t 71 = lul i171

< Hp t Hp = 2 . Hp < H$ ,

and by the inductive hypothesis H$ = Hp.Hence Iu t 71 < Hp < H,, and so u+7<

H,.

Define an ordering << on H, X H, as follows: (a, 7)<< (u’,7’)* (a t 7 < u’ + 7’)or (u + 7 = u’ + 7’and u

< u’) .

It is easy to see that << is a well ordering. Let tp(H, X Ha,’<<)= y. Clearly N, < y. In fact y = H,. For suppose, on the contrary, that H, < y. There is than an order isomorphismf : y + H, X H,, and let f ( H 3 = (u,7).Consider

s = ((flu, v ) E H, x N,; (fl, v) << (u,7)) . Put u t 7 = [, so [ < H,. If ( p , u)<< ( 0 , 7 ) then p t u < u t 7 = [, so p , v < t . Thus S C ([ + 1) X (C; + I), so IS1< I[ t 11’. Since [ < Ha,also It t I1 < H, so by the inductive hypothesis I[ t 1I2 < H,. Thus IS1 < H,. Yet f restricted

to H, is one-to-one and onto S, so I S1 = Ha. This contradiction shows that y = H,. Hence H, X K, 4 Ha,as required.

Corollary A.3.4. If K

IfK

K,

h are infinite cardinals then

t h = *h=max(rc,h). ~

< A then K’

= 2‘.

hoof. Suppose K < h. Then A < K ih< h -hi= 2. h< h2 = h, So K 4A = h. Also h < K . h < h2 = h SO K . h = A. And 2’< K h < h‘< (2’)‘= 2” = 2’, SO K = 2‘.

’

The definitions of sum and product of two cardinals generalize to the sum and product of an arbitrary family of cardinals as follows. Given a family

192

Appendix, Cardinal and ordinal numbers

Ap. 3

( K ~ i; E I ) of cardinals, define C ( K ~ iE ; I ) to be the cardinality of U{Ai; i E I} where the A i are pairwise disjoint sets with [Ail = ~ iand , define I l ( ~ ii;E I ) to be the cardinality of the Cartesian product X(Ai; i E I ) of sets A i where Mil = K ~ There . is the classical theorem of Konig connecting these two notions.

Theorem A.3.5 (Konig). Let ( K ~ i; E I ) and (hi;i E I ) be two families of cardinals with always 1 Q ~i < hi. Then C(Ki;iEI)< rI(hi;iEI).

Proof. For each i in I , choose two sets A l , Bi such that Ai C Bi, Mil = ~ i , IBil = hi where the Ai are pairwise disjoint. Choose bi from Bi - A i . Put A = U{Ai; i E I}and B = X(Bi; i € I ) . Define a function F : A -+ B by, for a in A with a E A i ,

Then F is one-to-one and so MI < IBI. Suppose in fact MI = IBI, so there is a one-to-one and onto map G : A +B. Then B = U{G[Ai];iE I}.For each i inI, put Ci = { g ( i ) ; g E G [ A i ] }then ; Ci C Bi. And Ci # Bi since lCil < IG[Ai]I = Mil < IBil. Choose xi from Bi- Ci and define f in B by f ( i ) = x i for all i. Then f 4 G[A], contradicting that G maps onto B. Hence MI f IBI, so IAl< IBI. Thus C ( K ~i E ; I ) < I&; i E I ) as claimed. We shall now review the concept of cofinality, particularly as applied t o cardinal numbers. For any ordinal number a, the subset C of a is said t o be cofinal in a if for all p with fl < a there is y in C with y 2 p.

Definition A.3.6. The cofinality of the ordinal a (denoted by a' or cf(a)) is the least ordinal fl such that there is a mapping of p onto a cofinal subset of a. The ordinal a is regular if a' = a ; otherwise a is singular. It is clear from the definition that a' is always a cardinal, and a' < a. If a is a successor ordinal then a' = 1, if a is a limit ordinal then a' 2 a.

Lemma A.3.7. If a is any ordinal then a' is the least ordinal fl such that there is a strictly increasing map f' : fl + a with f [p] cofinal in a. Proof. This is trivial if a'

= 0, 1 so we

may suppose a' 2 o.By the definition

Ap. 3

Cardinal arithmetic

193

of a’, there is a map g : a’ -+ a with g[a’] cofinal in a.To prove the lemma, it will suffice to find a strictly increasing map f : a’ a with f [a‘] cofinal in a. Define the values f(y) for y with y < a’by induction as follows: f(y) is the least ordinal [ with E < a such that [ >g(6) and [ > f(6) for all 6 with 6 < y. Since y < a’,neither (g(6); 6 < y) nor { f(6); 6 < y) is cofinal in a so there is such a [. Then f is strictly increasing, and f [a’]is cofinal in a since g[a’] is. -+

Theorem A.3.8. For all a, the co.finality a’ is regular. Proof. We need to show that a” = a’. It suffices to show that there is a mapping of a” onto a cofinal subset of a, for then a‘ < a“. Since anyway a” < a‘, this would give the result. By Lemma A.3.7 there are increasing maps f : a” a‘ and g : a’ Q with f[a”] cofinal in a’ and g[a‘] cofinal in a. Then the conipositiongf is increasing, g f : a” a. Alsogf [a”]is cofinal in a. For given /3 with p < a , there is y in a’ withg(y)>/3and6 i n a ” w i t h f ( 6 ) a y . But thengf(6)>g(y)>/3,sogf[af’] is indeed cofinal in a. -+

-+

-+

Theorem A.3.9. Let K be an infinite cardinal. Then K ’ is the least cardinal h such that K can be expressed as a sum of h cardinals all less than K . If K is singular then K can be expressed as the sum of a strictly increasing sequence of K ’ cardinals all less than K . Proof. Given K , by Lemma A.3.7 there is a strictly increasing function f : K ‘ + K with f [ ~ ‘ ]cofinal in K . For [ with [ < K ’ , put A t = f([+ 1)- f(5). Then K = U{At; < K ’ } and the A t are pairwise disjoint, so K = C(IAtI; [ < K ’ ) . And IAtl < If([ t 1)1
= E(K,,;

U<

K’)<

1/31 . K ’ <

K

.

Thus inductively we may choose cardinals h, for T with T < K ’ such that

Appendix. Cardinal and ordinal numbers

194

Ap. 3

< u < K ' } and A, 2 U{ A, + 1 ; v < T } , since always U{ A, + 1; 7 < K ' . Then (A,; 7 < K ' ) is a strictly increasing sequence of cardinals below K , and K = Z(h,; 7 < K ' ) . h, E { K,;

7

v < T} < K when

It follows from Theorem A.3.9 that if K is any infinite successor cardinal then K is regular. For suppose K = A+. By Theorem A.3.9, write K = Z(K,; u
= z(K,,;u < K ' )

< A . A = h2 = x < K

;

hence K ' = K so that K is regular. Thus for the infinite cardinal K to be singular, it must be that K = H, for some limit ordinal a. It is easy to see that then K' = a'. A regular cardinal H , with a a limit ordinal is said to be weakly inaccessible; the existence of a weakly inaccessible cardinal cannot be proved from the axioms of ZFC set theory. We shall conclude this section with a few remarks about cardinal exponentiation. Cantor's theorem showed that always 2K> K . This result can be strengthened as follows.

Theorem A.3.10. I f K is infinite then (2K)' > K . Proof. Write 2K = Z(A,; u < ( 2 K ) ' )where always 1 < A, < 2K. If (2K)' < K , from Konig's Lemma (Theorem A.3.5) we would obtain the contradiction 2K = qx,; u < ( 2 9 ' ) < n ( 2 K ;(5 < (2K)') < (2K)K= 2K . = 2 K ;

and hence (2K)' > K The Continuum Hypothesis is the statement 2'O = HI;the Generalized Continuum Hypothesis (GCH) is the statement Va(2'" = Both statements are consistent with and independent from the axioms of ZFC. Assuming the GCH, or at least particular instances of it, enables one to evaluate K' for infinite K and h, as in Theorem A.3.11 below. In ZFC alone apart from Theorem A.3.10 very little can be proved about the value of K'.

Theorem A.3.11. Let K , X be infinite cardinals. Suppose 2e = 0' for all infinite cardinals 0 with 0 < max(K, h). m e n K

ifh
K+

ifK'
h+ i f K < h .

Cardinal arithmetic

Ap. 3

195

Proof. Suppose X < K ’ . I f f i s a functionf: A - + K , thenf[A] is not cofinal in K and sof[A] C a for some a with a < K . Hence A~ = U { ’a;Q < K } . And if a < K then

< (2l”l)h = 2101 . A = (la1 A)’ < K . Hence K h = lhKl < z ( l % l ; Q < K ) < K K = K and SO K h = K . Now suppose K ’ < h < K , so K is singular. By Theorem A.3.9 there is a strictly increasing sequence ( K , , ; u < K ’ ) of cardinals below K with K = Z ( K ~ ; u < K ’ ) . By Konig’s Lemma, h K = x(K,,; u < K ’ ) < U(K,,+~; u < K ’ ) < K ~ < ’ K . Also K h < (2K)h= 2K . = 2K = K ’ , so K < K~ < K + and hence K~ = K i Finally, if K < A by Corollary A.3.4, K’ = 2’ and so K’ = A’. [$I = lap

’

‘

REFERENCES

[ 11 H. Bachmann, Transfinite Zahlen, Ergebnisse der Mathematik und ihrer Grenzgebiete, Vol. 1 (2nd ed.) (Springer, Berlin, 1967). [ 2) F. Bagemihl, The existence of an everywhere dense independent set, Michigan Math. J. 20 (1973) 1-2. [ 3 I J.E. Baumgartner, Improvement of a partition theorem of Erdos and Rado, J. Combinatorial Theory Ser. A 17 (1974) 134-137. [ 4 ] J.E. Baumgartner and A. Hajnal, A proof (involving Martin’s axiom) of a partition relation, Fund. Math. 78 (1973) 193-203. [ 5 ] J.E. Baumgartner and K. Prikry, On a theorem of Silver, Discrete Math. 14 (1976) 17-21. [ 6 ] N.G. de Bruijn and P. Erdos, A colour problem for infinite graphs and a problem in the theory of relations, Nederl. Akad. Wetensch. Proc. Ser. A 54 = Indag. Math. 15 (1951) 371-373. 171 C.C. Chang, A partition theorem for the complete graph on w w , J. Combinatorial 7lieory Ser. A I2 (1972) 396-452. [ 8 ] R.O. Davies, An intersection theorem of Erdos and Rado, Proc. Cambridge Philos. SOC.63 (1967) 995-996. [ 9 ] B. Descartes, Solution to Advanced Problem No. 4526,Arner. Math. Monthly 61 (1954) 352-353. [ 101 F.R. Drake, Set Theory. Studies in Logic and the Foundations of Mathematics, Vol. 76. (North-Holland, Amsterdam, 1974). [ 1 1 1 B. Dushnik and E.W. Miller, Partially ordered sets, Amer. J. Math. 6 3 (1941) 600-610. [ 121 W.B. Easton, Powers of regular cardinals, Ann. Math. Logic 1 (1970) 139-178. [ 131 P. Erdos, Some set-theoretical properties of graphs, Univ. Nac. Tucumbn. Revista A 3 (1942) 363-367. [14] P. Erdos, Some remarks on set theory,Proc. Amer. Math. SOC. 1 (1950) 127-141. [15] P. Erdos, Graph theory and probability, Canad. J. Math. 1 1 (1959) 34-38. [ 161 P. Erdos and G. Fodor, Some remarks on set theory - V, Acta Sci. Math. Szeged 17 (1957) 250-260. [ 171 P. Erdos and G. Fodor, Some remarks on set theory - VI, Acta Sci. Math. Szeged 18 (1958) 243-260. [ 181 P. Erdos and A. Hajnal, On the structure of set-mappings, Acta Math. Acad. Sci. Hungar. 9 (1958) 111-131.

References [ 191 P. Erdos and A. Hajnal, On a property of families of sets, Acta Math. Acad. Sci.

197

Hungar. 12 (1961) 87-123. (201 P. Erdos and A. Hajnal, Some remarks on set theory - IX. Combinatorial problems in measure theory and set theory,Michigan Math. J. 11 (1964) 107-127. [21] P. Erdos and A. Hajnal, On chromatic number of graphs and set-systems, Acta Math. Acad. Sci. Hungar. 17 (1966) 61-99. [22] P. Erdos and A. Hajnal, On decomposition of graphs, Acta Math. Acad. Sci. Hungar. 18 (1967) 359-377. [23] P. Erdos and A. Hajnal, On chromatic number of infinite graphs, in: P. Erdos et al., eds., Theory of Graphs (Proc. Colloq., Tihany, 1966) (Academic Press, New York, 1968) pp. 83-98. [24] P. Erdos and A. Hajnal, Unsolved problems in set theory, in. D.S.Scott, ed., Axiomatic Set Theory (Proc. Sympos. Pure Math., Vol. 13, Part I, Univ. California, Los Angeles, Ca., 1967) (Amer. Math. SOC.,Providence, R.I., 1971) pp. 17-48. I251 P. Erdos and A. Hajnal, Unsolved and solved problems in set theory, in: L. Henkin et al., eds., Proceedings of the Tarski Symposium (Proc. Sympos. Pure Math., Vol. 25, Univ. of California, Berkeley, CA, 1971) (Amer. Math. SOC.,Providence, R.I., 1974) pp. 267-287. [ 261 P. Erdos and E.C. Milner, A theorem in the partition calculus, Canad. Math. Bull. 15 (1972) 501-505;Corrigendum, ibid. 17 (1974) 305. [27) P. Erdos and R. Rado, Combinatorial theorems on classifications of subsets of a given set, Proc. London Math. SOC.Ser. 3, 2 (1952) 417-439. ' [28] P. Erdos and R. Rado, A problem on ordered sets, J. London Math. SOC.28 (1953) 426-43 8. [29] P. Erdos and R. Rado, A partition calculus in set theory, Bull. Amer. Math. SOC. 62 (1956) 427-489. [30] P. Erdos and R. Rado, A construction of graphs without triangles having preassigned order and chromatic number, J. London Math. SOC.35 (1960) 445-448. [31] P. Erdos and R. Rado, Intersection theorems for systems of sets, J. London Math. SOC.35 (1960) 85-90. t [32] P. Erdos and R. Rado, Partition relations and transitivity domains of binary relations, J. London Math. SOC.42 (1967) 624-633. [33] P. Erdos and R. Rado, Intersection theorems for systems of sets (II), J. London Math. SOC.44 (1969) 467-479. [34] P. Erdos, L. Gillman and M. Henriksen, An isomorphism theorem for real-closed fields, Ann. of Math. Ser. 2, 61 (1955) 542-554. [ 351 P. Erdos, A. Hajnal and E.C. Milner, On the complete subgraphs of graphs defined by systems of sets, ActaMath. Acad. Sci. Hungar. 17 (1966) 159-229. [36] P. Erdos, A. Hajnal and E.C. Milner, On sets of almost disjoint subsets of a set, Acta Math. Acad. Sci. Hungar. 19 (1968) 209-218. [37] P. Erdos, A. Hajnal and E.C. Milner, Set mappings and polarized partition relations, in: P. Erdos e t al., eds., Combinatorial Theory and its Applications I (Proc. Colloq., Balatonfured, 1969) (North-Holland, Amsterdam, 1970) pp. 327-363. [38] P. Erdos, A. Hajnal and R. Rado, Partition relations for cardinal numbers, Acta Math. Acad. Sci. Hungar. 16 (1965) 93-196. [39] P. Erdos, EL'. Milner and R. Rado, Intersection theorems for systems of sets (III), J. Austral. Math. SOC. 18 (1974) 22-40.

198

References

[40] G. Fodor, Proof of a conjecture of P. Erdos, Acta Sci. Math. Szeged 14 (1952) 219-227. 1411 C. Fodor, Some results concerning a problem in set theory, Acta Sci. Math. Szeged 16 (1955) 232-240. [42] G. Fodor, Eine Bemerkung zur Theorie der Regressiven Funktionen, Acta Sci. Math. Szeged 17 (1956) 139-202. 1431 A.A. Fraenkel, Einleitung in die Mengenlehre. Die Crundlehren der Mathematischen Wissenschaften in Einzeldarstellungen, Vol. 9 (3rd. ed.) (Springer, Berlin, 1928). 1441 F. Calvin, On a partition theorem of Baumgartner and Hajnal, in: A. Hajnal, ed., Infinite and Finite Sets, Vol. 11, (Colloq., Keszthely, 1973. Colloq. Math. Janos Bolyai, Vol. 10) (North-Holland, Amsterdam, 1975) pp. 71 1-729. 1451 F. Galvin and A. Hajnal, Inequalities for cardinal powers, Ann. of Math. Ser. 2, 101 (1975)491-498. 1461 F. Calvin and J. Larson, Pinning countable ordinals, Fund. Math. 82 (1974) 357-36 1. 1471 F. Calvin and S. Shelah, Negation of partition relations without CH, Notices Amer. Math. SOC. 18 (1971) 666. [48] F. Galvin and S. Shelah, Some counterexamples intthe partition calculus, J. Combinatorial Theory Ser. A , 15 (1973) 167-174. [49] K. Godel, The Consistency of the Axiom of Choice and the Generalized Continuum Hypothesis with the Axioms of Set Theory. Annals of Mathematics Studies No. 3 (Princeton University Press, Princeton, N.J., 1940). [ S O ] L. Haddad and G. Sabbagh, Calcul de certains nombres de Ramsey gdn&ralisds, C R . Acad. Sci. Paris S h . A-B, 268 (1969) A1233-A1234. 1511 L. Haddad and G. Sabbagh, Nouveaux rdsultats sur les nombres de Ramsay g h d ralisds, C.R. Acad. Sci. Paris S6r. A-B, 268 (1969) A1516-A1518. [52] A. Hajnal, Some results and problems on set theory, Acta Math. Acad. Sci. Hungar. 11 (1960) 277-298. 1531 A. Hajnal, Proof of a conjecture of S. Ruziewicz, Fund. Math. 50 (1961/62) 123- 128. [54] A. Hajnal, Remarks on the theorem of W.P. Hanf, Fund. Math. 54 (1964) 109-113. [ S S ] A. Hajnal, On some combinatorial problems involving large cardinals, Fund. Math. 6 9 (1970) 39-53. (561 A. Hajnal, A negative partition relation,Proc. Nut. Acad. Sci. U.S.A. 6 8 (1971) 142- 144. [ 571 W. Hanf, Incompactness in languages with infinitely long expressions, Fund. Math. 53 (1963/64) 309-324. 1581 J.M. Henle and E.M. Kleinberg, A combinatorial proof of a combinatorial theorem, Acta Math. Acad. Sci. Hungar. 26 (1975) 3-7. 1591 I. Juhisz, Cardinal Functions in Topology. Mathematical Centre Tracts Vol. 34 (Mathematisch Centrum, Amsterdam, 1971). 1601 J.L. Kelly, General Topology (van Nostrand, New York, 1955). 161] E.M. Kleinberg, Infinitary combinatorics, in: A.R.D. Mathias e t al., eds., Cambridge Summer School in Mathematical Logic (Cambridge, England, 1971. Lecture Notes in Mathematics Vol. 337) (Springer, Berlin, 1973) pp. 361-418. [62] A.H. Kruse, A note on the partition calculus of P. Erdos and R. Rado, J. London Math. SOC.40 (1965) 137-148.

References

199

[63] C. Kuratowski, Sur une caractdrisation des alephs, Fund. Math. 38 (1951) 14-17. [64] D. Kurepa, On the cardinal number of ordered sets and of symmetrical structures in dependence of the cardinal numbers of its chains and antichains, Clasnik Mat.Fiz. Astronom. Drustvo Math. Fiz. Hrvatski Ser. II 14 (1959) 183-203. I651 J.A. Larson, A short proof of a partition theorem for the ordinal mu, Ann. Math. Logic 6 (1973) 129-145. [66] D. L k i r , On a problem in the theory of aggregates, Compositio Math. 3 (1936) 304. [67] S. Mazur, On continuous mappings on Cartesian products, Fund. Math. 39 (1952) 229-23 8. [68] E. Michael, A note on intersections, Proc. Amer. Math. SOC. 13 (1962) 281-283. [69] E.C. Milner, Some combinatorial problems in set theory, Thesis, London, 1962. [70j E.C. Milner, Transversals of disjoint sets, J. London Math. SOC.43 (1968) 495-500. [71] E.C. Milner, Partition relations for ordinal numbers, Canad. J. Math. 21 (1969) 317-334. [72] E.C. Milner, A finite algorithm for the partition calculus, in: W.R. Eames et al., eds., Proceedings of the Twenty-Fifth Summer Meeting of the Canadian Mathematical Congress (Lakehead Univ., Thunder Bay, Ont., 1971) (Lakehead Univ., Thunder Bay, Ont., 1971) pp. 117-128. [73] E.C. Milner and R. Rado, The pigeon-hole principle for ordinal numbers, Proc. London Math. SOC. Ser. 3, 15 (1965) 750 -768. [ 741 M. Morley, Partitions and models, in: M.H. Lob, ed., Proceedings o f t h e Summer School in Logic (Leeds, 1967, Lecture Notes in Mathematics Vol. 70) (Springer, Berlin, 1968) pp. 109-158. [75] E. Nosal, On a partition relation for ordinal numbers, J. London Math. SOC.Ser. 2 8 (1974) 306-310. [76] S . Picard, Solution d’un problbme de la th6orie des relations, Fund. Math. 28 (1937) 197-202. S . Piccard, Sur un problbme de M. Ruziewicz de la thdorie des relations, Fund. Math. 29 (1937) 5-8. K. Prikry, On a problem of Erdos, Hajnal and Rado, Discrete Math. 2 (1972) 5 1-59. K. Prikry, On a set-theoretic partition problem, Duke Math. J. 39 (1972) 77-83. F.P. Ramsey, On a problem of formal logic, Proc. London Math. SOC.Ser. 2, 30 (1930) 264-286. F. Rowbottom, Some strong axioms of infinity incompatible with the axiom of constructibility, Ann. Math. Logic 3 (1971) 1-44. S. Ruziewicz, Une gknkralisation d’un thkoreme de M. Sierpinski, Publ. Math. Univ. Belgrade 5 (1936) 23-27. R.A. Shore, Square bracket partition relations in L, Fund. Math. 84 (1974) 101-106. W. Sierpinski, Sur une ddcomposition d’ensembles, Monatsch. Math. Phys. 35 (1928) 239-242. W. Sierpinski, Sur un problbme de la th6orie des relations, Ann. Scuola Norm. Super. Pisa Ser. 2, 2 (1933) 285-287. [86] W. Sierpinski, Sur un problbme de la thdorie des relations, Fund. Math. 28 (1937) 71-74. [87] W. Sierpinski, Sur les suites transfinies finalement disjointes, Fund. Math. 28 (1937) 115-119.

200

References

[ 881 W. Sierpinski, Sur quelques propositions concernant la puissance du continu, Fund. Math. 38 (1951) 1-13. [ 891 W . Sierpinski, Cardinal and Ordinal Numbers. Polska Akademia Nauk, Monografie Matematyczne Vol. 34 (2nd ed.) (P.W.N.-Polish Scientific Publishers, Warsaw, 1965). 1901 J.H. Silver, Some applications of model theory in set theory, Thesis, Berkeley, 1966. See alsoAnn. Math. Logic 3 (1971)45-110. [ 91) J.H. Silver, The independence of Kurepa’s conjecture and two-cardinal conjectures in model theory, in: D.S.Scott, ed., Axiomatic Set Theory (Proc. Sympos. Pure Math., Vol. 13, Part I, Univ. California, Los Angeles, CA, 1967) (Amer. Math. SOC., Providence, R.I., 1971) pp. 383-390. [92] J.H. Silver, On the singular cardinals problem, in: Proceedings of rhe fnternational Congress of Mathematicians (Vancouver 1974) Vol. 1 (Canadian Mathematical Congress, Montrdal, 1975) pp. 265-268. 1931 E. Specker, Teilmengen von Mengen mit Relationen, Comment. Math. Helv. 3 1 (1957) 302-314. [ 941 A. Tarski, Sur la ddcomposition des ensembles en sous-ensembles presque disjoints, Fund. Math. 12 (1928) 188-205 and 14 (1929) 205-215. [95] N.H. Williams, A partition property of cardinal numbers, Dissertationes Math. (Rozprawy Mat.) 90 (1972) 1-26. [96] A.A. Zykov, On some properties of linear complexes, Mar. Sbornik N.S. 24 (66) (1949) (Russian) 163-188.

INDEX OF AUTHORS

Bachmann, H., 184,185 Bagemihl, F., 70 Baumgartner, J.E., 15,157, 174 Bernstein, F., 189 de Bruijn, N.G., 125,132

Henriksen, M., 3 Juha'sz, I., 26 Kelly, J.L., 184 Kleinberg, E.M., 47,63 Konig, J., 192 Kruse, A.H., 156 Kuratowski, C., 109 Kurepa, D., 24

Cantor, G., 189, 194 Chang, C.C., 153,168 Davies, R.O., 142 Drake, F.R., 61 Dushnik, B., 24 Easton, W.B., 15 Erdos, P., 3 , 7 , 1 5 , 1 7 , 2 3 , 2 4 , 2 6 , 3 6 , 4 0 46,53,60,61,64,66, 70, 71, 78, 81, 83, 102,112,113,116,121, 125,128, 129,130,132,141,142,144,147, 157, 161,165,173,174 Fodor, G., 11,61,64,66,67,69, Fraenkel, A.A., 184

71

Galvin,F., 15,59,109,161,164,174 Gillman, L., 3 Godel, K., 59, 184 Haddad, L., 159,165 Hajnal, A., 7,15,17,23,24,26,36,46,47, 53,60,61,64,70, 71,74,78,81,83, 102, 105, 109, 113, 116,121, 128, 130, 141,161,165,173,174 Hanf, W., 44 Hartogs, F., 189 Henle, J.M., 63

Larson, J.A., 161, 168 Lka'r, D., 64 Mazur, S., 142 Michael, E., 142 Miller, E.W., 24 Milner, E.C., 7, 9, 15,70, 130,141, 144, 147,154,155,165,168 Morely, M., 61

Nosal, E., 161 Piccard, S., 64 Prikry, K., 15, 101,105,174 Rado, R., 24, 2 6 , 3 6 , 4 0 , 4 6 , 5 3 , 7 1 , 7 8 , 8 3 , 102, 113, 125,142,144,147,154, 155, 157,174 Ramsey, F.P., 24, 33, 34 Rowbottom, F., 47,61 Ruziewicz, S., 64, 70 Sabbagh, G., 159,165 Schroder, E., 189

202

fndex of authors

Shelah, S., 5 9 , 6 0 Shore, R.A., 5 9 , 6 0 Sierpinski,W., 1 , 2, 3 , 2 4 , 4 0 , 6 4 , 108,184 Silver, J.H., 10, 1 5 , 6 3 Specker, E., 113,159,168

Tarski, A., 1 , 4 Turin, P., 64 Tutte, W.T., 112 Zykov, A.A., 112

INDEX OF DEFINITIONS

adjacent vertices, 110 almost disjoint, 1 almost homogeneous, 175 Axiom of Choice, 185

connected set, 130 contain (of indexed family), 141 Continuum Hypothesis, 194 cover number, 130

beth, x branch of tree, 27

decomposition, x degree of disjunction, 1 delta family, 142 domain, ix

canonical partition, 36 of [ W ( n ) ] * , 161 Canonization Lemma (Lemma 2.4.2), 36 Cantor normal form, 189 cardinal number, 190 ,beth, x , limit, x ,regular, x, 192 , singular,x, 192 , strong limit, x ,strongly inaccessible, x ,successor, x ,weakly inaccessible, x, 194 cardinality, 190 Choice, Axiom of, 185 chromatic number, 111 circuit in graph, 110 closed sequence, 12 closed set, 11 cofinal set, x, 192 cofinality, 192 colouring by q colours, 121 v-colouring, 111 colouring number, 111 complete bipartite graph, 110 complete graph, 110 concatenation of sequences, ix

edge of graph, 110 family, ix ,delta, 142 ,indexed, ix , weak delta, 147 ,(01,141 ,(A,
204

Index of definitions

height, of ramification system, 28 , of tree, 27 homogeneous, almost, 175 , sequence, 83 ,set, 25 inaccessible cardinal, strongly, x ,weakly, x , 194 independent set, 110 indexed family, ix initial ordinal, 190 Konig’s Lemma (Theorem A.3.5),192 length of sequence, ix level in tree, 27 lexicographic ordering, 41 limit cardinal, x limit ordinal. 186 Milner-Rado paradox, 155 node of tree, 27 normal form, Cantor, 189 order, of element in tree, 27 , of ramification system, 28 ,of set mapping, 64 ordinal number, 186 ,finite, 186 ,initial, 190 ,limit, 186 , successor, 186 ordinary partition symbol, for cardinals, 24 , for ordinals, 153 partition, x , canonical, 36, 161 , disjoint, x partition symbol, ordinary, 24, 153 ,polarized, 83 ,polarized with alternatives, 84 , square bracket, 53 pin, 160 polarized partition symbol, 8 3 ,with alternatives, 84 Ramification Lemma. 30

ramification system, 27 ,height of, 28 , order of,28 Ramsey’s Theorem, 33,34 range, ix regressive function, 11 regular cardinal, x, 192 restriction, of function, ix , of sequence, 27 set, 184 , cardinality of, 190 , connected, 130 , independent, 110 ,stationary, 11 , transitive, 186 ,unbounded, 11 , well founded, 186 set mapping, 64 ,order of, 64 , type of, 78 shape, 169 similar, 189 ,sequences, 161 singular cardinal, x, 192 square bracket partition relation, 5 3 stationary set, 11 Stepping-up Lemma, 34 strong limit cardinal, x strongly inaccessible cardinal, x subgraph, 110 , spanned by H, 110 successor cardinal, x successor ordinal, 186 transfinite induction, 186, 187 transfinite recursion, 187 transitive set, 186 transversal, 5 q-transversal, 15 tree, 27 ,branch of, 27 , full, 27 ,height of, 27 ,level in, 27 ,node of, 27 type, of set mapping, 78

Index of definitions unbounded set, 11 vertex of graph, 110 weak delta family, 130

weakly inaccessible cardinal x, 194 well founded set, 186 Zermelo-Fraenkel set theory, 184

205

INDEX OF NOTATION

f L

membership, ix, 184 subset relation, ix proper subset, ix power set, ix union of (the members of) x, ix intersection of (the members of) x, ix set difference, ix ix ordered pair, ix sequence, ix length of sequence, ix concatenation of sequences, ix domain, ix range, ix function from A into B, ix set of functions, ix value of function, ix restriction of function, ix image off on X,ix indexed family, ix Cartesian product, x in the same class of A, x cardinality, x, 190 K-size subsets of X,x
.

dom(A) ran(A) f:A+B AB f(x)

fr x

fIXl

(Aj;iEI)

X (Aj ; i E

I)

a = b(mod A)

1x1 1x1

[XI
Index of notation order type, x X

subsets of X of order type 01, x supremum, x cardinal successor, x n-fold cardinal successor, x cofinality, x, 192 cardinal beths, x Generalized Continuum Hypothesis, x, 194 least infinite ordinal, x, 186 degree of disjunction, 1 constructible universe, 10 19 ordinary partition relation, 24 ordinary partition relation, 24 24 24 ordinal sequences of length o, 27 in ramification system, 27 in ramification system, 27 in ramification system, 27 in ramification system, 27 first point of difference, 48 48 square bracket partition relation, 53 square bracket partition relation, 53 partition relation, 6 1 67 for set mappings, 7 1

polarized partition relation, 83

polarized partition relation with alternatives, 84 complete graph, 110 complete bipartite graph, 110 110 colouring number, 11 1 chromatic number, 111 121 130 14 1 14 1 delta family, 142

201

208

Index of notation 142 147 number of cardinals < K , 147 ordinary partition relation, 153 set of sequences of length n, 157 162 set of finite sequences, 168 175 176 179 179 Zermelo-Fraenkel set theory, 184 equality symbol, 184 pair set, 184 ordering of t h e ordinals numbers, 186 ordinal successor, 186 similar, 189 initial ordinals, 189 infinite cardinals, 190