Combinatorics of Finite Sets (Dover Books on Mathematics)

Ii

Combinatorics of Finite Sets Ian Anderson University of Glasgow

DOUER PUBLICATIONS, INC. Mineola, new York

Copyright Copyright © 1987 by Ian Anderson

All rights reserved under Pan American and International Copyright Conventions.

Published in Canada by General Publishing Company, Ltd, 895 Don Mills Road, 400-2 Park Centre, Toronto, Ontario M3C 1 W3. Published in the United Kingdom by David & Charles, Brunei House, Forde Close, Newton Abbot, Devon TQ12 4PU.

Bibliographical Note This Dover edition, first published in 2002, is a corrected republication of the work as published by Oxford University Press, Oxford, England, and New York, in 1989 (first publication: 1987).

Library of Congress Cataloging-in-Publication Data Anderson, Ian, Ph, D. Combinatorics of finite sets / Ian Anderson. p. cm

a corrected republication of the work as published by Oxford University Press, Oxford, England, and New York, in 1989 (first publication: 1987)"- -T.p. verso. Includes index. ISBN 0-486-42257-7 (pbk ) 1. Set theory. 2. Combinatorial analysis. I. Title.

QA248 .A657 2002 511.3'22 - dc2l 2002019228

Manufactured in the United States of America Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501

To Margaret Grace

Preface

The past quarter century has seen the remarkable rise of combinatorics as a distinctive and important area of mathematics. Combinatorial topics have found a place in many university degree courses, and, since the founding of the Journal of Combinatorial

Theory in 1966, there has been a flood of publications in the combinatorial area. Some branches of combinatorics are already well established, with their own unified body of theory and applications: examples are graph theory, with its topological flavour, coding theory and design theory, with their algebraic connections, and enumeration theory, concerned with the techniques of counting. These areas are

already supplied with many textbooks at both undergraduate and postgraduate levels. The purpose of the present work, however, is to advertise another area of combinatorics, where a body of theory, at one time very scattered and ad hoc, is gradually being moulded into an elegant unity. Without attempting to be exhaustive, the book is intended to be a readable introduction to what is, for the author, a fascinating subject. The origins of the theory can perhaps be traced back to 1928, when Sperner published a simple theorem which has had repercussions far beyond his wildest dreams. Sperner's theorem simply asserts that if

you want to find as many subsets of an n-element set as possible, subject to the condition that no subset is contained in another, then you cannot do better than to choose all the subsets of size [in]. This theorem has been reproved and generalized to such an extent that it has given rise to a whole branch of the theory of partially ordered sets

(or posets) called Sperner theory. Although we shall go some way down this path, we shall not restrict ourselves to this area. Instead, we shall use Sperner's theorem as a springboard and a signpost, discovering where the ideas involved in its various proofs lead to. We shall be led to consider the structure of the set of subsets of a finite set viewed as a poset, its chain decompositions and its antichains, its rank levels, and the inclusion relations between these levels. This will take

us, for example, to the famous Kruskal-Katona theorem which answers the following question: given r subsets of S, all of size k, vii

viii

I

Preface

what is the least possible number of sets of size k - 1 contained in them? The answer to this question involves a nice interplay between two different orderings, namely the partial ordering of the subsets of S by inclusion, and the total ordering of the subsets of S of a given size by what we shall call the squashed ordering, which is a variation on the more familiar lexicographic or alphabetical ordering. Anyone writing a book in this area is faced right at the start by a fundamental problem. Many results for the poset of subsets of a set

can be extended to more general posets. Sometimes one of the available proofs for sets easily extends to more general posets; sometimes, however, the nicest proof for sets does not generalize. So should the results be presented in their most general form (thereby

sometimes losing out on clarity) or should they be presented for subsets of a set (thereby losing out on generality but perhaps gaining in clarity)? An example of this problem arises in connection with the Kruskal-Katona theorem. Several proofs are available, but a more difficult proof due to Clements and Lindstrom establishes the result in

the more general context of the poset of divisors of a number (or subsets of a multiset). Since we include the Clements-Lindstrom theorem in Chapter 9, there is strictly speaking no need to include a

separate proof of the Kruskal-Katona theorem. However, in the simpler context of sets, the Kruskal-Katona theorem has such an elegant theory surrounding it that it would be almost criminal to omit the simpler case. So we present both proofs. On many other occasions we prove results in more than one way because the different proofs

illustrate different ideas and different techniques. On the whole, I have taken the view that I should present results in their simplest forms, concentrating mainly on sets and multisets. Accordingly, the reader will not find a discussion of, say, geometric lattices, although their `prototype', the poset of partitions of a set, is discussed. In a few

places we look at posets more generally; the final chapter, for example, discusses extensions of the theorem of Dilworth concerning chain decompositions of a general poset. In searching out the material for this book I was greatly helped by several survey articles, two of which deserve special mention: the first

is that by Greene and Kleitman (1978), and the second is the more recent one by D. B. West (1982). As will be seen from a glance at the extensive list of references at the end of the book, a number of more

recent results have been included. Inevitably some of the results presented here will have been improved upon by the time this book appears in print, but in a sense this does not matter for the aim of the book is not to provide an exhaustive survey but to present some of

Preface

I

ix

the ideas and techniques which go to make up the subject. Inevitably, also, my choice of material will not meet with the approval of all, but a number of interesting results not in the text have been included in the exercises at the end of each chapter with hints or outlines of their solutions at the end of the book. A number of people have helped and encouraged me in the writing

of this book. In particular, I gladly acknowledge the helpful comments of Professor George Clements and Dr Hazel Perfect. I should also like to thank the Oxford University Press for encouraging me to write, and the University of Glasgow for granting me a period of study leave during which the final compilation of the book was accomplished. Glasgow December 1985

I.A.

Contents Notation

xv

1. Introduction and Sperner's theorem 1.1 A simple intersection result 1.2 Sperner's theorem

1.3 A theorem of Bollobas

4

Exercises 1 2. Normalized matchings and rank numbers 2.1

1

2

Sperner's proof

7

2.2 Systems of distinct representatives

10 12

2.3 LYM inequalities and the normalized matching property 2.4 Rank numbers: some examples

13 17

Exercises 2 3. Symmetric chains 3.1 Symmetric chain decompositions 3.2 Dilworth's theorem 3.3 Symmetric chains for sets 3.4 Applications 3.5 Nested chains 3.6 Posets with symmetric chain decompositions Exercises 3 4. Rank numbers for multisets 4.1 Unimodality and log concavity 4.2 The normalized matching property 4.3 The largest size of a rank number Exercises 4

23 27 30 32 36 43 45 53

56 60 63

68

5. Inter secting systems and the ErdOs-Ko-Rado

theo rem 5.1

The EKR theorem

70

5.2 Generalizations of EKR 5.3 Intersecting antichains with large members

73 77

5.4 A probability application of EKR

79

5.5

Theorems of -Milner and Katona xi

81

xii I Contents

5.6 Some results related to the EKR theorem Exercises 5 6. Ideals and a lemma of Kleitman 6.1 Kleitman's lemma 6.2 The Ahlswede-Daykin inequality 6.3 Applications of the FKG inequality to probability theory 6.4 Chvatal's conjecture Exercises 6

83 85 87 90 97 103 108

7. The Kruskal-Katona theorem 7.1

Order relations on subsets

7.2 The 1-binomial representation of a number 7.3 The Kruskal-Katona theorem

7.4 Some easy consequences of Kruskal-Katona Compression Exercises 7 8. Antichains 8.1 Squashed antichains 8.2 Using squashed antichains 8.3 Parameters of intersecting antichains Exercises 8 9. The generalized Macaulay theorem for multisets 9.1 The theorem of Clements and Lindstrom 7.5

112 115 119 124 126 127 131 135 139 143

9.2 Some corollaries 9.3 A minimization problem in coding theory

145 151 154

9.4 Uniqueness of maximum-sized antichains in multisets Exercises 9

157 160

10. Theorems for multisets 10.1 Intersecting families 10.2 Antichains in multisets 10.3

Intersecting antichains

Exercises 10

11. The Littlewood-Offord problem 11.1 Early results 11.2 M-part Sperner theorems 11.3 Littlewood-Offord results Exercises 11

162 168 172 174

176 178 183 185

Contents I xiii

12. Miscellaneous methods 12.1 The duality theorem of linear programming 12.2 Graph-theoretic methods 12.3 Using network flow Exercises 12 13. Lattices of antichains and saturated chain partitions 13.1 Antichains 13.2 Maximum-sized antichains 13.3 Saturated chain partitions 13.4 The lattice of k-unions Exercises 13 Hints and solutions

187 192 194 198

201 203 205 212

212 214

References

241

Index

249

Some notation used in the text

(n)

ad e(k)d Vsi Cs/I

(k)S Ni <S

Fk(n) 2C(P)

0 i(m) dk(P)

binomial coefficient (n-choose-r) the shadow of s/4 the shadow of s/I at level k

the shade of si the compression of s/I the set of complements of members of s/I the set of k-subsets of S the number of members of rank i the squashed ordering the first n k-sets (or k-vectors) the set of antichains of the poset P the empty set the number of divisors of m the size of the largest k-union in P

xv

1 Introduction and Sperner's theorem

1.1

A simple intersection result

The main theme of this book is the study of collections s4 of subsets of a finite set S, where s4 is described in terms of intersection, union, or inclusion conditions. An amazing richness and variety of results will be discovered, developed, and extended in various directions.

Although our main initial theme will be a study of a theorem of Sperner which could be said to be the inspiration of all that follows, we get into training by first of all asking what must surely be one of the simplest questions possible.

Problem Let 4 be a collection of subsets of an n-element set S (or an n-set S) such that A, fl A; 0 0 for each pair i, j. How big can IsiI be? The answer, and more besides, is given by the following theorem.

Theorem 1.1.1 If s4 is a collection of distinct subsets of the n-set S

such that A; fl A; # 0 for all A,, A; E 4, then I4 2n '. Further, if I si I < 2"-', s4 can be extended to a collection of 2"-' subsets also satisfying the given intersection property.

Proof If A E 4 then the complement A' = S - A is certainly not in A since A fl A' = 0. So we immediately obtain IdI -- 22" = 2"-' This bound cannot be improved upon since the collection of all subsets of {1, . . . , n} containing 1 satisfies the intersection condition and has 2"-' members. Now suppose IsiI <2"-'. Then there must be a subset A with A f si and also A' f si. We can then add A to the collection s4 unless there exists B E 4 such that A fl B = 0. But then B c A' and so we could add A' to 4. If the resulting collection has fewer than 2"-' members, repeat the process.

This example pinpoints some key questions. Given a property involving union, intersection, and inclusion, how large can a collec-

tion 4 of subsets of S be if s4 satisfies the property? Can we 1

2 I Combinatorics of finite sets

characterize those collections which maximize I4I? These are the sort of questions which we shall study. 1.2

Sperner's theorem

We now consider the property: if A;, A; E s4, then A. A;. A collection of subsets of S with this property is called a collection of incomparable sets, or an antichain, or sometimes a clutter. It is an antichain in the sense that its property is the other extreme from that of a chain in which every pair of sets is comparable. Theorem 1.2.1 (Sperner 1928) an n-set S. Then

Let s4 be an antichain of subsets of

IS'I

This result is clearly best possible since the subsets of size [n/2]

form an antichain. The original proof given by Sperner will be analysed in Chapter 2, but we start here by giving a simple elegant

proof due to Lubell which is pregnant with generalizations and extensions. Altogether we shall give three different proofs, not just because they exist, but because each in its own way presents us with

ideas which can be developed to suit a wider range of ordered structures.

Proof of Theorem 1.2.1 (Lubell 1966) First of all note that there are n! permutations of the elements of S. We shall say that a permutation jr of the elements of S begins with A if the first IA I members of r are

the elements of A in some order. Now the number of permutations beginning with A must be IAI!(n - JAI)!. Also, no permutation can begin with two different sets in s4, since one of these sets would contain the other; therefore permutations beginning with different sets in s4 are distinct. Thus Z IAI !(n - JAI )! _- n!

Aes4

If we let pk denote the number of members of 4 of size k, we have 7, k!(n - k)!pk -_ n! whence

k

(n) k

Introduction and Sperner's theorem 1 3

Thus ICI =

n k

Pk = [n/2]

Pk

n

k

l

( n [n/2]1

([n/2])

Pk

k (nl

n

n/2]l.

(1.2)

k)

0 Note that Lubell's proof actually gives a stronger result than Sperner's theorem. The inequality (1.1) is called a LYM inequality after Lubell, Yamamoto (1954), and Meschalkin (1963), whose work includes similar results. Instead of giving an upper bound for EkPk, the inequality gives an upper bound for the weighted sum EkPk

nk).

Note that

Pk/(nk)

is the proportion of all those subsets

of S of size k which are in d; the LYM inequality asserts that the sum of these proportions is at most unity. The special case where Pk =(k) if

k = [n/2], Pk = 0 otherwise, shows that the bound can be attained.

We have thus shown that the maximum size of an antichain of

subsets of an n-set S is ([n2J) . Can we identify all the antichains which are as big as this? The first inequality in (1.2) shows that we can

attain the bound ([n2]) only if Pk = 0 whenever (k) < (E',) Therefore, if n is even, there is only one maximum-sized antichain,

namely the collection of all n/2-subsets. If n is odd, all sets in a maximum-sized antichain must be of size J(n - 1) or J(n + 1). We now show that there can in fact be no mixture of sizes; a maximum-sized antichain consists either of all subsets of size z(n - 1) or of all subsets of size J(n + 1).

Theorem 1.2.2 If n is even, the only antichain consisting of n

[n/2])

subsets of the n-set S is made up of all the n/2-subsets of S. If

n is odd, an antichain of size ([n/2]) consists of either all the J(n - 1)-subsets or all the '(n + 1)-subsets.

Proof (Lovasz 1979) The case of even n has been dealt with. Suppose now that n = 2m + 1 and that

.

is an antichain of size

(n ). Note that in Lubell's proof the only way of finishing up with m

4 1 Combinatorics of finite sets

equality in (1.2) is to have equality at each stage, so in particular every permutation must contribute a member of s4 with which it begins.

Now 4 must consist only of sets of size m or m + 1. Suppose that X, Y are subsets of size m, m + 1 respectively and that X c Y. If X = {x,, . . . , x,,,} and Y = {x,, ... , x,,,, xm+,} then since any permutation beginning with x,, . . . , must begin with a member of s4, X or Y must be in 4. Our aim is to prove that s4 consists of all m-sets or all (m + 1)-sets. Suppose that s4 contains some but not all of the (m + 1)-sets. Then we can find sets E, F such that JET = IFI = m + 1, E E s4, F f s4. By

relabelling the elements of S if necessary we can suppose that E = {x,, ... , xm+,} and F = {x;, ... , x,,,+r} for some i. Since E E S4 and F f s4 there must be a largest integer j < i with '{x;, . . . , x,,,+;} E

s4. Then E*={x1,...,xm+;}ES4 and F*={x;+,,...,x,,,+i+1}f s4. We now have an impossible situation. Since E* fl F* c E* where E* cs4, we must have E* fl F* s4. However, E* fl F* c F* where IE* fl F*1= m and IF*l = m + 1, so by an earlier part of the proof one of E* fl F* and F* must be in s4. This contradiction shows that our assumption must have been false and so s4 indeed consists only of sets all of the same size. 0

1.3 A theorem of BollobSs

As another example of how the permutation approach of Lubell's

proof can be used to obtain elegant proofs of results obtained originally by more complicated arguments, we now give a generalization of Sperner's theorem due to Bollobds (1965). The result was also independently proved by Katona (1974), Tarjan (1975), and Griggs,

Stahl, and Trotter (1984). This repeated discovery of results by authors working independently is a frequent occurrence in this area of mathematics!

Theorem 1.3.1 (Bollobas 1965) Let A,, . . . , Am, B1, . . . , Bm be subsets of an n-set S such that A; fl B; = 0 if and only if i = j. Let a; = IA; I and b; = I B; I. Then m

1 1.

a;

1

Introduction and Sperner's theorem

1

5

Proof Consider each of the n! permutations of the elements of S, and say that a permutation r contains A followed by B if all the elements of A occur in jr before all the elements of B. If a particular

permutation n contains A, followed by B; and also contains A; followed by B;, then A, fl B, = 0 (if A. ends before B; begins) or A; fl B; = 0 (if A, ends after B; begins), and either of these contradicts the hypotheses. So, for each permutation jr, there is at most one i for which r contains A. followed by B,. However, given i, the number of permutations containing A; followed by B; can be found as follows. Choose the a, + b, positions to be filled by the elements of A, and B;;

this can be done in (a, + b,) ways. Then place the a, members of A, in

some order in the first a, of the chosen positions and then the b, members of B, in some order in the remaining b, positions; this can all

be done in a;!b;! ways. Finally order the remaining n - a, - bi elements of S and place then in the remaining places of the permutation; this can be done in (n - a, - b;)! ways. Thus the number of permutations r containing A, followed by B; is (aj +n bi

l)a`!b`!(n - a; - b;)! =

n!

(ai+bi)

Summing over all i we now obtain

I

n - gin! (a; + b;l a,

J

as required. Note that Sperner's theorem follows on taking B, = A,!, the comple-

ment of A;, for the condition A, n B; = 0 becomes A; n A; = 0, the condition A; n B, 0 0 becomes A; n A, * 0, i.e. A, A;, and the conclusion yields 1

1

(n) k (k)

i

a,-

J

(a, a bj)

Theorem 1.3.1 has been generalized in a number of ways. Frankl (1982) and Kalai (1984) weakened the condition to A; n A; * 0 for

1 --i < j --m and obtained the same conclusion. Lovasz (1977)

6

Combinatorics of finite sets

generalized the theorem to subspaces of a linear space. As a recent application, we now apply Theorem 1.3.1 to the following generaliza-

tion of the Sperner situation. Suppose we are given m chains of subsets of an n-set S which are incomparable in the sense that no member of one chain is contained in a member of any other chain. How large can m be? In the case where all the chains have k + 1 members, let f (n, k) denote the largest possible value of m. Then Sperner's theorem corresponds to the case k = 0 and asserts that f(n, 0) =

n

([n2])

Theorem 1.3.2 (Griggs et al. 1984) Let f (n, k) denote the largest value of m for which it is possible to find m chains of k + 1 distinct subsets of an n-set S such that no member of any chain is a subset of a member of any other chain. Then

f(n, k) =

n-k l

(Kn

- k)l2]).

Proof Suppose that we have m chains

Aj,ocA,,ic...cA,.k (i=1,...,m) satisfying the conditions of the theorem. In Theorem 1.3.1 take A; = A;,o and B; = S - A,,k. Then a; = jA;,oj and b; = n - IA;,kJ. It is clear that IA;,kI , a; + k, so b; , n - k - a,; thus `a` a,

b`I

` `n

-k)_(,( n-k a,

n

k)l2]1'

Now we certainly have A, fl B, = 0. Also, if we had A; fl B; = 0 for some i O j we would then have A;,o c A;,k, contradicting the hypotheses. So the sets A; and B; satisfy the conditions of Theorem 1.3.1, and we have

m= 1 i=1

( i=1

n-k l a; +b;1 (n - k)l2]a;

n-k \[(n - k)121)'

Thus n -k l f(n, k) , \[( n - k)l2]I'

To complete the proof we exhibit ([( k)/2]) chains with the required properties. Consider the [(n - k)/2]-subsets X of

Introduction and Sperner's theorem 1 7

{k + 1,

... , n). There are ([(nn k)/2]) such subsets. For each

such X take the chain

XcXU{1}cXU{1,2)c...cXU{1,2,...,k}. These chains have the required properties.

Further generalizations of Sperner's theorem will be discussed later, particularly in Chapter 8 and in the study of the LittlewoodOfford problem in Chapter 11. Exercises I 1.1

1.2

Prove that if A1, . . . , Am are distinct subsets of an n-set S such that for each pair i, j, A, U A10 S then m , 2"-1

Can we have equality in Theorem 1.1.1 without all the A, having a common element?

1.3

Show that if .si is an antichain of subsets of an n-set with JAI , h < Zn for all A E Si, then I.iI _- (h).

1.4 1.5

1.6

How many antichains of subsets of S are there if (a) ISI = 2, (b) ISI = 3? (This will be followed up in Chapter 3.) Show that the number of pairs X, Y of distinct subsets of an n-set S with X c Y is 3" - 2" A collection 9D of subsets of an n-set S is called a cross-cut if

every subset of S is comparable with (i.e. contains or is contained in) at least one member of R. Suppose that 90 is a minimal cross-cut (i.e. R is a cross-cut but no proper subset of

I is a cross-cut). Show that 1601 1.7

Let x1, .

. .

, x" be real numbers, Jx11 % 1 for each i, and let I be

any unit interval on the real line. Show that the number of linear combinations D,7=1 six, with s, = 0 or 1 lying inside I is at

most ([n?2]). (Hint: First explain why we may assume that each

x1>0. Then associate with each sum Y,(2e1-1)x1 the corresponding set of indices i for which (2s,-1)x1>0.) (Erdos 1945) 1.8

Let A1, ... , Am, B1, ... , Bm be subsets of an n-set S such that

8 1 Combinatorics of finite sets

A, c B; for each i and A; = B1 whenever i *j. Show that m

1

;_, (n - IB; -A;I

<1.

lA,j

1.9

Let .s4 = {A1i ... , Am} be a collection of subsets of an n-set S and let a, = lA11. Call a subset B; of A; an own-subset of A; if B.

Al for all j * i. Prove that if every member of s4 has an

own-subset then 1

m

,_1 (n - a; + b;\

t

b,

1

J

(Tuza 1984)

where b; =1B11.

1.10 Let s4 = {A1i ... , Am} and let r(s4) =min{IBI : B fl A; * O for

each i}. Call s4 critical if r(s4 - {A,}) < r(s4) for all i. Show that if s4 is critical with r(s4) = t + 1, then m

1

M(IAii+t)

-- 1.

J

Deduce that, if lA, I = k for each i and s4 is critical with r(s4) = t + 1, then s4 can have at most 1.11

k + t) members.

(Tuza 1984) s4 = {A1, ... , Am} is a completely separating system of subsets of S = {x1, . . . , if for any two elements x, #x; of S there

exist sets Ak and A,, in s4 such that x, E Ak, x, I Ak, x, I A,,, x; eA,,. The dual s4* of s4 is the collection s4* = {X1, ... , of subsets of {1, ... , m} given by X, = {k:x, EAk}. Show that s4 is completely separating if and only if A* is an antichain. (Cai 1984) 1.12

(i) If s4 is an antichain of subsets of S, define the blocker b(s4)

of s4 to be the collection of all minimal subsets of S which intersect every member of s4. Clearly b (.s4) is an antichain. Show that b(b(s4)) = s4.

(ii) Similarly define the antiblocker c(s4) of s4 to be the collection of all maximal subsets of S having at most one element in common with each set in s4. Give an example to show that the assertion that c(c(s4)) = s4 is not always true.

Introduction and Sperner's theorem 1 9

1.13 An antichain si is a Menger antichain if the maximum number of disjoint sets in .RQ is equal to the minimum size of a member

of the blocker b(sf). Verify that

{a, b, c} {c, d, e} {b, d, f} {a, e, f} {c, f} {b, e} {a, d}

form a Menger antichain si but that b (si) is not a Menger antichain. 1.14

Menger's theorem in graph theory asserts that if x and y are non-adjacent vertices of a graph then the maximum number of

vertex-disjoint paths from x to y is equal to the minimum number of vertices whose removal from the graph separates x from y. Re-express this result in terms of Menger antichains.

(For further details see, for example, Woodall (1978) or Seymour (1977).)

2 Normalized matchings and rank numbers

2.1

Sperner's proof

The aim of this chapter is to present Sperner's original proof of his

theorem and to study a property of sets, called the normalized matching property, which arises in that proof. In doing so we shall discuss one of the basic theorems of combinatorics, the so-called marriage theorem, and we shall finish by having a look at the lattice of partitions of a set, about which some interesting conjectures have both risen and fallen. The basic idea behind Sperner's proof is the following. Suppose that si is an antichain of subsets of an n-set S, and suppose that Si has p; members of size i, 0 , i , n. If there is an i < 2(n - 1) for which p; > 0, we consider the smallest such i and show how to replace the p, sets of size i by p; sets of size i + 1, in the process preserving the antichain property. By repeating this process we `push up' all the sets in si of size <2(n - 1) replacing them by sets of size n/2 if n is even and size 2(n -1) if n is odd. We then consider sets of size greater than n/2 and push them down. In this way we finish up with an antichain

sil of the same size as d but with all sets having size [n/2]. The theorem will then follow immediately.

The entire process described above depends upon the following simple lemma about the shade and the shadow of a collection of k-sets.

Definition Let The collection

be a collection of k-subsets of an n-set S, k < n.

V ={DcS:IDI=k+1,BcDforsomeBEP.8} is called the shade of B. Thus V90 consists of all subsets of S which can be obtained by adding an element to a set in 90. Similarly the shadow A91 of R is defined to be

A ={DcS:IDI=k- 1,DcBfor some BeB). 10

Normalized matchings and rank numbers 1

11

Thus the shadow of y8 consists of all subsets of S which can be obtained by removing an element from a set in -04.

Lemma 2.1.1 (Sperner) Let R be a collection of k-subsets of an n-set S. Then

I..I_k+1I.I

ifk
and

ifk>0. Proof Consider the pairs (B, D) with B E 6-4, D E V90, B c D. We use a common combinatorial technique and count these pairs in two

different ways. For each B there are n - k elements of S not in B which can be added to B to obtain a set D; therefore the number of pairs (B, D) is (n - k) I Q1. However, each D is of size k + 1 and so has k + 1 subsets of size k (not all of which are, of course, necessarily in 58). Therefore the number of pairs (B, D) is at most (k + 1) IVRI. Thus

(n-k)160I _- (k+1)IVP1 as required. The proof of the second part of the lemma is similar and is left as an exercise. Corollary 2.1.2 If k--12(n-1), IVPI=1,64I. Ifk,2(n+l),

Proof of Sperner's theorem Suppose that the antichain s4 has p; members of size i, and suppose that the smallest value of i for which p, > 0 is io < 2(n - 1). By the corollary above we can find p,0 sets of size io + 1 containing them, none of which can be in s4. Therefore we can replace the p,, sets in s4 of size io by an equal number of sets from their shade. We can repeat this process until an antichain is obtained

with the same number of sets as s4 but with no member of size , 2(n - 1). We then proceed in the same way as above, replacing all members of s4 of size , 2(n + 1) by an equal number of sets of size [n/2]. The resulting antichain a, consists entirely of sets of size [n/2], so we have finally Is'I = I4i1 -- ([n/2])

12

1

Combinatorics of finite sets

Note that the conclusion of Lemma 2.1.1 can be written in the form

(k1)Ck) The lemma therefore asserts that the proportion of sets of size k + 1 in V is at least as big as the proportion of sets of size k which are in 64. This property of sets is known as the normalized matching property. We shall see later that the analogous property holds for

divisors of an integer, where inclusion is replaced by divisibility and size by the number of prime factors. It will also be shown that the

normalized matching property is intimately related to the LYM property discussed in Chapter 1. 2.2

Systems of distinct representatives

The normalized matching property is reminiscent of a fundamental theorem in combinatorics due to Konig and P. Hall which is often

known by its popular title of the `marriage problem'. Given a . . . , A, of an n-set S, we define a system of distinct representatives (s.d.r.) for A1, . . . , A, to be a system of distinct elements a1, . . . , a, of S such that a; E A, for each i. For example, the sets (1, 2, 3} {2, 4) {1, 4) (2, 5) possess an s.d.r. since we can select 3, 2, 1, 5 respectively to `represent' the sets. However, the sets (1,2,3) (2,41 (1,31 11,2) {2,31 cannot possess an s.d.r. since, for example, the four sets {1, 2, 3}, (1, 3}, {1, 2), {2, 31 have only three distinct elements in their union, which is not enough to represent four sets. It is clear that if an s.d.r. exists then, for each k, any k of the sets must have at least k elements in their union. Perhaps surprisingly, this obviously necessary condi-

collection of subsets A1i

tion turns out to be sufficient.

Before stating the theorem, we remark that the title `marriage problem' arises from the following interpretation. Imagine that r men are each asked to make a list of the ladies they would be willing to marry. Then (assuming that the ladies have no say in the matter) it is possible to marry each man to a lady on his list if and only if the lists possess an s.d.r.

Normalized matchings and rank numbers 1 13

Theorem 2.2.1 (Konig 1931, P. Hall 1935) The sets A,, ... , A,. possess an s.d.r. if and only if, for each m , r, the union of any m of the sets A; contains at least m elements.

Proof (Halmos and Vaughan 1950) Since the condition is clearly necessary, we prove the `if part, and proceed by induction on r. The result is clearly true for r = 1, so we assume it to be true for r = k % 1

and prove it for r = k + 1. So let A,, ... , Ak+i be given, and put S = Uk' A;. We consider two cases. (a) Suppose that any r , k sets A; contain at least r + 1 elements in their union. Choose any a, E A, to represent A,. Then any r of the remaining k sets contain in their union at least r elements other than a, and so by induction hypothesis A2, ... , Ak+, possess an s.d.r. which does not contain a,. We therefore obtain an s.d.r. for all the A;.

(b) Suppose that there is a value of r , k for which there are r sets A; with exactly r elements in their union. Suppose that these sets are labelled A,, . . . , Ar. By induction hypothesis, they possess an s.d.r., say a, E A,, . . . , ar E Ar. Now consider the remaining sets Ar+D

. . .

,

Ak+l They will possess an s.d.r. of elements of S -

{a,, . . . , ar} unless there exist h of them, for some h, which contain less than h elements of S - {a,, . . . , ar} in their union. However, in

that case these h sets, along with A,, ... , Ar, would constitute a collection of r + h sets with less than r + h elements in their union, contradicting the hypotheses of the theorem. So these remaining sets Ar+1,

. .

,

Ak+j do possess an s.d.r. from S - {a,, ... , ar} which,

along with a,, ... , ar, constitute an s.d.r for all the sets A. We shall make use of this theorem while studying the normalized matching property in the next section. For a quantitative version of

the theorem, concerned with how many s.d.r.s exist for a given collection of sets, see Exercises 19 and 20 at the end of the chapter.

2.3 LYM inequalities and the normalized matching property

It turns out that the normalized matching property and the LYM property are closely related; this section will be devoted to a study of that relation, describing results which are largely due to Kleitman. To set the scene for the discussion we extend the context from that of subsets of a set to a general partially ordered set with a rank function.

14 1 Combinatorics of finite sets

Definition A partially ordered set (or poset) P = (S, which an order relation = is defined, satisfying

is a set S on

(i) x--x forallxES

(ii) if x--y and y< x then x= y (iii) if x , y and y -- z then x _- Z. Examples are (a) the set of subsets of a set under inclusion (b) the set of divisors of a number under divisibility (c) the set of partitions of a set under refinement.

We say that x and y are comparable if x _- y or y -_ x. If x , y and x *y we write x
Associated with a poset P with a rank function r are the rank numbers or Whitney numbers No, N1..... defined by Ni = number of elements of P of rank i.

For example, in the poset of subsets of an n-set, Ni = (n). i

Not the same as a saturated chain, a maximal chain in a ranked poset is a chain containing one element of each rank. It is clear that a maximal chain must be saturated, but not vice versa. Finally among these definitions, we record that an antichain in a poset P is a set of elements of P, no two of which are comparable. In all that follows, we assume that all posets under discussion are finite, i.e. have a finite number of elements.

Theorem 2.3.1 (Kleitman 1974) Let P be a poset with a rank

function and with rank numbers Ni. Then the following three conditions are equivalent.

Normalized matchings and rank numbers 1 15

(1) P has the LYM property, i.e. if si is an antichain in P with p, members of rank i, then

(2) P has the normalized matching property, i.e. if 58 is a set of members of P of rank k and if VR, the shade of 9, is the set of all elements of P of rank k + 1 which are comparable with at least one member of 1,10, then Nk+1

Nk

(3) There is a regular covering of P by chains, i.e. there exists a non-empty collection ' of maximal chains in P such that, for each k, every element of rank k occurs in the same number of chains of IV. Proof (1) (2) Suppose that P has the LYM property, and let be a set of elements of P of rank k. Let denote the set of all elements of P of rank k + 1, excluding the elements of V9. Then 98 U is an antichain, and the LYM inequality gives Wk

+Nk+1 - IV90I Nk+i

1

I R LI V J Y J

Nk

Nk+1

(2)=>(3) Suppose that P has the normalized matching property. Define a new poset P' by the following construction. For each j, take II,,; Ni copies of each element of P of rank j, and then say that in P' each copy of x is less than each copy of y precisely when x < y in P. The poset P' has Il, Ni elements of each rank. Now consider any r elements of P' of rank k. They arise from at least r/II,#k Ni distinct elements of rank k in P, and by the normalized matching property of P these are comparable with at least Nk+i

r

Nk ll,*k Ni

r Hi k+1 Ni

elements of rank k + 1 in P. These r elements of P' are therefore comparable with at least r elements of P' of rank k + 1. It follows from the marriage theorem (Theorem 2.2.1) that the elements of P' of rank k can be paired with comparable distinct elements of rank

16 1

Combinatorics of finite sets

k + 1. Repeating this pairing at each rank level and putting the matchings together, we obtain a collection of II; Ni maximal chains such that each element of P of rank k lies in exactly iii#k Ni chains. (3) (1) Let T be a collection of maximal chains forming a regular covering. Each element of rank k occurs in precisely I T I IN,. of the chains. If d is an antichain with Pk elements of rank k, then these Pk elements together make an appearance in (pk/Nk) 119chains. Considering each k in turn and noting that, since .si is an antichain, the chains so obtained for different values of k must be distinct, we obtain kNk whence

pk

<

k Nk

We now present a far-reaching consequence of this important theorem. One of the earliest extensions of Sperner's theorem was obtained by Erdos, who proved that if . is a collection of subsets A. of an n-set S such that no k + 1 of the sets form a chain of length k, then I si I cannot exceed the sum of the k largest binomial coefficients

(1). i

Clearly, the case k = 1 reduces to Sperner's theorem. We shall

now see that this is an immediate consequence of the following result

of Kleitman, which exploits the existence of a regular covering of chains.

Theorem 2.3.2 (Kleitman 1974). Let P be a poset with rank function r and with the LYM property, and let T be any regular covering of P by chains. Then, if A is a real-valued function defined on the elements of P, and R is any subset of P, A(X) XER N.(x)

_ max E A(x). CeV xECf1R

Proof. For any chain Cc- T consider f (C) = EXEcnR A(x) We have

If(c)=z CEIB Z A(x) = z A(x) z 1= 2XER A(x) C XECf1R xeR

CE 8

1191

r(X)

XEC

since each of the Nr(X) elements of P of rank r(x) occurs in the same number of the 116 1 chains. Thus the average value of f (C) taken over all chains C in T is EXER {;,(x)/N,(X)}. Thus the maximum value of f (C) must be at least that value.

Normalized matchings and rank numbers 1 17

Taking A(x) = N,(x) we immediately obtain the following corollary.

Corollary 2.3.3 Under the hypotheses of Theorem 2.3.2

'RI _ max Z Nr(x) CE` xeCnR

Consider now the special case where P is the set of subsets of an n-set S and R is a collection of subsets of S no k + 1 of which form a chain. Then, if C is any regular covering of P by chains and if C E C, then c fl R can contain at most k sets; thus ExEcnR N,(x) can be at

most the sum of the k largest binomial coefficients. We have thus proved the following theorem.

Theorem 2.3.4 (Erdos 1945) If s4 is a collection of subsets of an n-set such that no k + 1 members of 4 form a chain, then k-1

141

n ([(n + 0/21)'

We close this section with an important general consequence of Corollary 2.3.3. If R is an antichain in the ranked poset P, then each C fl R can contain at most one element and so IRI can be no greater than the largest rank number. Definition A ranked poset P is said to have the Sperner property if the size of the largest antichain in P is the largest rank number. We have proved the following corollary.

Corollary 2.3.5 If the poset P has any of the three equivalent properties described in Theorem 2.3.1, then P has the Sperner property.

2.4 Rank numbers: some examples In the poset of subsets of an n-set, the rank numbers Nk are given by Nk =

(k).

These rank numbers have the property that (k n 1) < (k) if k -- 12,n

18

1

Combinatorics of finite sets

and ('k)

> (k +

1) if k =

Zn.

Thus the rank numbers first increase with

rank and then decrease. This property is known as unimodality.

Definition A sequence No, N1, . . . , N is unimodal if there exists j N -_N such that N; N, -> i

o

.

.

.

Nn.

The poset of subsets of an n-set S = {x,, . . . , xn} can be considered as the poset of divisors of a square-free number m =p, ... pn (where the p; are distinct primes) under division, by means of the correspondence

where h; = 0 if x; A and h; = 1 if x; E A. For example, the subset {x2i x4, x5} of S = {x1,

... , x5} corresponds to the divisor pZp4p5 of

m = pl ... p5. Under this interpretation the rank of a divisor of m is the number of prime factors of m. When this system is generalized the divisors of m =pi' . . . pR where the p; are distinct primes and the k; may exceed unity, are numbers of the form pi' . . . pn' where 0 , h; , k; for each i. Under divisibility the set of divisors of m form a poset with rank function r given by r(pi' ... pn^) = h, + ... + hn = total number of prime factors of m counted according to multiplicity.

Example m =p2g3r where p, q, r are distinct primes. The divisors of rank 4 are pg2r, p2 qr, p2g2, q3r, pq3. The rank numbers are No = 1,

N1 = 3,

N2 = 5,

N3 = 6,

N4 = 5,

N5 = 3,

Note that these form a unimodal sequence as in the square-free case. That this is true for the divisors of any number will become obvious in the next chapter. Sometimes the divisors of a non-square-free number are described in terms of multisets. The idea here is that the divisor p2g2r of p2g3r can be thought of as a `set' {p, p, q, q, r} where each distinct element

has a multiplicity, namely the number of times it occurs in the factorization. Corresponding to the number m = p;' ... pn^ is the multiset l,Pl,

k1

) P1,P2) . . ,P2, k2

,Pn,

,Pn}' k.

Normalized matchings and rank numbers 1 19

Posets of multisets will be discussed more fully in the next two chapters. It will be found that the LYM property holds. Partition Another familiar poset with a rank function is the poset r l,

of all partitions of a set

{1,

... , n} ordered by refinement. A

partition of S is a disjoint union of non-empty sets whose union is S. Each partition of S consists of disjoint blocks. An order relation can be defined on the set of partitions of S by writing 2r1 -- ?r2 if every block in partition r1 is wholly contained in a block of Jr2 (i.e. if every block of Jc2 is a union of blocks of 2r1), in which case we call r1 a refinement of ;r2-

Example S = {1, 2, 3, 4}. The partitions of S include Jr1 = {1, 31 U {2} U {4}

Jr2={1)U{2,3,4} Jr3={1}U{2}U{3,4}. Here Jr3 ' Jr2 but ir1

Jr2.

A rank function r can be defined on 11 by r(z) = n - J;rJ, where IJrl is the number of blocks in Jr. Therefore in the above example r(ir1) =1, r(Jr2) = 2, r(7r3) = 1. The partition with rank 0 is {1} U {2} U U {n}. The reader should check that in this example the rank numbers are No = 1, N, = 6, N2 = 7,N3 = 1. The rank numbers associated with partitions are known as Stirling numbers of the second kind. We have

Nk = number of partitions of an n-set of rank k = number of partitions of an n-set into n - k blocks = S(n, n - k)

where, in general, the Stirling number S(n, m) is the number of partitions of an n-set into m blocks. We shall prove that, for fixed n, the numbers S(n, m), m = 0, . . . , n - 1, form a unimodal sequence. Theorem 2.4.1

(i) S(0, 0) =1; S(n, 0) = 0 if n ,1 (ii) S(n + 1, k) = S(n, k - 1) + kS(n, k) n

(iii) S(n + 1, k)/=k-1 = E (')sj, k - 1). l Proof (i) S(0, 0) = 1 can be taken as definition. S(n, 0) = 0 for n ;1 is obvious.

20 1

Combinatorics of finite sets

(ii) Consider the S(n + 1, k) partitions of {X1....

,

into k

is is a block or blocks. In any given such partition, either in a block of size . 2. The first possibility arises in S(n, k - 1) ways. However, a partition with in a block of size -- 2 can be obtained from a partition of {x1, ... , into k blocks by adding xn+1 to any

of the k blocks; thus there are kS(n, k) such possibilities. Thus S(n + 1, k) = S(n, k - 1) + kS(n, k). (iii) A partition of {x1, . . . , x,,,

into k blocks can be obtained

by first choosing some j, k -1-- j -- n, then choosing n - j of the elements x1, . . . , x,, to form a block of size n -j + 1 with and then partitioning the remaining j elements into k -1 blocks. Thus

S(n + 1, k) i=k-1 = E ( nn- j)S(j, k -1) =

()sq, k -1).

!=k-1 I

(Note that the summation can equally well be written from j = 1 to

0

j=n.)

Theorem 2.4.2 For each n the numbers S(n, 0), S(n, 1).... form a unimodal sequence. There is a number k(n) such that

S(n, r - 1) < S(n, r) if r , k(n),

S(n, k(n)) , S(n, k(n) + 1), S(n, r) > S(n, r + 1) if r > k(n). Further, k(n) = k(n - 1) or k(n) = k(n - 1) + 1. Proof We proceed by induction on n, assuming that the theorem is true for n and considering the case n + 1. First of all, suppose that k -- k(n). Then, by Theorem 2.4.1(ii),

S(n + 1, k) -S(n + 1, k - 1)

=S(n, k-1)+kS(n, k)-S(n, k - 2) - (k - 1)S(n, k-1) = S(n, k - 1) - S(n, k-2)+k(S(n, k) - S(n, k - 1)) + S(n, k - 1) > 0 by the induction hypothesis. Next suppose that k -- k(n) + 2. Then, by Theorem 2.4.1(iii),

S(n+1,k)-S(n+1,k-1)=> (')(so,k-1)-S(j,k-2))<0 j=1 j since k - 2 , k(n) % k(j) for each j -- n. We have therefore proved that S(n + 1, k) increases strictly up to k = k(n) and decreases strictly from k = k(n) + 1. It follows that the

Normalized matchings and rank numbers 1

21

numbers S(n, k) are unimodal with peak at k(n + 1) = k(n) or k(n) + 1. How large is k(n)? It was shown by Harper (1967) that k(n) - n /log n.

More recently, Canfield (1978b) and Jichang and Kleitman (1984) have shown that, for sufficiently large n, k(n) lies around the root of the equation

(x+2)logx=n-2.

We now turn our attention to the normalized matching property. Does IIn possess it? It turns out that the answer is `no' if n , 20, and a

proof of this is outlined in the exercises. It was thought for many years, nevertheless, that the partition poset had the Sperner property, i.e. that the maximum size of an antichain in II is equal to the largest Stirling number S(n, k(n)). This conjecture was publicised by Rota (1967) and much work was done on the problem. However, thanks to the work of Canfield (1978a) it is now known that the conjecture is false for sufficiently large n. Since Canfield's paper, Shearer (1979) and Jichang and Kleitman (1984) have reduced the estimate for the

smallest value of n for which the conjecture fails. We finish this chapter by giving an outline of Shearer's argument. First, though, we introduce some notation. A partition of {1, ... , n} into k, blocks of size 1, k2 blocks of size 2, ... , kr blocks of size r, where necessarily E, ik; = n, is called a partition of type 1k12k2

... rk'.

Consider now the set T of all partitions of (1, .

. .

, n) of type

m'" (2m)''2(3m)'`' into k(n) + 1 blocks where

h, + h2+ h3= k(n) + 1 and

mh, + 2mh2 + 3mh3 = n.

Choose the h; as follows:

h, = 3 {5(k(n) + 1) - 2m h2 = 3 {m

- 0}

- (k(n) + 1) + 20} h3=3{m-(k(n)+1)-0},

where 0 is chosen as 0, 1 or -1 so as to make each h; an integer. The value of m remains to be chosen.

22

1

Combinatorics of finite sets

Now if we have a partition into k(n) blocks which is related under the ordering , to a partition in ', it must be of one of the following types: (i) mhl-2(2m)h2+1(3m)h3 mh,-1(2m)h2-1(3m)h3+1 (ii) (iii) mhl-1(2m)h2(3m)h3-1(4m)1 mh,(2m)h2-2(3m)h3(4m)1

(iv)

(v) mh,(2m)h2-1(3m)h3-1(5m)1

(vi) mh'(2m)h2(3m)h3-2(6m)1.

If there are altogether less than 116 1 partitions of these types, then the partitions in 16 and the remaining partitions with k(n) blocks will form an antichain of size greater than S(n, k(n)) as required. So we have to show that, for sufficiently large n, there are less than 116partitions of types (i)-(vi). Now the number of partitions of type (i) is, by Exercise 2.15, n!

(m!)h,-2((2m)!)h2+1((3m)!)h3(h1 - 2)!(h2 + 1)!h3!

and the number of partitions in 16 is n! (m!)h'((2m)!)h2((3m)!) 23h1!h2!h3!

so that the ratio of these two numbers is h1(h1- 1)(m!)2 (h2 + 1)((2m)!)

-

h1(h1- 1) (h2 + 1)(2m l)

If we carry out a similar calculation for each of the other types (ii)-(vi) we find that we have to prove that

h1(h1-1) + (h2 +

+ h1h3

h1h2

1)2m3 (m ) (h + 1)(3m) m + h2(h2 -1) +

(4m) m h2h3

+ h3(h3 -1) < 6m

4m

(m)

(2m )

Now take m = [nl ak(n)]. Since k(n)

(2.1)

(3m)

n/log n we have m

log n/a.

We choose a a little greater than unity to ensure that each h, is positive. We now have

h1=3(5-2a)lo

h2_"h3_"3(a-1)1 gn

nn. g

Normalized matchings and rank numbers 1 23

We can now use Stirling's approximation n! = (2 rn)ln" a-" to show that each of the six terms on the left-hand side of (2.1) tends to zero

as n -* -. Their sum is therefore eventually less than unity as required. Detailed numerical work shows that if a = 1.06 then it is sufficient to take n > 4 X 109. (Jichang and Kleitman (1984) reduced this estimate to 3.4 x 106.) Exercises 2 2.1

Complete the proof of Lemma 2.1.1.

2.2

Let si be an antichain of subsets of an n-set S, and suppose

that si contains a; (n) sets of size i, 1 _- i _ n. Let 6-4 = {B c i

S : A c B for some A E 4 }, and suppose that 98 has fl, (n) sets i

of size i. Show that fle+1 % 3, + a;+1 for each i < n. 2.3

2.4 2.5

2.6

Let 4 be a collection of subsets of an n-set S, such that A E 4, A c B ' B e si. Show that the average size of the sets in si4 is at least ?n. Show that if k < Zn then all the k-subsets of an n-set S can be paired with distinct (k + 1)-subsets containing them. Let I S I = mn, S = A 1 U ... U A" = B1 U ... U B" where IA, I = I B, I = m for each i. Show that there exists a permutation a of 1, ... , n such that A, n B,,(I) 0 for each i. Deduce that S can

be split into m disjoint sets of n elements each of which is an s.d.r. for the As and for the Bs. Show that if P is a regular poset in the sense that, for each k, each element of rank k covers the same number of elements of rank k - 1 and is covered by the same number of elements of rank k + 1, then P has the normalized matching property. (Baker 1969)

2.7 2.8

Obtain the rank numbers for the poset of divisors of 60. Prove that, if si is a collection of subsets of an n-set such that

there are no sets A, B e 4 with B c A and I A- B I < k, then IdI is at most the largest sum of the form E;

n

a+ki

(Katona 1972a) 2.9

Prove that, if si is a collection of subsets of an n-set such that

24 1

Combinatorics of finite sets

A., A; E si, A; c A;

lA1- A; < r, then I.siI is at most the sum

of the r largest binomial coefficients

(12). i

Give two different

proofs: (a) imitate the proof of Sperner's theorem given in this chapter; (b) use Theorem 2.3.2 or its corollary. (Erdos 1945) 2.10

Let si,, ... , s4, be t collections of subsets of S = (1, ... , n) such that if i 0 j then there is no set in ,91 which is a proper subset of a set in siJ. Use Theorem 2.3.2 to prove that I,si,d +

... + lsi j --max(2"' t([nl21)).

(Daykin, Frankl, Greene and Hilton 1981) 2.11

Verify, by listing all possible partitions, that S(5, 4) = 10 and S(5, 3) = 25.

2.12 Prove that (a) S(n, 2) = 2"-' - 1 and (b) S(n, 3) = 2.13

2(3"-' - 2" + 1).

Show that k"-k _- S(n, k) , (k -1)k "-k whenever k
2.14 Prove that 11n does not have the normalised matching property

as follows. Consider n even. Let X be the set of all partitions of

the n-set S into blocks of size 2n; then 1XI = 2(nJJ2). Verify that each partition in X covers 2S(n/2, 2) partitions into three

blocks, covering ('2)S(n/2, 2) such partitions altogether. The partitions of X and the remaining partitions into three blocks form an antichain, so if nn has the LYM property, then the LYM inequality implies S(n, 3) -_ 2S(n, 2)S(n/2, 2). Use Exercise 2.12 to show this false if n , 20. (Spencer 1974) 2.15

Prove that the number of partitions of {1, ... , n} of type lk,2k

... rk, is

n!

IIm_1 (m!)km(km)!

2.16 This exercise is to construct a ranked poset which does not

have the Sperner property. A graph G has n vertices, a, b, C1, ... , cn_2; a and b are adjacent, and each of a and b is adjacent to each of the c;, but no two c; are adjacent. Let P be the set of partitions of the vertex set V into blocks which are such that the subgraphs induced by the vertices in the blocks

Normalized matchings and rank numbers 1 25

are connected. Then P is a poset under refinement. If r is a partition define r(r) to be the number of blocks in r; then r is a rank function. Verify that the elements of rank k in P are as follows: (i) partitions in which k - 1 of the c, are blocks on their

own, the other (n - k - 1) c, being with a and b in another block; (ii) partitions in which k - 2 of the c,, form blocks on their own, a is in a block with j c,s and b is in a block with

n-2

n - k -j c,'s for some j. Verify also that there are (k -1) partitions of type (i) and n-k (n

1

lk - 2/ j k) = 2- k(k - 2) partitions of type (ii). Next note that we can obtain an antichain by taking all the partitions of type (i) for some k, and all those of type (ii) for some k2
of (k -1) occurs when k is approximately in, and the maximum value of 2n-k(k

- 2)

occurs approximately when

k = 3n. Choosing k, and k2 appropriately we obtain an antichain of size approximately

(n 2(n -22)) +

22n,3(n

) 3n -22)

However,

Nk=(k-1)+2"-k(k-2) so that the maximum Nk is attained when k is approximately the value of Nk being smaller than the size of the antichain. In 3n,

particular, choose n = 12; in this case max Nk = N5 = 15 570 whereas the antichain obtained by choosing k, = 6 and k2 = 5 (Dilworth and Greene 1971) has 15 612 members. 2.17

Show that if IXI = n and Y = {1, ... , k} then the number of surjections ('onto' maps) f : X -3 Y is k!S(n, k). Exercise 2.17 to show that if [k], = k(k - 1)(k 2). .. (k - i + 1) then k" = E, S(n, i)[k],. Illustrate with n = 3.

2.18 Use 2.19

Suppose that A,, ... , A, possess an s.d.r.. Show that there

26 1 Combinatorics of finite sets

exists j such that, for any a e A;, there is an s.d.r. in which A; is represented by a. 2.20 Show that, if A,, ... , A, possess an s.d.r. and IA;I ; k for each

i, then there are at least k! different s.d.r.'s if r % k, and at (M. Hall 1948) least k!/(k - r)! s.d.r.s if k > r.

3 Symmetric chains

3.1

Symmetric chain decompositions

The problem of generalizing Sperner's theorem on subsets of a set (or

divisors of a square-free number) to multisets (or non-square-free

numbers) was posed in 1949 as a prize problem by the Dutch Wiskundig Genootschap. The solution, due to de Bruijn, Tengbergen, and Kruyswijk (1951), is not only very simple but exhibits a structural aspect of the poset of divisors of a number which turns out to be of great usefulness. The aim of this chapter is to describe this structural property and to give a number of varied applications. The basic idea in their solution was to decompose the poset into chains with certain symmetry properties. If P is a ranked poset with

rank function r, we say that the elements x1, ... , x,, of P form a symmetric chain if (i) x;+1 covers x, for each i < h (ii) r(x1) + r(x,,) = r(P)

where r(P) is the largest rank in P. Note that (i) requires that the chains are saturated and (ii) requires that each chain finishes as far above the middle rank as it started below, i.e. it is `symmetrically positioned' about the middle rank. In the case of subsets of an n-set S, A,,. . . , A,, form a symmetric chain if (i) A; c Ai+1 and JAt+1I = 1 + IA,l for each i < h (ii) IAII + IAhI = n.

In the case of divisors of a number m, the divisors d1, ... , d,, form a symmetric chain if (i) d; divides d;+1 and d;+1/d; is prime for each i < h (ii) r(di) + r(d,,) = r(m), where r(d;) is the number of prime factors of d; counted according to multiplicity. de Bruijn et al. showed that the set of divisors (subsets) of a number m(set S) can be partitioned into symmetric chains. Here are two examples to illustrate the idea. Example 3.1.1 Subsets of {1, 2, 3, 4). 27

28 1 Combinatorics of finite sets

(3, 4) {2, 4}

{4} c {1, 4} {1, 2, 4} {3} c {1, 3} c {1, 3, 4} f2)1-- {2, 3} c {2, 3, 4}

0c{1}c{1,2}c{1,2,3}c{1,2,3,4}. Example 3.1.2 Divisors of 180 = 22325.

2.5 22.5 32. 5

32

1

5

3.5 2.3.5 22.3.5

3

2.3 2.32

2

22

22.3

2.32.5 22.32

22.32.5

In terms of multisets, Example 3.1.2 partitions the set of subsets (or more strictly submultisets) of the multiset {2, 2, 3, 3, 5} as follows: {2, 5} c {2, 2, 5} {3, 3} c {3, 3, 5) (5) c {3, 5} c {2, 3, 5} c {2, 2, 3, 5}

{3} c {2,3}c{2,3,3}c{2,3,3,5} 0c{2}c {2, 2)c{2,2,3}c{2,2,3,3}c{2,2,3,3,5}. Theorem 3.1.1 (de Bruijn et al. 1951) The set of divisors of a number (or subsets of a set) can be expressed as a disjoint union of symmetric chains.

Proof We shall use the notation of divisors of m, and proceed by induction on the number n of distinct primes dividing m. If m = p" then 1, p, p2, ... , p" is a symmetric chain containing all the divisors

of m; thus the theorem is true for n = 1. Suppose now that the theorem is true for n = k, and consider a number with k + 1 distinct prime factors. Such a number can be written as m = m,p" where m,

has k distinct prime factors and where p 4. ml. We show how to construct symmetric chains for m from symmetric chains for m,. Let d,, ... , d,, be one of the symmetric chains in the chain decomposition

of mand consider all the divisors of m of the form dps, 1 -- i -- h, 0 -- 0 = a. We put these divisors of m into symmetric chains by the following procedure. Write them in a rectangular array as shown, and

Symmetric chains 1 29

then peel off chains as indicated:

d1 d2

... dh_2

dh-1

dh

dip dzp

... dh-zP

dh-lp

dhP

d1p 2 d2p2

...

dh-1P

dh_zpz

dipa-1 d2pa-1

dh-2P«-i

dlpadzpa

dh-2Pa The outer layer gives the first chain tl

d1, dz,

z

dhpz

CI-1

dhpa-1

dh-iP dh-1pa

... , dh-I, dh, dhP, dhP 2, .

dhP a

. .

dhPa

,

which clearly satisfies condition (i) and is symmetric since

r(d1) + r(dhp a) = r(d1) + r(dh) + a = r(m1) + r(p a) = r(m). Similarly the second layer gives a symmetric chain, and so on. Finally, every divisor of m will be obtained in this way from some symmetric chain for m 1.

Example 3.1.3 We can use Example 3.1.1 and the construction given in the proof of the theorem to obtain a symmetric chain decomposition of the set of subsets of {1, 2, 3, 4, 5}. For example, starting with the chain {3} c {1, 3} c (1, 3, 4) we consider the rectangular array {3} {1, 3} {3, 5} {1, 3, 5)

{1, 3, 4} {1, 3, 4, 5}

and obtain the chains

{3}c{1,3}c{1,3,4}c{1,3,4,5} and {3,5}c(1,3,5). Having introduced the idea of symmetric chains, we can now give our third proof of Sperner's theorem. Proof of Theorem 1.2.1 Partition the poset of subsets of the n-set S

into symmetric chains. Now the number of chains must be

n ,n/2]) since each chain must contain exactly one subset of size

30 1

Combinatorics of finite sets

[n/2]. However, if si is an antichain then s4 can contain at most n one member of each chain; so Isif - ( \[n/2]

3.1.1 Rank numbers for multisets

Consider a symmetric chain decomposition of m = pi' ... p;' and put

K = k, + ... + k,. Since each chain starting at rank h ends at rank K - h, we must have N,, = NK_,,. Also, if h < JK then all chains containing a divisor of rank h must contain a divisor of rank h + 1, and so we must have N,,+, , N,,. Thus we have the following theorem.

Theorem 3.1.2 If No, ... , NK are the rank numbers for the poset of divisors of m = pi' ... where K = E; k;, then (i) No -- N,

...

N[Knj;

.1....

N[K,7

NK

and

(ii) N,, = NK_,,

for each h.

In the next chapter we shall find out where we have strict inequalities in (i). Meanwhile we note that the N, form a unimodal sequence and that the largest value taken by any Ni is N[K/21. The proof of Theorem 1.2.1 given above therefore generalizes directly to give the following theorem.

Theorem 3.1.3 If si is an antichain in the poset of divisors of m = pi' ... p;,, then 141 --N(Kr2, where K = E, k,. Thus the poset of divisors of m possesses the Sperner property. 3.2

Diiworth's theorem

Before looking further at symmetric chains it is of interest to compare Theorem 3.1.1 with the theorem of Dilworth concerning chains in a general poset. It is clear that if a poset can be expressed as a union of w chains then any antichain in P can have at most w members, one from each chain. Thus maximum size of an antichain in P -- c(P)

(3.1)

where c(P) denotes the minimum number of chains into which P can

be decomposed. Dilworth's theorem asserts that there is in fact

Symmetric chains

1

31

equality; c(P) is the largest size of an antichain in P. If there is a symmetric chain decomposition, then we acquire some extra information, namely that c(P) is the largest rank number, but we emphasise

that Dilworth's theorem holds in any poset, even one without any rank function. There are now many different proofs of Dilworth's theorem in the literature. The proof we now present is one of the simplest, and is due to Perles (1963). Theorem 3.2.1 (Dilworth 1950)

In any poset P, the maximum size of

an antichain is equal to the minimum number of chains in a chain decomposition of P.

Proof (Perles 1963) In view of (3.1), we have to show that the minimum number of chains in a chain decomposition of P is less than or equal to the size of a maximum antichain. We use induction, and assume that the required result is true for all posets Q with IQI < IPI. Let m be the maximum size of an antichain in P. We now consider two cases.

a(i) Suppose first that there is an antichain of size m which is neither the set of all maximal elements nor the set of all minimal elements of P. Let W be such an antichain, and define

P+ = {x E P: x- y for some YEW} P- = {x EP:x -- y for some y c- W).

Since there is at least one minimal element not in W, P+ * P. Similarly P- * P. Thus IP+I < IPI and IP I < IPI, so we shall be able to apply the induction hypothesis to them. Observe that P+ U P- = P, for if there were an element of P not in P+ U P- then we could add it

to W to obtain a larger antichain. Also, P+ fl P-= W. Now, as already mentioned, P+ and P- are, by induction, the union of m chains, say P+ = U" , C, and P- = U?=1 D;. The elements of W, being the minimal elements of P+ and the maximal elements of P-, are the minimal elements of the chains C, and the maximal elements of the chains D.. We can therefore glue the chains C, and D; together in pairs to form m chains which form a chain decomposition of P.

(ii) Next suppose that there are at most two maximum-sized antichains, one or both of the set of all maximal elements and the set of all minimal elements. Choose a maximal element a and a minimal element b with b _ a. Then P - {a, b } contains an antichain of size m - 1 but none of size m, so that by the induction hypothesis it can be

32 1

Combinatorics of finite sets

decomposed into m - 1 chains. These chains, together with {a, b }, give a decomposition of P into m chains.

3.2.1 The marriage problem

Dilworth's theorem will be considered further in Chapter

13.

Meanwhile we close this section by showing how it can be used to give an alternative proof of the marriage theorem (Theorem 2.2.1).

Proof Let A...... A. be m , n subsets of S = {x,, .

. . , x,, } such that, for each k -- m, any k sets A; contain at least k elements in their union. Define a poset as follows: take objects u,, ... , u and

v,, ... , vand define an order relation by

u;;v; u;

u;

if and only if x;EA;

ifi0j

v,4v; ifi*j. It can be seen that u; is really representing x; and v; is representing A;. Now {u1, ... , U.1 is an antichain in P, and we now show that it is a

maximum-sized antichain. Suppose, after a suitable relabelling if necessary, that {U1.... , Ur, v,, ... , v5} is an antichain. Then none of x,, ... , Xr is in any of A,,

.

. .

, As, so the only possible elements of

Thus thessets A...... A.have --n - r elements in their union. However, they have ;s elements in their union, so we must have s _ n - r, i.e. r + s , n. Therefore n is indeed the maximum size of an antichain. It now follows from Dilworth's theorem that P can be decomposed into n chains. Each chain must contain exactly one u;. Since the v; are all in different chains, there corresponds to each v; a u; in the same chain with u; -- v,; thus to each A; there corresponds an xi with x, E A;. 3.3 Symmetric chains for sets

Having looked at symmetric chains in a more general setting, we are

now going to return to the particular case of subsets of a set and examine the structure of the chains more closely. Suppose we wish to find a symmetric chain decomposition of the set of subsets of a given

set. The construction given in the proof of Theorem 3.1.1 is a step-by-step construction, altering the chains at each step. This makes

for a tedious construction if the given set is large. Is there a neater way of writing down the chains without having to find the chains for

Symmetric chains 133

all the smaller sets in the process? A positive answer to this question has been given by Leeb (pers. comm.) and, independently, by Greene and Kleitman (1976b). we associate a Corresponding to each subset A of S = {x1, ... , sequence of n parentheses. If x; E A we put a right parenthesis in the

ith place; if x, I A we put a left parenthesis in the ith place. For example, if S = (1, ... , 9) and A = {2, 3, 4, 7, 81 we represent A by Having done this we can now `close' the brackets as far as possible. In the example given, we obtain

()))(())(

Elements of S which correspond to right parentheses which are paired

in the closing process are called the basic elements of A. In the example 2, 7 and 8 are the basic elements of A. The unpaired right parentheses represent non-basic elements of A. Note that all unpaired

left parentheses must occur to the right of all the unpaired right parentheses, for otherwise further closing of brackets would be possible. These unpaired left parentheses represent unpaired elements of the complement A'. Now define A- to be the set of basic elements of A, and A+ to be the union of A with the set of unpaired elements of A'. Then IA+I = IAI + IA'I - number of paired elements of A' = ISI - number of paired elements of A

=ISI - IAI so that Al I+ IA+I = ISI Now form a chain containing A by starting with A-, the set of basic elements of A, and adding one by one the non-basic elements of A in order, and then adding one by one the unpaired elements of A' in order. The chain therefore starts with A-, contains A, and ends with A+. Since Al I+ IA+I = ISI, the chain is a symmetric chain containing A. In the example above, the basic elements of A are 2, 7, 8, and the elements 3, 4, 9 are added in turn to give the chain

{2,7,8}c{2,3,7,8}c{2,3,4,7,8}c(2,3,4,7,8,9).

(3.2)

The reader could check that this is one of the chains which the proof of Theorem 3.1.1 would construct! In terms of parentheses, the basic elements determine all the closed parentheses uniquely and leave all the unpaired parentheses to be

34 1 Combinatorics of finite sets

fitted in. The chains are constructed by first taking all the unpaired parentheses as left ones (giving A-), then changing the first of them to a right parenthesis, then the second, and so on. The chain (3.2) is, in terms of parentheses,

Observe how the closed parts remain unchanged while the remaining parts change step by step from left to right parentheses. If two subsets of S have the same basic elements, they will have the same closed parts and hence will give rise to (and be contained in) the

same chain. The set of all chains obtained in this way forms a symmetric chain decomposition of the set of subsets of S; every set

lies in a chain, and the chains are disjoint since they are each determined by basic elements and each subset has a unique set of basic elements.

We can now see that the chains formed in this way are precisely those obtained in Theorem 3.1.1. For small values of n we could find the chains explicitly and verify this; therefore we proceed with an induction argument. Suppose that the chains of Theorem 3.1.1 for an

n-set are the same as those of the parenthesis construction, and consider how the chains for an (n + 1)-set are formed. The first new

chain in the de Bruijn construction is obtained by extending an to its final set:

existing chain by adjoining

A, A, U

A2 U

...

A,, A,, U

This corresponds to appending `(' to each string of parentheses representing the sets A,, . , Ah and then changing it to `)' to represent A,, U The resulting chain is precisely a chain of subsets of {x,, ... , with the same basic elements as those of the original chain. The second new chain formed from A, c ... c A,, is obtained by appending to each of A,,. . . , A,_,, i.e. appending `)' to the strings of parentheses representing A,, ... , Ah_,. The fact . .

that A,, is omitted corresponds to making the last non-basic element, say x;, of Ah no longer available. This is achieved by forming a closed

pair of brackets, `(' at the jth place and `)' at the (n + 1)th place;

Symmetric chains 1 35

therefore the resulting chain consists of all subsets of {x1, . . . , whose basic elements are precisely those of the original chain together with Thus again the new chain consists of subsets with the same basic elements. The chains possess some elementary properties which are worth recording. Theorem 3.3.1 Let A, c ... c A, be one of the symmetric chains of subsets of the n-set S constructed as above. (i) If A;+1= A; U {x,,,;}, then m; < m;+1 for each i, and the m; are alternatively odd and even. (ii) The number of basic elements in each of the A. is 2(n + 1 - r). (iii) For each i, there is a set B. such that Ae_1 c B; cA;+1 and with B; in a shorter chain.

Proof (i) That m; < m;+1 is immediate from the construction. Note

also that any block of closed parentheses must contain an even number of elements, so that m;+1 = mi + (even number) + 1. (ii) We have

n = (number of paired elements) + (number of unpaired elements) = 2(number of basic elements) + (number of unpaired elements).

Now the chain is constructed by adding one unpaired element at a

time, so that the number of unpaired elements must be r - 1. Therefore

n = 2(number of basic elements) + r - 1. (iii) The sequences of parentheses representing Ai_1 and A;+, differ in two places. We have

Ai-1:... (XX(...

A1:...)XX(...

A;+1: ... )XX)... where XX denotes a (possibly empty) closed part. Consider the subset B; represented by

B,:... (XX).... It clearly satisfies A;_, c Bi c A;+1. Also, since a further bracketing is now possible in the sequence for Bi it follows that B; has more basic

36 1 Combinatorics of finite sets

elements than A;, and so, by part (ii), the chain in which it lies is shorter.

There are yet more equivalent ways of constructing the same symmetric chain decompositions as those described above, and we briefly mention two of these. If we look at, for example, the 2-subsets of (1, . . . , 4) in lexicographic order (alphabetical, with 1 in place of a, 2 in place of b, etc.), and assign to each the first available 3-set containing it we obtain {1, 2)-{1, 2, 3} {1, 3)-{1, 3, 4} {1, 4}-{1, 2, 4) (2, 3)-{2, 3, 4) {2, 4) -none left {3, 4)-

Note that this is precisely the pairing between 2-subsets and 3-subsets in the symmetric chain decomposition of Example 3.1. Aigner (1973) has shown that this always holds, so that the chains can be built up rank by rank. Another approach is due to White and Williamson (1977). To find out which set covers the set {a,, . . . , a,,) in its chain, work out a; - 2i

for each i (taking ao = 0) and define t to be the largest interger i, 0 < i , h, for which a; - 2i is minimal; then the set covering {a,, ... , a,,} in its chain is {a...... a,,, I +ar}. For example, to find

the set covering {2, 3, 7, 8}, work out 0 - 2.0 = 0, 2 - 2. 1= 0,

3-2.2=-1, 7-2.3=1, and 8-2.4=0; take t=2 to obtain the set {2, 3, 4, 7, 81 since 1 + a2 =1 + 3 = 4. 3.4

Applications

We now give some applications of the symmetric chain structure.

3.4.1 The number of antichains Given an n-set S, how many antichains of subsets of S are there? We shall accept as antichains two rather dubious candidates, namely the empty antichain which contains no subsets and the antichain consist-

ing of only the empty set. This is to enable us to formulate an equivalent version of the question posed above. We show that antichains of subsets of S are in one-one correspondence with increasing functions from 9'(S), the set of subsets of S, to

Symmetric chains 1 37

{O, 1). Here we say that a function f : 91(S)--- {O, 1} is increasing if

f (X) --f (Y) whenever X c Y. Given an antichain d, we can define f : P(S) - {O, 1} by f (B) = 0 if B is a subset of a member of d and f(B) = 1 otherwise. To check that this f is increasing we need only check that if B c D and f (D) = 0 then f (B) = 0. However, if f (D) = 0 then D c A for some A E .saf ; then B c A also, so f (B) = 0. It is clear that under this correspondence the sets in the antichain

are precisely the maximal sets on which f takes the value zero. Conversely, given an increasing function f : 6P(S) -3 {0, 1), the maximal sets on which f takes the value zero will form an antichain. Thus the required correspondence has been exhibited. Note that the empty antichain corresponds to the function f which takes the value unity everywhere.

The number of antichains is therefore the number of ways of defining an increasing function f : 9'(S)-+ {0, 1}. We therefore consider the problem of constructing as many increasing functions as we

can. Suppose that we decompose the set of subsets of S into symmetric chains, and that we attempt to construct f by defining it on the sets in one chain at a time. More precisely, suppose that we have

succeeded in defining an increasing function f consistently on all chains of size less than k and that we now attempt to extend the definition off to a given chain of size k. If k = 2, any chain of size k is of the form X c Y. Here there are at most three possible choices of values for f(X) and f(Y): we could have f (X) =f(Y) = 0, or f (X) =f(Y) = 1, or f (X) = 0 and f (Y) = 1. Certainly the choice f (X) =1, f (Y) = 0 is not allowed. Thus there are at most three possibilities. If k > 2, consider any two sets X, Y in the chain under consideration with IYI = IXJ + 2. We then have

..XcZaY... for some set Z. Now by Theorem 3.3.1(iii) there is a set U, X c U c Y, with U in a shorter chain, and hence with f (U) already defined. It follows that if the chain X,CXZC.. . CXk is being considered, then, for each i = 2, . . . , k - 1, there is a set U; such that X_, c U c X+, and with f(U;) already defined. If f(U) = 0 for all i, 2 -_ i , k - 1, then we must take f (X _,) = 0 for each i , k -1 and we are possibly free to define f on Xk_, and Xk. As before, there are at most three possible choices here, f(Xk_,) =1, f(Xk) = 0 not

being allowed. So again there are at most three possible ways of

38 1 Combinatorics of finite sets

extending f. If f (U,) = 1 for all i, 2 -- i , k - 1, then similarly there is only possible freedom of choice at f (XI) and f(X2), and again there are at most three possible ways of extending f. Finally, if f is not constant on the U,, let j be the first i at which f (U,) = 1. Then f (X,) = 1 for all i > j and f (X,) = 0 for all i < j -1, so we are only free to define f on X _1 and X j. As before there are at most three possible choices.

Thus as each chain is considered in turn, there are at most three ways of extending the definition of f consistently to that chain. The total number of possible increasing functions f is therefore at most 3'

where m is the number of chains. Since m =

we therefore

have the following theorem.

Theorem 3.4.1 (Hansel 1966) The number of antichains of subsets of an n-set is at most

0

3([.2J).

By using a much more complicated argument based on the same ideas, Kleitman and Markowsky (1974) later reduced this upper bound to 2(1+0(log n/n))((:OJ)

More recently, Korshunov (1981) has published a complicated asymptotic formula for the number of antichains. In the case of even n, the estimate is

p (n/2). )(2-' + n22 n5 _ n X 2 -n-4)

2(-"12) ex p

Even in the case of n = 6 this gives remarkably good results: 7 996 118

instead of the exact value 7 828 354. The first few exact values are tabulated below. n

Number of antichains

5

3 6 20 168 7581

6 7

7 828 354 2 414 682 040 998

1

2 3

4

Symmetric chains 1 39

3.4.2 A storage problem The following problem arises in the mathematical theory of information storage and retrieval. Given a collection 99 of subsets of an n-set S, find the shortest sequence T of elements of S such that every set in 90 appears at least once as a subsequence of consecutive terms of T. In the case where A is the set of all subsets of S, the problem is to

find as short a sequence as possible containing every subset as a subsequence of consecutive terms. If n = 5, for example, and S = {1, 2, 3, 4, 5), the sequence

1234512413524 satisfies the requirements.

A trivial upper bound for the length of the shortest sequence is obtained by considering the sequence obtained simply by writing down each subset one after the other. Since there are 2" subsets of average size 2n, the length of the resulting sequence is 2In2". However, an improvement on this can be achieved by exploiting the existence of a symmetric chain decomposition of the poset of subsets.

Theorem 3.4.2 (Lipski 1978) If S is an n-set, there exists a sequence of length (asymptotically) E(2/r)2" which contains every subset of S as a subsequence of consecutive terms.

Proof. We consider first n even, say n = 2k, and we write S = T U R

where ITI = I R I = k. Then both T and R have symmetric chain decompositions of their posets of subsets into m = ([k2]) symmetric

chains: P(T) = 01 U ... U Pm, P(R) = yi, U ... U RV,,,. Corresponding to the chain 4,, Oi : {x,, ... , xi} c {x1i .. . , xj, xi+1} c ... c {x...... x,,}

(h !j)

we associate the sequence U = x1x2 ... x,,. Then every subset of T occurs as an initial part of one of the sequences U1, . . . , U,". Similarly let V1, ... , V. be sequences corresponding to the chains `Y1, ... , V. such that every subset of R occurs as an initial part of some V. If we

let V denote the sequence obtained by writing V in reverse order, then every subset of R occurs as a final part of one of the V . Now consider the sequence V1 UI V, U2 .

.

. V1 U. V2 U1 V2 U2 ... V2 Um ... V. U1 ... V. Um.

If A is any subset of S we can write A = E U F where E c T and F c R. Now E occurs as the initial part of some Ue and F occurs as the

40 1 Combinatorics of finite sets

final part of some 11f: thus A occurs in the above sequence as part of YfU,. Thus the sequence contains every subset of S as required. The 2 length of the sequence is F. 2m2(k + 1) = 2(k + 1)( \[k/2]1 However, (k2ir)"22k

(tk12k

by Stirling's approximation, so that

2m2(k+1) =k

. 22'`.

k=42".

This estimate for the length of the sequence can be halved by using Exercise 3.10, thus proving the theorem. The proof for n odd is similar.

3.4.3 A probability result related to Sperner's theorem Another application of the symmetric chain structure leads to an elegant probabilistic version of Sperner's theorem. We investigate the

following problem. Suppose that two subsets A, B are chosen independently and at random from the set of subsets of an n-set S, and suppose that the underlying probabilities of each being chosen are known. Then what can be said about the probability that A C- B?

We shall show that, no matter what the underlying probability distribution is, the probability that A c B is at least 1 Lemma 3.4.3 Let p1, ... , p,, be a probability distribution on an m-element set T. Then the probability that two elements of T chosen at random are identical is ,1/m, irrespective of the p;. Proof The required probability is E7=1 p; . Now we clearly have M

1

(P' so

P+-2 r=1

0

l

- M/

m '=1

P'+m.1 2,0 m

whence 1 m ,2 YIPI -m m,_1Pi-m =1

2

1

m

m=m

1

0

Symmetric chains

1

41

Lemma 3.4.4 If {q(x)} is a probability distribution on a poset P and P is partitioned into m chains, then the probability that two elements of P chosen at random lie in the same chain is --1/m, no matter what the values of q are.

Proof In Lemma 3.4.3 take T to be the set of chains. The required probability is the probability of choosing two chains at random and obtaining the same chain twice. The probability p, of choosing the ith chain is given by pi = E q(x) where the summation is over all x in the ith chain. The result is immediate from Lemma 3.4.3.

Note that the probability that two elements chosen at random lie in the same chain can be written as

Y,

xeP

q(x)( Ey

x,ym

q(y)) _ Z q2(x) + 2 x

same chain

E q(x)q(y)

x
x,yin

same chain

Therefore the result of Lemma 3.4.4 can be written as

I q2(x) + 2 x

:

q(x)q(y) - 1 m.

x
x,yin

same chain

Suppose now that we can find two chain decompositions of P into m chains with the property that no pair of elements of P in the same chain in one of the decompositions lies in the same chain in the second decomposition. (Such a pair of chain decompositions are said to be orthogonal.) We can then obtain an inequality as above for both the chain decompositions, and on adding them together and dividing by 2 we obtain

Z q2(x) + Z* q(x)q(y)

m

where E * denotes the sum over all x, y such that x < y and x, y occur in the same chain in one of the two decompositions. Since

T, * q(x)q(y) -- Z q(x)q(y) x
we therefore have T, q2(x) + Y, q(x)q(y) % 1 . x x
We can now prove the following theorem.

(3.3)

42 1

Combinatorics of finite sets

Theorem 3.4.5 (Baumert, McEliece, Rodemich, and Rumsey 1980) If two sets A, B are chosen independently according to a probability distribution defined on the subsets of an n-set S, then the

probability that A c B is at least 1/(,

Proof Take m =

n

n/2]

The required probability is

Z q2(A) + A q(A)q(B) A AcB

where q(A) is the probability of choosing A. Thus, by (3.3), the probability is -1/m-provided of course that an orthogonal pair of chain decompositions exists! This is now proved.

Theorem 3.4.6 (Shearer and Kleitman 1979) If n - 2, there exist two orthogonal chain decompositions of the poset of subsets of an n-set

into ([',2]) chains. Proof If n = 2 or 3 it is easy to find such decompositions (see Exercise 3.11). Therefore we now suppose that n - 4. First of all take the symmetric chain decomposition constructed earlier in this chapter. Then obtain another chain decomposition by replacing each set by its complement. Certainly 0 and S will lie in the same chain in both decompositions, but we show that no other pair of sets have that property. If A, B with A c B lie in the same chain in both decompositions then A and B must have the same basic elements

and their complements A' and B' must also have the same basic elements. Also, the string of parentheses representing A' is obtained from the string representing A by replacing all left parentheses by right ones and vice versa. We now study the strings for A and B and show that if A and B have the same basic elements, then A' and B' cannot have the same basic elements except when A and B are 0 and S. Suppose first that A is not the smallest set in its chain. Then with A

and B in the same chain and A c B, we must have the following strings of parentheses:

where the C, respresent (possibly empty) closed parts. Since A is no, the smallest set in its chain, there is at least one ')'before C,. Now the

Symmetric chains 1 43

strings representing A' and B' are obtained by interchanging left and right parentheses, and it is clear that this interchange applied to the string for A will result in the reversed bracket between C, and C;+, closing with a previous bracket whereas this closing will not take place in the case of B. Thus A' and B' do not have the same closed parts. A similar argument applies if B is not the last set in its chain. We are left with the case when A and B are the first and last members of their chain. We then have

A:Co(C1(... B: CO)C1)... If there is a non-empty C, to the left of a in A then on reversal there will be a closing of brackets in A which will not occur in B. If there is a non-empty C, to the right of a `)' in B then there will, on reversal, be a closing of brackets in B which will not occur in A. Thus,

unless all C, are empty (in which ase A = 0 and B = S), A' and B' will have different closed parts and hence different basic elements.

The two symmetric chain decompositions are therefore almost orthogonal. To make them orthogonal, we remove 0 from its chain in the first decomposition and add it to any chain which does not contain

any of the n - 1 sets other than S or itself in its chain in the other decomposition. This is possible since there are more than n chains if

0

n>3. 3.5

Nested chains

In obtaining a generalization of Sperner's theorem, Lih (1980) considered collections si of subsets of an n-set S which are not only antichains but which have the further property that every member of sQ has a non-empty intersection with a given k-element subset T of S. The case k = n, i.e. T = S, clearly reduces to Sperner. Lih showed that .sal has the Sperner property, so that .sz2) s max{no. of subsets of size i which intersect T}

=ma x{(i) -\n i

i

kl

since the number of i-subsets of S which do not intersect T is (n

kl i

J

It is not hard to check that the maximum occurs when i = [Zn] (see

44 1 Combinatorics of finite sets

Exercise 3.12), so we finally obtain

(,n

In-kl

f \[n/2]J - \[n/2]1

An alternative approach to proving that s:f is Sperner has been given by Griggs (1982). By relabelling the elements of S = {x1,

. . .

,

if necessary, we can suppose that T = {x1,

. . .

,

xk}. Let

C(n, k) denote the collection of all subsets of S which intersect T. Then Griggs showed that, although C(n, k) need not have a decomposition into symmetric chains, it does nevertheless have the next best thing, a decomposition into nested chains.

Definition Two chains in a ranked poset P are nested if they are saturated and if the chain containing the element of least rank in their union also contains the element of greatest rank. Also, P is called a nested chain order if P can be partitioned into pairwise nested chains.

Example 3.5.1 Let S = {1, 2, 3, 4) and T = {1, 2). Then C(4, 2) consists of all subsets of S which contain 1 or 2. A decomposition into pairwise nested chains is shown: {1} c {1, 2} c {1, 2, 3) c {1, 2, 3, 41 {2} c {2, 31 c {2, 3, 4)

{1,3}c{1,3,4} (1, 4) c {1, 2, 4)

(2,41 Note how these nested chains are related to the symmetric chains exhibited in Example 3.1.1. This gives us a hint of the proof of the following theorem. Theorem 3.5.1 (Griggs 1982) C(n, k) is a nested chain order (and so has the Sperner property).

Proof Consider the symmetric chain decomposition given by de Bruijn et al. (1951). If the first member of a chain intersects T then every member of that chain will intersect T, and so the whole chain is in C(n, k). If none of the members of a chain intersect T, then ignore that chain. Consider next a chain in which the first member does not intersect T, but at least one other member of the chain does intersect T. Since each member of a chain is obtained by adding an element to the previous one, the elements being added in the order of the x., and

since T = {x...... xk}, it follows that if an element of T is added at

Symmetric chains 1 45

some stage in the chain, it must be added at the very first stage. Thus every member of the chain except the first is in C(n, k). In conclusion, therefore, C(n, k) is the union of chains each of which is one of the symmetric chains or is a symmetric chain minus its first member. Thus C(n, k) is a nested chain order.

Corollary 3.5.2 Any antichain sd in C(n, k) has at most n

\[n/2])_(n-k)

[n ] Proof 4 can contain at most one member of each of the nested chains but the number of chains is the number of subsets of size [in] which intersect {1,

3.6

n l ( \[n - k/' l ... , k), i.e. is ([n/2/ - n/21

Posets with symmetric chain decompositions

A ranked poset is called a symmetric chain order if it possesses a decomposition into symmetric chains. In this section we look for some partial answers to the following question: which posets are symmetric

chain orders? Our first result is really just a reworking of Theorem 3.1.1.

Definition If P, Q are posets with rank functions r, r' respectively, then the direct product P x Q = {(p, q) : p E P, q E Q} is a poset with (p,, q1) - (p2, q2) 'P, ,P2 in P and q, s q2 in Q. P X Q has rank

function p defined by p(p, q) = r(p) + r'(q).

Example 3.6.1 The poset of divisors of pagb (where p, q are distinct primes) is the direct product of the two chains {1, p, p2, ... , p°} and {1, q,

... , qb}.

Theorem 3.6.1 If P and Q are symmetric chain orders, then so is P X Q.

Proof Let P = C, U ... U Cand Q = D, U ... U D be symmetric chain decompositions of P and Q. Choose any C, and D;, say C,: cO
46 1 Combinatorics of finite sets

and

Di: do < d1 <

... < dh,

where r(co) = r,

r(ck) = R,

r + R = r(P)

r'(do) = s,

r'(dh) = S,

s + S = r'(Q).

and

We imitate the proof of Theorem 3.1.1 and consider the array E0--> codo cido...

Ck_edo ck-1do Ckdo

E,--,, cod, c,d1..

ck-2d1 Ck-ld1

ckdl

Ck1dh

Ckdh

code

.

..

from which we read off chains Eo, E1i etc. We show that these chains are symmetric (they certainly cover P X Q). Consider Ei : (co, di) < (c1, d) < .

. .

< (Ck-i, d1) <

. . .

< (Ck-j, dh)

where the rank of (co, d) is r + j + s and the rank of (ck_j, dh) is

r+k-j+s+h=R+S-j, so that

rank of (co, d1) + rank of (Ck_j, dh) = r + R + s + S = r(P) + r'(Q)

=P(PxQ). This completes the proof.

We now turn our attention to the posets L(m, n) defined by

L(m, n) = {(x1i ... , x,,,):x, integers, 0-- x1' x2 with order relation

...

x,,,

n)

defined by

x=(x1,...,x.,,)--y=(yl,...,ym)pxr--yifor each i. The posets L(m, n) are ranked posets, with rank given by r(x) _ x, +. - - + xm. As an example, consider L(3, 2). The element of maximum rank in L(3, 2) is (2, 2, 2) with rank 6. There are 10 elements altogether, and they can be placed into two symmetric chains as follows:

(0,0,2)<(0,1, 2)<(0,2, 2) (0, 0, 0) < (0, 0, 1) < (0,1,1) < (1, 1, 1) < (1, 1, 2) < (1, 2, 2) < (2, 2, 2)

It has been conjectured that each L(m, n) is a symmetric chain order.

Now there are two necessary conditions on the rank numbers of a

Symmetric chains

poset which have to be satisfied; they are (i) the rank numbers Ni are unimodal, (ii) Ni = N,_; for each i,

where r = r(P) is the largest rank in the poset P. Since (xl, ...

has rank j if and only if (n - x,,,

. . .

1

47

,

, n - x,) has rank mn - j,

condition (ii) is certainly satisfied. It is also known that the rank numbers satisfy (i). At present, the conjecture has been confirmed only for L(3, n) by Lindstrom (1980) and for L(4, n) by West (1980). We give the proof for the first of these. Theorem 3.6.2 (Lindstrom 1980) L(3, n) is a symmetric chain order.

Proof We shall consider the cases of n odd and n even separately. First we consider n odd, n = 2k + 1. The case k = 0 corresponds to L(3, 1) which is trivially symmetric as it consists of one chain (0, 0, 0) < (0, 0, 1) < (0,1,1) < (1, 1, 1). We therefore use an induction argument, assuming that L(3, 2k - 1) is symmetric (k a 1) and considering L(3, n) where n = 2k + 1. Let S(n) denote the set of all

(a, b, c) in L(3, n) with a = 0 or c = n (or both). The remaining elements of L(3, n) have 1, a , b c , n - 1 and form a poset isomorphic to L(3, n - 2) = L(3, 2k - 1) under the correspondence (a, b, c) a L(3, n) H (a - 1, b - 1, c - 1) E L(3, n - 2). By the induction hypothesis, L(3, n) - S(n) has a symmetric chain decomposition. We then show that S(n) has a symmetric chain decomposition as well, by exhibiting k + 1 chains C,, i = 0, ... , k, which partition S(n):

C1:(0,i,i)<(0,i,i+1)<...<(0, 1,n-i)<(0,i+1,n-i) < (0, i + 1, n - i + 1) < (0, i + 2, n - i + 1) < ... < (0, 2i, n) < (0, 2i + 1, n) < (1, 2i + 1, n)

<(1, 2i+2, n)<(2, 2i+2, n)<.. . < (n - 2i - 1, n, n) < (n - 2i, n, n). These chains C, are symmetric and partition S(n).

We next consider n even. We let S(n) be as before, and we obtain

S*(n) by adjoining to S(n) all elements (1, 2i + 1, n - 1), i = . . . , k - 1, where n = 2k. It can now be checked that S*(n) can be partitioned into chains C and C*, i = 0, . . . , k - 1, defined by

0,

C: (0, 0, n) < (0, 1, n) < ... < (0, n, n)

48 1 Combinatorics of finite sets and

C*: (0, i, i) < (0, i, i + 1) < ... < (0, i, n - i - 1) < (0, i + 1, n - i - 1) < (0, i + 1, n - i) < (0, i + 2, n - i) < (0, i + 2, n - i + 1) < ... < (0, 2i + 1, n - 1) < (1, 2i + 1, n - 1) < (1, 2i + 1, n)

< (1, 2i+2, n) < (2, 2i+2, n)<(2, 2i+3, n) < ... < (n - 2i -1, n, n) < (n - 2i, n, n). Each chain C; contains an element (1, 2i + 1, n - 1) of S*(n) - S(n);

this element is in T(n) = {(a, b, c) E L(3, n); a =1 or c = n -1 or both}. We can also check that T(n) - {(1, 2i + 1, n - 1)) is the union of symmetric chains D;, i = 1, . . . , k - 1, given by

D;:(1,i,i)<(1,i,i+1)<...<(1,i,n-i-1)<(1,i+1,n-i-1) < (1, i + 1, n - i) < ... < (1, 2i, n - 1) < (2, 2i, n - 1) < (2, 2i + 1, n - 1) < (3, 2i + 1, n - 1) <.. . < (n - 2i, n - 1, n - 1) < (n - 2i + 1, n - 1, n - 1). Now S(n) U T(n) contains all elements (a, b, c) in L(3, n) with a = 0 or 1 or c = n - 1 or n. Therefore L(3, n) - (S(n) U T(n)) is isomor-

phic to L(3, n - 4) under the correspondence (a, b, c) - (a - 2, b 2, c - 2). Thus we can use an induction argument, obtain a symmetric chain decomposition of L(3, n) - (S(n) U T(n)) from the induction hypothesis, and combine it with the symmetric chain decomposition obtained above for S(n) U T(n).

The proof for L(4, n) given by West (1980) is more complicated. Although it is not known whether L(m, n) is symmetric in general, it has recently been shown by Stanley (1980) that L(m, n) is always Sperner. His proof uses deep algebraic methods. 3.6.1 Symmetric chains and the normalized matching property

We now show that, along with the conditions (i) and (ii) on the rank numbers which are necessary for a symmetric chain decomposition to exist, a sufficient condition which ensures that P is symmetric is, remarkably, that P has the normalized matching property. Before we

prove this, we have to describe a generalization of the marriage problem to the context of common systems of distinct representatives. Definition

If A1, .

. .

, A. and B1, .

. .

, Bm are two families of sets, a

common system of distinct representatives (c.s.d.r.) is a set of m

Symmetric chains

1

49

elements which is simultaneously an s.d.c. for the A; and for the B; (in some order).

Example 3.6.2 The sets A, = {1, 2, 3}, A2 = {1, 4), A3 = (3, 4) and the sets B, = {3, 4}, B2 = {2, 4}, B3 = {3} have a c.s.d.r. consisting of 2, 3, 4, for we can represent A, and B2 by 2, A3 and B3 by 3, and A2 and B, by 4. Theorem 3.6.3 (Ford and Fulkerson 1958) The families A,, ... , Am and B1,. .. , Bof subsets of the n-set S possess a c.s.d.r. if and only if

(,U

A;) n (;U B,) ;JEJ +JFJ -m

for all subsets E, F of {1,

.

.

.

(3.4)

, m}.

Proof (Perfect 1969) The key idea behind Perfect's elegant proof is

to show first that the existence of a c.s.d.r. is equivalent to the existence of an s.d.r. for a related family of sets. Suppose that S = {x,, .. . , let T = S U {1, ... , m), and define m + n subsets Y,of Tby

Y;=A; (j=1,...,m) YX={x)U{i:xEB;} (xES). Now any s.d.r. for the Y is uniquely described by an injection a: {1, ... , m) - {1, ... , n} and a permutation 0 of {1, ... , m), which are given by xa(;)EA;

(i=1,...,m)

/3(i)EYX,(;, (i=1,...,m), the remaining Y; being represented by x;. The second of these relations is equivalent to x«(t) E Bp(i)

so that we obtain xa(;) E A. n Bp(;) for each i = 1, ... , m. Clearly a different s.d.r. for the Y would involve a different a or a different 0 and so would give rise to a different c.s.d.r.. Conversely, it is easy to check that any c.s.d.r. yields an s.d.r. for the Y. Now by the marriage theorem the Y possesses an s.d.r. if and only

if IU,EKYI-- IKI whenever Kc{1,...,ml write K = I U J where I c {1, . . . , m} and J c {x,, ... ,

If we the

50 1 Combinatorics of finite sets

condition is

(uA1) 1EJ whenever I c 11, .

. .

,

U

(U Y) J=J

(3.5)

III + IJI

m} and J c {x,, ... , x,,}. However,

; Y)=(UA;)nJ (uA1)n(u and

UY;=JU{i:B;nJ*0) ;EJ so that, since IS n TI + IS U TI = ISI + ITI, expression (3.5) can be rewritten as

U

A,

UY

(uA1)nJ iEJ

III+IJI,

{i:B;nJ*o}I,1I1.

iel

(3.6)

Finally we show that (3.6) is equivalent to the Ford-Fulkerson condition (3.4). Suppose that (3.6) holds for all I, J. Given subsets E,

F of {1,...,m}, define J by

U(B,:jEF) and

take I = E. J is the set of elements not in any B;, j e F, so that if B, n J * O then j f F. Then (3.6) yields

I (n (JU B;)I + m Conversely, suppose that (3.4) holds for all E, F. Given I, J, define E and F by E = I and F= {i : B; n J = O}. Then lEE

JEF

tEE

so that (3.4) gives iEJ

=III-(m-IFI) _III-l{i:B,nJ*Q}I, i.e. (3.6) holds.

0

We now use Theorem 3.6.3 to obtain a sufficient condition for a poset to be a symmetric chain order.

Symmetric chains

1

51

Theorem 3.6.4 (Anderson 1967b, Griggs 1977) Let P be a ranked poset with normalized matching property whose rank numbers are unimodal and satisfy Ni = N,_; where r is the rank of P. Then P is a symmetric chain order.

Proof We construct a symmetric chain decomposition step by step.

Suppose that all elements of P of ranks t, k < t < r - k have been placed in symmetric chains. We then have Nk+l pairs (u, v) of elements of P with r(u) = k + 1, r(v) = r - k - 1, u < v, and u in the same chain as v. Let H denote the set of all such pairs. For each ai E P

of rank k, let A. = {(u, v) E H: ai < u}, and for each b. E P of rank r - k let B. = {(u, v) E H: v < bi}. Then we can extend the symmetric chains to include all elements of P of ranks k and n - k if we can find a c.s.d.r. for the sets AJ, i = 1, . . . , Nk, and the sets Bi, i = 1, ... , Nk. We therefore only have to show that the sets A, and Bi satisfy the condition (3.4) of the Ford-Fulkerson theorem. Therefore let E, F be any two subsets of {1, . . . , Nk}. We assume that IEl + IFI > Nk since otherwise (3.4) is trivial. Now, by the normalized matching property, U AJ

iEE

Nk+1 IEI

Nk

Nk+'IFI %U BJ

Nk

so that

(uA1)n(uB1)I=IUAiI+ UB' JEF !EE jEF !EE

l

l

UA,)U(UB') JEP

iEE

Nk'(IEI+IFI)-Nk+j

-_Nk+1 (IEI +IFI -Nk) Nk

IEI + IFI - Nk.

Thus the Ford-Fulkerson condition is satisfied, and the chains can be extended to include all elements of ranks k and n - k. Repeating the process at each rank level gives a construction of the chains for P. If n is even the whole process begins with the elements of the middle rank in place of the pairs (u, v), and if n is odd the first step is to pair off

the elements of rank 2(n - 1) with those of rank 2(n + 1), which is possible since the normalized matching condition reduces to the Hall condition between the two ranks in this case.

52 1 Combinatorics of finite sets

Quantitative results in this area are also possible: here we ask how many symmetric chain decompositions does P have? Recall from

Exercise 2.20 that if the sets A,, ... , A. possess an s.d.r. and if I A; I , t for each i, then there are at least f (m, t) different ways of choosing an s.d.r. for the A;, where

fm,t

ifm - t, -{ t!t(t-1)... (t-m+1) _

if m
(3.7)

Using this result we find that the Perfect proof of the Ford-Fulkerson theorem yields the following theorem.

Theorem 3.6.5 (Anderson 1985) Let A,, ... , Am and B1,. .. , Bm be two collections of subsets of an n-set S which satisfy condition (3.4). Suppose also that IA;I = t for each i and that every element of S is contained in at least u of the B;. Let p = min(t, 1 + u). Then there

are at least p! different c.s.d.r.s for the A. and the B;. (Here two c.s.d.r.s are considered different if at least one of the sets is represented by different elements in the two c.s.d.r.s.)

Proof In Perfect's proof, I Y I , t for each i =1,

. . .

, m and I YX

1 + u for each x E S, so the result follows from (3.7).

Corollary 3.6.6 The poset of subsets of an n-set can be partitioned into symmetric chains in at least g(n) ways where g(n) = 3!4! ... (In + 1)!n if n is even and g(n) = 3!4!

... (n 2 1)!(n,21)! n

if n is odd.

Proof If n is even, then, in the notation of the proof of Theorem 3.6.3, IAII = n - k and each element is in k + 1 sets B;. Since

mint, 1 + u) = min(n - k, k + 2) = k + 2, there are at least (k + 2)! ways of extending the chains to ranks k and n - k. At k = 0, 0 and S can clearly be added to any of n chains. The case of odd n is similar, except that at the start there are at least {(n + 1)/2}! ways of pairing the sets of ranks (n - 1)/2 and (n + 1)/2. This lower bound can be increased by a more detailed argument, but it is certainly an improvement on the trivial lower bound of n!, which is the number of symmetric chain decompositions obtained using the construction of de Bruijn et al. (1951) by considering each of the n! possible orderings of the elements of S.

Symmetric chains 1 53

Exercises 3 3.1

Show that, in a symmetric chain decomposition of the set of

subsets of an n-set, there are (k) n + 1 - 2k, and hence of size i. 3.2

- (k n 1)

n

chains of size

l

n

[(n + i)/2]) - `[(n + i +D/21)

chains

Define µ(A) = (-1)IA1. Show that if (SI = n then

IAE (A)

3.3

whenever .sf is a convex set of subsets of S, i.e. if A E Si, B E s4, A c D c B D E,4. (de Bruijn et al. 1951) Use symmetric chains to show that if .i1i ... , s4, are r disjoint antichains of subsets of an n-set then

I_ Us$,,2 =1

n

i.1 ([(n +j)121)'

3.4

(Kleitman 1965) Show that, if A1i ... , A. are distinct k-subsets of an n-set and

3.5

k - h - n - k, then there exist distinct h-subsets B1, ... , B,,, such that A; f1 B1=0 for each i=1,...,m. Show that the number of antichains of subsets of an n-set is 32(121).

3.6

3.7

Use Dilworth's theorem to show that if A1, ... , A. are any n sets then we can find ?n" of them such that the union of no two is a third. (a) Show that the maximum size of a chain in a finite poset P equals the minimum number of disjoint antichains into which P can be decomposed.

(b) Deduce that if r, s are positive integers then a poset of rs + 1 elements possesses a chain of size r + 1 or an antichain of size s + 1 (or both).

(c) Deduce that, given a sequence of rs + 1 real numbers, there is either an increasing subsequence of length r + 1 or a decreasing subsequence of length s + 1. 3.8

Show that a subset A of S = {x1,

... ,

(Mirsky 1971)

is a set of basic

elements if and only if, for each i, at most half of x1i

. . .

, x1 are

in A. A zigzag path between two points with integer components is a path made up of straight line segments joining a

54 1

Combinatorics of finite sets

point (x, y) to either (x + 1, y + 1) or (x + 1, y - 1). Given a set of basic elements, construct a zigzag path from (0, 0) as follows: if x, is a basic element go from (0, 0) to (1, -1), but if it is not a basic element go to (1, 1); at the (i + 1)th stage, go from the point (i, v,) so far reached to (i + 1, v; - 1) if x; is basic, but to (i + 1, v, + 1) if x; is not basic. Hence show that the number of zigzag paths from (0, 0) to a point on x = n which do not go below the x-axis is ([n/2)). Show also that the

number of zigzag paths from (0, 0) to a particular (n, p) is n

n

k)

k -1

where k = 2(n -p) is integral. (Greene and Kleitman 1976b)

3.9 Show that (2/nrc)"2 2" is a lower bound for the storage problem of Theorem 3.3.2. 3.10 Obtain the upper bound of Theorem 3.4.2 by showing that

the average size of the greatest member of a chain in a 3.11

symmetric chain decomposition of the set of subsets of an n-set is 2n +O(nln). Find orthogonal chain decompositions as required in the proof of Theorem 3.4.6 for n = 2 and n = 3.

3.12 Show that, if n and k are fixed, dm

(n) m

maximum value when m = [in). 3.13

('em

k) takes its (Lih 1980)

By imitating the first proof of Sperner's theorem show that if .s1 is an antichain in C(n, k) then

E (In-1 IAnTI -- k. n-1 )

AEaL

-11

Also extend this result to the poset C(n, k, h), consisting of those members of C(n, k) which intersect T in ,h elements. (Griggs 1982) 3.14

Show that L(m, n) has (m

+n

members.

) 3.15 An intersecting family is a collection of subsets of a set no two of .which are disjoint. A maximal intersecting family is one n

which cannot be extended to a larger one. Show that the

Symmetric chains 1 55

number of maximal intersecting families of subsets of an n-set

is at most the number of antichains of an (n - 1)-set, as follows.

(i) For any maximal intersecting family 4 define g(4) _ tA c

{1,...,n-1}:A=BfI{1,...,n-1} for some BE,s4, and there does not exist C E s4 such that c fl {1, ... , n - 1} c A}. Prove that g(4) is an antichain. (ii) Prove that if A c {1, . . . , n - 11 then

(a) AU{n}E4 if and only for some Bcg(4) (b) A E Si if and only if B c A for some BE g(4) and there is no C E g(s4) with c fl A = 0. (iii) Observe that (a) and (b) show that g is one-one. (Erdos and Hindman 1984)

4 Rank numbers for multisets

4.1

Unimodality and log concavity

In this chapter we shall take a closer look at the rank numbers in the poset of divisors of a number (or subsets of a multiset). We shall use the notation of divisors of a number and throughout the chapter we shall assume that m pktpk2 2 ... p;'

(4.1)

where the p; are distinct primes and k; a 1 for each i. The rank of m is then r(m) = k, + . . . + k and N;(m) is the number of divisors of m of rank i. We have already seen from the symmetric chain decomposition of the poset that, if r(m) = K and A = [1K], N,(m) = NK-i(m)

(4.2)

for each i, and No(m) _ N,(m) , ...

Nx(m); Nx(m) ..... NK(m).

(4.3)

In the case of subsets of an n-set, or square-free m, we have N; = ('f), and N; < Ni,, whenever i < In, so that strict inequality occurs throughout (4.3) if n is even and everywhere except at

(?(nn

1))

n

if n is odd. It is natural to ask where strict inequality occurs in (4.3) when m is not square free. This question has been

((n + 1)

answered by Clements (1984a, b) and Griggs (1984).

Theorem 4.1.1 Let m be as in (4.1), with k, = k2 --... , k,. Let K = E; k and put L = L(m) = min{[ZK], K - k,}. Then

No... >NK. Note that this tells us that if k, ,1K the rank numbers increase strictly to Nx and there is a unique maximum rank number if K is even and two equal maximum rank numbers if K is odd. However, if k, > 1K 56

Rank numbers for multisets 1 57

there will be a `plateau' of 2k, - K + 1 equal maximum rank numbers NK_k,, ... , Nk,.

Example 4.1.1 m = p4g3r2. Here K = 9, L = min(4, 5) = 4, k, <21K. The rank numbers are 1, 3, 6, 9, 11, 11, 9, 6, 3, 1.

Example 4.1.2 m = p5g2r. Here K = 8, L = min(4, 3) = 3, k, > ZK, 2k, - K + 1= 3, so that there should be a plateau of three maximum rank numbers. The rank numbers are

1,3,5,6,6,6,5,3,1. Proof of Theorem 4.1.1 (Clements 1984a, b) We first prove that N;_,
N,(m) - N,-I(m) = {Ni(m') + N;-,(m') +. . . + N,_k,(m')} - {N,_,(m') +. . . + N,_,_k,(m')} = N,(m) - N1_1-k,(m'), so we have to show that N,_,_k,(m') < N,(m').

(4.4)

It will help to have an idea of how large i - 1 - k, is, so we next show

that if i , L(m) then i - kr -- L(m').

(4.5)

There are two cases to check in proving (4.5). First suppose that L(m') = [ZK(m')]. Then

iEL(m)=> i 2K(m)'i-kr-i kr -- [ZK(m')] = L(m') as required. Next suppose L(m') = K(m') - k,. Then

i;L(m) 'i;K(m)-k, => i-krK(m)-k,-kr ' i - kr K(m') - k, = L(m').

[2{K(m')+k,}]-kr

58 ( Combinatorics of finite sets

So (4.5) is proved. Thus, by the induction hypothesis, N`-1-k,(m ') < N,_k,(m').

To prove (4.4) it is therefore sufficient to show that N;_k,(m') N,(m'). Since i - k, -- L(m'), this will be true by (4.2) provided that i -- K(m') - (i - k,), i.e. provided that i -- i{K(m') + k,} = ZK(m). (4.4) is proved. This in turn is true since i < L(m) -We next show that N,(m) = N,+1(m) for all i such that L(m) , i < [ZK(m)]. Clearly we can assume that L(m) = K(m) - k1. Again, we proceed by induction and assume that N,(m') = N,+1(m') for all i such that L(m') - i < [ZK(m')]. We then have that N,(m') - N,(m') for all i, j such that L(m') -- i -- j. Now if L(m) -- i < [ZK(m)] we have ZK(m). So

i-k,-K(m)-k1-k,=K(m')-k1-L(m') that N;_k,(m') - Ni+1(m'). However, Nt+1(m) - N;(m) _ Nt+1(m') - N;_k,(m'), so we have Ni+1(m) --N,(m). However, so

N;(m) , Ni+1(m) since i < [jK(m)], so we conclude that N;(m) = N,+1(m) as required.

For completeness we record the number of maximum rank numbers as a corollary. Corollary 4.1.2 In the notation of Theorem 4.1.1, if k1 > jK there are exactly 2k1- K + 1 equal rank numbers of maximum size. 4.1.1 Logarithmic concavity

In attempting to prove that a sequence

is unimodal it

is

sometimes easier to prove a stronger result, namely that the sequence is log concave.

Definition The sequence (a,,) of positive numbers is log concave if a`-1a,+1 = a2

For each i. Similarly, {b,,) is log convex if b,-1b,+1= b;

for each i.

If we write (4.6) as a;+11a, , a;lai_1 it becomes clear that if ai+1- a, is certainly unimodal. Also note that we then a, - a;_1, so that

Rank numbers for multisets

1

59

can write (4.6) as 2(log a1_1 + log a1+,) -- log ai

so that the sequence {log a;} is concave in the sense that in the cartesian plane the point (i, log a;) lies on or above the straight line segment joining (i - 1, log a;_,) and (i + 1, log a;+,).

It is trivial to check that, for fixed n, the binomial coefficients n i

are log concave:

I.

nl)\i+\ilZ_

It is also fairly straightforward to show that, for fixed n, the Stirling numbers S(n, k) are log concave (see Exercise 4.1) We now show that the rank numbers for multisets are also log concave. This is a special case of results of various authors (see Exercise 4.7) but we give a simple direct proof. Theorem 4.1.3 The rank numbers for multisets are log concave, i.e. N; _,N;+, , N; for each i, 0 < i < K.

Proof (Anderson 1968) Observe that N;-,N;+, is the number of

ordered pairs (a, b) of divisors a, b of m with r(a) = i - 1, r(b) = i + 1, and that similarly Ni' is the number of ordered pairs (c, d) of divisors c, d of m with r(c) = r(d) = i. Let M be any divisor of m2. The theorem will be proved if we can show that M can be expressed in the form cd in at least as many ways

as in the form ab. To prove this for all M dividing m2 we use induction on n, the rank of m. The result is trivial if n = 1, so we suppose that m has rank n -- 2 and that the result is true for all M arising from an integer of rank less than n. Let M be any divisor of m2 of rank 2i. First suppose that m 4-M. Then there is a prime p which divides m

to a higher power than it divides M. Let m = m,p. Then M I m; and

r(m,) < n. However, the numbers of representations of M in the forms ab and cd remain unchanged on replacing m by m,, so that the result follows by induction. The condition m 4. M is certainly satisfied if i < in, since if m I M

then r(m) , r(M), i.e. n , 2i. Suppose next that i = in, i.e. r(m) = r(M). If m * M then m 4.M and the result follows as before. If m = M then the numbers of representations of M in the forms ab and cd are and N}., and the result follows since Ni,,. precisely

60 1 Combinatorics of finite sets

It remains only to consider those M for which i > In. Now there is a one-one correspondence between representations of divisors of m2 of degree 2i and representations of divisors of m2 of degree 2(n - i). For if ab = cd = M then

m m_m m_m2

b.a c.d M'

where m2) r(M = 2(n - i).

So if the result is true for i < Zn then it is also true for i > In.

El

Corollary 4.1.4 If i <j then N,N; , N;+,N;_,.

Proof Ni

N,+,

N,+1

N,+2

N _j ..

N

0

4.2 The normalized matching property

We next show that the poset of divisors of m has the normalized matching property. As in the case of log concavity, there are several methods of proof. As well as the proof given here we shall present one in Chapter 12 which treats the poset as a direct product of chains. Theorem 4.2.1 Let 6-4 be a set of divisors of m of rank i, and let V a denote the shade of R. Then Iveol

> I_I

N,+j(m) - N,(m) Proof (Anderson 1968) Let 1,641 = s and ,Vat = t. We have to prove

that tN,(m)=sN;+,(m). Now tN;(m) is the number of ordered pairs (a, v) with a e VQ, v I m, and r(v) = i, while sN,+,(m) is the number of ordered pairs (b, u) with b E 90, u I m, and r(u) = i + 1. As in the proof of Theorem 4.1.3 it is sufficient to prove that any divisor M of m2 of rank 2i + 1 can be expressed in the form av in at least as many ways as in the form bu.

We proceed by induction on the rank n of m. The case n =1 is

Rank numbers for multisets 1

61

trivial so we suppose that m has rank n , 2 and that the result is true for divisors of numbers of rank less than n. We let M be a divisor of m2 of rank 2i + 1. The case m 4. M is dealt with as in the previous

proof: m = m,p where M is effectively a divisor of m;. So now suppose that m I M, say M = mg where r(g) = 2i - n + 1. Suppose that the representations of M in the form bu are

M=mg=blu,=... =brur. Since b; I m and u; I m it follows that g I u; and g I b,, so we can write b, = ge,, u; = gw,. Then m/g = e,w and

r(e,) = 2 {r(m) - 1}. g

If we now consider a symmetric chain decomposition of the poset of divisors of m 1g, we see that, since

r(e) = 2 fr(m)g-1}, we can associate with each e, a multiple f of rank one more, namely the next number in the chain containing e,, to obtain numbers f such that e, If I m/g, r(f) =1 + r(e). We then have m

=fz, where z,=w,.

g

e, m f

g

Thus

M=mg=gfi.gz1=...=gf,.gzr where each gf E V90 and each gz, I m. This gives r representations of M in the form av as required. Corollary 4.2.2 Let 64 be a set of divisors of m of rank i, and let A A denote the shadow of R. Then

Proof Let 1V'= {m/b : b E 90}. Then 1.6h'I =1y81 and JA Applying Theorem 4.2.1 to 12', we have

I0'64'I JI i

N-,+1

N-+

= Iv94'1.

62 I Combinatorics of finite sets

so that IA-0A i

I-I

N,-1

Ni

0

We also record the fact that, since the poset of divisors of m has the normalized matching property, it must also have the other properties

equivalent to it, as described in Theorem 2.3.1. We thus have the following theorem.

Theorem 4.2.3 If D is an antichain of divisors of m then

I

1

deD Nr(d)

11

`1

As an application of this theorem, we establish a result of Kleitman and Milner (1973) concerning the average rank size in an antichain. The proof given here is essentially due to Grimmett (1973). As in Theorem 4.1.1, we let L(m) denote the largest value of i such that No < N1 < ... < Ni holds. Theorem 4.2.4 (Kleitman and Milner 1973) Let D be an antichain in the poset of divisors of m = pi' . . . pn^, and suppose that ID I , Nk for some k , L(m). Then EdeD r(d) % k ID 1, i.e. the average rank size of the divisors in D is at least k.

Proof Let q denote the average rank size of the divisors in D, and suppose that q < k with a view to obtaining a contradiction. Let f(j) =1/N for each j = 0, ... , L(m). Then, since N;_1Nj±1-- N, for each j, 1

1

N -1) +N + 1) - Ni -1 + Nj+1

2(

1

\v2

N,-1N;+1

=N=2f(), so that f is a convex function on 0, 1, ... , E; k;. By Theorem 4.1.1 it is strictly decreasing on {0, . . . , E, k. ). We can extend f to a convex function f : [0, E, k;] - R by linear interpolation; this extended f is also strictly decreasing on [0, L(m)]. If we let d, denote the number of members of D of rank i, we have DI

=1,

Rank numbers for multisets 1 63

so that, by the convexity of f,

.f (:

I I'i f (i).

The LYM inequality of Theorem 4.2.3 therefore yields TDI

DI

aDf(r(d)) = I1

Z df(i);f(

Ld i)

=f(4) >f(k) = Nk Thus ID I
4.3 The largest size of a rank number

In the case of subsets of an n-set the largest rank number is

([',2J) . Using Stirling's approximation n! - (2nn)' n"e-" we obtain (,n/21n

\,nl112n

If m =p'. .. p1', the largest rank number is

N(K,21(m) where

K = kl + ... + kr = r(m). If we let r(m) denote the total number divisors of m, we have r(m) = No(m) + ... + NK(m) = (1 + k,) ... (1 + kr) since any divisor of m is of the form p;' ... p;`', where there are 1 + k; choices 0, 1, ... , k; for h,. In this section we

of

obtain an estimate for the size of N[K,21 in terms of the k,. If we write

A(m) = 3 > k,(k, + 2)

we find that we can show that the size of the middle-rank number is essentially a multiple of r(m) (4.7) (A(m))`2 Note that in the square-free case where each k, = 1,

t(m)

2"

(A(m)}"2 - nin

The first lemma below shows that A(m) is closely related to the standard deviation of the distribution of ranks.

64 1 Combinatorics of finite sets

Lemma 4.3.1 Let a2 be the variance of the distribution of ranks among the divisors of m. Then

a2=L zk1(k;+2). Proof The power of p; occurring in a divisor of m as in (4.1) is a random variable taking the values 0, 1, . . . , k; with equal probability. Its average value is therefore 2k, and its variance is 1

k'

l+k. M=0

(m-2k,)2

=

1 1) l 1+k,JEm , -k,Em+4k?(k;+ m 2

1

=6k;(2k;+1)-Zk?+4k? = 2 k;(k1 + 2).

The lemma now follows since the variance of a sum of independent distributions is the sum of the variances. Since A(m) = 4a2, we obtain information about A(m) by studying the variance a2. Now since 62

(K - 2i)2N, i)2N, _ = i(m) =0I (K 2 2i(m t=0 )

where A = [ZK], we have to estimate shall make use of the following lemma.

(4.8)

(K - 2i)2NV. To do this, we

Lemma 4.3.2 (Reduction formula) x-r

(K - 2i)2Ni

a`Z-r 1=0

A-r-1

x-r-l

(K - 2i)N,

(K - 2i)2NL + 8Nx_,_1

1V21_,_1 i=0

i=0 .i-r-1

+ 16N2;,-,-1 Z N, + 3(2r + 5)2Nx_,. f=o

Proof

A-r-2

A-r

Nx_r Z (K - 2i)2N;

Nx_,

(K - 2i)2N1 + 2(2r +5)2Nx_,

i=O

where the first term on the right is, by two applications of Corollary

Rank numbers for multisets 1 65

N2Z

4.1.4,

d-r-2 (K - 2i)2Ni+2

i=0 x-r

Na2_,_, 2 (K - 2i + 4)2N, i=o

A-r-1

Nx-r-i , (K - 21 + 4)2N, + (2r + 5)2N3_r i=0 A-r-1 N2 A-r-I = (K - 2i)2N, + 8Nx_r_1 Z (K - 2i)N, i=o

i=o

A-r-1

0

+ 16Na2_,_1 E Ni + (2r + 5)2N31_,. i=0

Lemma 4.3.3 There is a constant C > 0 such that

r(m) a
Proof As in (4.8), x

2r(m)Nx(m)a2 = N,L(m) Y, (K - 2i)2Ni(m). i=o

Using the reduction formula of the previous lemma repeatedly, for r = 0, 1, ... , A -1, we obtain A

A-r

x

A-:

2r(m)Nx(m)a2 _ 8 E Nx_i fi-O E (K - 2i)N,} + 16 2 Nx_t(E N,) t=1

:=1

i=0

A-1

+ 3 Z (2t + 5)2Nx_,.

(4.9)

r=0

However, A-1

A-1

(2t + 5)2N3 _, t=0

25N34 + Z Nx_i(7tNA_,)2 t=1

x-1

25N3+ 49

NA_i(NA_t +. t=1

.. + Nx_1)2

x-1

NX_i{?r(m)}2

25N3A + 49 t=1

z

25r(m)(NA + 2 NA_t) < 25r3(m).

(4.10)

66 1 Combinatorics of finite sets

We also have

1=1

N,_,(

iO

N) -- ii(m) r=1

NIL-,

ir(m)( NA.. )2 1=1

(4.11)

< x3(m).

Finally, a reduction similar to the above (see Exercise 4.6) leads to x

NA 2 (K - 2i)N, -- 4i2(m), i=O

so that we have

E Nx_r(E (K - 2i)N,) , 4i2(m)NA_r) < 4i3(m). r=1

(4.12)

r=1

:=o

Combining (4.9)-(4.12), we therefore obtain

2t(m)Nx(m)a2, (constant)r3(m) whence

t(m) CNA(m)

0

as required.

NA(m)l v(m) - 0 as K -* c, and

Note that, since or - as K so we can write

NA(m) = o(r(m)).

(4.13)

We now use symmetric chains to obtain a lower bound for a. In the

next lemma we assume that the poset of divisors of m has been decomposed into symmetric chains.

Lemma 4.3.4 Let H(m) denote the number of ways of choosing three divisors of m from the same symmetric chain. Then

H(m)

2i(m)' Proof We shall use the identity \x32)-(3)-x2.

Rank numbers for multisets 1 67

We have, by (4.8), x-1

1 oz. 2i(m) > (K-2i-1)2N,

i=o

2sm+3 J 2t1m) E

{(K-2i+1)(K-2i-1)}N;

{ )

=

1

2r(m)

R)N)}(K+1)

H(m)

since, for i ;1, there are Ni - Ni_1 chains of size K + 1 - 2i. Lemma 4.3.5 There is a constant C' > 0 such that

o > C' T(M)

Nx(m)

Proof Suppose that the divisors of m are in s = NA(m) symmetric chains of sizes x1, ... , xs where x1 + ... + xs = r(m). Then H(m)-E

(3i)

6x,(x;-1)(x;-2).

Now under the constraint E; x; = r(m), H(m) has its minimum value essentially when all the x; are equal, i.e. when each x, = r(m)/s. It follows from (4.13) that H(m) % {6 - o(1))s(r 13 = {6 - o(1)} t3(m)

s/

Nx(m)

and the lemma follows from Lemma 4.3.4.

We can now state our main theorem of this section, justifying the estimate (4.7) for NA(m).

Theorem 4.3.6 (Anderson 1969) Let m = p; ... p;, A = [3K], r(m) = (1 + k1), .. (1 + k,), and A(m) = 3 E; k;(k, + 2). Then there exist positive constants cl and cz such that cl {A(m)}l1z

Nx(m)
r(M) (A(m)) 112

Proof The theorem follows immediately from Lemmas 4.3.1, 4.3.3, and 4.3.5.

68 1 Combinatorics of finite sets

Observe that in the case where k1= ... = k = 1 and K = n,

A(m)=3.3n=n so that r(m)l{A(m)}112 = 2"/n1'2. Thus Theorem 4.3.6 agrees with the estimate for NA(m) in the square-free case. By using complex variable methods, an asymptotic formula can be

obtained for NA(m). It has been shown by Anderson (1967a) that, under reasonably general conditions, Nx(m)

(

(2)1'2

)

.n{A(m)}la

Exercises 4 4.1

Show that, for fixed n, (a) the binomial coefficients (k) and (b)

the Stirling numbers S(n, k) are log concave. (In (b), use induction and Theorem 2.4.1(ii).) 4.2

Verify Theorems 4.1.1 and 4.1.2 for m =p7 q 3 r 2.

4.3 4.4 4.5

Extend Theorem 2.3.4 to multisets. Extend Exercise 2.2 to multisets. Show that if N;-1-- Ni and R is a collection of divisors of m of

rank i, then IAA- IM -_ N,-N;_1.

4.6

(a) Prove the reduction formula x-r

x-r-1

x-r

NA_, Y, (K - 2i)N, , Nx-r-1 Z (K - 2i)N; + 2Nx_r I Ni i-O

i-O

i=o

+ 2r{N21_, + Nx-r_1Nx-r}.

(b) Use (a) to prove that Nx E' Lo (K - 2i)N, , 4r2(m).

(c) Let J(m) denote the number of ways of choosing two divisors of m from the same chain of a given symmetric chain

decomposition. Show that J(m) -_2 E o (K - 2i)N,, and use (b) to deduce that J(m) < C r2(m) Nx(m)

(d) Verify that, if m is square-free,

J(m) - 2

(Anderson 1968)

Rank numbers for multisets 1 69

4.7

In the product H = F X G of two ranked posets (defined in Section 3.6) the rank numbers hk of H are given in terms of the rank numbers fk, gk of F, G by hk = I fpgk-p' pck

Suppose that the fk and the gk are log concave. Show that the hk are then also log concave as follows.

(i) We have to prove that G fpgk-l-pfmgk+1-m -- E fm pgk-mgk-p'

pck-1

mck+1

pck

mck

(ii) If p -- k, verify that both of the expressions f n fp - fm-lfp+1 and gk-mgk-p - gk-m+lgk-p-1 are %0 or --0 according as m - p or m %p, so that their product is always O. Deduce that fm9k-m+lfpgk-p-1 +fm-lgk-mfp+lgk-p -fmfpgk-mgk-p + f m-lf p+lgk-(m-1)gk-(p+1)'

(iii) Sum over all p and m to prove the inequality of (i). (iv) Deduce Theorem 4.1.3. (Hsieh and Kleitman 1973, Hoggar 1974)

5 Intersecting systems and the Erdos-Ko-Rado theorem

5.1

The EKR theorem

The main result to be discussed so far, Sperner's theorem, is concerned with antichains of subsets of a set. We now turn our attention to a special class of antichains which have the extra property that no two members of the antichain are disjoint.

Definition A collection of sets A1, . . . , A, is called an intersecting family, or is said to have the intersection property, if A, fl A10 0 for all i, j. The following result due to Erdos, Ko, and Rado (1961) was published in 1961.

5.1.1 (Erdos-Ko-Rado (EKR) theorem) Let Am) be an antichain of subsets of an n-set S which has the intersection property. Suppose also that there is a number k -- Zn such that 1A1I - k for each i. Then

Theorem {A,

,.. .

,

l m' rnk -1 - ' Before proving this result we observe that we need only prove it in the special case where JA, I = k for each i, for if some of the sets A; have fewer than k elements, we can replace each of them by a set of size k containing it (e.g. by the k-set in the same chain of a symmetric chain decomposition), meanwhile preserving the antichain property. The intersection property is also automatically preserved since each set is replaced by one containing it. It follows that we need only prove the following theorem.

Theorem 5.1.2 Let A1, ... , Am be m k-subsets of an n-set S, k _-21n, which are pairwise non-disjoint. Then

70

Intersecting systems and the Erdos-Ko-Rado theorem 1

71

Note that the upper bound on m given by this theorem is best possible; it is attained when the A, are precisely the k-subsets of S

which contain a chosen fixed element of S. Note also that the condition k _- In is essential since if k > In any two k-subsets are non-disjoint, so that m could be as large as (k). The case k > In will be considered in Section 5.3. The first proof of Sperner's theorem in this book involved a study

of all permutations of the elements of S, estimating how often the initial elements of such a permutation can constitute one of the sets Ai. We shall make use of similar ideas in proving Theorem 5.1.2, this

time looking at cyclic orderings of the elements of S. There are (n - 1)! such cyclic permutations. We say that a cyclic permutation contains A if the elements of A are consecutive in the permutation.

Proof of Theorem 5.1.2 (Katona {1,

1972b)

Suppose that S =

... , n}. For each cyclic permutation a of 1, ... , n and each set

Ai, define

1

f (a, Ai) = fO

if a contains Ai otherwise

We count &,J (a, Ai) in two different ways. First,

f(a, Ai) i,a

(

E'a l

where E' is over all cyclic permutations a which contain Ai. Now each a has n k-sets of consecutive elements, and, by symmetry, each

of the (k) k-subsets of S will occur as consecutive elements equally

often when all of the (n -1)! cyclic permutations a are considered. So we must have

n.(n-1)!- n! (k)

so that

Ef(a,Ai)=m.

(k) n!.

,a

(k) However, if we can prove that any one cyclic permutation a can contain at most k of the sets Ai, it will follow that

Jf(a,A;)=2: I 1-_Zk=(n-1)!k. i, a a i (A; in a)

a

(5.2)

72 1 Combinatorics of finite sets

Comparing the two estimates (5.1) and (5.2) we shall then be able to conclude that

m. n!

l

(k_-

k.

(n-1)!,

/

min\k/-\k-1/' It therefore remains to prove the following lemma. Lemma 5.1.3 If a is a cyclic permutation of 1, ... , n and the subsets A. of (1, . . . , n j are as in Theorem 5.1.2, then a can contain at most

k of the sets A. Proof Suppose that A,, appears as consecutive elements x 1 , . . . , xk of a. Then, owing to the intersection property of the sets A;, the only sets of k consecutive elements of a which can constitute other sets are the k - 1 sets A. beginning with x; (2 -- i -- k) and the sets B. ending with x; (1 - i - k - 1). But at most one of A. 1, B; can be in the collection if 1 , i - k - 1, and, since A1= Bk, the collection can therefore contain at most 1 + (k - 1) = k of these sets. The above proof is due to Katona (1972b); the original proof was not so elegant. It will be possible to make further use of the ideas of Katona's proof in obtaining more general results in the following sections. Another proof of the EKR theorem will be given in Chapter

7 where it will be shown that it is an easy consequence of the Kruskal-Katona theorem. It is natural to generalize the intersection property. Suppose we replace the condition A, n A10 0 (i.e. all intersections of two sets are non-empty) by the condition all intersections of h sets are non-empty (where h , 2 is a given integer). For this more general situation the following result holds.

Theorem 5.1.4 (Frankl 1976a) Let d be a collection of k-subsets of (1, ... , n} such that Al n... n Ak * 0 for all choices of

Suppose that kh - n(h - 1). Then IsiI_(k-1/'

Intersecting systems and the Erdos-Ko-Rado theorem 1 73

Clearly the case h = 2 reduces to the EKR theorem. The theorem

in fact follows from the following result in the same way that Theorem 5.1.2 followed from Lemma 5.1.3. Lemma 5.1.5 If a is a cyclic permutation of 1,

. . .

, n and the sets A;

are as in Theorem 5.1.4, then, if kh , (h - 1)n, at most k of the sets A. can be contained in a. 5.2

Generalizations of EKR

Just as the original version of Sperner's theorem can be strengthened to a LYM inequality, so can the EKR theorem be strengthened to a LYM version. The first stage in this strengthening is to adapt Lemma 5.1.3.

Lemma 5.2.1 If {A1, . . . , A,,,} is an antichain of intersecting subsets of S = {1, ... , n}, if 1A;1, 2n for each i, and if a is any cyclic permutation of 1, . . . , n which contains A,,, then a contains at most jA,,I of the sets A;.

Proof. Follow the proof of Lemma 5.1.3. Let A,, appear in a as the consecutive elements x;, ... , xk. Then any other set A. contained in a must begin at one of x2, ... , xk or end at one of x1, ... , xk_1. The sets need not all be the same size, but the antichain property ensures that no two can begin at the same element of a or end at the same element. As in the proof of Lemma 5.1.3, the antichain can contain at most k = l A,iI of these sets.

Theorem 5.2.2 (Bollobas 1973). Let {A,,. . . , A.) be an intersecting antichain of subsets of an n-set S such that IA,I _- Zn for each i. Then

"`

1

( n-1 \1A;1 -1)

<1.

Proof (Greene, Katona, and Kleitman 1976) We imitate the proof of Theorem 5.1.2, but this time define 1

to

if a contains A. otherwise.

74 1 Combinatorics of finite sets

Again we estimate Ei,a f(a, A.) in two different ways. First we have

Ef(a, Ai) _a E 1i i.a

1

(A; in a)

IAiI

Consider the inner sum where a has been fixed. Choose j such that A;

is the smallest of the Ai contained in a. Then, by the lemma, there are at most IA; I terms in the inner sum, each --1/ IA; I. Therefore the inner sum is at most

IA;I.j

=1,

and we have

I f(&,A.)-- E 1 = (n - 1)!. i,a

(5.3)

a

However,

Y, f(a,Ari)=Y, 1 Y 1=E 1n. (n-1)! i IAiI

i.a

i IAiI

a

(Ain a)

(A1) i

as in the proof of Theorem 5.1.2, so that i,a

i

1

(n

nI

IAiI / n l \IAiI/

1)!

i (n -1 l \IAiI - 1)

Comparing (5.3) and (5.4) we obtain

c

1

0

(n-1

IAil -1

Theorem 5.1.1 is clearly a consequence of this result since, if n

IAiI , k , ?n for each i, then (n -1) % (IAil 11) for each i. Note also that Theorem 5.2.2 implies that if IAiI < k for some i, we have strict inequality: m < Corollary 5.2.3 (Bollobas 1973) Let 4 be an antichain of subsets of an n-set S such that A E s4 A' E Si. Then 1 &(A)1-<

Aed

Intersecting systems and the Erdos-Ko-Rado theorem 1 75

where

B(A)=min

n-1 jj

(n-1

t(IAl -11'

n-1 l

2

I41

([n/2]

-1).

Proof We first note that if sad contains any sets of size Zn, these sets must occur in complementary pairs. Form a subcollection Oh of a by putting into 98 all those sets A e s4 with JAI < in, and one set from each complementary pair of sets of size Zn in 4. Then R must be an

intersecting family, for if X, Y E 04 and x fl Y = 0, then X c Y', contradicting the antichain property. Since 9(A) = (IAI -11) for each A E -'V, we can apply Theorem 5.2.2 to -04 to obtain

The remaining members of d are precisely the complements of the sets in 90. Applying Theorem 5.2.2 to their complements, we obtain

Z

1

nJA- 1)i _ 1)

AE.r4_99

-

((n

AE

1

99

(5.6)

0(A)' 1.

The required result now follows by adding (5.5) and (5.6). Finally, note that

0(A)( n-1 ) [n/2]-1 so that 141 n-1

_

1

A

b(A)

--- 2

0

\[n/2]-1) Corollary 5.2.4 (Brace and Daykin 1972) Let si be an antichain of

subsets of an n-set S such that if A, B e 4 then A fl B * 0 and

76 1 Combinatorics of finite sets

AUB#S. Then

IsiI`(,(n -1 l n - 2)/211.

Proof Let si' = {A': A E d}. Then d U d' is an antichain satisfying the conditions of Corollary 5.2.3 so that 1

AesfUja'

9(A)

<,.

Thus, by (5.7), 2I.I

n-1

,

E

1

<2,

([n/2]-it whence Isil

n-1 l

0

(n - 2)/211.

The conditions of Corollary 5.2.4 can be rewritten as follows: for

all A, B Ed, A f1 B * o, AFB' # Q, A' fl B # O, A' fl B' * O, i.e.

all four regions defined by A and B in a Venn diagram are non-empty. The collection si is in this case called 2-independent. More generally, s is k-independent if, for all choices of k sets A,, . . . , A,. in si, all 2' regions of their Venn diagram are non-empty. We conclude this section with a few remarks about alternative paths through the theorems. Instead of proving Theorem 5.2.2 by altering the choice of function f (a, A,), we could derive it from Theorem 5.1.2. This can be achieved by `moving up' all sets of smallest size, replacing them by sets of size one more, and using the following lemma (see Exercise 7.14).

Lemma 5.2.5 Let . ICI --

be a set of k-subsets of an n-set such that

`k - D. Then

/n-11 Iv.

I=Ink1 I4I 1k-11

However, we cannot prove this lemma until we have met the Kruskal-Katona theorem in Chapter 7. The argument in fact enables a LYM inequality to be derived from Theorem 5.1.4 also.

Intersecting systems and the Erdos-Ko-Rado theorem 1 77 5.3

Intersecting antichains with large members

The EKR theorem was concerned with intersecting antichains whose

members are all of size --k -

Zn.

In this section we remove this

restriction on size and show that

is then at most ([n/2]

+ 1). By considering the set of all subsets of size [n /2] + 1 we see that this bound is the best possible. Theorem 5.3.1 (Greene et al. 1976) Let d = (Al, ... , A,n} be an antichain of subsets of an n-set S such that A; fl Al * 0 for all i, j. Then 1

1

l II - 11 n

AE.i

IAI=n2

+E

1

n kI>n/2 IAI)

=1.

Ae.sB

Proof The proof follows that of Theorems 5.1.2 and 5.2.2, but involves yet another choice of the weight function f(a, Ai). Define

n-IA;I+1

if IAJ -- 12n and a contains Ai

IA iI

0

otherwise.

Then

i.a

f(a, A;) _

t

{l Y, f(a, A,)} a

A;ina

n--IAiI+1 Al

i IA;Icn2

=

' 1+ Iiof

Aiina n

-IAI+1 i

IA,I IA,Isn/2

=n! 2

n.

21

i or IA,I>n2A,ina

+ n

n

IAiI)

IA,In2].) +n!

IA,I>n2 (IAiI/

1

(IAr-

IAjI>n2

(IAI)

78 1 Combinatorics of finite sets

However, i,a

a

f(a,A,)}

i

Am n a

n-IA1I+1+ A;ina

IAi I

1. /z

IA

A;ina

}

If we can show that right-hand side of this expression is

n=n.(n-1)!=n! then the required result will follow on comparing the two expressions for El,a f (a, A,). Thus it remains to prove the following lemma.

Lemma 5.3.2 Let the sets Ai be as in Theorem 5.3.1, and let a be any cyclic permutation of the elements of S. Then

2 n-IA1I+1

IiAl

A;ina

A;ina

Proof Suppose first that the first sum is empty. We have then to

prove that the number of sets Ai contained in a is at most n. However, by the antichain property at most one A; can begin at any given element of the cyclic permutation a, so the result is trivial.

Next suppose that there is a set A, of sd contained in a, with IAjI = r -_ 2n. Choose such an Aj with r as small as possible, and without loss of generality suppose that A, = {1, . . . , r} and S = {1, . . . , n}. Any other A, contained in a must consist of consecutive elements of a beginning with some s, 2 , s , r, or ending with some s - 1, 2 , s , r. By the antichain property, at most one A, can begin with any such s, and at most one Al can end with any such s - 1. However, note that if Ai, begins with s and Ail ends with s - 1 then, by the intersection property, at least one of them must contain more than 2n elements. Since r is the minimum size of an A,, the sum of the weights of Ail and Aj2 is therefore at most

1+n-r+ln+l r

r

The total sum of the weights of the sets contained in a is therefore at most

n

r+1+(r-1)n r

r

1=n.

Intersecting systems and the Erdos-Ko-Rado theorem 179 Corollary 5.3.3 (Brace and Daykin 1972; Schonheim 1974b) If Si is an intersecting antichain of subsets of an n-set, then

([n/2]+1). Proof LetA E si. If IAI , Znthen (JAI 1)

n -_ \[n/2] -1) ` \[n/2] + 11'

+ 21 so that (,n,) If IAI > Zn, then IAI -_n [--2

( A[n/2j +n 1 ). The result

J

now follows from Theorem 5.3.1.

This result is a special case of a more general result due to Milner:

if si is an antichain such that IA; fl A; I , t for all A., A; E Si, then n

The case t = 0 is just S erner's theorem, and

] p + t + 1)])' the case t = 1 is the above corollary. We prove Milner's theorem as Theorem 5.5.1.

IAI

-_ ([12(n

5.4 A probability application of EKR

Suppose that Y1, ... , Y are independent variables, each taking the value unity with probability p and the value zero with probability 1- p, where p , 2. We use EKR to prove the following theorem. Theorem 5.4.1 (Liggett 1977) Let al, ... , a be non-negative real numbers such that E, a; = 1. Then, with the Y, as above,

P(aIY1+...

%2)%p.

Proof Suppose first of all that no partial sum of the a; is J. Let Mk denote the number of subsets A of {1, ... , n} of size k such that E;EA a; > 2. Then clearly Mk +

n k).

Note that, for any fixed k,

no two sets A contributing to Mk can be disjoint, for otherwise the sum of the a; over all i in their union would be greater than unity. Thus by the EKR theorem we must have

Mk_(k-1) if k,2n.

80 1

Combinatorics of finite sets

Now

P (i=1 aiY zMkpk(1-p)"-k. k=1 However, on writing k = h + 1, k

_p)"-k =P

\k -1/Pk(1

n-l n

1)Pk(1-P)"-1-h

( h

=P(P+1-p)"-I =p,

so that we have " 1) P(rY%2

n

n

_P=>Pk(1_P)n-k(Mk-\k k=1 _

pk(1 _ p)n-k(Mk-\n

k

k

+lE Pn-i(1_pyM"-;=

pk(1

k

1

-j - J) 1

_ p)n-k(Mk-(-111 n-1

+ I pn-i(1- py ((n) i
= kE12 pk(1- p)n-k r

_ M - \n j 1//

Mk_ (n

+,E Pn-i(1-py\\nn kE

n

(n

n

- 11)1 -

1

11

-M'1

(pk(1-p)n-k-pn-k(i-p)k)(Mk-\k-111

We want to show that the expression on the right is Mk

O. Since

\k -11 we therefore have to show that pk(i - p)n-k , pn-k(1- p)k for all k
However, this is true since p

21, i.e. p % 1- p.

Finally, if a partial sum of the a; is equal to 2, alter the a; fractionally to avoid this and then apply a limiting procedure.

Intersecting systems and the Erdos-Ko-Rado theorem 1 5.5

81

Theorems of Milner and Katona

At the end of Section 5.3 we proved that if .i is an antichain of subsets of an n-set such that IA; f A;I -- 1 for all A;, A; E 4, then Is' I

-- (1I1'

This result is a special case of a more general result

due to Milner, and it is our aim in this section to state and prove Milner's result.

Theorem 5.5.1 (Milner 1968) If 4 is an antichain of subsets of an n-set S such that IA, fl A; I -- k for all A;, A; E 4, then

Jai

n

rn+k+ll L

2

J

In the proof of this result we shall make use of a valuable result of Katona concerning the shadow of a collection of k-sets. Recall that the shadow of a collection of k-subsets of an n-set S is AR

for some BE-64}.

In proving Sperner's theorem we showed very simply that k ,0

+111.

In Chapter 7 we shall study the size of A in much greater depth, but at this stage we are going to study I A 931 in the special case where R is

an intersecting family. It turns out that in this case we always have IA9 1=1.1B, 1. The proof of this fact will make use of a shift operator which will shift the sets away from a particular element of S.

Lemma 5.5.2 Let s4 be an intersecting collection of k-subsets of S = {1, ... , n} and, for 1 <j _ n, define {A

S;(A)=(A-{1})Uj)

if1EA,jfA,(A-{1})Uj)fsi otherwise

for each A E s4. Then S;(-4) = {S;(A):A E A}

is also an intersecting collection of k-subsets of S, and

IiS,(4)I -- l 4I.

82

1

Combinatorics of finite sets

Proof We first check that S;(A) fl S;(B) * 0 whenever A, B E . . The only case which is non-trivial is when S,(A) _ (A - (11) U {j}, S,(B) = B, 1 E B, j I B. However, in this case (B - { 11) U { j } and A are both in .sB and so must have an element x in common; we must have x:* 1

andx*j, soxES,(A) andxeS,(B). We next check that IASiI -- IAS1(si)1. If X E AS; (a) but X 4t Asi then X c (A - { 1 }) U { j ) for some A, X A, so that j E X, 1 O X, Y = (X - {j}) U {1} E 04 and YI AS;(si). We thus have

XEAS;(S:Q)-e,i=> Y=(X-{j))U(1)Eesaf-OS;(si) so that lose - AS1(si)I - ILS;(.i) - As1l, i.e. Iosdl - IAS,(,i)I. Theorem 5.5.3 (Katona 1964) If si _ {(A,, ... ing family of k-subsets of S = {1, ... , n}, then

, Am} is an intersect-

IosiI-IS41. Proof If k , 2(n + 1) the result is immediate from the symmetric chain decomposition of the set of subsets of S; so now suppose k , First consider k = 2. If m = 1, IOsil = 2 > Isil, so now suppose m -- 2. ?n.

If no member of S is in more than two members of s/i, and if A1= {a, b}, A2 = {a, c} E Si, the only other possibility for another set in si is {b, c); therefore IA.s'I = 3 -- I41. If there is an element a E S occurring in A1, A2, A3, then a must be in every set A, (for if a A;, Al needs three members to enable it to intersect each of A,, A2, A3); therefore IA, I =1 + ICI > Isil. Suppose now that k > 2. If we repeatedly apply the shift operators S;, j = 2, . . . , n, to sQ, the number of sets containing 1 must decrease

and after a finite number of applications the shifts must therefore cease to make any change. We therefore eventually obtain a new collection s/i° of k-subsets of S, with S,(s/i°) = s4° for each j _- 2, Isi°l = Isil, and IO.sll°l , IOs/I. The theorem will be proved for si if it can be proved for .sO. Therefore we can now assume that si has the property that S,(s/i) = si for each j - 2. Thus we can suppose that, for

all AEs/iand j-2, 1EA,jfA=>(A-{1})U{j}E.sB.

(5.8)

Suppose that the sets A; are now relabelled so that 1 E A; if and only if

i , m°i and let B; = A. - {1} for each i , m°. Then we claim that the B; are pairwise non-disjoint. Since I A; U A,, I -- 2k - 1 < n, there exists some w r: S, w f A; U A,,. Thus on applying (5.8) with j = w, we find that B,, U {w} E .s/i, so that by the intersection property

IB;fB,,I=IB;l(B,,U{w})I= IA; fl(B.,U{w})I-1.

Intersecting systems and the Erdos-Ko-Rado theorem 1 83

Thus the sets B,, i < mo, form an intersecting family of (k - 1)-subsets of the (n -1)-set 12, ... , n). If we put 0-4 = {B,), we therefore have by induction = mo. Finally, the collection V = {A, : i > mo) is also an intersecting family I

I

I

of subsets of 12, ... , n), so by induction we have

I A'6 I = I T I =

m - mo. Adjoining the element 1 to the sets in A,64 we thus have

IoM>IA-041+IA19I>CAI+I19i=Is4I. We now use Katona's result to prove Milner's theorem.

Proof of Theorem 5.5.1 Let t = [2(n + k + 1)]. We are given an antichain s4 of subsets A; of an n-set satisfying IA,, fl A,2I , k, and we

have to prove that I.

I , (n). t

If IA,I = t for all i, then the result is trivial. Next consider the case when I A, I , t for each i, but IA, I < t for at least one i. Let j be the smallest size of an A,, j < t, and suppose, after relabelling if -- t. The idea of the necessary, that IA1I = ... = IA,I =j < IAh+1 ,

proof is now to replace A1, . . . , Ah by sets of size j + 1 in their shade. Now the sets in the shade of A1, . . . , A,, are the complements of the sets in the shadow of X = {A;, ... , Ah}. The condition IA,, fl A,2I -- k implies that IA,, U A,2I -- 2j - k, so that IA;; fl A72I % n - 2j + k, where j + 1= 2(n + k + 1), i.e. n - 2j + k , 1. So by Katona's theorem (Theorem 5.5.3) IACI -- I YCI = h. Thus we can replace A1i ... , Ah by h sets of size j + 1, and clearly this replacement does not destroy the

antichain or the intersection property. If we keep repeating the process we eventually obtain 1,41 sets of size t, so that IdI n t

as required.

Finally, consider the case when some of the sets A. have size greater than t. As in the previous case, replace all those sets of size less than t by an equal number of sets of size t. Then replace each set of size greater than t by the set of size t in the same chain in any chosen symmetric chain decomposition. (Since t > Zn, this is possible.) (The)resulting collection of t-sets has I.s4I members, so again 1,41 t

5.6 Some results related to the EKR theorem

We consider briefly a few topics which are variations on the EKR

84 1 Combinatorics of finite sets

theme. If the condition A, n A, * 0, i.e. IA, n A,I ;1, is replaced by IA ; fl A; I , t, then with each IA! , k, we might expect to be able to prove that Is'l -- (k

t) This was indeed proved for sufficiently large

n by Erdos, Ko, and Rado in their 1961 paper, but it is not necessarily true for small values of n. More recently, Frankl (1976b) and Wilson

(1984) have between them shown that the result is true for n 31 no(k, t) where no(k, t) = (t + 1)(k - t + 1). Another possible direction in which to consider extending the EKR theorem is the following. Suppose that si is a collection of subsets of an n-set such that IAI -- Zn for each A E si, and with the property that si does not contain t + 1 simultaneously disjoint pairs of subsets. How

large can Is4l be? Clements (1976) proved the following result, a proof of which is outlined as Exercise 7.12.

Theorem 5.6.1 (Clements 1976) Let si be a collection of subsets of an n-set S, n even, where IAI , Zn for all A E A and where s$ does not possess t + 1 simultaneously disjoint pairs of subsets. Then -(n-11+t.

IS4I

A number of authors have obtained analogues of the EKR theorem for structures other than subsets of a set. Hsieh (1975) obtained an analogue for subspaces of a finite vector space, Livingston (1979) for ordered sets, Frankl and Furedi (1980) for integer sequences, Moon

(1982) for Hamming schemes, Rands (1982) for t-designs, and Stanton (1980) for Chevalley groups. The more obvious problem of analogues for multisets will be discussed in Chapter 10. Finally, we remark that equality can occur in Theorem 5.1.2 only

when Al n ... n Am 0 0. An upper bound for m in the case Al n... n Am = 0 is given by the following theorem. Theorem 5.6.2 (Hilton and Milner 1967) Let s4 = {A1i ... , Am} be an intersecting collection of k-subsets of an n-set, k--'2n, such that

Aln... nAm=Q. Then

IsiI-(k-1)1(n kk

1

1).

0

The proof uses a shifting argument similar to that in the original proof of the EKR theorem.

Intersecting systems and the Erdos-Ko-Rado theorem 1 85

Exercises 5 5.1

Let A1,

. . .

,

Am be subsets of an n-set S such that if A, fl A; _

0 then A; U A; = S.

Show that m--2n-'+ ([('/2])

5.2

Give an example of an antichain s4 for which equality holds in Corollary 5.2.4.

5.3

Prove Corollary 5.2.4 directly from EKR and Sperner, taking the cases of n even and n odd separately. Let 1-- g -- h -_ n, g + h -- n, and suppose that Al, ... , Am are

5.4

subsets of an n-set S such that g , jA;J , h for each i and A. fl A; 0 0 for all i, j. Use EKR to show that m En

n-1

I-8 (i-1).

5.5

(Hilton 1974)

Let f be a non-negative real function defined on the nonnegative integers. Show that if si _ {Al, . . . , Am} is an intersecting antichain of subsets of {1, . . . , n} with 1A11, in for each i, then Em lfflA, 1)

omaax

f(k)\k (Greene et al. 1976)

5.6 5.7

Use Lemma 5.2.5 and Theorem 5.1.2 to prove Theorem 5.2.2. Verify that if 4 satisfies the conditions of Corollary 5.2.4 then the collection obtained from s4 by choosing any one set in 4 and replacing it by its complement also satisfies the conditions.

Hence obtain an alternative proof of 5.2.4 via Milner's theorem. 5.8

(i) By applying EKR to the sets C - A1, show that if C is a k-set and Al,_ , Am are h subsets of C with no i, j such that A; U A; = C, then, if k = 2h, m `

h kl

k k

\hl

(ii) Suppose now that 4 is a collection of subsets of an n-set containing no two sets whose union is also in 4. Suppose also that h ; in. By considering, for each k, h , k , 2h, and each set C of size k in s4, the number of saturated chains between ranks

h and 2h through C and containing no smaller set in 4, show

86 1 Combinatorics of finite sets

that

2h Pk h 2----l (l (n) k k

k-h

5.9

where Pk denotes the number of sets in si of size k. (This will be used in Chapter 12) (Kleitman 1976b) The set of all q-subsets, where q = [2(n + k + 1)], is an example of a collection ,sQ which satisfies the conditions of Milner's theorem and which gives equality: I.s6tl = (n). In the case when q n + k is odd, give another example which gives equality.

6 Ideals and a lemma of Kleitman

6.1

Kleitman's lemma

In this chapter we shall consider collections q of subsets of an n-set S with the property

AEI, BcA=> BE0). We call such a l an ideal of S or a downset or a simplicial complex. The terminology `downset' conveys the idea very clearly, and leads naturally to the corresponding idea of an upset: Oil is an upset or filter if

AE°1l,AcB=> BE°ll. Clearly, °ll is an upset if and only if °ll' = {A': A e °ll} is a downset. There is a simple connection between upsets and antichains. If Si is an antichain of subsets of S, then

si+={XcS:XDAfor some AEs4} is an upset, in which the members of si are the minimal members of 4', and similarly

sB-={XcS:Xc- Aforsome AES9} is a downset in which the members of si are the maximal members of saI-. Conversely, if °ll is an upset, then the set of minimal members of °ll is an antichain. One of the most useful results concerning ideals is due to Kleitman (1966); appropriately, it appeared in the first issue of the Journal of Combinatorial Theory. Theorem 6.1.1 (Kleitman's lemma) downset on S, where ISI = n, then

If °ll is an upset on S and 0 is a

IOU I.1.91,2"j°lln21.

There is a natural generalization to divisors of a number (Anderson 1976). An upset of divisors of m is a set 01l of divisors of m such that if

aE°Iland aIbI mthen bEV. 87

88 1

Combinatorics of finite sets

Theorem 6.1.2

is a downset of divisors of

If ' L is an upset and then r

IqtI.1-91-II(1+k,).1°un.9I.

Since the proof of Theorem 6.1.2 is not much more difficult than that of Theorem 6.1.1, we now prove the more general result. In the next section we shall see that they are in fact consequences of a much more general result due to Ahlswede and Daykin (1978).

Proof of Theorem 6.1.2 We use induction on n = E; k;. If n = 1 or 2

the result is trivial, so we proceed to the induction step. Writing pr = p and k, = s, we have m = m'p = m"p' where p does not divide m". Partition it and 2 as follows: all =gtpU°IlP

2=!@PU!@P

where all,, and P consist precisely of those members of °11 and 2 then respectively which are divisible by p'. Now if hp' E h, hp, .

. .

,

hp'-' are

all in 2p; therefore 12P 1 --s 1 2P I. Similarly

IRiPI `s IOuP1 Thus m'

(S I°PI - (iPI)(I2 I -s 1!@,.1) --0 whence

S IOPI

I-,5)PI +S 1°ipI . I - ' D pI -- I°itPI .

1!@,,

I

PI

I9P1.

(6.1)

Using the induction hypothesis for divisors of m', we have IMPna,pI r(m,)
I9PI

where i(m') = (1 + k1) ... (1 + kr_1)k,.

Now let °IP, P denote the sets of divisors of m" obtained by dividing each member of °llp, OP by p'. Then, since 1% n 9P 1= 'Up" n 2'n

the induction hypothesis applied to m" gives IMPnop l

Ol1I

IsPI=IOil I. I2PI.

(6.3)

Ideals and a lemma of Kleitman 189

Thus

IIn2Ir(m)=I°upn!@pIx(m)+°itpn!@plr(m)

=(s+1)lIpnCpl

r(m')

-(s+1) by (6.2) and (6.3) I9pl+I0UPI

IVA Opl

I_qPI +S 1q'p1 I2PI+S

N PI

I41 + II&PI

N PI - 0,61 + IOUPI

PpI by (6.1)

= (I upI + Iq'PI)(I@'I + i.,@pl)

0

= IOLI 0I

As described in the exercises, there are other equivalent forms of Theorem 6.1.1. For example, we have the following theorem. Theorem 6.1.3 If sz and a are ideals of subsets of the n-set S, then Isil.I6ol_2"Is:¢n9BI-

6.1.1 Applications of Kleitman's lemma

We now give two simple applications of Theorem 6.1.1. The first of these is to give a very simple proof of a result which had been proved in several different ways by various authors (Daykin and Lovdsz 1976, Schonheim 1974a, Seymour 1973, Hilton 1976). The following proof is due to Anderson (1976) and Kleitman (in Greene and Kleitman 1978).

Theorem 6.1.4 If .si is a collection of subsets of an n-set S such that

ifA,BesatthenAnB*Q andAUB*S, then

I4I1__2n-2.

Proof Define °1 to be the collection of subsets of S consisting of all the sets in d and all their supersets, and define to be the collection

of subsets consisting of all sets in s1 and their subsets. Then s:Q q ;,W n 2, and Kleitman's lemma yields IsiI2"-I/II_qI.

90 ( Combinatorics of finite sets

However, V is a collection of intersecting subsets, so, by Theorem 1.1.1, IVI , 2- -1. Similarly (Exercise 1.1), Ill , 2n-1. Thus Is4I 2n ` 2n-1.2n-1

as required.

This result is clearly best possible, since there are . . . , n} which contain 1 but do not contain n.

2n-2 subsets

of

{1,

Kleitman originally proved his lemma while investigating the following problem. If si is an intersecting family of subsets of an n-set

S, then, as we recalled above, Isil ,

So suppose now that dl, ... , .s/k are disjoint intersecting families of subsets of S; then 2n-1

how large can I U d l be?

Theorem 6.1.5 (Kleitman 1966) If d,, . . . , silk are disjoint intersecting families of subsets of an n-set S, then IU Sill 2n-2"-k.

Proof The case k =1 reduces to Theorem 1.1.1, so we proceed by induction on k. As explained in Theorem 1.1.1, s4k can be extended to a collection sik of 2n-1 intersecting subsets. Let S = {X: X 0 dk}; then T is a downset. Consider next = Uk 1 sQ;. If 90 is not an upset, we can adjoin any missing superset to an appropriate saf; to obtain an upset 2 which contains B and which, by the induction hypothesis, has at most 2" - 2n-k+1 members. Applying Kleitman's inequality to 19 and 9, we obtain

IikUdil=lRus¢kl,l99 ncl+l&ikl,l-nWl+2"-1

1

2n

ICI . I(6I + 2'

-1

<

2n .

(2n - 2n-k+1) 2n-1 +

2n-1

=2n-1__2n-k+2n-1=2n-2n-k

Again, the bound obtained is best possible: if S = {1, ... , n}, take A to be the collection of all subsets of S containing i but none of 1, ... J-1. -1. Then 14,1=2 n -'and the bound is attained.

6.2 The Ahlswede-Daykin inequality

Since Kleitman proved his inequality in 1967, a succession of generalizations of his result have been obtained. In this section we

Ideals and a lemma of Kleitman

1

91

choose one fairly general theorem and deduce from it a number of inequalities, including Kleitman's, as special cases. In the exposition we shall write

(S) for the set of subsets of S, and, for any collections

4, R of subsets of S, define

.siv 58={EcS:E=AUB for someAEsf andBER} .s4A 4={EcS:E=Af1B forsomeAE. and B E.04). Theorem 6.2.1 (Ahlswede and Daykin 1978) If a,

/3,

(p, b are

non-negative functions defined on -6%(S) and satisfying

a(A)P(B) -- qv(A U B)6(A fl B)

(6.4)

for all A, B c S, and if si, R are any two collections of subsets of S, then

(. a(A))(Z /3(B)) -- (EZ AEJR

BEa

Er4v

99

(E))(FY,

.5 (F)).

(6.5)

Ec4A

Note If we use the abbreviation

f(sfI) =AES4 E f(A) we can rewrite (6.5) more compactly as

a(Si)/(94)

rp(sB v a)S(.si n a).

(6.6)

Proof of Theorem 6.2.1 We use induction on n = ISI, and first consider the case n = 1, where we can take S = {1}. Condition (6.4) becomes a(0)0(0)

rp(o)b(o)

a(o)f(S)

(P(S)S(o)

a(SWO)

q9(S)6(0)

a(S)f3(S)

-V(S)b(S).

(6.7)

If si or OA has only one member then (6.5) follows immediately. For example, if . = {f} and R = {Q, S} then the left-hand side of (6.5) is a(o){fl(O) + /3(S)) while the right-hand side is {qp(O) + gq(S))b(Q) so that (6.7) gives the required result immediately. Now suppose that 4 = 99 = {0, S}. In this case we have to prove that

{a(P) + &(S)) {0(0) +/3(S)} -- 0)(0) + qp(S)){8(P) + 8(S)}. (6.8)

92 1

Combinatorics of finite sets

If any of a(0), f3(0), q,(0), 6(0) is zero, (6.8) is immediate; therefore, after suitable scalar multiplication, we can suppose now that a(O) _ /3(0) = cp(O) = 6(0) = 1. The conditions (6.7) become /3(S)

a(S),6(S) -- 99(S)6(S)

a(S) -- q'(S),

cp(s),

(6.7')

and (6.8) becomes

(6.8')

{1 + C(S)}{, + /3(S)} -- {1 + qq(S)}{1 + 6(S)}.

Now (6.8') is certainly true if p(S) = 0, so we can assume now that cp(S) > 0. Since 6(S) -- a(S)(3(S)1cp(S), (6.8') will follow if we can prove that a(9 S(S)

{1 + a(S)}{1 + /3(S)) -- {1 + cp(S)}{cp 1+

a(S) + /B(S) -- cp(S) +

I

a(S)S(S)

(6.9)

q? (S)

However, (6.7') gives {cp(S) - a(S)}{q4(S) - S(S)} , 0, which yields (6.9). Therefore the proof of the case n =1 is complete. We now consider the induction step. Assuming that the theorem is true for sets of size m, we consider S with ISI = m + 1. Let

S={1,...,m+1) and define T={1,...,m} and W={m+1}.

In an attempt to use the induction hypothesis for m-sets, we write any A c S as A = A' U A" where A' = A fl T and A" = A fl W. Define a,, #,, q71, 61 on °,P(T) by

a,(C) =AE.I a(A) A'=C

(C) _

fl,(C) _Be-% E /3(B) B'=C

6,(C) _ 1 6(F).

Es gv

Fesdn95

E'=C

F'=C

Then

a(s/2) = E a(A) = E E a(A) _ Y, al(C) = al(P(T)) CQT Aes4

AeE4

A'=C

CQT

and similarly

P(R) = /,(,'(T)),

qq(s/ v 1V) = 991(9(T)),

6(sd A S8) = 6,(,9(T)).

Suppose for the moment that we can prove that al(C)PI(D) -- qq,(C U D)61(C fl D)

(6.10)

Ideals and a lemma of Kleitman

93

for all C, D c T. We can then apply the induction hypothesis to obtain

_(p(.4 vO8)6(SiA6B) as required. Therefore it only remains to prove (6.10).

Choose any C, D c T and let E = C U D, F = C fl D. Define a2, $2, P2, 62 on 1(W) = {Q, {m + 1}} by

a2(R)-10(RUC) if RUCEM otherwise

102(8) _ g22(R) _

(O (R U D)

if RUDE otherwise

{0(R U E) if RUE E.1v_1V

62(R) = l0

otherwise (R

U F) if RU FE otherwise.

A

Then it is easy to check that

al(C) =Ae.d Z a(A) = RcW E ZAea(A) _ E a2(R) = a2(9'(W )) d RcW A'=C

A'=C A"=R

and that similarly 6,(D) = SZ(J'(W)), qp,(E) = q?2(9'(W)), 81(F) 62(J'(W)). If it can be shown that a2(R)f32(Q) = q)2(R U Q)62(R fl Q) (6.11) for all R, Q c W, it will then follow, on using the induction hypothesis

for n = 1, that al(C)$1(D) = a2(9'(W))f32(P(W )) q,2(9'(W))b2(9'(W )) = gg,(E)61(F),

thus proving (6.10). We complete the proof by proving (6.11). Note that the left-hand side of (6.11) is zero unless R U C E si and Q U D E A, in which case it equals a(R U C)/3(Q U D). Now (R U C) U (Q U D) = (R U Q) U E

so that (RUQ)UEESfvP, and (RUC)fl(QUD)=(RfQ)UF so that (R fl Q) U F E Si A A. Therefore the right-hand side of (6.11)

94 1

Combinatorics of finite sets

is then cp(R U Q U E)b((R fl Q) U F). However, (6.4) gives a(R U C)fl(Q U D) , rp(R U Q U E)6((R n Q) U F), so that (6.11) is true.

0 Ahlswede and Daykin have pointed out that many previously proved inequalities are contained in Theorem 6.2.1 as special cases. If

we take a = ig = 9' = S =1 we immediately obtain the following corollary.

Corollary 6.2.2 (Daykin 1977) subsets of S, then I.t1.1,A1--I

If si, 90 are any two collections of V001.JAAOhl.

Corollary 6.2.3 Kleitman's lemma 6.1.1.

Proof See Exercise 6.8. Corollary 6.2.4 (Marica and Schonheim 1969) Let si - 90 = {A - B : A E si, B E 64). Then, for any collection s4 of subsets of S,

Proof Let sV = {A' : A E si}. For any collections si, A of subsets of S we have, by Corollary 6.2.2,

ISi v@'I.J4Ao'I = 1(s4 V

.

Id A Oh'l

=!4'ARl.j,4A@'I [email protected]. Now take R = 4 to obtain I si - si I .. 1.

-ql

Before leaving the Marica-Schonheim result we note that various generalizations of it have been obtained. For example, Ahlswede and Daykin (1979) have shown that, if A,,... , Am are distinct sets and if B,, . . . , B are sets such that each A, contains at least one B; as a subset, there are at least m distinct differences A; - B, (see Exercise

6.16). Also, Daykin and Lovasz (1976) have shown that any non-

trivial boolean function takes at least m distinct values when evaluated over m distinct sets.

Ideals and a lemma of Kleitman 1 95

For our next corollary, take a(A) = ,u(A)f (A), #(A) =,u (A)g(A), cg(A) = f(A), and 6(A) = u(A)f(A)g(A), where f,g are decreasing functions (f is decreasing if f (A) --f(B) whenever A c B) and where u is a log-supermodular function, i.e.

u(A)u(B) -- p(A U B)u(A fl B) for all A, B. We then have

a(A)P(B) = u(A)u(B)f(A)g(B) u(AUB),u(Af1B)f(Af1B)g(Afl B)

=(p(AUB)6(Af1B) so that, by Theorem 6.2.1,

t Z u(A)f(A)j{ Y, u(B)g(B)) AES!

BE96

{E

SBv96 EE

u(E)j{ FE SA/.9 u(F)F)g(F)}. (6.12)

We thus have the following Corollary.

Corollary 6.2.5 (The FKG inequality of Fortuin, Kasteleyn, and Ginibre (1971)) If u is a non-negative log-supermodular function

and if f and g are both decreasing (or both increasing) functions defined on _13A(S), then

I i(E)j{ ZFSS u(F)f(F)g(F)}. { E u(A)f(A)I{ BSS E u(B)g(B)} = { EGS ASS

We choose here s? = °rli = P (S) in (6.12), and we also note that if f, g are increasing then the proof easily adapts (interchange cp and S). If, instead, we take si, 96 in (6.12) to be downsets and f,g to be their characteristic functions,

XE4 fm _(10 ifotherwise

g(X)

IifXE-04 0 otherwise.

Then f,g are clearly decreasing and we obtain the following corollary. Corollary 6.2.6 Let 4, °O8 be ideals of subsets of an n-set S. Then

{AE u(A)I{ Z u(B)} --- { E u(E)}{FEY, rsi

. p(F)j.

11

96 1

Combinatorics of finite sets

The special case µ = 1 yields Kleitman's lemma in the form of Theorem 6.1.3 since, if .s and a are ideals, Si A .68 = a fl go.

We now consider some consequences of Theorem 6.2.1 for an for each

sequences. A sequence {a,,} is log-convex if

n>1. and Corollary 6.2.7 (Seymour and Welsh 1975) Let decreasing sequences of non-negative real numbers and let

be be

log-convex. Then

(ak)(akbkck) (akbk)(akck). Proof Let dk = akl (k) and let S be any n-set. For A c S put µ(A) = dlA1, f(A) = bjA1, g(A) = CIAI. Then f and g are decreasing functions and so, if we can show that y is log-supermodular, the FKG inequality will yield

(IAQS dIAlbIAI)(ASS dlAicl`'1) -_ (A-S d1A1)( AQS dIAlblAIcIAI) (k)dkckl

{G) dkbkJ f 1

f k=O

()}{

(n)d, k=o

k

bkckl

\k

akbk) (I akck) k=O ak) \kI akbkck) as required. To prove that µ is log-supermodular we have to show that aIAl

AI)

alel` B{)

alAUBl

alAnBL

(JA U B1) (

Now IAl + IB I _ IA U B I + IA fl B I, so by the log-concavity of the

binomial coefficients we have (cf. Corollary 4.1.4)

(,AI)(IBI) __

(IA U BI)(!A fl BI)

However, alAlaIBl = alAualalAnBI, so the required result follows.

Ideals and a lemma of Kleitman 1 97

As a special case we have the following corollary.

Corollary 6.2.8 (Chebyshev) If and quences of non-negative real numbers, then (n

iti bi )(n-

are decreasing se-

n bic). n i.,

c')

Proof Take a; = 1 in Corollary 6.2.7.

11

It should be clear by now what a powerful result Theorem 6.2.1 is. Further, although all the results in this section have been presented in the context of the lattice of subsets of a set, they could be presented in the more general setting of a finite distributive lattice L, with U

and fl replaced by v and n . We then have, for example, the following form of the FKG inequality which will be used in the next section.

Theorem 6.2.9 (FKG) If (S, <) is a finite distributive lattice and if

f, g are both increasing (or both decreasing) on S and µ is a non-negative function on S such that 14 (x)µ (y) _ µ(x v y)µ (x A y) for all x,y E S, then

fEf(X)tl(X)ItEg(X)IA(X))--IE[t(X)Ify,il(X)f(X)g(X)IXEs

XEs

XES

XES

We remark briefly that a partially ordered set is a lattice if every pair of elements x,y have a join x v y (with x, x v y and y; x v y)

and a meet x A y (with x A y-- x and x A y; y). A lattice

is

distributive if x A (y v z) = (x A y) v (x A z) for all x, y, z.

6.3

Applications of the FKG inequality to probability theory

During the past few years a number of authors have used the FKG inequality to obtain results of a probabilistic nature. This is hardly surprising since the inequality itself was discovered in an investigation into correlation problems in statistical mechanics. We can see to begin

with that Kleitman's inequality can be expressed in terms of probabilities:

n-91

98

1

Combinatorics of finite sets

IqtI 2"

ICI

Iou n !@I

2"

2"

P(xEcU). P(xE9)%P(xEv ng) i.e.

P(xE2)-P(xE!@IxE'4L) where the conditional probability P(A I B), the probability of A occurring given that B occurs, is defined by P(A I B) =

P(A n B) P(B)

Kleitman's lemma can therefore be interpreted as saying that the condition x E Oh tends to pull x `up', making it less likely for x to be in the downset !D. In this section we attempt to give the flavour of these applications

by presenting proofs of two different results, first a result originally proved by Graham, Yao, and Yao (1980) and second a proof of a conjecture known as the XYZ conjecture. Suppose we are given two chains a, < a2 < ... < am and b, < ... < b,,, i.e. we are given two linear orders. Let C =A U B where A = {a;)

and B = {b;}. A linear extension .t of C is a mapping of C onto {1, ... , m + n} such that x
preserves the separate rankings in A and B. The set L of linear extensions of C has

(m+n m

)

members, all of which can be considered

initially to be equally likely. Now suppose that P,, P2, P3 are subsets of the set of rankings of the members of C, each defined by a number

of conditions of the form a; < b;. We then state the following theorem. Theorem 6.3.1 (Graham et al. 1980) With P1, P2, P3, L as above,

P(AEPi IAeLnP2nP3).P( eP,IA,ELnP2). This result asserts that the extra information given by P3, that some

a;'s are ranked before some b;'s, strengthens the chances of the conditions of P, occurring; intuitively, P3 reinforces the view that all the a, are in general of smaller rank than the b..

Ideals and a lemma of Kleitman

99

Before proceeding to the proof, we note that the result is not in general true if the a, and b, are not totally ordered.

Example 6.3.1 (Graham) Assume nothing about the orderings of a, and a2, and of b, and b2; therefore L is the set of all rankings of the

four elements a,, a2, b,, b2. Let P, be defined by a,< b,, P2 by a2 < b,, and P3 by a2 < b2. Then we can check that there are eight rankings satisfying P2 and P3, of which five satisfy P1, so that

P(AePi IAELf1P2f P3)=8. However, there are 12 rankings with a2 < b,, of which eight satisfy P,, so that

P(A EPI1),ELf1P2)=3>8. Proof of Theorem 6.3.1 (Shepp 1980) Let T denote the set of all (

m+n ) m-subsets of {1, ... , m + n}. If X= x, <x2 <... M

xm}

and Y= f y,
X v Y = {max(x,, y,), ... , max(xm, ym)};

it is easy to check that these are both in T. Since, for any three real numbers a, b, c,

min(a, max(b, c)) = max(min(a, b), min(a, c)), it follows that X A (Y V Z) = (X A Y) V (X A Z) for all X, Y, Z E T. Thus (T, <) is a distributive lattice. We now exhibit a one-one correspondence between members of L and members of T. Given X E T, say X = {x, < x2 < ... < xm }, we can construct a member 6(X) of L as follows. Place a; in position x;

for each i, and then place b; in the jth remaining position. For example, if m = 4, n = 3, and X = (1, 4, 5, 71 we obtain the ranking a,b2b2a2a3b3a4 since, for example, x2 = 4 requires that a2 is in the fourth position. Conversely, any linear extension gives a unique X. Next observe that if a; is in position x; there are x, - i `gaps' to be filled by b;s before a;, so that a; will be ranked before b, if and only if j > x; - i, i.e. if x; _ i + j - 1. Thus each condition in the definitions of P,, P2, P3 corresponds to a condition of the form x; _- k;,;.

100 1

Combinatorics of finite sets

Thus, corresponding to P,, P2, and P3 we obtain subsets P of T defined by a number of inequalities of the form x; , k,,;. Define functions p, f, g on T by 1

ifXEP2

'u (X) = 0 otherwise f(X)_11

ifXEP,

0

otherwise

1

ifXEP3

g(X) = {0 otherwise.

Then f, g are decreasing functions, for if X < Y and f (Y) =1, then YEP,, and hence X EP,, since x; < y, and y, < k,,; imply x; k,,;. To show that y satisfies U(X)14(Y) -- µ(X A Y)µ(X V Y) for each X, Y, note first that it is trivial if µ(X) = 0 or µ(Y) = 0, so that we can now suppose that µ(X) = µ(Y) = 1, i.e. X, Y E P2. Now each condition on X, Y is of the form x; k, y, -- k, but then min(x,, y,) -- k and max(xf, y) -- k, that so X A Y and X v Y are also in P2 and so

µ(X AY)=µ(XvY)=1.

The conditions of Theorem 6.2.9 have now been shown to be satisfied so we can apply that theorem to obtain

IT nP,nPPIIT nP2nP3I\ ITnP2I JTnP1nP2nP3I (m + n)!

(m + n)!

(m + n)! *

(m + n)!

i.e.

P(P,nP2nL)P(P2nP3nL)-- P(P2nL)P(P,nP2nP3nL) which is the required result. The second application of the FKG inequality is also due to Shepp.

Let x...... x be random variables about which some partial information is known in the form of a set G of inequalities x; < x,. It was conjectured by Rival and Sands (see Shepp 1982) that P(x,<x21 G) -- P(x,<x21 G, x1<x3)

since, if it is known that x, <x3, there is all the more reason to expect that x, <x2. This intuitively reasonable conjecture was known as the XYZ conjecture, and it became a theorem 2 years later when Shepp applied the FKG inequality to obtain a proof. It was pointed out by Shepp that an analogous conjecture P(x, < x2 < X41 G) -- P(x, < x2 < X41 G, x, < x3 < x4)

Ideals and a lemma of Kleitman

1

101

is in general false. To see this, take n = 6 and G = {x2 <x5 <x6 <x3i X1 <x4}; then

P(x1<x2<x4I G) =

4 15

>1=P(x1<x2<x4'G,x1<x3<x4). 4

Theorem 6.3.2 (Shepp 1982) Let G be a set of inequalities of the form x, < -.; relating to n variables x1, ... , xn. Then

P(x1<x2I G)-P(x1<x2I G,x1<x3) or, equivalently,

P(x1 <x2, G)P(xl <x3, G) - P(G)P(x1 <x2, x1 <x3, G).

Proof Let S denote the set of ordered n-tuples x = (x1i ... , xn) . . , N). N is a large integer which will later tend to infinity. Define an order relation - on S by

where each x, E {1, .

x -y pxl %Y1, xi - x1-y1-Y1 (i = 2, ... , n). It is straightforward to show that x A y has ith component (x A y), = min(x; - x1, y, - y1) + max(x1i YI)

and that x v y is the n-tuple whose ith component is (x v y), = max(x, - x1, y, - y1) + min(x1, YI)

It is also easy to check that x A y and x v y are both in S. We now show that (S, <) is a distributive lattice. To do this we have to check

that xA(yvz)=(x Ay)v(xAz) for each x, y, z. Put w=yvz. Then (x A (y v z)), = min(x, - x1, w, - w1) + max(x1, w1) = minx, - x1, max(y, - YI, z, - z1)) + max(x1i min(y1i z1))

whereas

((x A y) v (x Az)),=max((x Ay),-(x Ay)i, (x Az),-(x Az)1) + min((x A y)1, (x A z)1) = max(min(x, - x1, y, - Yi), min(x, - x1, z, - z1)) + min(max(x1, Yi), max(x1, z1)).

However, for any three numbers a, b, c we have min(a, max(b, c)) = max(min(a, b), min(a, c))

(6.13)

102 1 Combinatorics of finite sets and

max(a, min(b, c)) = min(max(a, b), max(a, c)),

(6.14)

so that all we need do is replace a, b, c by x; - x1, y; -Y,, z, - z, in (6.13) and replace a, b, c by x,, y,, z, in (6.14). Thus (S, <) is a distributive lattice. For an application of FKG we next introduce the characteristic functions f, g of the events x, -- x2 and x3: ifx, -- x2 0 otherwise

f(x)-{1

1

g(x) =I 0

ifx, -- x3 otherwise.

It is easy to check that both f and g are increasing functions. For example, if x
To prove that µ satisfies the condition of Theorem 6.2.9 it is sufficient to show that µ(x) = µ(y) =1 ' (x n y) = µ(x v y) = 1. Therefore suppose µ(x) =y(y) =1. Then x and y both satisfy the inequalities of G. If x; < x; is one of the inequalities of G, we then have x; < x; and y; < y, so that

(x A Y): = minx; - x,, Y, -Yj + max(x,, yi) min(x, - x,, yi - y,) + max(x,, Y,) = (x A y);, and similarly (x v y); -- (x v y),. We can therefore apply Theorem 6.2.9 to f, g, and p to obtain EXES!^'(x)

isl

EXEs

'

µ(x)f(x)g(x)

ISI

i Ex ESf(x)1L(x) EXESg(x)P(x) ISI

ISI

P(G)P(x, -' x2, x1 -' x3, G) = P(x, -- x2, G)P(x, -- x3, G).

Up to

this

stage we have been considering n-tuples x =

(x,, ... , x; E {1, . . . , N). These are not permutations since it is possible to have x; = x, and i : j. However, as N --* - the probability that x; = x, for some i * j tends to zero, and the theorem therefore

follows. We remark that we had to deal with n-tuples rather than

Ideals and a lemma of Kleitman

1

103

permutations so as to obtain a distributive lattice. We note also that the complicated order relation on S was needed to give functions satisfying the conditions of the FKG theorem. 6.4

Chvdtal's conjecture

In 1974 Chvatal made an interesting conjecture concerning intersecting families in ideals. Recall that a collection OF of subsets of an n-set S is an intersecting family if A fl B * 0 for all A, B e S. Among the intersecting families are those of a very simple type, called stars.

Definition 6.4.1 A collection 04 of subsets of an n-set S is a star if there is an element x E S such that x E B for all B E 9B. Chvatal conjectured that if we are looking for as large an intersecting family as possible in an ideal, then we can restrict our search to the stars. Chv6tal's conjecture Let si be an ideal of subsets of S, let w(si4) denote the maximum size of an intersecting family in 8, and let s(4)

denote the maximum size of a star in 4. Then w(4) = s(4). Note that

s(s4)=max I {A es4:x eA}l. xeS

It is clear from the definitions of w(s4) and s(s4) that we certainly have w(si) , s(si). Chvdtal conjectures that there is equality. There are some special cases where the conjecture follows from earlier

results in this book. If 4 consists of all subsets of S, then the conjecture follows from Theorem 1.1.1. If 8 consists of all the k-subsets of S and their subsets, k
Example 6.4.1 Suppose that the ideal s4 consists of the following subsets of {1, . , 5}: 0, {1}, {2}, {3}, {4}, {5}, {1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {1, 5}, {3, 5}, {4, 5}, {1,2,3}, {1,2,4}, {1, 3, 5}. Then 4 is the union of the following pairs of disjoint sets,

104 1 Combinatorics of finite sets

together with 0: (1) and {4, 51, (2) and (1, 3, 5), {3} and (1, 2, 4), {4) and {1, 2, 3), (5) and {1, 4), (1, 3) and {2, 4), {1, 2) and {3, 5}, {1, 5} and {2, 31.

It had been shown earlier by Erdos, Herzog, and Schonheim (1970) and by Marica (1971) that if .s4 is an ideal, then there is a bijection 0 : si --> s4' (where si4' = {A' : A E 4}) such that A c 0 (A) for all A E si. Berge's result is that there is a bijection 0 as described, with the additional property that - ((4 (A))') = A' for all A E si. For if A is paired with (O(A))' then certainly A and (4(A))' are disjoint (since (A))'))' = A. We A c 0(A)) and, further, (4(A))' is paired with therefore prove the following theorem.

Theorem 6.4.2 Let 4 be an ideal. Then there exists a bijection 0 : s4 -- si' such that A c 0(A) and 0((0 (A))') = A' for all A e si4. Proof (Daykin, Hilton, and Mikl6s 1983). We use induction on 141.

The case (s41=1 is trivial, and so we consider the induction step, assuming that the result holds whenever s4 is replaced by an ideal of smaller size.

Let E be a subset of S of minimum size such that A U E E X for some A E si, and let J= {A E Si : A U E e 4'}. Note that A fl E = 0 for all A E!@, since otherwise there is an unnecessary element in E, contradicting the minimality of E. We shall now show that s4 - is an ideal. Since 0 0, we shall be able to apply the induction hypothesis to s4 -!@.

So let X E si4 - .9, Y c X. Then X E 4 but X U E f si'. Certainly Y E Si since .s4 is an ideal, so suppose that Y E!@. Then Y U E E .s3' so that, since s4' is an upset, we must have X U E E .s4'. This contradicts X U E f .s.4', so the assumption that Y E must be false. So Y E Si 9 as required, and si - is an ideal.

-

Applying the induction hypothesis to si - @, we obtain a bijection : s4 - 9 --> (s4 - .°0)' such that A c p(A) and ip((ip(A))') = A' for all A E Si -!@. Now define 0 on .s4 by

O(A) -

JAUE ifAe V(A)

if A e 4 - e.

Then trivially A c 4 (A) for all A E s4. Further, if A E si - . $((0(A))') = 0((V (A))') =1V ((tV (A))') = 4' since (tp(A))' E 4 - 2. However, if A E 2, then

0((O(A))') = 4((A U E)')

Ideals and a lemma of Kleitman 1105

where (A U E)' E A. If we can show that (A U E)' E 2 then we shall have

4((O(A))')=4((AUE)')=(AUE)'UE=((AUE)nE')'

=(AnE')'=A' since A n E = 0. It therefore remains to show that if A E 2) then (A U E)' E 2. Certainly (A U E)' E A, so suppose that (A U E)' E

A-2. Then (AUE)'UEIA', i.e. (AUE)nE' f.s4, i.e. AnE'f

A. However, A n E' c A where A E A, so that A n E' E s 1 since s:4 is

an ideal, and we have a contradiction. Thus (A U E)' E

and the

proof is complete.

An immediate consequence of this theorem is the following corollary.

Corollary 6.4.3 If A be an ideal, then w(A)<12 IAI.

Proof At most one set from each disjoint pair can be in any given intersecting family.

We note in passing that this corollary can be obtained in a different way by using Kleitman's lemma (see Exercise 6.17). Some of the partial results concerning Chvatal's conjecture involve the maximal members of the ideal A; these are called the bases of A.

Definition 6.4.2 If

BEA=>A=B.

.

'

is an ideal, A E .s is a base if A c B,

Theorem 6.4.4 (Schonheim 1976) If the bases of the ideal A have a non-empty intersection then Chvatal's conjecture is true for . 4, and

w(A)=21AI.

Proof Suppose that the element x is in all the bases of A. Write = U It where x is in every set in 98 and in no set in 16. If A E 99 then A - {x.} E T, so we have I R I , 16 1. However, if D E then D c B for some base B (which contains x) so that D U {x} c B, and D U {X) E R. Thus 116 1 , I Oh1. It follows that I RI = I' I, and 1,% I = 2 IAI. Finally observe that R is a star, so 1991 -_ s(sf) w(A); thus .

w(s4) ? 21.41. It now follows from Corollary 6.4.3 that w(.s) = 2 1,41 S(A) and = I°S( = w(.si).

Since these results were proved in the mid 1970s, Chvatal's conjecture has been proved for some further classes of ideals. For

106

Combinatorics of finite sets

example, Stein (1983) proved the conjecture for ideals in which all

but one of the bases form a simple star, i.e. a star in which the intersection of all the members is equal to the intersection of any two of its members. Another partial result has been obtained by Mikl6s. Recall from Corollary 6.4.3 that w(sf) -_ [2 Isil]. Mikl6s's result is that if w(4) = [2141] then Chvatal's conjecture is true for 4. We need a preliminary lemma. Lemma 6.4.5 Let si be an ideal of subset of S. If there is an x E S such that s(x, 9) =1 {A e 4 : x E A} 1= 2 1.s41, then x is in every base of sB. If there is an x E S such that s(x, si) = (14 -1), then x is in 2 all but one of the bases.

Proof Let B1, . . , Bp be the bases of 4. First suppose that s(x, 4) = 1A1. If there is a base B1 with x* B,, then s(x, Si .

{B1}) = 2 21sf1 > [2 1sf - {B1}11, contradicting Corollary 6.4.3. There-

fore x is in every base. Next suppose that s(x, 4) = 2 (1,41-1). Since s(x, .4) # 21.41, it is clear from the proof of Theorem 6.4.4 that x o (l, B. However, if x 0 B; and x B; (i *j), then s(x, 4 {B,, B;)) = 2(1s4) - 1) > [2 I s4 - {B,, B;}1],

contradicting

Corollary

6.4.3. Thus x is in all but one of the bases. Theorem 6.4.6 (Mikl6s 1984) Let si be an ideal of subsets of an n-set S such that w(A) = [21s41], and let B1, ... , B,, be the bases of s4.

(i) If 141 is even then M = f, B; # 0 (so that by Theorem 6.4.4 Chvatal's conjecture is true for 4), and every intersecting family 69 c si of maximum size arises in the following way: take a maximum intersecting family 980 of subsets of M and let 98 = {A E si : B c A for some B E 98a}. (ii) If 1si 1 is odd and R c si is an intersecting family of maximum

size then there is either a base set B, such that 98 c si - {B,} or a set Bp+1 c S, Bp+1 I d, such that 98 U {Bp+1} is an intersecting family and .4 U {Bp+1} is an ideal (i.e. either 1981= 21d - {B,}1 or 198 U {Bp+1}1= 2 is U {Bp+1}1, in either of which cases Chvatal's conjecture follows from (i)).

Proof We consider only the case where 141 is even. It is clear that any family 98 as described in the statement of the theorem is an intersecting family. We now show that it is of maximum size by proving that 1981= 21si1. If A E 4 then the symmetric difference A + M = (A - M) U (M - A) is a disjoint union Al U M1 for some

Ideals and a lemma of Kleitman

1

107

A, c A, M, c M. However, Al c B; for some i, so that A, U M, c B; and A + M E s4. Also, (A + M) + M = A, so s4 consists of pairs (A, A + M). Now either M n A or its complement in M must be in Ro by maximality, so there is a B E -04a such that B c A or B c M + A; therefore at least one of each pair (A, A+ M) is in 04. Thus 1-64I = 21s4I. However, 190 1, 21,41 by Corollary 6.4.3, so we must have

A=2114 1 We next use induction to prove that m = (1, B; : 0, and assume the result holds for ideals of size less than 1,41. Imitating the proof of

Theorem 6.4.2, we can choose E c S of minimum size such that A U E E s4' for some A E s4 and define 2= {A E d: A U E E sV} to find that .s4 - 2 is an ideal. Provided that 4 - 2 * 0, we shall apply the induction hypothesis to s4 - 2; if -9 = s4 then s4 = P(S) and the result is trivial. Now we are given that .9 has an interesting family 9 of size 2 1s4 9 is a union of disjoint pairs (Exercise 6.21), so lii n Si _- 2 J!@ I. If 2 n gI < I_qJ then 9 - .9 would be an intersecting family in s4 2 n with 19 - 21 >2 I s4 - 21, contradicting Corollary 6.4.3. So 91= 2 X21, whence IS - 21= 21s4 - !@ 1. We therefore have w(s4 b) = 2 1 s4 - 2 J, and we can apply the induction hypothesis to . - 2, I

concluding that the intersection N of the bases of .s4 - 2 is nonempty. Since all intersecting families of maximum size in s4 - 2 must

contain N, N E °F - 0. We next show that N n E' * 0. Suppose that

NnE'=0, i.e. NcE. Since AnE=O for all AEI, AnN=O

n -91= 1.9 1 * 0, we can choose for all A E 2. However, since A E S n g and for this A, A E S and N 2E f, whence A n N* 0, 1

which is a contradiction. Therefore we must have N n E' 0 0 after all. Let N n E' = M; then M will turn out to be the intersection of all the bases of s4.

If X E M, i.e. x e N n E', then x is in every base set of s4 - 2; it

therefore follows as in the proof of Schonheim's theorem that s(x, . - 9) = 2 Id - qJ. However, is the union of pairs A, B with A n B = 0 and A U B = E' (Exercise 6.21), so that x occurs in half and thus s(x, s4) = Is41. It therefore follows from the sets of Lemma 6.4.5 that M is in every2 base of .s4. Since if x f M, s(x, s4) < 1 s41, M is the intersection of all the bases of s4; this 2 intersection is therefore non-empty. Since - 2 is a maximum-sized intersecting family in s4 - !2, the

induction hypothesis tells us that 9 - 2 can be obtained by taking a maximum intersecting family of subsets of N and then taking all supersets in s4 - 0. We can then add half the sets in 2 to obtain the intersecting family 9 of size 12 1 s4 1. If M = N then f is as described in

108 1

Combinatorics of finite sets

the theorem. Next suppose M 0 N, i.e. N n E 0 0. Consider any partition M = M, U M2; we shall next show that M, or M2 lies in 9 - 9. Suppose that neither is in 9 - !@. Note that

M,U{M2U(NnE)}=MU(NnE)=(NnE')U(NnE)=N and

M,n{M2U(NnE)}=M,n(NnE)c(NnE')n(NnE)=0 so that M, and M2 U (NnE) are complementary subsets of N. Since 9 - contains a maximum intersecting family of subsets of N, it must contain one of M, and M2 U (NnE). Similarly it must contain one of M2 and M, U (NnE). Therefore if 9 - 2 contains neither M, nor M2 it must contain both H, = M, U (NnE) and H2 =M2U (N n E). However (Exercise 6.22), there is a pair of sets in such that one of them intersects M in M, (and hence does not intersect H2) while

the other intersects M in M2 (and hence does not intersect H,); therefore ° can contain neither of the sets of this pair, contradicting the observation made earlier that half of the sets of 2 can be added to

the sets of 9 - 2 to form 9. Thus .° - 2 must, after all, contain M, or M2, so that 9 contains a maximum intersecting family of subsets of M, say 98. Thus 9 a {A E sc4 : B c A for some B E M}. However, such a collection was shown to have cardinality z Is$I = 191, so we conclude that F = {A E S4 : B c A for some B E 90) as required. Exercises 6 6.1

Prove the following two alternative forms of Kleitman's lemma.

(a) If s4, R are ideals of subsets of an n-set, then ICI is. MI 2" Isin,r'.6I.

are upsets of an n-set, then I si . 190 1 , 2" I. n I. , Let si4 be an ideal of subsets of S. Show that the average size of

(b) If . 6.2

members of si is ,I ISI. 6.3

For 1 < k < in construct k disjoint intersecting families of subsets of an n-set S whose union contains

(.

k-1 n i=1

)

1 -1

sets, such that none of the families can be increased without destroying disjointness. (Note that g(n, k) < 2" - 2"-', so that

Ideals and a lemma of Kleitman

the situation

is

1

109

different from the case k =1 where any

intersecting family can be extended to one of size 2n-1 ) (Kleitman 1966) 6.4

Let 9, 2 be families of subsets of the n-set S such that

AE P,BE°2 Show that I 9'I 1l2 +

19-11/2

-- 2'

as follows. Put

X={XcS:XDAforsome AE9'andX-B for some BE 1}

I={XcS:XDAfor some AE9'but

all BE°-}

{X cS:X AforallAE9'butXcBforsomeBE Z} X={XcS:X Afor all AEOP andX Bfor all BE2}. Apply Kleitman's lemma to °U = W U .0 and 2 _ .0 U X to prove that Ij 1 1 / 2+1 ,0 1 "- -2n- ', and note that 9' c .0, °oL c J0. (Seymour 1973) 6.5

6.6

Use 6.4 to prove Theorem 6.1.4. Let sY be a collection of subsets of an n-set S. Let

Si+={BcS:BcAfor some AEs4}, s4`={BcS:BcA for some A E d},

16(si) = sz2+ U si-, mid(st) = 4+ fl sv. Show that

(i) sr = (mid(s4))+ (ii) T(4) = 16(mid(.s )) (iii) ISi+I + I.si-I %2(2" jmid(.s1)j)112

(iv) (szZ) , 2(2" Imid(s4)I)v2 - Imid(.s4)I)1rz (v) if 0 -- i _ Zn and Isatl -- 2"-2i then I16(sz2)I %

2n+1-;

- 2"-21

(Frankl and Hilton 1977) 6.7

Verify that

(i) if si, R are upsets then 4 v = si fl R, (ii) if 4, R are downsets, then Si AR =,4 fl R. 6.8 6.9

Deduce Kleitman's inequality from Corollary 6.2.2.

Prove FKG (6.2.9) in the special case where the lattice is {1<2< ... < n) by expanding the inequality

Z ((i) -fU))(S(i) -S())µ(i)pU)

O.

110 1 Combinatorics of finite sets

that µ,, µ2 are probability distributions (i.e. EA pi(A) =1) such that µ,(A)µ2(B) , µ1(A U B)µ2(A fl B) for all A,B c S, and that h is an increasing function. By choosing a(A) = µ1(A), #(A) = 112(A)h(A), qq(A) = µ1(A)h(A) and S(A) = 142(A), prove Holley's result that

6.10 Suppose

E µ,(A)h(A) ':A92(A)h(A) A (Holley 1974) 6.11 Show that Holley's result in 6.10 implies FKG, by taking µ2 = µ and 141 = 8l1/EA 8(`4)µ(A). 6.12 A polarity is an order-inverting bijection a : P(S) 01(S) such

that a-'= or, where order-inverting means A c B a(B) c a(A). For any sB c 07'(S) let .si+ denote the collection of all supersets of members of si, and .4- the set of all subsets of members of ssi.

(i) Show that I(a(si))+I = Isi I. (ii) Use Kleitman's lemma to show that I. I , I. fl (a(s/I))+I and Isi) _- Is- fl (a(sh))-I. (Daykin, Kleitman, and West 1979) 6.13 Let .s', 98 be two complement-free collections of subsets of an n-set S such that, for all A E si and B E 98, A B and B A. Prove that Id) + ICI 12^-' as follows. (i) Show that if Si = s 1 U .s&2i sit fl .s12 = Q, 6-D4 = 1 U R,nA=0, SiU S42fl982=0,'6 =5711V-041, ' 2 U 19, then 15fl iU I = 0, so that I I , 2"-'. (ii) Next show that .si fl ;:82 = si fl T =902n % = QJ, so that

I.4U6B2U'9I=IaI+I

I+I'WI.

(iii) Use Marica-Schonheim to prove that I 16 I . Isi,I = I. (Hilton 1976) (iv) Deduce that I s i I = 199 1 = I J I -_ 2R-'. 6.14 Give an example of collections si, R satisfying the conditions 2"_' of 6.13 and with Is$I + 1601= 6.15 Use 6.13 to give another proof of Theorem 6.1.4.

... , Am are distinct subsets of S and , B are such that each A; contains at least one B1, then

6.16 Prove that if A1, B1,

. . .

there are at least m distinct differences A, - B1. Deduce the Marica-Schonheim result (see hints). (Ahlswede and Daykin 1979) 6.17 Use Kleitman's lemma to prove Corollary 6.4.3. (Schonheim, private communication)

Ideals and a lemma of Kleitman

6.18

1

111

Prove Chvatal's conjecture in the case when 4 has only two bases.

6.19

Let sf, 60 be two intersecting families of subsets of an n-set S such that A e .s.4, B E y8 =>A U B * S. Prove that Isil + +RI 2"-' as follows. (i) Extend .sB, PA to intersecting families W, 98* of size

2"-', and let

={X:XEsrnR*),

g_go*-

D.

(ii) Show that E E 5S ei E' e S.

(iii) Apply Berge's theorem to the ideal' _ {D': D E .9), and hence show that the members of can be put into pairs with the union of the two sets in any pair being S. (iv) Then use (ii) to extend the pairing in (iii) to the whole of W U P*. (Hilton 1978) 6.20 (i) Show that if PA is an intersecting family of 2' subsets of an

n-set S, then' is an ideal. (ii) Use (i) and Corollary 6.4.3 to give another proof of Theorem 6.1.4. 6.21

(Schonheim 1974a)

With .9 and E as in the proofs of Theorems 6.4.2 and 6.4.6, prove that if A c-!@ then B = (A U E)' is also in !@, and (B U E)' = A. Deduce that is the disjoint union of disjoint

pairs A, B with A=(BUE)' andAUB=E'. 6.22 Show that, if M = Ml U M2 as in the proof of Theorem 6.4.6,

then one of the pairs Dl, D2 of 6.21 is such that D, n M = M;. 6.23 Prove Theorem 6.1.4 by considering I d U (d - .s) I and using (Daykin and Lovasz 1976) Corollary 6.2.4.

7 The Kruskal-Katona theorem

7.1

Order relations on subsets

In the original proof of Sperner's theorem, use was made of the following fact: if si is a collection of k-subsets of an n-set S and if Ad denotes the shadow of .si, i.e.

Asi={BcS:IBI=k-1, BcAforsomeAEs4), th en

k

k+111 This estimate for the size of Asaf was obtained very easily, and was sufficient to yield the powerful normalized matching property, but it is possible to improve upon it by using deeper arguments. The aim of this chapter is to discuss such an improvement, obtained independ-

ently by Kruskal and Katona in the 1960s. Before we state their result, it will be helpful to have a look at some ways of putting an ordering on the set of k-subsets of S = {1, . . . , n). 7.1.1 Lexicographic order

Here we write A <,., B if the smallest element of the symmetric difference A + B = (A fl B') U (A' fl B) is in A. As usual, A' denotes the complement S - A of A. This ordering is a total ordering of the k-subsets of S. For example, if the 3-subsets of (1, . . . , 5) are listed in lexicographic order we obtain the first column below:

123 124 125 134 135 145 234 235 245 345 112

45 35 34 25 24 23 15 14 13 12

The Kruskal-Katona theorem 1 113

This is just the `dictionary' order, with 1 replacing a, 2 replacing b, and so on. Note also that, since A + B = A' + B', A
qB'
Here we use the dictionary idea again, but consider the ordering of 1, ... , n to be reversed. Imagine therefore that the alphabetic order is reversed. We write A
543 542 541 532 531 521 432 431 421 321

345 245 145 235 135 125 234 134 124 1 2 3.

7.1.3 Squashed ordering

We write A <s B if the largest element of A + B is in B, so that the squashed order is clearly the reverse of the antilexicographic order. Under the squashed order all subsets containing the element m , n come after all the subsets consisting only of elements less than m. The

114 1

Combinatorics of finite sets

3-subsets of {1, .

. .

,

5) in squashed order are

123 124 134 234 125 135 235 145 245 3 4 5.

Although the squashed order appears less natural than the lexicographic order, it turns out that it plays a particularly important role in the investigation of the size of the shadow of a collection of k-sets. Roughly speaking, if we want to minimize ji4 over all choices of . of a given size, we want to squash the Idl sets so that they are made up from as few elements as possible, and this corresponds to choosing

the first Idl sets in the squashed ordering. This is essentially the kernel of the Kruskal-Katona result. In our investigations it will sometimes be convenient to represent subsets by sequences of Os and is. For example, if n = 5, the subset {1, 2, 4} of {1, ... , 5} could be represented by the sequence 11010.

The ith digit is 1 if the element i is in the set, and is 0 otherwise. However, for reasons which will become clearer shortly, it is convenient to represent subsets by reverse sequences; for example {1, 2, 4) would be represented by the sequence 01011. Here the ith digit from the right is 1 precisely when i is in the set. This representation will be particularly useful when we come to study multisets; for example, we will be able to represent the divisor

2.3

. 7 by the reverse sequence 1021, the sequence of of 22. powers in the factorization 71.5° . 32.21. The reason for representing by reverse sequences is the following.

We have A <s B q largest element in one but not the other is in B < *first place in representing reverse sequences where they differ has

0 for A and 1 for B q sequence for A is before the sequence for B in the lexicographic ordering of the sequences, where (x1, . . . , x,,)
The Kruskal-Katona theorem 1 115

listed in their squashed order above are represented by the following sequences which are in lexicographic order:

00111 01011 01101 01110 10011 10101 10110 11001 11010 11100. 7.2 The 1-binomial representation of a number

In order to shed further light on the squashed ordering of sets, we now prove the following theorem.

Theorem 7.2.1 Given positive integers m and 1, there exists a unique representation of m in the form

m=(1t)+(a, 1) +...+(t`)

(7.1)

1

where at > a,_ I> ... >a, , t ,1. (This representation of m is called the 1-binomial representation of m.)

Proof We can easily find such a representation of m. First choose at

as the largest integer for which

( fit) ,

M. Then choose a,_1 as the

largest integer for which (tat-1)

m - (It). If a,_1--at then we

would have m, (a) + (1 -at 1) _ (1

1

1

at),

contradicting the maxi-

of at. Therefore a,_1 < at. Continuing this process we eventually reach a stage where the choice of a, for some t . 2 actually gives equality,

(a`)=m-(lt)I r

1)-...-\t+,

116 1 Combinatorics of finite sets

or we get right down to choosing al as the integer such that

(ai)<(ai)

(a2)(al, 1)

in which case we have

so that

m=(a')+...+(a,). We now show that such a representation of m is unique. If l = 1 there is clearly only one representation, namely m = (T) so we use ,

induction and suppose that the representation of any m is unique if I = k - 1. Suppose then that we have two different representations t m=(k)+...+(a)=(k)+...+(br)

for some t .1, r

1. If ak = bk we then obtain two representations for

m - (k) with l = k - 1; by the induction hypothesis these representations are identical, so that a, = b; for each i. Next suppose that ak # bk. Without loss of generality we can assume that ak < bk. We then obtain

m=(k)+(kk-1)+...+(tt) (k)+(a,, -

)+...+(ak-(k-t))+...+(ak-1 +1)

=(ak+1)-1 k

on using the identity (,k) +

We thus obtain m < (ak

+

k-11)

k

1)

,

(n0 k)=(nk 1).

(7.2)

(kk) .< m, which is a contradiction.

Thus, by induction, the l-binomial representation of m is unique.

0

The Kruskal-Katona theorem 1 117

Example 7.2.1 Take 1= 4 and consider m = 26. The largest x for which (4x) --- 26 is x = 6. We have

26=(4)+11

-(4)+(3)+1 (1) + (1) + (2). 4

3

2

What exactly is going on here? The representation is closely related to squashed ordering. Suppose that (7.1) holds. In the squashed

ordering of 1-sets the first (1`) sets are precisely the 1-subsets of {1,

... , a,). The next (1a)

are those obtained by adjoining at + 1

}...

, a,_1 . The next1-2 ('') are those to the (1-1)-subsets of {1 obtained by adjoining (a1_1 + 1, at + 1) to the (1- 2)-subsets of

(1,

... , a,_2) and so on until the final

(a`) are the union of

{a,+1 + 1, ... , a, + 1} with the t-subsets of 11, I-set in the squashed order is

... , a,). Thus the mth (7.3)

Example 7.2.2 Take 1= 3. 9=

=

(31)

(34)

+

(21)

2

+ (1), so the 9th set is {5, 4, 21.

Z

+ (2), so the 5th set is {5, 2, 1).

4 = (43), so the 4th set is (4, 3, 2).

Example 7.2.3 In which place in the squashed order is (a) (2, 3, 5) among the 3-sets and (b) {2, 3, 5, 7} among the 4-sets? Solution

(a) The set {2, 3, 5) is rewritten as (5, 3, 2) and compared with

118 1

Combinatorics of finite sets

(7.3). We see that a3 + 1= 5 and a2 = 3, so the set is the mth set where

m=(3)+(2)=7.

(b) Write the set as {7, 5, 3, 2} and compare with (7.3) to obtain

a4+1=7, 1a3+1 (= 5, a2 = 3. Thus the set is the mth set where

m=(4)+(3)+(3)=22. The reversal of the ordering of the elements so as to compare with

(7.3) suggests that we could work with the reverse representing sequences instead.

Example 7.2.4 In which place in the squashed ordering of 4-sets is the set {3, 4, 5, 6}? Solution This set is represented by the sequence 111100, so we have

to find where 111100 occurs in the lexicographic ordering of sequences of length 6 with four 1s. However, 111100 is clearly the last sequence in the lexicographic ordering, so it is in the place number

(6) =15. 4

Clearly, in general, if the sequence consists of r is followed by s Os, then its place in the lexicographic ordering of those sequences with r r+s is is ). r

Example 7.2.5 In what place in the squashed ordering of 4-sets is the set (2, 3, 5, 7)?

Solution We have to place 1010110 in the lexicographic ordering of sequences with four is. It is preceded by all such sequences which begin with 0; there are (41) of them. The other sequences preceding it

begin 10 .... Therefore we now consider the position of 10110 among

those sequences with three is. It is preceded by all the

(4)

3sequences

beginning with 0. The other sequences before 10110 begin with

.... so we have to place 110 in the ordering of sequences with two is. By the remark after the previous example, its position is

10

The Kruskal-Katona theorem 1 119 (23).

Thus the required position number is

(4)+(3)+(2)=22.

In general, we peel off is from the left until a sequence consisting of a string of is followed by a string of Os is obtained.

7.3 The Kruskal-Katona theorem Consider the first m l-sets in the squashed order, as described before

Example 7.2.2, when m is given by (7.1). Note that the (1-1)-sets contained in them are all (1

a,1)

,a,}

(1-1)-subsets of { 1, .

all (I`-2) (1- 2)-subsets of {1,

a,_1), unioned with {1 +a,}

.

all (t a` 1) (t -1)-subsets of 11,

a,}, unioned with {1 +a,, 1 +a,_1i

Thus there are

(')')

. . .

C)

such (l -1)-sets, and they are precisely the first (I

,

1 +a,+1).

a,l) +... +

( a`) (1-1)-sets in the squashed order. (Care has to be taken if t-1 t = 1, for then the representation is not as in Theorem 7.2.1 (t - 1 = 0), but on rewriting

(all)

to give

(0)

as

(a01)

it can be combined with

(at+i+ 1) and the process continued until the repre-

sentation is as in Theorem 7.2.1.) We are therefore in a position to understand what the Kruskal-Katona theorem stated below essentially says: to minimize the shadow of a collection of m 1-sets, where m is given, take the first m I-sets in the squashed order. Theorem {A1,

. . .

,

7.3.1 (Kruskal 1963, Katona 1966a) Let si = A,,} be a collection of I-subsets of S, and suppose that the

120 1 Combinatorics of finite sets

1-binomial representation of m is

m=(')+(l

11

+...+

ttl

where a,>a,_1>...>a,=t.1. Then

Iall)+\1,

21+...+\ta`11

Note that this result has nothing really to do with n = JSJ; all that matters is how many sets are in . (for recent work on the case where U, A; = S, see Mors (1985)).

Various proofs of Theorem 7.3.1 have been given. The original proofs of Kruskal and Katona are quite complicated, but since they were published a number of authors have given shorter proofs; for example, we mention Daykin (1974a), Hilton (1979) and Frankl (1984). Hilton's proof is by induction, the main step being to show

that, for any i E {1, ... , n} and any positive integers b, c with

b + c = a, the `earliest' possible collection .si of a k-sets with minimum shadow and containing b sets without i and c sets with i is given by sat = Fk(b)` U {{i} U Fk_1(c)`}

where F(x)` denotes the first x I-sets in the squashed order which do

not contain i. Later we shall give a proof involving the idea of compression in the more general context of multisets (the generalized Macaulay theorem of Clements and Lindstrom). However, the proof we now present is due to Franki and depends on a shifting process;

we begin by trying to replace an element of a set by 1 if 1 is not already present. This shifting idea goes back to the original proof of

the Erdos-Ko-Rado theorem, where the shifting is away from n; recall that a shifting away from 1 was used in Section 5.5.

Lemma 7.3.2 Let .si be a collection of l- subsets of S = {1, ... , n and, for 1 < j , n, define S;(A) _ Also define Then

{(A - {j}) U {1}

oif j e A,

A, (A - {j}) U {1}

therwise. S;(si) = {S1(A): A E Si).

A(Si(d)) c S;(Ad).

si

The Kruskal-Katona theorem 1

121

Note It follows immediately that 1A.s1I ,1A(S,(saf))1.

Proof of Lemma 7.3.2 We have to show that if A E 4 then AS,(A) E- S;(A4). First suppose that A = S;(A). If B E AS;(A) then B E AA and so A = B U {i} for some i , n. We show that B = S1(B) where (this is important!) S, is the shift operator for Ash, not 4; it will then follow that B E S,(As4). If i = j then j 0B so that B = S,(B).

If i=1 and jEB then (B-{j})U{1}=A-{j}EAsi, so S1(B)=B. If i =1 and j B then S1(B) = B. If i # 1 or j, B = S1(B) unless possibly when j E B and 10 B. But then j E A and 1! A and, since S;(A) = A, we must have (A - {j}) U {1) E 4. Thus (B - {j}) U {1} E AsQ, and so S,(B) = B.

Next suppose that A # S,(A). Then j E A, 10 A, S,(A) = (A {j}) U {1}. If B E AS,(A) and 10 B, then B = A - {j} and so S1(B) =

B. If B E MS;(A) and 1 E B, then B* = (B - {1}) U {j} cA and so B* E Asi. If B 0 Ash, then S1(B*) = B so that B E S,(Asz1). If, finally, B e i 4, note that j 0 B so that B = S1(B), whence B E S3(As4). Lemma 7.3.3 If S,(s4) = si for all j -- 2, and if

s4o={AE4: 1 A},

s4T={A-{1}:1EAEs4},

then

TI' lAsiol.

I

Proof If E E As$o then E U {i) E 4Bo for some i > 1 so that, on applying S ; with j = i, E U {1} E s4. Thus E E Asio=> E E s4T.

Proof of Theorem 7.3.1 (Frankl 1984) If we repeatedly apply the shift operators S,, j = 2, . . . , n, to si, the number of sets containing 1 increases, so that after a finite number of applications the shifts must therefore cease to make any change. We have then obtained a new collection s4* of the same size as sQ, with S;(si*) = 4* for each j , 2, and with I Asi l -- j As4 * j. The theorem will therefore be proved for s4 if we can prove it for W. Therefore we now suppose that s4 has the property that S,(s4) = s4 for each j 2.

Now the theorem is true for l =1 and any n = jS(, so we now assume that the theorem is true whenever n < p and whenever I< k and n = p, and consider the case of a collection 4 of k-subsets of a p-set T. We have (

where ak>...>as?s;l.

l

(1)

122 1 Combinatorics of finite sets

Suppose, if possible, that

{sitI<(k-1)+...+\ss-11' Then, since I.sQ = 141 - IStI, Is'01

>Yk)-\k-1I1+.+f(aS)

-\s-11j

(akk 1)+...+(ass 1)

so that, by the induction hypothesis,

j°sf

jak-1)+...+(as-1

°\k-1/

s-11'

Thus I41I % 1A401 > IdtI, which is a contradiction. Therefore we must have

k-1)+.+\ss-11' but then, by the induction hypothesis,

°atl

=(k-2)+.+(as.,--21)

so that a,

0

as required. Corollary 7.3.4 Let si be a collection of k-sets and, for r < k, let

°(')ai={B:IBI=r, BcA for some AES4}. Then if

Isil = (7)

+- - -+

("),

ak>...>a, t:1, a A(r).Jja(akr )+...+t-a,+r)'

The Kruskal-Katona theorem 1 123

The Kruskal-Katona theorem is a very powerful result, but, owing to the complicated nature of its statement, it is sometimes not very

suitable for applications. Instead, the following variation due to Lovasz is often more convenient. In it we extend the definition of the binomial coefficient (k") to (k), where x is a positive real number, by f x)

x(x-1)... (x-k+1)

\k)

k!

These numbers satisfy the usual recurrence relation (k) + (k x 1) _

(x

k 1); further, if (n k

< m < (k), we can find a real number

x such that m = (k). Theorem 7.3.5 I

(Lovasz 1979) If s4 is a collection of k-sets and

- k) (x , k), then lAsil % (k X 1)

Proof (Frankl 1984) As in the proof of Kruskal-Katona, we can

suppose that S1(s4) = d for each j , 2. Again, we use double Id l induction on k and m = j4 j. If j s41 j < (k - 1) then iT)

- (k -1) = (x k 1), so that, by the induction hypothesis, x 1)' However, Lemma 7.3.3 then gives ISQTI (k 1) (k

(k)

1

X

which is a contradiction. Therefore we must have l.siil % (k -1), but so that then, by the induction hypothesis, 1Ddt1 -- (k

- 2)

(kx

as required.

(k x 1)

0

This Lovasz version of Theorem 7.3.1 leads to a simple proof of the following result from which Lemma 5.2.5 can be deduced.

124 1 Combinatorics of finite sets

Corollary 7.3.6 If sad is a collection of k-subsets of S, and if I41

(k) then

Ii siI

(k u 1)

I sBI.

(k) Proof Let 1,41 = (k), x -- u. Then, by Theorem 7.3.5, X

loSi I , (k

= k

x

1)

1)

ICI .

(k -1)

ICI

((u k)

(Xk)

since X

l

k-1) (x)

k

x-k+l

k is a decreasing function of x.

7.4 Some easy consequences of Kruskal-Katona 7.4.1 The Erdos-Ko-Rado theorem

As our first example of the power of the Kruskal-Katona theorem we

give the elegant deduction of the EKR theorem due to Daykin (1974b) and Clements (1976) independently.

Theorem 7.4.1 (EKR) Let 4 = (A 1 i ... , A) be a collection of k-subsets of an n-set S, k -- ?n, such that A, fl Al # 0 for each pair i, j. Then

m-(k1)

Proof (Daykin 1974b, Clements 1976) Suppose that m > (k Then I si' I = m > (n

1)-

- k) where, as usual, s4' = {S - A : A e d). The

condition A; fl A14 0 is equivalent to A;

A;, so the sets in A(k)s4'

The Kruskal-Katona theorem 1 125

must be distinct from the sets in .i; thus ICI + IO(k)J

But I,&(k)s'I

I

(7.4)

(k).

(n k 1) by Theorem 7.3.1, so that ICI+IA(k'SJ11 >`k1)+(nk 11-(k)'

contradicting (7.4). 7.4.2 Subadditivity of Fk(n)

Let Fk(n) denote the first n k-sets in the squashed order. The following result will be used frequently in later chapters. Theorem 7.4.2 (Clements 1974) For any positive integers n,, n2, IAFk(nl + n2)I -- I AFk(nl)I + I AFk(n2)I.

Proof We consider subsets of { 1 , 2, 3, ... } . Let h denote the largest

element occurring in the sets in Fk(nl), and let 64 denote the collection of the first n2 k-sets with least element greater than h. Then AFk(n,) and EP are clearly disjoint, so that IL (Fk(n,) U 1)I = I AFk(n,)I + I A,@I. However, @ is isomorphic to Fk(n2). It therefore follows from Kruskal-Katona that ILFk(nl + n2)I

I A(Fk(nl) U R)I = I EFk(nl)i + IA-04I

0

=IAFk(nl)I +IAFk(n2)I

7.4.3 Extensions of EKR

... ,

Suppose we are given a collection .ss = {A,, an n-set S, and an integer 1. Define

of k-subsets of

I,(.4)={BcS:IBI=1, BfA#Q forallAe. ). Thus I,(s/I) consists of all 1-subsets of S which intersect every member

of d. What can be said about the size of I,(sd)? Hilton (1977) observed that, since the condition B fl A # 0 is equivalent to B J A', it follows that if l < n - k then

Ih(SO +(') 1

126 1 Combinatorics of finite sets

where, as usual, si' = (A': A e si). Thus, if m

-

(nnk)+...+(p)

is the (n - k)-binomial representation of m,

(1)-(al k)-...-(p+l a(n-k))' We can therefore prove the following theorem. Theorem 7.4.3 (Kleitman 1968) Let s/I = {A1, ... , Am} be a collection of k-subsets of (1, ... , n), let l + k -- n, and let B1, ... , B. be all the 1-subsets of {1, ... , n) such that A, fl B; # 0 for all i, j. Then

eitheru -- (1n-1) 1orm<(k-1)' Proof (Hilton 1977) Suppose that m - (k

1). Then m -_ (n

- k)

so that, by the above remarks,

(n -

u=III(SO--()-(n 1

0

This result was conjectured by Hilton and Milner (1967) in the special case when k =1. Corollary 7.4.4 If .

k

,

A

are two collections of k-subsets of an n-set,

in, such that A fl B# 0 whenever A E., B E 61 then 1,41

(k-1) or

(k-1)'

The EKR theorem arises when si and

0 coincide.

7.5 Compression Let .sQ = {Al, ... , Am) be a collection of k-sets. Then the compression Cs/I of .s/I is defined to be the collection of sets consisting

of the first Isil k-sets in the squashed order. The Kruskal-Katona theorem asserts that ILs/J - ILCs1II. In fact we have seen that the shadow of the first so many k-sets consists of the first so many

The Kruskal-Katona theorem 1 127

(k - 1)-sets, so that the Kruskal-Katona theorem actually asserts the stronger result ACSIIcCA.I. More generally, if s& contains sets of different sizes, let d = d U U ... U .s , where d, = {A E d : JAJ = i}. We then define the compression Csaf of d to be Cs/I = U, Cs. ; thus compression is performed at each rank level. We also say that s/I is compressed if C.QQ = s/I. We have therefore seen in Section 7.2 that the following .c

result holds.

Theorem 7.5.1 If d is compressed, so is Ad.

Compression will play an important role when we extend the Kruskal-Katona theorem to multisets. It will also be important when we study antichains in greater detail in the next chapter. However, for the moment we content ourselves with the following simple observation. Theorem 7.5.2 If . is an ideal then so is Cd.

Proof Let s/ = U J , o . , where s/; = {A E sd : JAI = i}. Then Os/I; c ,s/,_1 for each i so that, on writing & = U, Os4,, we have

ACd=A(UCA)=UA(C4) c_ U CAsaf,

by Kruskal-Katona

cUCsA,_1cCs/I.

Exercises 7 7.1

7.2 7.3

(a) Find the 4-binomial representations of 23, 27 and 37. (b) How many 3-element sets are there in [tF4(23)? What is the 25th set in the squashed ordering of 3-sets? In what position in the squashed ordering of 5-sets is

(a) {1,3,5,6,8} and (b) {3,4,5,7,8}?

0

128 1 Combinatorics of finite sets

7.4

Show the following:

(i) if A <s B then B' <s A'; (ii) if . consists of the first m (in)-subsets of an n-set S, where n is even, then d' consists of the last in (4n)-subsets; Is, (iii) if m = 2 (n, 2) then U szQ'I = (n/2 7.5

Show that if A is a set of size k then its successor in the squashed ordering of k-sets is

B={1,...,b}U{a+1)U(A-{1,...,a}) where a is the least integer such that a E A and a + 1 j A, and b = IA fl 11, . . . , a } I - 1. Hence verify that the successor of {2, 3, 4} is (1, 2, 5). 7.6

Let m =

(ak) +. .

. + (a`) be the k-binomial representation of

m, let s be the smallest index for which as > s, and let

J.

7.7

Show the following: M = (k 1) + ... + (s (a) IOFk+,(M)I m and I AFk+,(M + 1)1 > m; (b) IAFk+,(M + m)I = m + I zFk(m)I By considering s = Fk(r)' U ({1) U Fk_,(s)'), prove that IAFk(r + s)I < IIFk-l(s)I + max(s, I AFk(r)I ) (Hilton 1979)

7.8

Show that if m = r + s, where the k-binomial representation of m is as in 7.6, then s _- IAFk(r)I a (k-1)++(t`

11)

a

(kk-1)+...+(t`

when a,>t

11)-1 whena=t. (Hilton 1979)

7.9

(a) Show the following:

(i) if v < (k n 1) then AFk((k) + v) I = (k n 1) +

IAFk-1(v)I

The Kruskal-Katona theorem 1129

(ii) if 0 _ ml _ m2 _ (k) _ m, + m2 then IAFk(ml)I +IoFk(m2)I --

(kn

1/+IdFk(m,+m2-(k))

(b) Let d be a collection of k-subsets of S = S, U S2 where

(kn)`I',I(k)+(Iki),

S1nS2=0,

and IS, I-IS2I=n, where, for each A E s4, A c S, or A c S2. Use (a) to show that

IoM -- (k

n

1) +

(k)) (Katona 1973)

7.10 Deduce the EKR theorem from Theorem 7.4.3. 7.11 Prove that I AFk(m + 1)I _ IAFk(m)l + (k - 1). 7.12

(i) Show that, if n is even and 2 (2) then there exists an antichain sio of m (Zn)-subsets of S =

... , n} such that the number of pairs A, B E ,s4o with AflB=Q ism-2(n/2) {1,

(ii) With m as in (i) show that every antichain sat of m

subsets of S each of size _Zn must have at least m n

i n/2) simultaneously disjoint pairs of members. 7.13

(Clements 1976) (iii) Deduce Theorem 5.6.1. Prove the following version of the Kruskal-Katona theorem.

Let .sf be a collection of k-subsets of an n-set S. Then IVs4l % IVAI where A consists of the last Isil k-subsets of S in the squashed ordering. 7.14 Deduce Lemma 5.2.5 from Corollary 7.3.6. 7.15 In investigating all selections of 12 numbers from {1, ... , 20}, the selections 1; a, < a2 < ... < a,2,-< 20 are considered in the lexicographic order of sequences a, . . . . a12. Show that 1, 2, 4, 5, 8, 10, 11, 12, 13, 15, 16, 18

is the 29 937th sequence in the lexicographic ordering. (Lehmer 1964)

130 1 Combinatorics of finite sets

7.16 Show that if d consists of the first m (n - k)-subsets of an n-set S,

k
(nk1J'

m=(k-1)-(nkk11)+1,

then

1(k)1

8 Antichains

8.1

Squashed antichains

The study of antichains has been very central to the subject matter of most of the previous chapters. Now that we have met the KruskalKatona theorem we are better equiped to study antichains and their structure at greater depth. Much of our time will be spent studying the parameters of an antichain. Definition 8.1.1 If .si is an antichain of subsets of an n-set S, the parameters p, are the numbers defined by p, = 14,l where sc, = {A E d: JAI = i}.

Many of the results obtained so far are really theorems about these

parameters; we summarize these results before going on to obtain some new ones. Sperner's fundamental theorem, in its LYM form, asserts that Pi i=o (n ) i

, 1.

The EKR theorem is concerned with intersecting antichains, and, in its LYM form of Theorem 5.2.2, asserts that if p, = 0 for all i > k, where k <_ in, then

If .i is an intersecting antichain, but there is no restriction on the size of the sets in A ., then, by Theorem 5.3.1, tnrz (i

f>n/2 (n)

n 1) 131

132 1 Combinatorics of finite sets

Also, Bollobds's result (Corollary 5.2.3) asserts that if the antichain .s& satisfies A e sxl

A' a Si then

E A l+ E p` ,2. ;.nn(n-1/ ,>n/2(n 1\ All these results put necessary conditions on the parameters which

must be satisfied if an antichain of a particular type with these parameters is to exist. This raises an obvious question: is there a necessary and sufficient condition on numbers p; for the existence of an antichain with these numbers as its parameters? In answering this question (in the affirmative) it is necessary first of all to discuss a special family of antichains which are, in a certain sense, compressed. Suppose that the non-negative numbers po, ... , p are given. We attempt to construct an antichain se with parameters p (. ... , p in the following way. Let g be the largest value of i for which p; * 0. Define . to consist of the first pg sets of size g in the squashed. order. This definition is satisfactory, provided that pg

, g). n

(8.1)

In the shadow of these pg g-sets there will be ng_, = IOFg(pg)I sets of

size g - 1, namely Fg_,(ng_,); none of these sets can be in .24*. Therefore define sflg_, to consist of the next pg_, sets of size g - 1 in the squashed order. This definition is satisfactory provided that there are pg_, sets available to be chosen, i.e. provided that Pg-1 + ng-1

-n

.

8.2) (8.2)

We then have pg_, + ng_, sets of size g -1 which are in s4g_, or are contained in sets in 4*, namely the sets of Fg-,(pg-, + ng_,). In the shadow of these pg_, + ng_, sets there will be ng_2 = I'Fg_,(pg-, + ng_,)i sets of size g - 2, namely Fg_2(ng_2), none of which can be in ,V. Therefore define . g-2 to consist of the next pg_2 sets of size g - 2 in the squashed order. This definition is satisfactory provided that there are pg_2 sets of size g - 2 available to be chosen, i.e. provided that Pg-2 + ng-2' (g n 2).

(8.3)

Continuing in this way we construct an antichain sat * = s& U ag_, U U .io with parameters p; = IAN, such that, for each i, the sets in as

,

together with the i-sets contained in larger members of si*,

constitute an initial segment of the sets of size i in the squashed order.

Antichains 1 133

The construction works provided that the inequalities (8.1), (8.2), (8.3), etc. all hold. We have therefore proved the following theorem.

Theorem 8.1.1 (Clements 1973, Daykin, Godfrey, and Hilton 1974) Let po, ... , p be non-negative integers, and let k and g be the smallest and largest values of i for which p; * 0. Define integers

nk,...,ngby

ng=0 ng-1 = IAFg(p(g,),I

ng-2 = IMFg-1(Pg-1 + ng-1)I nk = I AFk+1(pk+l + nk+1)I.

Then, provided that pi +n;----

(n) i

(8.4)

for each i = k, ... , g, there exists an antichain sd* with parameters p, such that, for each i, the sets in si* of size i, together with the sets of size i contained in larger members of .i*, constitute an initial segment of the sets of size i in the squashed order. We call sB* the squashed antichain with parameters po, ... , p,,. Example 8.1.1 Construct the squashed antichain on { 1, ... , 6) with parameters po =P1 =Ps =P6 = 0, p2 = 2, ps = 3, Pa = 1.

Take the first 4-set, namely {1, 2, 3, 4}. Its 3-subsets are {1, 2, 3}, {1, 2, 41, {1, 3, 4}, {2, 3, 4}, the first four 3-sets in the squashed order. Therefore take the next three 3-sets, namely (1, 2, 5), (1, 3, 51, (2, 3, 5). The 2-sets contained in the sets so far chosen are {1, 2), {1, 3}, {2, 3), {1, 4}, {2, 4}, {3, 4), (1, 5), {2, 5), (3, 5); therefore take the next two 2-sets, namely {4, 5) and {1, 6). Thus the squashed antichain consists of the following sets: {4, 5}, {1, 61, {1, 2, 5}, (1, 3, 5), (2, 3, 5), {1, 2, 3, 4}.

Solution

The importance of squashed antichains arises from the following result which says essentially that the conditions (8.4) are automatically satisfied if any antichain with parameters p; exists.

Theorem 8.1.2 (Clements 1973, Daykin et al. 1974) Let .sB be any antichain on 11, ... , n}; suppose that its parameters are po, . . . , p,,, and that k and g are the smallest and largest values of i for which p; 0 0. Then the conditions (8.4) are satisfied by the P.

134 1 Combinatorics of finite sets

Proof. Let 4 = d k U ... U safe where .si, = {A E .cf : JAS = i }. Since Is4g = pg,

Pg , (n), so that (8.4) holds for i = g. Now there are g

J0(dg)J sets of size g - 1 in the shadow of sag, and none of these can

be in . g_,, so that there are at least pg_, other sets of size g - 1, n I0(,4g)l, i.e. Pg-1 + namely Y the sets in dg-1. Thus Pg-, (g (g n JA(.s8g)J However, by the Kruskal-Katona theorem,

-1 -

1).

JA(dg)i IAFg(Pg)l = ng_,, so that pg-, + ng-1- (g n 1), i.e. (8.4) holds for i = g -1. Next, consider the pg_, + J0(sig)I sets of size g -1 in .44 or contained in larger sets of s$. There are ;pg-, + ng_, such sets, so by

the Kruskal-Katona theorem they contain at least

IAFg_,(pg_, +

ng_,)I = ng_2 sets in their shadow. Since none of these can be in d$g_2,

there must be at least pg-2 other sets of size g - 2, so that we must have Pg-2 + ng_2 = (g

n

2), and (8.4) is satisfied for i = g - 2.

Continuing in this way we find that (8.4) holds for each i. Corollary 8.1.3 If .9.f is an antichain on {1, . . . , n} with parameters po, ... , p,,, then there exists a squashed antichain Ssi with the same parameters.

Note that we have, in the process of studying squashed antichains, obtained the necessary and sufficient conditions we were looking for.

Theorem 8.1.4 The non-negative integers po,... , p are the para-

meters of an antichain on (1, ... , n) if and only if the conditions (8.4) hold.

Proof. Theorem 8.1.2 proves the `only if part; Theorem 8.1.1 proves the `if part.

This result can be presented in a slightly different form. First of all note that if one of the conditions (8.4) fails to hold, say for i = j, then all the subsequent ones, for i <j, will also fail to hold. Therefore we can replace (8.4) by the one condition pk + nk , (k). Further, if we rewrite JOF(m)) as b;(m), we obtain the following corollary. Corollary 8.1.5 Let po, ...

, p be non-negative integers and let k be

Antichains

1

135

the smallest value of i for which p, * 0. Then there exists an antichain on (1, . . . , n) with parameters p, if and only if Pk + bk+1(Pk+1 + 6k+2(Pk+2 + bk+3(.

))) - (J)

We finish this section by pointing out one important property of squashed antichains. It is implicit in the proof of Theorem 8.1.2 that the number of sets of size i contained in sets of A of size greater than

i is at least n;, while the corresponding number of sets for the squashed antichain is precisely nj. We therefore have the following theorem. Theorem 8.1.6 (Clements 1973) If A is an antichain on {1, ... , n} with parameters Pk, Pk+1, ... , Pg, where Pk * 0, pg 0 0, let E(k-1)(s/1)

denote the set of all (k -1)-subsets of (1, ... , n} contained in at least one member of 4. Then the minimum value of jz(k-1)(4)I, taken over all antichains 4 with these given parameters, is nk_1= IAFk(nk)I, attained when si is squashed. 8.2

Using squashed antichains

We begin this section by showing how improvements on the LYM

inequality can be obtained by an elegant use of the existence of squashed antichains. If A is an antichain with parameters po,

.

, Pn,

define

top (4) =

i>n/2 (n

bot (z2) = Z

i
P`

11

P`

(';i) ifnisodd

0

mid (si) = 1

P,n

2 rnm 1\

if n =2m

and let r(s4) = top(4) + mid(st), i (4) = bot(si) + mid(st). Theorem 8.2.1 (Daykin 1984) If 4 is an antichain with parameters Po,

, pn,

then either r(4) -1 or (s4) ,1.

136 1

Combinatorics of finite sets

Proof. Because of Corollary 8.1.3 we can assume that .s8 is squashed. First suppose that n is odd, n = 2m + 1, and let A be the last member of si of size m + 1 in the squashed order. If n E A then, for each i -- m, there is a subset B of A of size i which contains n, and so all the members of d of size i coming later in the squashed order than B must contain n. If we delete n from each of them we obtain an antichain of subsets of {1, ... , n -1), and the LYM inequality yields

p'

1,

(n-11

(s4) -- 1.

i.e.

\i-1

If, however, n f A, then no set in si of size --m + 1 can contain n, so

these sets form an antichain on {1, ... , n -1} and the LYM inequality yields r(sQ) ,1. Finally, if n = 2m, take A to be the `middle' set of size m in 4, and argue as above. Corollary 8.2.2 (Greene and Hilton 1979)

If pr = pn-; for each i then

T(.s:¢) + 1(.f) < 2.

Proof -r(,4) = (si), so both are --1. This generalizes Bollobas's result (Corollary 5.2.3), since p; = pn holds there. Note also that we can deduce EKR from Corollary 8.2.2 (see Exercise 8.2). We now present a theorem the truth of which was first conjectured

by Kleitman and Milner (1973); the proof will be another nice application of the existence of squashed antichains. Theorem 8.2.3 (Daykin et al. 1974) If there exists an antichain with parameters po, . . . , p,,, then there exists another antichain with parameters qo, . . . , qn where

q;

0

ifi
p;

ifi=2n

A +Pn-;

if i > In.

Proof Choose the largest h such that Ph + p,, -h > 0. If n is even and

h = In, then there is nothing to prove. If n is odd and h = I(n + 1) then, by Sperners's theorem, ph_1 + ph -- (h), so we can choose ph_1 +Ph sets of size h as the required antichain.

Antichains

1

137

We now consider the case h > Z(n + 1). We shall suppose that the theorem is true for all antichains s4 for which Zn = h < k, and attempt to prove the theorem for any antichain for which h = k. We therefore

suppose that we have an antichain s4 with parameters po,

, Pn,

with Pi = 0 for i < n - k and for i > k, but with pk + pn-k > 0. We shall suppose first, for technical reasons, that k , 2n + 3. The way the proof goes is to derive from s4 a succession of seven antichains, the last one having the required parameters q;.

Consider first all sets in s4 of size k. Their shadow consists of at least ILFk(Pk)I subsets of size k - 1, so we can choose ILFk(Pk)I of these subsets to replace the Pk k-subsets, giving us a new antichain

s41. Next take complements to obtain an antichain s42i and then obtain S43 from S42 in exactly the same way as s4l was obtained from s4. The antichain s43 has parameters r, as follows:

r,=0 ifi--k rk-1 =Pn-k+l + I

AFk(Pn-k)I

ri=p,,_, ifn-k+/1,
n=0 ifi=n - k. Observe now that s43 is a valid subject for an application of the induction hypothesis. We can therefore deduce the existence of an antichain s44 with parameters ti given by

t;=0 ifi'k tk-1=Pk-1 +Pn-k+1 + IAFk(Pk)I +

IAFk(pn-k)I

t;=p,+p,,_; if1n
if n is even.

Now recall the subadditivity result of Theorem 7.4.2, by which IIFk(Pk)I +

IAFk(Pn-k)i

IAFk(Pk +Pn-k)I.

We can therefore omit IQFk(pk)I + IAFk(Pf-k)I - IAFk(Pk + pn_k)I of

the (k - 1)-sets in J44 to obtain an antichain 45 with the same parameters as s44 except that tk_1 is replaced by tk-1 -Pk-1 +P.-k+1 + I AFk(Pk +Pn-k)I

138 1 Combinatorics of finite sets

Now use Theorem 8.1.1 to obtain a squashed antichain s46 with the same parameters as sd5. The tk_, (k -1)-sets in ' 6 are precisely the first tk_, (k -1)-sets in the squashed ordering, so if we replace the first IOF,k(pk +Pn_k)J of these sets by the first Pn_k +Pk k-sets, we

obtain an antichain whose parameters are the q; required by the theorem. The condition k . in + 3 is needed for the inequality ?n < i < k -1 to make sense. If k <21n + 3 the argment above can easily be adapted.

We next use the existence of squashed antichains to obtain results

about antichains which are complement free. We say that si is complement free if A E 9 A' (t s:Q. For such antichains we can obtain a slight improvement on the LYM inequality and then deduce an improvement on Sperner's bound for the size of such an antichain. Theorem 8.2.4 (Clements and Gronau 1981) If si is a complement free antichain on S = {1, . . . , n} with parameters po, ... , pn, then

Pi + ,# n (n)

P./2

l

1.

(Znn 1)

Proof If n is odd, this is just the LYM inequality, so we now assume that n is even. We can also suppose that pnf2 * 0, for otherwise we just have the LYM inequality again.

By Theorem 8.2.3, there is an antichain d, with parameters . . , q,, where q; = 0 if i < in, qn,2 = p,,,2i q; = p, + pn_, if i > in. Now let d2 = d1' = {A: A' E .4.Q,}. Then d2 is an antichain with parameters n = qn_;, and, by Theorem 8.1.3, we can find a squashed antichain s43 with the same parameters ri as . f2. Note that the largest size of a set in S13 is in. We now (and only now) use the fact that sd is complement free. At most half of the in-subsets of S can be in si, so we must have pn12

qo, .

2

In d3 the p,,,2 sets of size in are the first p,,,2 sets of size in

in the squashed order so that, since they all occur in the first half of the ordering, none of them can contain n. They are therefore subsets

Antichains

1

139

of {1, ... , n - 1} so that, by the normalized matching property,

n-

Qin I AFn/2(Pna)1 %

1 11

_1

Pnn = Pnn

(n Zn We can therefore now construct another antichain st, by replacing the

pn1 sets of size 2n in s43 by Pn/2 of the IAFn,2(p,,,2)l sets of size 2n -1 in their shadow. The antichain .s44 has parameters t; where

t;=0 for i-2n,

for i<

2n - 1. The ordinary LYM inequality for 44 yields E pr +Pn_t+Pnr2-1 +Pnn+11 +Pnn tcna-2

(ni ) which is the required result.

(inn

11

Note that the proof of Theorem 8.2.4 uses the fact that s4 is complement free only to deduce that si contains at most half of the subsets of S of size 2n. The LYM-type inequality of the theorem therefore holds for any antichain with this weaker condition.

As a corollary to the theorem, we immediately obtain the following.

Corollary 8.2.5 (Purdy 1977) If 4 is a complement-free antichain on

an n-set S where n is even, then 8.3

1-41= (inn J.

Parameters of intersecting antichains

Some of the results of the previous sections and chapters seem to

suggest that there is a close connection between antichains on (1, . . . , n - 1) and intersecting antichains on {1, ... , n}. For example, we can compare Sperner's theorem with the EKR result of Theorem 5.1.2, or the LYM inequality with Theorem 5.2.2. In this section we show that a direct connection does exist, and that there is a result for intersecting antichains corresponding to Corollary 8.1.5.

Consider for a moment the EKR result of Theorem 5.1.1. The bound given for the size of an intersecting antichain of sets of size _k is (nk -1), and we can attain this bound by taking all the k-subsets of

140

(

Combinatorics of finite sets

... , n} which contain one chosen element. Therefore the k-sets in the extreme case are just (k - 1)-sets in a (n - 1)-set, with the nth element adjoined to each. We are led to ask the following question: in general, given an intersecting antichain on (1, . . . , n), can we in some natural way correspond to it an antichain on {1, ... , n - 1) in such a way that new light is shed on the theorems that we have been {1,

proving?

Now if we are given an intersecting family .s of k-subsets of ... , n}, the squashed family corresponding to it does not have the property that all its members contain a particular element; therefore squashing down the order does not appear to be a useful idea. However, if we squash in the opposite direction, replacing the {1,

sets of sf by the last jsij k-subsets of S in the squashed order, then the resulting sets are likely to contain the element n, for the later part of

the squashed ordering consists precisely of the k-subsets which contain n. This suggests that we should imitate the approach to squashed antichains of Section 8.1, but this time squash to the end of the ordering.

We have seen (for example in Exercise 7.2) that if A < B in the squashed ordering, then B'
Form the squashed antichain with parameters q;, and then take complements of the sets of this squashed antichain. The resulting antichain has parameters p; and is called the antilexicographic antichain with parameters p; (recall that the antilexicographic order is

the reverse of the squashed order). Since the complements of its members have been squashed to the beginning of the order, its members have been squashed to the end of the order as was required. An alternative way of describing this construction is the following. Let r be the smallest value of i for which p, 0. Replace the p, sets of

s i of size r by the last p, r-subsets of { 1, ... , n } in the squashed order, i.e. by L,(p,). Then replace the p,+l sets of size r + 1 in Si by the last p,+l subsets of size r + 1 which do not contain any of the sets in L,(p,) and so on. In what follows we shall write S.i and ALs4 for the squashed and antilexicographic antichains corresponding to si (i.e. with the same parameters as s4).

Theorem 8.3.1 (Greene and Hilton 1979) Let 4 be an intersecting antichain on {1, . . . , n}. Then the antilexicographic antichain ALs4 with the same parameters as si is also intersecting.

Antichains

1

141

Before proceeding to the proof, we state and prove the following lemma.

Lemma 8.3.2 s4 is an intersecting antichain if and only if, for all i,

(i) A( 4' fl s/4, = o. (ii) A = A(')s4' U s4, U ... U s4, is an antichain. (Here, as usual, si _ {A E s4: JAS = i} and A()s4' = {X: JXI = i, X c A' for some A' E S4'). )

Proof Suppose that .s/I is an intersecting antichain. Since A fl B * 0 for all A, B E s4, A = B' whenever A E s4, B' E sd'. Thus (i) holds for i. Suppose (ii) does not hold. Then we can find A E SY;, B E A(')s4', j < i, A c B. However, B c D' for some D' E S4', so A c D', i.e. A fl D = 0 for some A, D E Si which is a contradiction.

each

Therefore (ii) holds. Conversely, suppose (i) and (ii) hold for all i. Taking i = n in (ii) immediately gives that .sg is an antichain. Suppose it is not intersect-

ing. Then A fl B = 0 for some A, B E .s4, and we can suppose that A E s/,, B E S4;, i + j -_ n. Then A c B' where B' E S4;,_,, so that

A E .s4, fl Aw4', contradicting (i). Therefore si is an intersecting antichain. Proof of Theorem 8.3.1 By Lemma 8.3.2, R, = A(').s4' U ,s/I, U ... U .4.

is an antichain for each i. Now if we construct AL(sB, U ... U . 4,) there will be at least JA(')sB'J sets of size i at the beginning of the

squashed order not containing any set in AL(,41 U ... U 4,), for otherwise ALQ, would not be constructible. Thus S(A(')sY') U AL(4, U ...

U .5i,)

is an antichain. However, the squashed antichain Ssf' has a shadow at

level i no greater than the shadow of s4' at that level, so that c S(A(')s4'); thus A'°(Ss4') U AL(sii, U ...

U s4,)

(8.5)

is an antichain. If we denote ALs4 by ', then %'= Ss/I' and (8.5) becomes

A(')(9'U(N,U...U'b,. condition

of

the

lemma. However, A("(Ss/4') fl AL(si, U U s4,) = 0. Thus At''W fl (9, = 0, i.e. 1v also satisfies condition (i).

Thus

satisfies

S(Aw'ws4') fl AL(s4, U ...

(ii)

U s4,) = O so that

0

142 1 Combinatorics of finite sets

We now show that if the intersecting antichain d is such that JAI , in for each A E d, then ALJi is intersecting for the very simple reason that all the sets in AL.1 contain n.

Theorem 8.3.3 (Greene et at.

1976)

Let s9 be an intersecting

antichain of subsets of {1, . . . , n} such that IAI , in for each A E sf. Then every set in ALS4 contains n. Proof If all the sets in

.

have size k for some k -_ in, then Is'l

by EKR, so that ALsi consists of the last Is'l -- (k \k -1l) 1) k-sets in the squashed order. However, the last (k -1) k-sets are precisely those which contain n.

If si contains sets of more than one size, let k -- in be the largest

size of sets in sd. Let V)(.sf) denote the collection of all those k-subsets of (1, . . . , n) which are in si or contain a member of si; this is just the shade of ss at level k. Then Vtk)(.s) is an intersecting

family (since d is), so we must have IV)(. )I 1(k -1) by the EKR

theorem. Now just as shadows are minimized by taking squashed antichains, shades are minimized by taking antilexicographic anti-

chains, so that (V(k)(ALd)I , VW(4)I - (k -1). Thus V(k)(AL.s ) consists of some or all of the last (nk -1) k-subsets of {1,

... , n}.

Since these all contain n, no set in ALsf is a subset of a k-set not containing n; hence every set in ALs1 contains n.

It follows that, if we are given an intersecting antichain on {1,

. . .

,

n} with parameters p, where p, = 0 for all i > in, there exists

another intersecting antichain with the same parameters in which every set contains n. We can remove n from each of its sets to obtain an antichain on {1, . . . , n - 1} with parameters q, where q, -p;+,. We therefore have the following theorem. Theorem 8.3.4 (Greene et al. 1976) Let po, . . . , p,, be non-negative integers with po = 0 and p, = 0 whenever i > in. Then there exists an intersecting antichain on {1, ... , n) with parameters pi if and only if there exists an antichain on {1, . . . , n -1} with parameters q, =p;+,,

0-_ i
Antichains

1

143

Exercises 8 8.1

Find the squashed antichain Ss4 and the antilexicographic antichain AL4 corresponding to si if (a) 4 consists of all the k-subsets of {1, ... , n}

(b) 4 consists of the subsets {1, 2, 5} of {1,

8.2 8.3 8.4

... , 5}.

{1, 3, 4},

{2, 3}, {2, 4},

Deduce EKR from Corollary 8.2.2 by considering 4 U X. Use Corollary 8.1.5 to prove Theorem 7.4.3.

State a theorem similar to Corollary 8.1.5 for intersecting

8.5

antichains. Give another proof of Theorem 5.2.2 using Theorem 8.3.4.

8.6

Deduce Purdy's theorem in the form IdI , ([n/2]

1) for any + complement-free antichain d of subsets of an n-set (n even or odd) f r o m Theorem 5.2.3 as follows. Let 4 = {A1, ... , A,) be a complement-free antichain on {1, . . . , n}. Define a collection 90 of subsets of {1, . . . , n + 1) as follows:

(a) if A.cA; putA;U{n+1}, A;, A;U{n+1}, A; in PA; (b) if At z) A; put A; U {n + 1}, A;, A! U {n + 1}, A; in PA; (c) if, for some i, A. C A; and Ak cJ Al for all j, k, put A; U {n + 1}, A; in 60.

Then show that (i) A e R

{1,

... , n + 1} -A a R, (ii) 60 is

an antichain, and (iii) I80I = 2 Is4I. Finally, apply 5.2.3. (Greene and Hilton 1979) 8.7

Show that, if all the parameters pi of an antichain are zero except for Pk and Pk+1, the following strengthening of the LYM inequality holds: pk + bk+c(Pllk+l) ll

( k)

(kl

8.8

(Greene and Kleitman 1978) Does there exist an antichain on {1, . . . , 8} with parameters p, if (po, . . . , p8) is equal to (i) (0, 0, 0, 21, 35, 0, 0, 0, 0), (ii) (0, 0, 9, 8, 30, 8, 0, 0, 0), and (iii) (0, 0, 0, 8, 30, 8, 0, 0, 0)?

8.9

Suppose that we have equality in Theorem 8.2.4, where n = 2m. (i) Show that the only parameters p, of s4 which need not be zero are pm_1, pm, Pm+1 (2m 1 >. (ii) Show that pm = 0 or M

144 1 Combinatorics of finite sets

(iii) Show that if ISI = 2m and 60 is a non-empty collection of (m -1)-subsets of S such that Al E 98; IA21= m -1, A, n A2 =

0 A2 E R, then 9 = (m - 1)S. Hence show that if p,n = 0 then or is also zero. (Note: (m - 1)S denotes the set of all (m -1)-subsets of S.) (For further results, see Clements and Gronau (1981).)

9 The generalized Macaulay theorem for multisets 9.1

The theorem of Clements and Lindstrom

In this chapter we shall consider the extension of the Kruskal-Katona theorem to multisets. Recall that the Kruskal-Katona result essen-

tially states that, for given m, the number of (k - 1)-sets in the shadow As4 of a collection s4 of m k-sets is minimized by taking Si to consist of the first m k-sets in the squashed order. In 1969 Clements and Lindstrom obtained a similar result for multisets, and it is the aim

of this chapter to present their work and to look at one or two applications.

We saw in Chapter 7 that the squashed ordering of sets is equivalent to the lexicographic ordering of the reversed (0, 1)sequences which represent the sets. We can identify the subsets with the corresponding sequences, which are in many ways easier to work with. In studying multisets we shall therefore take up this idea and work with n-tuples instead of subsets. Thus let us consider the multiset

to consist of all n-tuples or vectors x = (x1, ... , such that each x, is an integer satisfying 0 xi -- k,. We shall define the rank jxj of x to be IxI = x1 + ... + x,,. The vectors in S(k1, ... , of a given rank can be ordered lexicographically: if and b = (b1, ... , write a 1. As an example, consider S(2, 3, 4). The vectors of rank 3 in lexicographic order are

S(k1, ... ,

003 012 021

030 102 111 120 201 2 1 0. 145

146 1 Combinatorics of finite sets

We shall call a vector of rank k a k-vector rather than a k-subset. If 4 is a collection of m k-vectors of S(k,, . . . , kn) then the shadow Ad of 4 is given by

Ad E 4 for some i --n}. The natural extension of the Kruskal-Katona theorem would be that, for fixed k,, . . . , kn, k and m, the size of Ad is minimized by taking 4 to consist of the first m k-vectors in the lexicographic ordering.

This is in fact what Clements and Lindstrom proved, under the condition that k, , k2 --... < kn. Their result is known as the generalized Macaulay theorem, as it is related to an earlier result due to Macaulay (1927). One of the key ideas in this chapter will be that of compression. We have already met this idea briefly in Chapter 7. If 4 is a collection of

m k-vectors of S(k,, ... ,

the compression C4 of 4 is the

in the collection consisting of the first m k-vectors of S(k,, . . . , lexicographic order. More generally, if .4 = U, 4,, where 4, is the collection of i-vectors in 4, we define the compression C4 of d to be C.nd = U, C4,; thus compression is carried out at each rank level. Further, we say that 4 is compressed if C4 = A We can now state the main result of this chapter. Theorem 9.1.1 (Clements and Lindstrom 1969) Let k, -- k2 , .. . kn and let 4 be a collection of k-vectors of the multiset Then S(k,, . . . , A(Csd) Z- C(Asi).

Note that this immediately implies that jA(Cd)j -- jA.Rdj since IC-641 = J R j for all -64.

The proof of Theorem 9.1.1 will require a number of lemmas. The following lemma has already been proved for k, = ... = kn = 1. Lemma 9.1.2 If 4 is a collection of k-vectors of S(k,, ... , kn) and if A is compressed, then Ad is also compressed. . . , an) and b = (b,, ... , bn) have rank k - 1, that a < b and that b e Asst. We have to show that

Proof Suppose that a = (a .

The generalized Macaulay theorem for multisets

147

a e Ash. Since b e Ash, there exists j _ n such that (b,, . . . , b; + 1, ... , b,,) is in 4. Let i be the first suffix for which a, < b;. We

consider two cases. First suppose that j _- i. Then

(a,,...,a,+1,...,an)<(bl,...,b;+1,...,bn)E.s4. Since sl is compressed, (al, ... , a; + 1, . , an) E 4, so that a E Ad. Suppose next that j > i. If a, < k, for some r > i then (a,, . . . , a,. + so that (al,...,a,.+

1,...,an)E4 and aEA4. If a;
so that again ac Ad. Finally, if i < j and a = (b,, ... , b,_l, b; -1, k;+l, .. , k,,) then, since Iai = Ibl, we must have b = (bl, . . . , b;_,, b;, ... , k, 1, ... , kn) for some t < n. But then j must be t, and so (b,, ... , b;, ki+,,

... , kn) E 4; thus a e Ash.

We now introduce the idea of i-compression. Let 4 be a collection

of k-vectors of the multiset S(k,,

. . . ,

k,,); denote by 4;,d the

collection of those members of 4 whose ith component is d, and let C4i.d denote the first I sz2; dl k-vectors with ith component d. (Thus we are compressing in the restricted field of only those vectors whose ith

component is d; this involves a reduction in the number of effective components by 1 and will therefore enable an induction argument to proceed.) We say that s4 is i-compressed if CsY,.d=sii.d for each

d=0,...,ki.

Let us illustrate these ideas with an example. Consider 2-vectors in S(1, 1, 1, 1) and take 4 = {(0, 0, 1, 1), (0, 1, 0, 1), (1, 0, 0, 1)}.

We show that s4 is 1-compressed. First consider d = 0. The 2-vectors

in S(1, 1, 1, 1) with first component 0 are, in order, (0, 0, 1, 1), (0, 1, 0, 1), and (0,1,1, 0), and the two vectors in 4,.0 are precisely the first two of these. Next consider d = 1. The vectors with first component 1 are, in order, (1, 0, 0, 1), (1, 0, 1, 0), and (1, 1, 0, 0), and the only vector in .RQI., is the first of these. Thus 4 is 1-compressed. It can easily be checked by similar arguments that si is i-compressed for each of i =1, ... , 4. Note, however, that 4 is not compressed since (0,1,1, 0) < (1, 0, 0,1) and (0,1,1, 0) 4 4. Therefore being icompressed for all i is not the same as being compressed. Starting with any collection 4 of k-vectors of S(kl, . . . , kn), we

148 1

Combinatorics of finite sets

now define a sequence of collections .sill, A2, ... as follows. Define

sd1=si ss2 = union of all Cs4l d (d = 0, ... , k1) sin+1 = union of all Csfln.d (d = 0, ... , kn)

(d = 0, ...

.Yin+2 = union of all C.m2; a1

,

k1)

and so on cyclically:

szQrt+1= union of all Cs?md (d=0,

... , k,)

where r = m (mod n), 1-- r , n. If si' # .s4j+1, at least one member of s4' is being replaced by an earlier vector in the lexicographic order. Eventually no more replacements will be possible, so that we have the following lemma.

Lemma 9.1.3 There exists a positive integer p such that sip is i-compressed for all i =1, ... , n.

The next lemma gives us some more practice in dealing with i-compression. We suppose that k, --... -- kn.

... , an), b = (b,, ... , bn), jai= Ibj = k, a < b and bn = 0 or an = kn. Then, if b E Si where si is

Lemma 9.1.4 Let n--3, a = (at,

i-compressed for i = 1, 2, and n, it follows that a E S4.

Proof We shall find an increasing sequence of k-vectors from a to b such that any two consecutive members of the sequence agree in the

first, second or nth component. It will then follow that all the members of the sequence, and a in particular, are in .s . We deal first of all with the case when an = kn. If a, = b 1 then the sequence a < b suffices, so now suppose that a, < b 1. The first subcase which we consider is when a; > 0 for some i, 2 -- i _ n - 1. In this case we have a = (a1, a2, .

.

.

,

an) < (a, + 1, a'2,

.. . , an-1, an)

a2 + ... + where a'2, ... , a'n_, are chosen so that az + ... + an -I - 1, and so that (a2, ... , a;;-,) is as early as possible in the lexicographic

order.

If

a1 + 1= b,

we

have

a<

(b1, az..... an"-j, an) -- b as required. If a, + 1 < b1 and a!'> 0 for some i, 2 -- i , n - 1, repeat the process. Either b1- a, applications of the process will give a sequence as required, or at some stage we

The generalized Macaulay theorem for multisets

1

149

shall enter the second subcase where we have a< ... < a", a' < b1, a"n-1t k) a"2 - . = a"n-1 - 0. But then we have a" = (a",1a", (b n < 2 1, a", ... , kn - b1 + b as required. Note that the middle vector here is in S(k1, ... , kn) since 0 _- (k1- b1) + a' _ kn - b1 + al n We next consider the case when bn = 0. The above argument can be applied to the vectors b' = (k1- b1i ... , kn - bn), a'=(k,-

a,, ... , kn - an). The complements of the vectors in the sequence thus constructed from b' to a' form a sequence of vectors from a to b as required.

Before proceeding to the proof of the Clements-Lindstrom theorem, we need one further result concerning i-compression.

Suppose that Theorem 9.1.1 is true in n - 1 dimensions and that -8 is a collection of (k - 1)-vectors of S(k1,... , kn), k1--... _- kn, such that Osi c 99. Then A (,W) c 0A' for all j -1.

Lemma 9.1.5

Proof Since the lemma is true for j = 1, we use induction on j and consider the induction step from j to j + 1. We let Sk denote the set of all k-vectors of S(k1i ... , kn). Suppose then that (9.2)

A(sd') c _04'.

Our aim is to prove that A(.sJ'+1) c g'+' In proceeding from j to j + 1 we shall use the fact that S4;+1 = U d

C'((.'): d),

i =-j mod n,

(9.3)

so we must study AC(si')i d

If d > 0, the members of A((si')i d) are of two types: there are those whose ith component is d and those whose ith component is d - 1. First consider those with ith component d. They constitute e((`4')i d) n (Sk-1)i.d and, by (9.2), we have d((si')i.d) n (Sk-1)i.d C (N)i.d. Now (.i1)i d has only n - 1 effective components so we can apply Theorem 9.1.1 to it to obtain L(C(('1')i.d)) n (Sk-1)i d C C(0((si2')i d)) n (Sk-1)i d

c C((R')i d).

(9.4)

Next consider those members of O((.s '),.d) whose ith component is d - 1. Since different members of s4' with ith component d clearly give rise to different members of A(sat') with ith component d -1,

150 { Combinatorics of finite sets (9.2) gives I(. ))i dl = I(A('4'))i d-1I -- I(R')i d-1I

for d - 1, so that (9.5)

I C((-04')i d-01-

I

In (9.4) we dealt with those members of i(C((Si')i d)) with ith component d. The other members constitute 0(C((.B'), d)) n (Sk_i)i d_1 which, by Lemma 9.1.2, consists of the first IC((sa2')i d)j (Sk-1)i d_I. Since C((164')1 d_1) consists of the first IC((A')i d-I)I members of (Sk_1)i d-1, (9.5) yields

members of

(9.6)

e(C(('sf')i.d)) n (Sk_1), d-1 c C((W)i d_1).

From (9.4) and (9.6) we now obtain z(C((s')i d)) 9; C((14')i.1) U C((9)1 d-1)

for each d -1. Since, by (9.2), we also have 0(C((-4')i.o) c C(i((s4')i.o) c C((o')i o) we finally obtain, from (9.3),

e(sij")1 = U iC(('s')i d) C U C((1')i.d) = '+1. \\

d

d

Proof of Theorem 9.1.1 We consider a collection of k-vectors of the

multiset S(k......

where k1--... -- k,,. The theorem is easily

seen to be true when n = 2 (see Exercise 9.2) so we assume that the theorem is true for n - I dimensions and consider the induction step

from n -1 to n. By Lemma 9.1.3 there is a positive integer p such that '' = .s1" is i-compressed for all i =1, ... , n. Let IV = (Asi)". If we take 64 = 0.s in Lemma 9.1.5 we get AV c V. The proof now proceeds by showing that V can be altered to Cszi and IV to a subset of COszi in such a way that ACs/i c CO.st is obtained. First we dispose of the case when '' = Sk, i.e. when every k-vector so that Cs/i = Sk. Also, is in V. Then l'l = I sip l = I s$ I = I C. j

Sk_I = / Sk = A'' c W', so we must have W= Sk_1. Since 'I = IOsiI = ICAsil it follows that CAsa2 = Sk_I, and hence OCsaf = ASk = I

Sk_I = COs// as required.

We now can assume that '' # Sk. Let a be the first vector of Sk which is not in '', and let b be the last vector of V. If b a. If b = 0 then Lemma 9.1.4 applied to ' would give a e T', which is a

The generalized Macaulay theorem for multisets

1

151

contradiction, and so we must have b > 0. Define b* = (b,,

.

. .

,

bn - 1)

and also define

a = (al, ... , an-,, an -1)

if an > 0.

Since AV c W, we have b * E'W. Now all the k-vectors in the shade of b*, other than b, must come after b in the ordering; therefore the vector b* in the shadow of '' arises from b but from no other vector in V. Now alter ' and V as follows. Define

V*=(''-{b})U{a} ,*={(91V'-{b*})U{a*}

ifan>0 otherwise.

IV

We show that AV* c'W*. Since b* 0 A('V' - {b}), it suffices to prove that A{a} c W*. Certainly, if a > 0 then a* E A{a} and a* E 9M*. To deal with the other members of A {a), first note that an < kn, since otherwise Lemma 9.1.4 would give a E Y. If a, > 0 for some i _ n - 1

then (al, ... , a; - 1, ... ,

an) is

also in AV since (al, ... , a,-

1, ... , an + 1) precedes a in the ordering and a is the first vector of Sk not in V. However, AV c: W, so we have shown that A{a} c W. To

show that A{a} c IV* we must show that b* 0 A{a}. Therefore suppose that b* = (a,, ... , a, -1, ... , an). Then b* = (al, ... , ail, . . . , an + 1), contradicting the fact that a < b. Having started with N' and W, AV c V. we have shown how to change them to 9V'* and W* with AV* c'W*. Now ''* is icompressed for all i just as V was, so we can repeat the process, each time replacing the last vector in V by an earlier vector. After a finite number of steps we shall compress '' to Csi (since J'V = +Csaf I) and

W' to a set 'lt satisfying AC,4 c lt. Now CAs4 consists of the first JAsil members of Sk_, while, by Lemma 9.1.2, ACS consists of the first JACsiJ members of Sk_l. However, JAC. iI -_ 101ti _ (WI = JAdl, so that ACss E_ CAsd as required.

9.2 Some corollaries The generalized Macaulay theorem plays a role similar to that of the

Kruskal-Katona theorem. Some of its applications to antichains, leading to results similar to those of Chapter 8, will be discussed in

the next chapter, but meanwhile we give a few of the simpler consequences of the theorem in this section, following this with two

152 1

Combinatorics of finite sets

nice applications due to Clements in the closing sections of the chapter.

If 4 is a collection of k-vectors of S(k1i ... , k,,), define A'(.4) inductively by A1(4) = Ash and A'+1(64) = A(A'(s4)) for each r .1. Then A'(sj) is the shadow of s2 at level k - r. Corollary 9.2.1 For each positive integer r and each collection s4 of k-vectors in S(kl, . . . ,

A'(Cs4) c CA'(4).

Proof The case r = 1 is just Theorem 9.1.1. Assume that it is true for r and proceed by induction. Ar+1(Csj) = A(A'(Cd)) c A(CAr(Sj)) C

The next corollary concerns ideals.

If Si = saf 1 U S42 U

... is a

collection of members of S(k1, . . . , k.), where sz4, denotes the set of elements of 4 of rank i, we call si an ideal if A.R4, c ,R4,_1 for each i.

Corollary 9.2.2 If s4 is an ideal of subsets of a multiset then so is CS4.

Proof The proof is precisely that of Theorem 7.5.2.

The next corollary re-expresses Theorem 9.1.1 in terms of the shade V 4 of 4. Corollary 9.2.3 Under the conditions of Theorem 9.1.1,

V(Lsi) c L(Vs4) where Ls4 denotes the last tJ k-vectors of S(k1,

... , k,,).

Proof Consider the mapping which sends (x1, ... , . ... , k,, If si is mapped onto si', then C.R4' is mapped onto La and Asd' is mapped onto V4. Thus &(CsV) c C(Asd') becomes V(Ls/i) c L(ast).

x1,

Our next result extends the subadditivity property of sets (Theorem 7.4.2) to multisets.

Corollary 9.2.4 (Clements 1977b) Let Fk(r) denote the first r k-vectors of S(k1i ... , Then, if a + b -- I(k)SI, the number of

The generalized Macaulay theorem for multisets

k-vectors in S(k,,

. . .

1

153

,

IEFk(a + b)I , IAFk(a)l + I AFk(b)I

Proof (Kleitman: see Clements and Gronau (1981)) We introduce

the multiset S'(1, 1, k,, ... ,

and let Fk(r) denote the first r

k-vectors of S'. Let si = {(0, 0, x): x is a (k + 1)-vector of S} 93 = {(0, 1, x) :x E Fk(a)} CC = {(1, 0, x) :x E Fk(b)}.

Let (k)S denote the set of k-vectors of S. The first

1(k + 1)SI

(k + 1)-vectors of S' are those in si; the next a + b are the first a + b (k + 1)-vectors of the form (0, 1, y) where y is a k-vector in S. Thus

IAFk+,(a+b+I(k+1)SI)I =IA(k+1)SI +IIFk(a+b)I. (9.7) However, by Theorem 9.1.1, we also have

IAFk+,(a+b+I(k+1)SI)IA(siU!J9 U1,66) Io(k + 1)SI + I AFk(a)I + IAFk(b)l. (9.8)

Comparing (9.7) with (9.8) we finally obtain I LFk(a + b)I = IOFk(a)I + IOFk(b)I

0

as required.

The final corollary will be needed in the next section. As usual, we thus IxI = x, + ... + x,,. If A is any collection of vectors we define the weight w(A) of i by

write IxI for the rank of x = (xl, ... ,

w(s4) = E IxI. xea

So far we have considered the lexicographic ordering of the vectors in a collection . only when all the vectors in .s have the same rank. We can, however, order lexicographically any collection of vectors of and irrespective of their ranks. If x = (x,, ... , S(k,, . . . ,

y=(y...... y,,)wewrite x
x, 1. When we consider ideals we have two comparisons between vectors: their ordering as vectors and their possible inclusion relation as subsets of the multiset. If x is a subset of

y in the multiset, i.e. if x; s y, for each i, then certainly x , y in the lexicographic ordering, but conversely if x -- y it is not necessarily true that x is a subset of y.

154 1 Combinatorics of finite sets

Corollary 9.2.5 (Clements and Lindstrom 1969) If 4 is an ideal in S(k,, ... , and l.il = m, then w(.set) < w(Fm) where F,,, denotes the first m vectors of S(k,, ... , in the lexicographic order.

Proof By Corollary 9.2.2, Csi is an ideal, and clearly w(C4) = w(4), so we might as well assume that 4 is compressed. If si = Fm be there is nothing to prove, so suppose 4 # F m . Let a = (a,, ... , the first vector which is not in 4 and let b = (b,, ... , b,,) be the last vector in saf. Then a < b. Suppose for the moment that we can prove that lal > lbl. We can then delete b from s4 and replace it by a to obtain an ideal ,s4, with w(s/I,) > w(4) and with si, compressed. If .m2, # Fm we can repeat this replacement process, and after a finite number of applications we obtain 4 = Fm, and w(Fm) = w(s4) >

w(4). We therefore have to show that lal > lbl. First note that if lal = lbl then it follows from the facts that a < b, b E sat, and s/I is compressed (at each rank level) that a E 4, which is a contradiction. Therefore we now suppose that lal < lbl and hope for another contradiction in this case. Since a < b we have, for some i, a; < b; and a; = b; for all j < i. If

we had a;+, = ... = a = 0 we would then have a contained in b as subsets, so that, since b c- d where 4 is an ideal, we would have a E Si, which is a contradiction. Therefore we can now suppose that a, > 0 for some j > i. Put d = l b l - la l > 0, and let d' = d min(d, b; - a, - 1); then we have d'-- b;+, +. . . + b (if d' = d - b; +

a;+1>0 then b; + a; =

d'). Now consider the vector c = (c,, ... ,

where

c, = a,, ... , c;_, = a,_,, c, = b; - d + d' and where c;+,, ... , c are obtained from b;,,_,, . . . , b by subtracting a total of d' in such a way that c;+,, . . . , c are non-negative integers; this is possible since d' -_ b;+, + ... + b,,. We have c; , b; for each i, so that, since b E s/I and d is an ideal, c E s/I. However, lal = lcl and a
9.3 A minimization problem in coding theory In this section we use Corollary 9.2.5 to solve a problem which arises in coding theory. Suppose that we are to send a message which is in the form of a string of non-negative integers, the permitted integers being 0,1, . . . , m-1 where m = (k, + 1) ... (k + 1). Each of these

m integers is to be encoded as a code vector (x......

with

The generalized Macaulay theorem for multisets

1

155

0 , x, , k, for each i, i.e. as a vector in S(k......

Now during the transmission of the code vectors some errors may possibly arise; one of the x, may be received falsely with the result that the received vector will be interpreted as representing an integer different from that intended. Let us assume that at most one error will be made in the transmission of any code vector. Then each received vector will be either the one that was sent or one of its neighbours, where we say that two vectors in S(k...... are neighbours if they agree in all but one component. Now there are many ways of choosing which

vector should represent which integer, and this gives rise to the following question: how do we choose the vectors to represent the integers so as to minimize the likely error involved on receiving a message? If we let x represent the integer g(x) where, as x varies over . . , k.), g(x) varies over {0, 1, . . . , m -1}, a

the set S = S(k,, .

reasonable definition of the error associated with the assignment g would be e(g) = E AX, Y) 1g(X) - g(Y)I x
where f(x, y) is 1 if x and y are neighbours and 0 otherwise. The problem now becomes: which assignment g minimizes s(g)? For any integer 1 define h,(a, b) by 1

h,(a, b) _ tO

if a_1andb>1 otherwise.

Then m-1

m-1

Z 1=0

Y,

f(x, y) = Y. Z h,(g(x), g(Y))f(x, y) 1=0 x.yEs m-i

_ x.yeS E f(x, Y) Y, h,(g(x), g(Y)) 1=0 x#y

m-1

= Z*f(x, y) E (h1(g(x), g(Y)) +h,(g(y), g(x)) ,=0

(9.9)

where E * denotes the sum over unordered pairs (x, y) with x 0 y. Now if g(x) * f(x, Y) 1g(y) - g(x)I = E(g)-

156 I Combinatorics of finite sets

Thus

I I

M-1

E(g)

1=0 g(x)-l
g(x)1 YES

f(x, y)

f(x, y) - g(x).Qr f(x, y) g(Y) c1

m-1

M-1

EI

E E AX, Y)

1=0

1=0 g(x)c1

m-1

m-1

1=0

I=0

Ay)-rl

E

e(1)

where e(l) is the number of neighbouring pairs of vectors involving only those vectors representing 0, . . . , 1. Thus to minimize E(g) we have to maximize E!-o1 e(l). Ideally the maximum of this sum will occur when each e(l) is individually maximized, but that might be asking for too much. Surprisingly, we find that we can maximize each e(l) simultaneously!

If T is any subset of S = S(k1i .

.

.

,

let E(T) denote the

number of unordered pairs of neighbours in T. If each k; = 1 it is

natural to think of a pair of neighbours, differing in only one component, as an edge of an n-dimensional hypercube. We also use the word edge in the more general multiset situation to denote a pair of neighbouring vectors. Lindsey (1964) showed that, for given 1, the way to choose a subset T of S consisting of I vectors so as to maximize

E(T) is to choose T to consist of the first I vectors of S in the lexicographic order. Thus e(l) is maximized by representing each integer j by the (j + 1)th vector of S in the lexicographic order. Before formally stating and proving Lindsey's result, we point out a connection between E(T) and w(T), where w is the weight function of Corollary 9.2.5.

Lemma 9.3.1 If T is an ideal of vectors of S(k1, ... , E(T) = w(T).

then

Proof For fixed y = (y1, ... , yn) E T, the number of edges (x, y) with x < y is yl + ... + yn = l y 1. Thus

E(T) = YET I IyI = w(T). Theorem 9.3.2 (Lindsey 1964) Let F, denote the first I vectors of in the lexicographic order, and let T be any set of 1 S = S(k1, . . . , vectors of S. Then E(T) -- E(F,).

The generalized Macaulay theorem for multisets

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157

Proof (Clements 1971) We shall show that the theorem follows from Corollary 9.2.5. Suppose that T is a set of l vectors such that E(T) is

maximum; we shall show that T can be replaced by F without decreasing the number of edges. Since the theorem is trivially true when n = 1, we proceed by induction and consider the induction step from n - 1 to n. We recall the notation: T, i is the set of all those members of T whose first component is i. Thus let T be a set of I vectors of S such that E(T) is maximum. Let ai = T, il. We can, without loss of generality, suppose that ao = a,

.. ' ak, since we could if necessary replace T by a set for which this is true and without altering E(T); this can be achieved by permuting 0, 1, . . . , k, appropriately in the first component. If i --j we shall say that an edge (x, y) is an (i, j)-edge if x E T, i and y E T, ,. Let NN,j(T) denote the number of (i, j)-edges of T. Then k,

k,

E(T) = I Z N;,j(T). i=0j=i

(9.10)

Suppose now that T is replaced by the set T' consisting of the first a, members of S, i for each i; in other words, T is 1-compressed into T'. Then by the induction hypothesis Ni,i(T') , N,,i(T) for each i. Also, if i <j we have NV,j(T) -- aj, whereas Ni1,(T') = aj, since the edges contributing to Ni1j(T') are the pairs of vertices in T' coinciding in all

other components and there are min(ai, aj) = aj such pairs. Thus Ni.j(T) _ Ni1j(T') for all i --j, so that (9.10) gives E(T) , E(T'). Finally, T' is an ideal (see Exercise 9.5), so by Lemma 9.3.1 E(T) ' E(T') = w(T'). Thus, by Corollary 9.2.5, E(T) , w(T') w(F) = E(F). Corollary 9.3.3 The error e(g) is minimized by choosing the

assignment g given by g(x) = j where x is the (j + 1)th vector of

S(k...... 9.4

in lexicographic order.

Uniqueness of maximum-sized antichains in multisets

We now return to the problem of finding the largest antichains in a multiset. We saw earlier, using the symmetric chain decompositions, that if .s4 is an antichain of subsets of S(k,, ... , k where Ni is the number of subsets of rank i. The special case of each k; =1 is of course Sperner's theorem, and we recall that in that case the antichains 4 with 141 = NN have been identified: if n is even there is a unique maximum-sized antichain,

namely the set of all [n/2]-subsets, whereas if n is odd there are

158

1

Combinatorics of finite sets

precisely two, the set of all 2(n -1)-subsets and the set of all 2(n + 1)-subsets. At first sight it might have seemed possible to obtain

an antichain of size

((F1

1)) consisting of a mixture of sets of sizes

2(n - 1) and 2(n + 1), but we saw that such a mixture cannot give a large enough antichain.

Now consider the corresponding problem for multisets. Let k, k2--. . .--kn be given and let K = k, + ... + kn. Is it true that if K is even the only maximum-sized antichain consists of all subsets of rank 1K? Similarly, if K is odd, is it true that there are only two antichains of maximum size, namely all subsets of size 2(K - 1) and all subsets of size 2(K + 1)? If kn > ZK the answer is generally no. For example, Clements (1984b) has pointed out that in the multiset S(1, 2, 4) there is a maximum-sized antichain which is a mixture of sets of ranks 3 and 4:

{(0, 0, 4), (1, 0, 3), (0, 1, 2), (1,1,1), (1, 2, 0), (0, 2,1)}.

However, if kn , 2K we do obtain the result we hoped for.

Theorem 9.4.1 (Clements 1984b) Let sift be a maximum-sized antichain of subsets of the multiset S(k,, ... , Q, where k, -_ k2 .

__kn_-2K, K=k,+... +kn. Then

(i) if K is even, szf is the set of all subsets of rank 1K; (ii) if K is odd, 4 is either the set of all subsets of rank 12(K - 1) or the set of all subsets of rank 2(K + 1). Proof Let .l = [21K]. Since 2k, _-K we know, by Theorem 4.1.1, that

Na
NA>... >NK (K even)

and

No
NA+,>...>NK (K odd).

Let . have parameters p,; thus p, is the number of members of Si of rank i. By the LYM inequality we have ,

Ni

.1

and it follows from this that if p, > 0 for any i with N; < NA then

IsiI =Zp,=EN`N,
and of ranks 2(K - 1) or 2(K + 1) if K is odd. The theorem will

The generalized Macaulay theorem for multisets

1

159

therefore follow if we can show that 4 cannot consist of a mixture of sets of ranks Si 2(K 03 - 1) 'e and 2(K + 1). Now if it were possible to have an antichain _ U where consists of sets of 1,91 rank 2(K - 1) and 16 _31 consists of sets of rank 2(K + 1) and where + ICI = Nx, then w4 we = o since si is an antichain, so that IV -041 , J j. But, by the

normalized matching property IV!@l ? 191, and so we must have IV@l = I@1. We can therefore complete the proof of the theorem by excluding this possibility as follows.

Lemma 9.4.2 Let 130 be a collection of subsets of S(k,, . . . , k ), each of rank A, where k, _- k2 k , 2(K - 1) jai. = A, K odd. Then if 93 does not contain all of the A-subsets, IVI(>

... ,

Proof By Corollary 9.2.3,

IV,13BI ,

so we might as well

assume that consists of the last JAI A-subsets of S(k,, consider two cases.

. . .

, Q. We

(i) Suppose first of all that all members of are of the form x = (x,, ... , where x, ;1. (We are reverting to thinking of subsets as vectors.) For x = (x,, ... , E !30, write x` = (k, -

x,, ... , k -

and put 90` = {x` :x E-04 }. Then @` can be considered

as a collection of (K - A)-vectors of S(k, - 1, k2, . , Q. This multiset has rank K` = K - 1 where K` is even, k _ ZK`, and .

.

K - A> IK`; therefore, by Theorem 4.1.1, NK_A_, > NK_A and the normalized matching property gives I0OA`I -_

K x-1

N K_x

l_A`i _

I.

(ii) Suppose now that not all members of 36 have first component ?1. Let P° denote the set of members of OA with first component 0. Form the set by taking all those A-vectors of S(k,, ... , which have first component 0 but which are not in and deleting the first component; thus is a set of A-vectors in S(k2, ... , k,,). Similarly form ii; by taking all those Q. + 1)-vectors of S(k,, ... , with first component 0 which are not in V-' and deleting the first component; thus W is a set of (A + 1)-vectors in S(k2, . . . , Q. We then have

A9 c 2. If we now let N, denote the number of i-vectors in S(k2,

... ,

we then have

I 98j_19J°l+Na-,+...+Nx-k, and IV- 641 = j(V_6h)°j +Nx+... +Na+,-k,

160 I Combinatorics of finite sets

so that IV99I - IRI = (Nx - I-640I) - (Na-k, - I(V94)°I)

= (Nx - IR°I) - (Nz+, - IMO)°I) = ICI - 191

If W = 0 we have I V, -ICI = ICI > 0 as required; therefore suppose now that ' # 0. Since A % max(k,,, 2(K - k,)) we have N), > Na+1 by Theorem 4.1.1, so that

oW . whence

ICI = theorem, is proved.

Nx Nx+I IWI>IWi

ICI > 0. Thus the lemma, and hence the

0

Exercises 9 9.1

9.2

Verify Theorem 9.1.1 for the multiset S(1, 2, 3, 4) and k = 4 in the case when s1 = {(1, 0, 2, 1), (0, 1, 1, 2), (2, 0, 2, 0)) by listing all the members of Csi, As4, OCs/t and C&9. Prove the case n = 2 of Theorem 9.1.1 as follows. Consider the multiset S(k,, k2) where k, k2 and let .sd be a collection of

k-vectors. First suppose k k,. Then si will be a union of blocks of consecutive members of the list (0, k), (1, k - 1), (2, k - 2), ... , (k, 0). Show that if 99 is any such block other

than the collection of all k-vectors, and Rp is any block

9.3

containing (0, k), then IARI - IOhI . IAMOI - IAI. Show that this conclusion is also true if k > k,. Then show that S4 I 1,41-- IACs/l - IC,41, and deduce the theorem. by N,. Show that if Denote the rank numbers of S(k,, . . . , N, < m < N,,, then there exists an antichain si, I s4I = m, of the form F1,1(j) U {(1)S - AF,+,(j)}.

(Clements 1974)

9.4

Is it true that if a set of k-vectors is compressed then it is

9.5 9.6

i-compressed for all i? Prove that the set T' in the proof of Theorem 9.3.2 is an ideal. Show that if s4 is a collection of k-subsets of an n-set S with

The generalized Macaulay theorem for multisets 1161

IsiIn

1 }then k

1

n-1l

-11

(n-1

ICI

k 9.7

by using an idea in the proof of Lemma 9.4.2. Given a multiset S(k...... and a positive integer 1, let M denote the largest integer such that t -_ M implies I AF(t) % t. (i) Use the marriage theorem to show that if X1, . . . , X, are t _ M 1-subsets then there exist t distinct (1-1)-subsets Y1, . . . , Y such that Y c X, for each i. (ii) Show that if t > M then no such conclusion is possible.

(iii) Generalize to the problem of finding t distinct (1- k)subsets Z,, . . . , Z, such that Z; c X, for each i. (Clements 1970)

10 Theorems for multisets

10.1

Intersecting families

In this section we shall attempt to extend to multisets some of the intersection results already obtained for sets. We begin with the problem with which we started Chapter 1. Now this problem was concerned with an intersecting family ss of subsets of a set, and we are therefore faced immediately with the following question: how do we extend the concept of intersection to multisets? It has often been as vectors helpful to think of subsets of a multiset S(k, . . . , x = (x,, ... , with 0 _- x; , k; for each i. In the case k, _ ... _

k = 1 the non-disjointness of two sets A, B corresponds to their representing vectors x and y having the property that x; = y, = 1 for some i. If we think of this as saying that min(x,, y,) > 0 for some i, we obtain a natural extension of the intersecting property which we shall call property I, . is said Definition 10.1.1 A collection s & of vectors of S(k,, ... , to possess property I, if, for all x, y E S , min(x,, y;) > 0 for some i.

In terms of divisors of a number, property I, requires that if .1 is any collection of divisors of m with property I,, then g.c.d(a, b) > 1 for all a, b E .QQ. Here g.c.d. denotes the greatest common divisor.

There is, however, another natural way of extending the intersection property to multisets. To have x; = y; = 1 for some i in the case when k, =. . . = k = 1 is to have x; + y, > k; for some i. We therefore make the following definition. . , k,,) is said to possess property I2 if, for all x, y E sQ (x = y allowed!), x; + y, > k; for some i. In terms of divisors of m, s$ has property I2 provided that, for all a, b E .s4, ab does not divide m.

Definition 10.1.2 A collection s4 of vectors of S(k,, . .

With these two definitions we have obtained two

different

properties 11 and 12, both of which reduce to the same intersection 162

Theorems for multisets 1 163

property in the special case of k, _ ... = k = 1. We shall look at both of I, and I2, beginning with I. Recall that Theorem 1.1.1 asserts that an intersecting family of subsets of an n-set has at most 2"-' members, and that any maximal intersecting family (i.e. an intersecting family contained in no larger intersecting family) has exactly 2"-' members. Is there a similar result for multisets and property I,? We find that we can obtain a bound

f(k...... k") corresponding to 2", but it is not true that every maximal I,-family has as many as f(k...... kn) members.

Whether or not two vectors x, y satisfy the condition that min(x,, y,) > 0 for some i really depends only on which components x,,

y, are zero and which are non-zero, and not upon their actual numerical values. For this reason, the idea of the support of a vector is useful.

Definition 10.1.3 The support s(x) of the vector x = (x,, ... , xn) is given by s(x) = {i :x, * 0}.

Lemma 10.1.1 A collection si of vectors has property I, if and only if s(si) = {s(x):x E Si) is an intersecting family. Proof min(x,, y,) > 0 for some i ax, > 0, y, > 0 for some i' i E s(x) fl s(y) for some i<* s (x) fl s(y) * 0. It is clear that if s4 is a collection which is maximal with respect to I, and if si contains a vector with support J, then si will contain all vectors with support J. If J is any subset of {1, . . . , n}, the number of vectors in S(k1, ... , k"} with support J will be IIIE k,. We denote this number by w(J) and call it the weight of J. Thus

w(J)=[Tk,. IE

Theorem 10.1.2 (Erdos and Schonheim 1970) Let si be a collection of vectors of S(k,, ... , kn) with property I,. Then Id)

f(k,,...,k")where

f(k,, ... , k") = 1

max{w(J), w(J')}, .n)

J' being the complement of J in {1, ... , n}.

164 1 Combinatorics of finite sets Proof

ISiI=z2J {IXES/i:S(X)=JI +IXESi:S(X)=J'I) Y, max{w(J), w(J')) J

since J and J' cannot both occur as supports of vectors in 4. It is straightforward to construct a collection of s4 with property I, and such that IsQI = f(k,, . . . , k"). For any J with w(J) > w(T) put into 4 all vectors x with s(x) = J. For all J with w(J) = w(J') put into

s2 all vectors x with support J (or J') according as 1 E J (or J'). Observe that w(J)w(J') = k, ... k" and that, if J, fl J2 = 0, then w(J,)w(J2)' k, ... k". It follows that if w(J,) > w(J,) and w(J2) w(JZ) then J, and J2 are not disjoint. Thus s(s4) is an intersecting family, and 4 has property I. Example 10.1.1

Consider S(2, 2, 5): w({1, 2, 3 ) )=2 . 2 . 5=2 0>w ( 0 )

w({1, 3}) =10> w({2)) w({2, 3)) =10> w({1}) w({3}) = 5 > w({1, 2)) so that, if sf is any collection of vectors with property I,

1s41 ,2{20+20+10+10+10+10+5+5}= 45. An example of a collection of size 45 is, by the above construction,

the set of all vectors in S(2, 2, 5) with supports {1, 2, 3), {1, 3}, {2, 3}, {3}, i.e. all vectors with third component ;1. Unlike the situation for sets, where any intersecting family can be . . , 1) = 2"-', there exist maximal I,-collections of size less than f (k,, ... , k") which cannot be extended. For example, in S(2, 2, 5) considered in the example above,

extended to one of size f(1, .

the collection of all vectors x with x, :1 is maximal, but has size 36 < 45. It is clear that in S(k,, ... , k") we can similarly obtain a maximal I,-collection of size k,(k2 + 1) ... (k" + 1). We now show that, if k, _ ... , k", then every collection of size less than k,(k2 + 1)

... (k" + 1) can be extended.

Theorems for multisets

1

165

Theorem 10.1.3 (Erdos, Herzog, and Schonheim 1970). Let sat be a collection of vectors of S(k,, ... , k"), k, _ k2 _ ... = kn, with pro-

perty I,, which cannot be extended to a larger collection with property I. Then

Isil-_k,[I(k,+1). i.2

Proof Let s(x) denote the support of x and let 99 = s(4) = {s(x):x E 4}. Then R is, by Lemma 10.1.1, an intersecting family. Since 4 cannot be extended, neither can R, so that, by Theorem 1.1.1, 1,641 = 2n-'. Let y8' denote the set of complements of members

of 64. Then, since 1.121 = 2"-' and 64 cannot contain a set and its complement, R U 64' is a partition of the set of all subsets of {1, . . . , n), and we have Is4 l

=

2

w(J ) =

E

k ... k,,

( 10 . 1 )

.

Therefore we must prove that

Z k1...k",k,fi(1+k;). E OR,

w(J)

(10.2)

i;-2

If (1, 0, ... , 0) E st then every vector with x, ,1 will be in s4, and IsiI % k,(k2 + 1) ... (k" + 1) is trivially true; therefore we now suppose that (1, 0, ... , 0) f s4, i.e. {1) 0 R We then have {1} E 99'. Consider the partition of @' : R' = R( '1) U O3; where a('I) consists of those sets in 9D' which do not contain 1, and 9B; consists of those sets in which do contain 1. Then, from (10.1),

_ ICI

kl ..k"

..k"

kl

(10.3)

w(J)

w(J)

M(')

Now

k,...k" w(J)

w(J)

_

w(J)

lea

1E

IEl

= k,

w(J) = k, fl (1 + k1), g(2,

n)

i;w2

so that, by (10.3), proving (10.2) is equivalent to proving that

Zk,...k,Ew(J) Eaj

w(J)

EMj

166 1 Combinatorics of finite sets

1.9 k1w(I)

Z(I)

lee

where 2 is the set of all members of A,' with the element 1 removed. But we can rewrite this as IEe

w(I)

>w(I)

lee

k, E w(I) -- E w(J) IE9

(10.4)

JET'

where is a collection of subsets of {2, ... , n}. We now observe that 98' is an ideal, so that is an ideal of subsets of {2, . . . , n). Also, 1.74 has the property that X E 2 => X' IV@ where

X' denotes the complement of X in {2, .

. .

,

n}. We can therefore

make use of Theorem 6.4.1 which now tells us that if 2 = {D,,

. . .

,

Dr} then there is a permutation 0 of {1, ... , r) such that

D; c D,(,) for each i. Corresponding to each term w(I) in the sum on the left of (10.4) there is therefore a term w(J) in the sum on the right

with I c J, so that w(J) k2w(I), k2 being the smallest of the k 2 < i -- n. We therefore have

k, Y, w(I) -- Z k2w(I) ` 2 w(J), lee

lee

JEe'

thus proving (10.4).

0

We now turn our attention to I2i the second extension of the intersection property. In this case we obtain a result very similar to Theorem 1.1.1. The argument involves the following collections 1, of in S(k,, . . . , vectors x = (x,, . . . , 98, = {x : x, > Jk, } 902 = (x : x, = '2k,, X2 > ?k2)

2,={x:x,=1k,,...,x,-I =lk1-,,xr>zk1) ..78={x:x,=Jk,foralliZkrt}.

Theorems for multisets

It is clear that

1

167

= 901 U. . . U Oh,, possesses property 12.

Theorem 10.1.4 (Greene and Kleitman 1978)

Let si be a collection of vectors of S(k1, ... , k,,), k1, ... , k,,, possessing property I2. then

1,41-16-41=[21 ll(1+k,)]. ,

Proof Define S*(k1, . . . , to be S(k1, ... , k,,) if k; is odd for at least one i, and to be S(k1, ... , k,,) - {(lk1, ... , Zk )} if all the k, are even. Note that if x E S*, then X e 9947>x' = (k1 - x1, . . . , k E S* - R. If s4 satisfies the conditions of the theorem then, for any x, x and x` cannot both be in s4. Thus, if some k; is odd, 1.41-1-041=ZIS*(k1i...,k,,)1=2 fl (1+k,),

whereas if all the ki are even,

191-1R1=12 (fj (1+k,)-1). t

Having extended Theorem 1.1.1 to multisets, it is natural to turn next to Theorem 6.1.4 and to seek extensions corresponding to I, and 12. We make the following definitions. has Definition 10.1.4 A collection of vectors in S = S(k1i . . . , property U1 if, for all x, y in the collection, max(x,, y,) < k; for some i. Similarly it has property U2 if, for all x, y in the collection, x; + y, < k, for some i.

In the language of divisibility, a collection 9 of divisors of m has property U1 if the least common multiple 1.c.m.(a, b) of any two members a, b of sQ is less than m, and has property U2 if, for all a, b E Si, m does not divide ab.

Theorem 10.1.5 (Anderson 1976) If s1 is a collection of vectors of ... , with properties I, and U1, then

S(k1i

1 ` {f(k1, ... 1

,

II, (1+ k,)

where f is the function of Theorem 10.1.2.

Proof Imitate the proof of Theorem 6.1.4.

kn)}2

168 1 Combinatorics of finite sets

Theorem 10.1.6 (Daykin et al. 1979) If d is a collection of vectors of S(k,, ... , with properties 12 and U2, then

ICI`[Zrl,(1+k,)]2 f 1i

+ k,)

Proof Again imitate the proof of Theorem 6.1.4, using Theorem 10.1.4.

Neither of these bounds is best possible in all cases. An exact bound for S(k, k, . . . , k) and I, and U, has been found by Engel and Gronau (1985). For I2 and U2, see Exercise 10.3. 10.2

Antichains in multisets

In Chapter 8 we obtained a number of results concerning the parameters of antichains of subsets of an n-set. In this present section we shall see that many of the results for sets extend in a natural way to multisets. The underlying reasons for this include the facts that the LYM inequality and normalized matching property hold for multisets, and that the generalized Macaulay theorem can be used in much the

same way as the Kruskal-Katona theorem was used in Chapter 8. where k, , ... , k,,. Throughout this section we consider S(k...... First of all we consider squashed antichains. If we go through the proof of Theorem 8.1.1 we find that, on replacing (n) by i

Ni,

the ith

rank number of the multiset, and squashed ordering by lexicographic ordering, the proof goes through in exactly the same way. Similarly the proofs of Theorem 8.1.2 and Corollary 8.1.3 can be followed and suitably adapted. So, corresponding to any antichain in the multiset we can construct a squashed antichain with the same parameters and the property that, for each i, the vectors of rank i in it, together with

the vectors of rank i in the shadow of members of greater rank, constitute an initial segment of the lexicographic ordering of the vectors of rank i. The remaining results of Section 8.1 then hold also. Summarizing, we have the following theorem.

Theorem 10.2.1 (Clements 1973) There exists an antichain in

S(k,, ... ,

with parameters p, if and only if p, + n, , Ni holds for all i -- g, where g is the largest value of i for which p, * 0, and where {n,} is defined by ng = 0 and n,_, = IAF(p, + n,)I, where, as usual,

Theorems for multisets 1169

Fk(m) denotes the first m k-vectors in the lexicographic ordering. Further, if these conditions are satisfied, there is an antichain called the squashed antichain with parameters pi such that, at each rank level

the pi vectors, together with those vectors in the shadow of members of the antichain of greater rank, constitute an initial segment in the lexicographic ordering. This antichain minimizes i,

IL(')sij, the number of r-vectors contained in at least one member of ss, over all antichains si with parameters A. Our next observation is that the proof of Theorem 8.2.3 made use of squashed antichains, the Kruskal-Katona theorem and the subadditivity result of Theorem 7.4.2. Since subadditivity still holds in multisets (Corollary 9.2.4) and the Clements-Lindstrom theorem

takes over from Kruskal-Katona, the proof of Theorem 8.2.3 immediately yields the following theorem.

Theorem 10.2.2 (Clements 1977a) Suppose that there exists an with parameters pi, and let K = Ei ki. antichain in S(k,, . . . , Then there exists another antichain with parameters qi given by

ifi
0

qi= pi Pi+pK_i

if i>ZK.

0

Using this result we can now extend Theorem 8.2.4. Theorem 10.2.3 (Clements and Gronau 1981) Let d be an antichain where K = Ei ki is even, such that with parameters pi in S(k,, . . . ,

sil contains no more than w vectors of rank ZK, where w is the with first component number of vectors of rank ZK in S(kl,... , less than k1. Then

2

PK/2 Pi + i:.ici2 I(i)SI I(ZK - 1)SI

where I(i)SI denotes the rank number Ni.

Proof As in the proof of Theorem 8.2.4, there is a squashed antichain sd3 with parameters ri given by ri = 0 if i > ZK, rK/2 = PKn,

and ri = pi + pK_i if i < ZK. Now since PK12 , w, all the vectors of rank ZK in SC must have first component ,k, - 1; they can therefore be thought of as ZK-vectors in S' = S(kl - 1, k2, ... , k,,). Applying

170

1

Combinatorics of finite sets

the normalized matching property in S' to them, we find that IOFxn(PK12)1%

1(IZK

- 1)S')

K1)l

Pxn=PKn

We can therefore construct another antichain s94 by replacing the PKn sets of size 2K in 43 by PK12 of the sets in their shadow. The antichain s14 so constructed has parameters t; where t; = 0 for i a K/2,

tK/2-1=PKn-1+Pxn+Pxn+1 and t,=p,+pK_, for each i
E

°1 +PK-i +Pxn-1 +Pxn +PKn+1.11. K(2K - 1)S, (i)S Since I(i)S) _ I(K - i)SI for each i, we thus have £
Pr + i.K,,2 I(i)SI

Corollary

10.2.4 If

sB

is

PKn < 1. ((12K -1)SI

a

complement-free

antichain

in

S(k,, ... , kn) where K is even, then Is4) , J(ZK -1)SI. We now turn our attention to antichains which, in strong contrast, consist entirely of sets and their complements. In the case of ordinary sets, we have seen the result of Bollobas in Corollary 5.2.3, and we

have widened the class of antichains by considering the condition p; = p,, _j in Section 8.2. Gronau has extended some of these results to multisets.

is called a Definition 10.2.1 An antichain 4 in S = S(k,, . . . , c-antichain if x = (X,, ... , xn) E sd , x` = (k, -x,,.. . , kn - xn) E Si. in Section 10.1, we define subsets Following the construction of of S(k,, . . . , kn), k, , k,,, as follows. Let t be the smallest value of i for which k,+, is odd; if all k; are even, define t = n. Then define

... ,

!@o= {xES:x,
i, x;+, < 2'k;+,}

(1, i ; min(t, n -1))

9n={(Zk1,...,2kn)}if t=n. consists of the first half of the lexiThen 9 = !@0 U ... U cographic ordering of S - On. In the particular case when k, is odd, i.e. t = 0, we have the simpler situation where S = 20 U 20'. Note further that all vectors in have rank .2(k, +. . . + k;).

Theorems for multisets

1

171

If sat is a c-antichain then, apart from x = (21k, ... , 'k,,), which may or may not be in saf, the vectors in si can be put into disjoint by pairs of complementary vectors x, x`. Form the antichain . choosing from each such pair x, x` a vector of rank -_ZK, and then put sV = {x` :x E 4+}.

Lemma 10.2.5 Let ski be a c-antichain in S(kl, ... , kn), and let si+. Then (.si+)* denote the squashed antichain corresponding to

U

irrmn(t,n-1)

Proof From the way that squashed antichains are constructed it is

clear that (.i+)* c si*. Let y be the last vector of (si+)* in the lexicographic order, and let z be the first vector of 99 = s4* - (s4+ U fin)*. Then y
in the first half of the lexicographic ordering of S - 2,,, as was claimed by the lemma.

Theorem 10.2.6 (Gronau 1980) Let .sf be a c-antichain and let qi be

the parameters of W. Let w = min(t, n - 1). Then there exist non-negative integers q,,1, JK _- i _- K, 0 (i) q, = E j q,, j for each i

j _ w, such that

(ii) for each j = 0, ... , w,

if i,K-k, j+1

q,,j=0

/=1

and

Z

//<1.

}K-i
The numbers q,,1 are in fact given by q,,1= number of vectors of rank i in (sz1+)* fl !@j.

Proof Consider the following partition of (si')*: (s4+)* = (sB+)O U .

.

. U (s4+)W

where (.f')7 = (d +) * n 2j. Each (d +); is an antichain in -qi and (Exercise 10.8) each '9j is a LYM poset. Let qi,1 denote the number of vectors of rank i in (.si+); . Since any vector in JOj has rank less than K - J(k1 +. . . + kj+1), we have q,,1= 0 whenever i -- K - J E;±; k,. Also, since (d+)* consists only of vectors of ranks -_JK, q,,1= 0 whenever i < JK. The LYM inequality for the

172 1 Combinatorics of finite sets

antichain (d+); in Oi therefore gives

E

I(Pi ,1 qq,,,

Ei=i k1

.

{

In the case when k, is odd, i.e. t = 0, we obtain qi

(10 x)

1

.

}K--i<-}k, 1(i)%1

where the q, are the parameters of the antichain d', i.e. where q}K = ZP}x and where q; = p, for each i >

ZK.

Similarly

q' _- 1 I(Z)30I `

(10.6)

where the qi are the parameters of the antichain (s4-)`. But qi = qK_i, and ZK s K - i < K - Zk, i Zk, < i , ?K; therefore on adding (10.5) and (10.6) we obtain the following corollary.

Corollary 10.2.7 Let d be a c-antichain with parameters pi in where k, is odd. Then S(k,, . . . ,

I

Pi

+

Y,

}k,
P'

- 2

f(i)-q0I `

This reduces to Theorem 5.2.3 when k = 1.

10.3

Intersecting antichains

Recall that the Erdos-Ko-Rado theorem is concerned with antichains of subsets of an n-set with two further properties: the sets in the antichain are all of size --k, where k _- 4n, and they are pairwise non-disjoint. Having extended the concept of intersecting families to multisets in two different ways by means of the properties I, and 12, it is natural to look at possible extensions of EKR to these situations. In the case of property I2 we are able to obtain extensions. Lemma 10.3.1 Let .sB be an antichain with property 12 in the multiset Then S4 U sfl` with every vector in .s having rank -_ S(k,, ... , is a c-antichain, where Si' = {Xc:x E Si). 4K.

Proof Suppose X E .s and y` E d`, with x; -_ k, - yi for each i. We would then have xi + yi , k; for each i, contradicting the fact that d

Theorems for multisets

1

173

has property 12. Therefore s4 U s8` is an antichain. Clearly it is a c-antichain.

We now obtain an extension of Theorem 5.2.2

Theorem 10.3.2 (Gronau 1980) Let 4 be an antichain with parameters pi in S(k,, . . . , kn) where k, is odd. Suppose that 4 has property 12 and that all vectors in 4 have rank _ K. Then Pi I(K - i)-%I

1.

Proof Since 4 U 4c is a c-antichain by the lemma, we can apply (10.5), noting that (s4 U s4`)+ = 4`. This gives q` }K,GiK -}k, 10)201

<1

where the qi are the parameters of .9i`; thus

E

PK-i

1.

}K-i. K-}k, I(Z)20I `

Finally put K - i in place of i to obtain the required form of the result.

If we restrict this result to the case k, = ... = k = 1, so that s4 is an for each A E Si, the theorem gives Zn

intersecting antichain with JAI _-

E

i.,n/2

n

i.e. E

Pi

Pi

i-n/2 n

\n -iJ

, 1.

\i-11

We thus obtain theorem 5.2.2. We now return to the general case, no longer assuming that k, is odd, and obtain a generalization of the EKR theorem first stated by Greene and Kleitman (1978). Recall the definition of the sets 139i in Section 10.1. Theorem 10.3.3 Let s4 be an antichain of k-vectors in S(k,, . . . , k_ Then I d I is at most the number of k-vectors in -04 = -04, U ... U !*,. ZK.

174 1 Combinatorics of finite sets

Proof By Lemma 10.3.1, F = .4 U i

is a c-antichain, so we can

apply Theorem 10.2.6 to it to obtain, for each j < min(t, n - 1), q,,]

where the sum is over all i such that 2K , i < K - 2 Eil, k, and where q,,; is the number of vectors of rank i in (i,°+)* fl !@j = (st`)* fl 9;. Since only i = K - k contributes, we obtain grc_k,;, J(K - k)ljl for all

j for which gj can contain a (K - k)-vector. Therefore

I.c2

)

E; J(K - k)-9jl. But X e !@j and Jxl = i ax` E Rj+1 and jx`l = K - i, so that J(K - k)!@il = j(k)R1+1I. Thus jal _ la`l = Ej 1(k)Rj+1I = l(k)90l. Exercises 10 10.1

Let f (k1 i ... , k,) be as in Theorem 10.1.2, and suppose that kn > k1k2 ... kn_1.

(i) Show that f(k1, ... , kn) = kn(1 + k1) ... (1 + k, -J. (ii) Verify that in this case the bound of Theorem 10.1.5 is

(

kn

2n

) [1 (1 + k;), and exhibit an example of a collection

1 + kn i=1 sad for which

1,41 = kn2 - 12 I

10.2

n

1J (1 + ki). (1 + kn) i=1

Show that

f(k, k,

... , k) = 2

(fl)km1.n_`> i=O

i

E (n) kn-' + 1 ( n ) kn2. 2 n/2)

i
10.3

1

In the case where at least two of the k; are odd, exhibit a collection .s which attains the bound of Theorem 10.1.6.

10.4

Let ss be a collection of vectors of S(k1i

... , k,) such that,

for all x, y E Si, if min(x,, y,) = 0 for each i then max(xi, yi) = k;

for each i. Use the ideas of the proof of Theorem 10.1.2 to prove that, if not all the k; are 1, then Isil , f(k1i . . . , kn). (Li 1978) 10.5

Deduce Theorem 5.2.3 from Corollary 10.2.7.

10.6

Show that l(i)2;I -- l([2(K + 1)])gjl for each i -- [2(K + 1)].

Theorems for multisets 10.7

(i) Show that if d is a c-antichain in S(k,, ... , j.

I -_j(ZK)Sj = NKrz

if K = E, k,

is

1

175

then

even, whereas 141

2 E;=o 1('(K + 1))2;) if K is odd. (ii) Show that these bounds can be attained.

(iii) Deduce that IdI , 2

n

1)

`[ 2n] -1!

when k,, =

10.8

(Gronau 1980) In the proof of Theorem 10.2.6 it was asserted that !@, was a LYM poset. Prove this assertion.

10.9

Extend Theorem 10.3.3 to the case where each vector in si

n

has rank = k. 10.10 Let sQ be a collection of divisors of m such that if a, b E .ss

then a and b have at most l prime factors in common, counted according to multiplicity. Show that Id I -_ E;±o N. Re-express this result in terms of vectors in S(kl, . . . , kn). (Woodall, private communication) 10.11 Prove that y < z in the proof of Lemma 10.2.5.

11 The Littlewood-Offord problem

11.1

Early results

In 1943 Littlewood and Offord published the following result. Let . . , z,, be complex numbers such that {z,l a 1 for each i; then the number of sums E= 1 e,z, (E, = t1) lying inside any circle of radius r cannot exceed Cr' log n

z1, .

n1/2

(11.1)

for some constant C. Two years later Erdos showed, by a very nice application of Sperner's theorem, that the log term can be omitted, thus giving the best order of magnitude possible for the bound. In this chapter we shall explore variations on this problem, showing that some generalizations of Sperner's theorem are central to the development. After giving the basic ideas and the early results in this section, we shall discuss M-part Sperner theorems in Section 11.2 and then show their relevance to the Littlewood-Offord problem in Section 11.3.

There are, in fact, two equivalent formulations of the result of Littlewood and Offord described above. If E;'__1 z, is added to each

sum, the differences between sums remain unchanged, and the coefficients of the z; are now 8, = 0 or 2. If we scale everything by a factor of Z, we obtain the following: the number of sums

E=1 biz, (S, = 0 or 1) lying inside any circle of diameter r cannot exceed the bound (11.1). This alternative formulation with coefficients 0 or 1 instead of ±1 is sometimes more convenient.

We start our survey by considering combinations of real numbers, noting that the one-dimensional analogue of a disc of radius 1 is an interval of length 2. Theorem 11.1.1 (Erdos 1945) Let x1, ... , x,, be real numbers such that 1x,l =1 for each i, and let J be any interval of length 2 open at at least one end. Then the number of sums E=1 e,x, (e, = ±1) lying in J n

is at most ([n21) 176

The Littlewood-Offord problem

1

177

Proof Without loss of generality we can assume that all the xi are positive; for any negative xi can be replaced by -xi. We now associate to each sum E"=1 Eixi the set A = {i : Ei =1}. If Al and A2 are two such

sets, and A1cA2, then the corresponding sums would differ by at least 2 and so they could not both be in J. It follows that the sets A corresponding to sums in J must form an antichain; the result is therefore immediate from Sperner's theorem.

Given any interval of length 2r, open at at least one end, we can split it up into r intervals of length 2. Applying the above theorem to each of these intervals we obtain the following corollary. Corollary 11.1.2 If the x, are as in Theorem 11.1.1 and J is any interval of length 2r open at at least one end, then the number of sums E7=1 E;x;(E; = ±1) lying in J is at most r [n/2]).

This bound is not, however, the best possible, for we can replace it by the sum of the r largest binomial coefficients (n).

i

Theorem 11.1.3 (Erdos 1945) Let x1, ... , x be real numbers, Ixil ,1 for each i, and let J be any interval of length 2r, open at at least one end. Then the number of sums E" 1 Eix, (e, = ±1) lying in J is at most n

r

([(n

+ i)/2])'

Proof As in the proof of Theorem 11.1.1, assume that each xi is positive, and associate to each sum E;7.1 Eix; the set A = {i : Ei =1}. If

Al and A2 are two such sets and Al c A2 and (A2 - A11, r, then the corresponding sums will differ by at least 2r, so that at most one of them can be in J. The result now follows from Exercise 2.9.

We now return to the case of complex numbers. We shall use j rather than i as a suffix so as to avoid confusion with i where i2 = -1!

Theorem 11.1.4 (Erdos 1945) Let z1, ... , z be complex numbers with Iz;I % 1 for each j. Then the number of sums E;=1 E;z; (e; = ±1)

178 ( Combinatorics of finite sets

lying inside a circle of radius r is at most Cr2" nin

for some constant C.

Proof If I z I.1 where z = x + iy, x and y real, then either Ix I = or 2 lYI = 2 Thus either at least half of the z; = x; + iy; have Jx;1 2 or at least half of them have )y; I.2. Thus, on replacing each z; by iz; if necessary, we can assume that Jx;J --'l for at least half of the z;. Further, since z, can be replaced by -z; if x; is negative, we can suppose that at least half of the z, have real part x; - 2 Therefore suppose now that z,, . . . , z, have real part ,2, t ' In. Consider the 2' sums E , E ; z ; where E,+,, ... , E" have been given fixed values. We require that the sums E;=, E;z; lie inside a circle of radius r, and hence that the sums E;=, s,x; lie inside an interval of length 2r. By Corollary 11.1.2, the number of such sums is therefore at most

2r([ J)

2

_r1/2

for some constant C; this is because the sums

E;(2x;) lie inside a region of length 4r, and 2x; ,1 for each j. Thus for each choice of E,+,, ... , E,, we obtain at most Cr2'lt'2 sums lying inside the circle. But there are 2"-' ways of choosing , E". Therefore the total number of sums lying inside the E,+,, circle is at most Cr2'2"-'

tin

where C -- C' ' 2"2C since

Zn

Cr2"

C'r2"

= tin = nin t -- n.

In the case r = 1 the bound given by Theorem 11.1.4 is a constant

multiple of ([n2]) . In fact, we can obtain ([n,2]) itself as the best possible upper bound. This improvement, conjectured by Erdos, was

proved by Katona and Kleitman independently. Since the proof depends upon a generalization of Sperner's theorem, we now turn to such generalizations. 11.2

M-part Sperner theorems

In this section we present some generalizations of the fundamental theorem of Sperner. We consider the set S partitioned into M disjoint

The Littlewood-Offord problem

1

179

non-empty sets: S = X, U ... U X. A colourful interpretation is to think of the elements of S being coloured by M colours, X being the set of those elements coloured by the ith colour. A generalization of the antichain condition is then as follows: a collection s1 of subsets of

S has the property that if A, B E . f and A c B then B - A is not contained in any of the parts X. In terms of colours this property requires that there do not exist sets A, B E d with A c B and B - A monochromatic. The antichain condition is of course given by M = 1.

Rather surprisingly, although the condition with M = 2 is less stringent than the antichain condition with M = 1, the upper bound ,n/21 n 1

still holds in the case M = 2.

Theorem 11.2.1 (Two-part Sperner theorem) (Katona 1966b, Kleitman 1965). If IS I = n, S = X, U X2, X1 * 0, X2 * 0, X, fl X2 = 0,

and if ,4 = (A,, ... , Am) is a collection of subsets of S such that there are no i, j with A. c A; and A. - A; c X, or X2, then I si l

Proof The set of subsets of X, can be partitioned into symmetric chains, as can the set of subsets of X2. Each subset of S is uniquely expressible in the form E U F with E c X, and F c X2. If

C:El cE2c... cEP and

C':F,cF2c...cFq

are, respectively, symmetric chains of subsets of X,, X2, then the subsets of S of the form E; U F; can be displayed in a `symmetric rectangle' as follows:

E, UF, E,UF2

E2UF, EZUF2

EPUF, EPUF2

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

E,UFq EZUFq

EPUFF.

this way the whole of i (S) is partitioned into symmetric rectangles. Note that no two members of si can appear in the same In

row or column of any such rectangle; therefore the number of members of si occurring in any such p x q rectangle is at most

180 1 Combinatorics of finite sets

min(p, q). However, by Exercise 11.2, min(p, q) is just the number of subsets of S of size [n/2] in the rectangle. Thus, considering all of the

symmetric rectangles, the number of sets in si is at most the total number of subsets of S of size [n/2], i.e.

([n/2] ).

It would have been nice if there were a corresponding three-part Sperner theorem, but unfortunately the bound ([':2]) does not hold in this case.

Example 11.2.1 Take S = {1} U {2} U {3}, i.e. colour the three elements of {1, 2, 3} differently. Let 4 be the set of all 2-subsets of S, together with 0. Then s4 does not contain A, B with A c B and B - A monochromatic, but Isi = 4 > (3).

An example for general n is given in Exercise 11.1. If the bound

([':2]) is to be valid in a three-part theorem, some extra conditions have to be put on the set si. Several such conditions have been found.

For example, Griggs and Kleitman (1977) have shown that if S = X1 U X2 U X3 then the following additional condition on 4 will,

along with the absence of A, B such that A c B and B - A monochromatic, ensure that AAA -- ([n12]): s4 does not contain four sets A1i ... , A4 such that A, n X1= A4 n X1 A2 n X1= A3 n X1,

A1nX2=A3nX2=)A2nX2=A4nX2, andA1nX3=A2nX3=) A3n X3=A4nX3. If the n-set S is k-coloured, how large can a collection Si of subsets of S be which has the property that there are no A, B with A C B and B - A monochromatic? Let f (n, k) denote the largest possible such 141,

taken over all k-colourings of S; then f (n, 1) = f (n, 2) _ < f (n, 3). The question is: how large can f (n, k) be?

([':2])

Theorem 11.2.2 (Graham and Chung: described in Griggs, Odlyzko, and Shearer (in press)). As n -* oo, fl n, k) _- {1 + o(1)}kln(

n [n/2]

The Littlewood-Offord problem 1

181

Proof Let .sB be a collection achieving the bound f (n, k), and let nk = JXkI where Xk is the largest of the k colour classes X, S = X1 U ... U Xk. For each subset T of the n - nk other elements of S let .s4(T) = {A fl Xk : A e 4, A - Xk = T}. Then 4(T) is an anti-

chain of subsets of Xk and so, by Sperner's theorem, Isi(T)l ( nk ). Now nk % n /k, so, on using the approximation [nk/2]

([n/2])='--'

7Ln122

which we obtain from Stirling's approximation, we have n, k) = 1.41 = 2

2n-n,,(,

nk])

- 2--nk(

f(lrz

2 J

)

2nk{1 + 0(1)}

1'22n{l

=(-' Znk

+ o(1)}

kln( 22 )'a2n 1/2

k1/2([ n n,2]){1+o(1)}.

It has recently been proved that dk = lim

f(n, k)

n-.W

([n/2]) exists. In fact, we have the following theorem which we state without proof. Theorem 11.2.3 (Griggs, Odlyzko, and Shearer, to be published)

.

As

n -,oo, f (n, k)n1'?2-n -> ek

k

}1!2

g k)

where ek - (21

0

For our next results, we replace the antichain condition by a condition involving chains.

Theorem 11.2.4 (Sali 1983) If IS I = n and S = X1 U ... U XM where the X are non-empty and disjoint, and if a is a collection of subsets of S such that s4 does not contain I+ 1 different members A1, ... , A,+1 satisfying Al c ... c A,+1 and A,+1- Al c X for some i, n then I sil s MI [n/2]

182 1 Combinatorics of finite sets

Theorem 11.2.5 (Griggs}}1980b) Under the conditions of Theorem 11.2.4, 141-- 2M_2l([n/2]l.

Note that Griggs' result reduces to Theorem 11.2.1 when M = 2

and 1=1, but that Sali's result is stronger than that of Griggs if M > 4. Both have applications to the Littlewood-Offord problem. We shall now give a proof of Sali's result, making use once again of symmetric chains.

Proof of Theorem 11.2.4 The basic idea is to note that for each i the set of subsets of X has a symmetric chain decomposition; just as, in the proof of Theorem 11.2.1, the set of subsets of S was partitioned into symmetric rectangles (products of two chains), so here the set of subsets of S can be partitioned into `symmetric bricks' (products of m chains). Each brick will be of the form 9V = C, X C2 x ... X CM where C, is one of the symmetric chains of subsets of X,. If we can prove that the number of members of .sB in P4 is at most Mlw(R) where w(,Q) is the number of sets in R of size [Zn], then the theorem will follow on considering all possible bricks 9$.

Now we can think of R = C, x ... X CM as the set of vectors (x1, .. '. , xM) with integer components satisfying 1 -- x; -- k; where k; is

the size of C. Therefore R is isomorphic to the set of subsets of a multiset, or to the set of divisors of Il,pk'-1; thus 9$ itself has a symmetric chain decomposition. Recall the structure of the chains in

the symmetric chain decomposition presented in Chapter 3. Each divisor is obtained from the previous divisor in the chain by including one more prime; the primes are added not in any order but in such a

way that in each chain the prime p, is added (as many times as necessary) before p,+, is added. Each symmetric chain of subsets in OA therefore consists of M parts (some of them possibly empty) such that

in the ith part the sets (vectors) differ only in the ith component. There can therefore be at most 1 members of 94 in each part and hence at most Ml members of 98 in each chain; thus there are at most Mlw(9$) members of .si in 94, where w(98) is the number of chains in

94. But, by the symmetry of the chains, each chain in 98 contains exactly one member of size [in]. Thus each symmetric brick 98 contains at most Mlw(P4) members of .si where w(61) is the number of members of 9D of size [in]. Considering each 98 in turn, we find that

the total number of members of d is

,Ml(, n

.

The Littlewood-Offord problem

183

L'ittlewood-Offord results

11.3

Having obtained some generalizations of Sperner's theorem in the

previous section, we now apply these to the Littlewood-Offord problem. First of all we obtain the best possible improvement of Theorem 11.1.4 in the case r = 1.

Theorem 11.3.1 (Katona 1966b, Kleitman 1965) Let z1, ... , z be complex numbers with Iz;) % 1 for each j. Then the number of sums E;1=1 e,zj (ej = ±1) lying inside a circle of unit radius is at most

Proof By replacing zj by -zj if necessary, we can assume that each zj has real part %0. To each of the sums Ej e,zj we can associate two sets, namely A = {j : ej = 1, z; in the first quadrant} and B = (j : ej = 1, zj in the fourth quadrant). If two sums give rise to the same A but to two different sets B1, B2 such that B, c B2, then the sums differ by a sum of complex numbers all in the fourth quadrant and all of size 2; the sums therefore cannot both lie inside a unit circle. So the sets C = A U B corresponding to sums in a unit circle must satisfy the conditions of the two-part Sperner theorem 11.2.1; thus the number of sums is at most

0

Theorems 11.1.1 and 11.3.1, both with bound (n 2]), deal with sums of vectors in one and two dimensions respectively whose sums

he in a `unit ball' in that particular dimension. What about three dimensions? Since the two-part Sperner theorem does not extend to three parts, a new idea is needed. The ingenuity of Kleitman provides the idea we need; his argument in fact deals with any dimension!

Theorem 11.3.2 (Kleitman 1970) Let xj, ... , x,, be n vectors in d-dimensional space, with 14 % 1 for each i. Then the number of

sums E7-1 e,x, (e, = ±1) lying in a unit ball is at most

[nn/2] .

Proof The idea behind the proof is to imitate the construction of a symmetric chain decomposition of the sets of subsets of an n-set into

('2) chains; here we shall partition the set of linear combinations ]

184 1 Combinatorics of finite sets

e,x, into ([n2]) blocks such that no two linear combinations in the same block can both have sums lying in a unit ball. The theorem will follow immediately from this partition into blocks.

The blocks are constructed step by step, as were the chains. have been put into blocks as required. In the chain construction, each chain for the case n - 1 gave rise to two chains (one possibly empty) for the case n, the chains being of sizes one more and one less than the size of the chain from which they arose. We now mimic this construction procedure. Consider any one of the blocks, say U = {w,, . . . , wk) where the w, are combinations of the form E"=,1 Ex;, no two of which lie in a unit ball. From U we construct first of all U' = {w, +x" : i = 1, . . . , k -1); U' is just a translate of U. Now choose from U the vector w; with the largest component in the direction of and consider the blocks

Suppose that all the combinations of x......

V,= U' -

V2= U"U

where V= { w, - x" : w, E U). These blocks are of size I Ul - 1 and I UI + 1 respectively, and we can check (Exercise 11.3) that no two

members of either block can both lie in a unit ball. Now the construction of the blocks is exactly parallel to that of the chains in the symmetric chain construction: each chain of size k gave rise to a chain of size k -1 and to another of size k + 1, and the blocks grow in exactly the same way. It follows that the final number of blocks

must be the same as the final number of chains, i.e. ([n/2]). The proof of the theorem is therefore complete.

Before we apply Sali's result 11.2.4 to obtain our final theorem of this chapter, let us introduce some notation. Let gd(n, r) denote the maximum number of sums E , E,x, (e, = 0

or 1) lying inside a ball of diameter r, where the x, are in

d-dimensional space and where 1x1J .1 for each i. We have therefore seen that

gd(n, 1) = ([n2])

for all d .1, and that n gl(n, r) = Z

Further results have been obtained by Kleitman (1976a) for r = N/2,

The Littlewood-Offord problem 1 185

V/3 and V5 in two dimensions (see also Exercise 11.7 for another related type of problem). We now prove the following theorem. Theorem 11.3.3 (Sali 1983) n Sd(n,r) = 2d-'[rdln](

[n/2]/

Proof (essentially Griggs 1980b) Let x,, . . . , x be n vectors each of length =1 in d-dimensional space. Without loss of generality we can assume that each has a non-negative first component, so they can be placed into 2d-' sets according to the orthants in which they lie. Let B be an open ball of diameter r, and let si = {E c {1,

... , n} : E;EEx; E B}.

Note that the sum of any k vectors x, in the same orthant must be

%k1dl'2 (since, for example, all vectors of length ,1 in the first orthant have component ?1/d12 in the direction of the unit vector (1/dl'2)(1, 1,

... , 1)); thus, if El c E2 and all the vectors x;, i E E2 -

El, are in the same orthant, then the difference between the sums L.,ie6, x, and E,EE2 x, is at least IE2 - E,I/d"2. Thus if E,, E2 E si and

E, c E2, we must have IE2 - Ell , [rd?2]. We now apply Theorem 11.2.4 with S as the set of sums E^=, E;x M = 2d `, the set X, consisting of the sums in the ith orthant, and I [rd12] to obtain I.iI = 2d-'[rd12]

0

Exercises 11 11.1

Show that there is no three-part extension of Theorem 11.2.1 by considering the following example. Let X, = {1, . . . , n - 2}, X2 = In -1}, X3 = {n}, S = X, U X2 U X3, n odd, and define

sd={AcS:JAI =lAnX,l=2(n-3)} 99 ={BcS: IBnX,l =1(n -3), BnX2l=IBnX3l =1}

={CcS:ICnX,l=1(n-1), CnX2l=0, ICnX3l=1} 2 ={DcS:IDnx,I =2(n-1), Dnx2I=1, IDnx3I=o}. 2( n - i l Show that if .F = dU 9OU U 2 then I9I =

\1(n - 1)/

186 1 Combinatorics of finite sets

n (12(n

1)), and that there do not exist F1, F2 E °.!r such that

F1 c FZ and F2 - Ft c X; for some i. 11.2

(Katona 1966b)

Verify that in the proof of Theorem 11.2.1 min(p, q) is the number of sets of size [Zn] occurring in the p x q symmetric rectangle.

Verify that in the proof of Theorem 11.3.2 any two vectors in V2 are distance a2 apart. 11.4 Given a colouring of a set S, we say that a chain A, c A2 c ... c A, of subsets of S is monochromatic if I = 1 or A, - Al is monochromatic. Show that if S is two-coloured then 9'(S) can 11.3

be partitioned into ([nn/2]) monochromatic chains. (Griggs, Odlyzko, Shearer) 11.5 With f(n, k) as in Theorem 11.2.2, show that f(n, n) = 2n-1 11.6 A two-colouring of the elements of S is balanced if the number of elements of one colour differs by at most 1 from the number of elements of the other colour. Suppose that S has been given

two different two-colourings, one arbitrary and the other inducing a balanced colouring on each block of the first. Let 4

be a collection of subsets of S which contains no A, B with A c B and B - A monochromatic with respect to the first colouring and balanced with respect to the second. Show that (,nn/2). (Greene and Kleitman 1976b) I.i{ -11.7

(i) Suppose that the two-dimensional vectors x1, x2 are such that 1x,j . 1 for each i, and that the angle between x1 and x2 is greater than ;r/3. Show that no two of 0, x1, x2 and x1 +x2 can be less than distance 1 apart. (ii) Let 1x,I =1, i =1, ... , n, and let x, and x2 be at an angle 0, jrl3 < 0 < 2jr/3. Show that the number of sums E, e,x, (s, = 0 or 1) in any disc of diameter 1 is at most ([I(nn

1

2+ 1)]) -

2([1(n -1)]). (Kleitman 1976a) 2

12 Miscellaneous methods

12.1

The duality theorem of linear programming

Some problems involving subsets of a set can be solved using the duality theorem of linear programming. Suppose that we are considering the following problem: maximize P =

c;x; = c'x

where c and x are n x 1 column vectors and c' is the transpose of c, subject to the conditions

x-0 and Ax--b where A is an m x n matrix and b is an m x 1 vector. Here we write x _- y to mean x, , y, for each i. We define the dual problem to be m

minimize Q =

by, = b'y

subject to the conditions

y - 0 and A'y - c. We observe that if x is any solution of the original problem and y is

any solution of the dual problem then Ax , b and A'y , c so that x'A' _ b', whence

b'y-x'A'y-x'c=c'x. Thus every solution of the original problem is at most as large as every solution of the dual problem, so that max P , min Q. The duality theorem asserts that we have equality.

Theorem 12.1.1 (The duality theorem) If either of the above problems has a solution then both of the problems are solvable, and

maxP=min Q. A proof of this theorem can be found in almost any book on linear programming (e.g. Ziants 1974). We shall give two illustrations of the 187

188 1

Combinatorics of finite sets

use of this theorem in the text, and another two in the exercises. The

first illustration is the original proof due to Kleitman and Milner (1973) of a result which we have already proved in a different way (Theorem 4.2.4). Theorem 12.1.2 (Kleitman and Milner 1973) Let s4 be an antichain

on an n-set and suppose that Is'I , (k) where k _- in. Then EA... JAI , k (4j, i.e. the average size of the sets in si is at least k.

Proof We first of all note that we can suppose that Isil _ (k), for we can ignore the I s4I - (k) largest sets in si. Let x; (instead of the usual p;) denote the number of i-sets in st; then (generously) (12.1)

and, by the LYM inequality, n

xi

(12.2)

ito (n) Our problem is to minimize E;'=0 ix;, i.e. to maximize

P=

n

2 (-i)x;, i=o

subject to the conditions x_-0, (12.1) and (12.2). Now (12.1) and (12.2) can be rewritten as 1

0

1

...

(n) -1

x

(kn

)

The dual problem is therefore to minimize Q =Y1- (k)Y2

subject to the conditions y = 0 and Y1

(1)

-y2 . -J (0 , j , n).

(12.3)

Miscellaneous methods 1 189

The idea is now to exhibit a specific example of a vector y % 0 satisfying (12.3) for which Q = -k(k). We shall then have min Q

-k(k) so that max P -k(k); the minimum value of Eno iz; will then be at least k(k) = k Is4l as required. Making an inspired guess, we therefore consider

Y2=k+n-2k+1'

yi=(k)(Y2-k),

For this y we certainly have y, - (k)y2 =

-k (k), so we have only to

check that this y satisfies the conditions y -_ 0 and (12.3). Certainly

y ? 0 since k , 2n forces y2 > k which in turn makes y, > 0. It therefore remains only to prove (12.3). Now (k)(Y2-k) Y,

-Y2 +i =

Y2 +i

(k)-1

=

(y2-k)-(k

(j) If we define

k-i

ip(n, k, i) =

(k)/(n) -1 we then have

p(n, k, k -1)=n

-2k+1=y2-k

so that n Y1

V,

1

{y,(n, k, k - 1) - yi(n, k, j)}, (12.4)

and we have to prove that the expression on the right is non-negative.

190 1 Combinatorics of finite sets

Now, for fixed n and k, lp(n, k, i) is an increasing function of i, i _ In

(see Exercise 12.2); therefore if j < k both brackets on the right of (12.4) are non-negative, whereas if k --In both brackets are O. Therefore (12.3) holds for all j _ Zn. But it then follows that (12.3) hold also for j > Zn, for if j > Zn then n - j < In so that

l-y2+j> 1 -y2+(n-j)-_ 0. jl (n _j)

(YI_

Thus (12.3) is satisfied and the theorem is proved.

The normalized matching property for products

As our second example of the use of the duality theorem we show that the normalized matching property of two ranked posets is, under certain conditions, preserved by their product. Let F and G be two ranked posets. The product poset F X G is defined to have elements (x, y) with x E F, Y E G, and to have the order relation defined by (x, y) _ (x', y') ax _ x' in F and y < y' in G. We can define a rank function on F X G simply by defining the

rank of (x, y) to be the sum of the ranks of x and y in F and G respectively. (Recall Exercise 4.7.) Theorem 12.1.3 (Harper 1974, Hsieh and Kleitman 1973) Let F and G be ranked posets whose rank numbers are log concave, and where both F and G possess the normalized matching property. Then F X G also has the normalized matching property.

Proof (Hsieh and Kleitman 1973) Let fp and gp respectively denote the numbers of elements of rank p in F and G. Then the number of

elements of rank k in F X G is Ep fpgk-p. Let X be a given set of elements of rank k in F X G, and, for fixed p, let xp,k_p denote the proportion of the fpgk-p elements of the form (x, y), r(x) = p, r(y) = k - p, which are in X. (Here, for convenience, we use r for the rank in either F or G.) Let Y = VX denote the shade of X in F X G, and let yp,k+1-p denote the proportion of elements of F X G of rank

k + 1 of the form (x, y), r(x) = p, r(y) = k + 1- p, which are in Y. We then have xp.k-p > O,

Yp,k+1-p' 0.

(12.5)

Let g be any element of G. By the normalized matching property of F, the proportion of (k + 1)-ranked elements of Y of the form ( , g) is at least as great as the proportion of k-ranked elements of X of the

Miscellaneous methods 1 191

form ( , g). Thus, on averaging over all g E G of fixed rank k - p, we obtain yp+1,k-p'xp,k-p Thus Yp+1,k_p-xp,k-p%0

forallp.

(12.6)

Yp,k-p+1 - xp,k-p % 0

for all p.

(12.7)

Similarly we obtain

To prove that F X G has the normalized matching property we have to show that

H _ E p xp,k-Afg k-p - E p p,k-p+llpgk-p+l ` O. upfpgk-p

Epfpgk-p+l

If H is ever positive under conditions (12.5)-(12.7) then it is in fact unbounded, for these conditions permit the replacement of the xs and ys by any scalar multiple. Therefore if H is ever positive, the linear programming problem of maximizing H under these conditions is not

solvable. Thus, by Theorem 12.1.1, we can prove that H _ 0 by showing that the dual problem has a solution. Now the dual problem

is to minimize a zero linear combination of S1i s2, ... , t1, t2, .. . subject to the conditions sp ? 0, tp % 0, and

f

-fpgk-p

SP + tp

(12.8)

FjPfPgk-p

and

SP-1 '{' tp ` fpgk-p+l Ep fpgk-p+1

(12.9)

The theorem will therefore be proved if we can show that conditions (12.8) and (12.9), together with sp 0 and tp -- 0, are feasible. We do this by exhibiting a solution. Take Sp

and

- Erspfrgk-r F,-r

Er.pfrgk-r+1 Erfrgk-r+1

tP = Ercp frgk-r+1 - L/r

Erfrgk-r

These clearly satisfy (12.8) and (12.9), so we need only check that they are --0. To show that sp . 0 we have to show that

i

Erfrgk-r+1> Ercpfrgk-r+1 Erfrgk-r Ero9pfr9k-r

192 1 Combinatorics of finite sets

Er

frgk-rek-r' Erp frgk-rek-r

(12.10)

Erfrgk-r where 0i = gi+1/gi. Now we use the given information that the gi form a log concave sequence, i.e. gi-1gi+1 -_ g? for each i, i.e. 0, , 0i_1, i.e. 0, is a decreasing function of i, i.e. 0k-i is an increasing function of i. The truth of (12.10) therefore follows from the following lemma, and tp , 0 follows similarly from the log concavity of the f . Lemma 12.1.4 Let 0. be an increasing non-negative function of i and let each pi be non-negative. Then, if f is defined by

f(k) =

Ei-Qkpie

f is an increasing function of k.

Proof f (k) , f (k + 1) gpo0o + ... + PkOk < ek + I(po +

... + Pk) <*

E'= O (0k+1 - 0i)pi = 0. But 0, < t1k+1 for each i -_ k.

12.2

Graph-theoretic methods

Sometimes the language of graph theory is helpful in the study of subset problems. Our first example is a nice application of the idea of a tree (a connected graph with no cycles) and the fact that in a tree the number of vertices is always one more than the number of edges.

Theorem 12.2.1 (Bondy 1972) If .i = {A1, . . . , is a collection of n distinct subsets of an n-set S, then there is an (n -1)-subset S' of S such that the sets A, fl S' are all distinct.

Proof We define a graph G with vertices labelled A1,

. .

.

, A,,, and

with Ai and Al joined by an edge labelled x if and only if A, = A; U {x} or Aj = Ai U {x}. Suppose we can show that there are at

most n - 1 distinct edge labels in G, i.e. there is some y E S which does not appear as an edge label. We then assert that S' = S - {y } has the required property, for two sets Ai fl S' and Ai fl S' can be equal only if one of Ai, A; is obtained from the other by removing y and this possibility has been ruled out by the non-appearance of y as an edge label.

We have therefore only to show that there are at most n - 1 edge labels in G. If G is a tree or a disjoint union of trees, then G has at

Miscellaneous methods 1 193

most n - 1 edges and so the required result is immediate. Therefore suppose that G has a cycle C. It is clear that every label appearing on an edge of C must occur at least twice on C, since every element removed from a set at some stage of the cycle must be replaced. It follows that if we remove an edge from C the resulting graph G' has the same set of edge labels as G. We can therefore remove edges from cycles until no cycles are left, leaving the set of edge labels unchanged in the process. But the graph we finish up with, having no

cycles, must be a tree or a disjoint union of trees, and we have already observed that such a graph can have at most n - 1 labels. It follows therefore that the original graph also has at most n -1 labels.

0 For generalizations of Bondy's result, see Frankl (1983).

As our second example of the use of graphs, we consider an extension of Exercise 5.1. In that exercise we showed that if A,, ... , are subsets of an n-set S such that A, fl A; = O A; U A; = S, then m_-2 n-1 + (11 nn

- 2)j).

Now let .si be a collection of

subsets of an n-set S such that, if is any subcollection of si, then nAE98 A = 0 4' UAE99 A = S. Can we find the best possible upper bound for I.sil?

Theorem 12.2.2 (Li 1977) Let sd be a collection of subsets of an n-set S such that, for any subcollection R of .si, (_J,E, A = 0 4' UAEJO A = S. Then 1.41 _

2"-1 + 1.

This upper bound can be attained (see Exercise 12.1). The proof of

the theorem will make use of the graph-theoretic concept of a covering: a covering T of a graph G is a set of vertices of G such that every edge of G is incident with at least one vertex of (C. The degree of a vertex is the number of edges incident with that vertex. A star is a graph in which there is a vertex which is incident with every edge, i.e. a graph which has a covering consisting of only one vertex. Lemma 12.2.3 Let G be a graph with n vertices, each of degree -1. Then G has at most 2n-'+ 1 coverings.

Proof We use induction on n. The removal of an edge from G cannot decrease the number of coverings, so we assume that every

edge is incident with a vertex of degree 1, and hence that every connected component of G is a star. If G is connected then G is a

194 1 Combinatorics of finite sets

star, and any covering contains the common vertex or consists of all the other vertices; the number of coverings is therefore 2n-1 + 1. If G is not connected then we can write G as a disjoint union, G = G1 U

G2, where G1 has k , 2 vertices and G2 has n - k , 2 vertices for some k, 2 , k ` n - 2. We can apply the induction hypothesis to G1 and

G2 to find that G1 has at most 2k-1 + 1 coverings and G2 at most 2n-k-1 + 1 coverings. Since a covering of G is obtained by combining a covering of G1 with a covering of G2, it follows that there are at 1)(2n-k-1 + 1) coverings of G. But this number is most (2k-1 + 2n-2+2k-1+2n-k-1,+-1, 2n-1+1. 0

Proof of Theorem 12.2.2 Let 4 satisfy the conditions of the theorem. Construct a graph G with vertices labelled 1, ... , n and with vertex i joined to vertex j by an edge if and only if the set {i, j} has non-empty intersection with every set in A Then every member of 4 is a covering of G, so the theorem will follow from the lemma provided that we can show that G has no vertex of degree 0. Consider therefore any i E { 1, . . . , n } . Since i f U (A e 4: i 0 Al, we have

U {A E 4: i - A) (1 {A E s$ : i 4tA}

S, so that, by the conditions of the theorem, 0, i.e. there exists some j such that j E A

whenever i f A. It follows that {i, j) has non-empty intersection with each A E Si; thus i is joined to j in G, and i has degree .1.

A related problemI^IAEq is to obtain the largest possible value of Isil A = 0 UAEa A = S whenever 'B is a under the condition subcollection of si consisting of k sets, k being a given fixed positive integer. (In contrast, 64 could be of any size in Theorem 12.2.2.) We have seen that the bound obtained in Exercise 5.1 for the case k = 2 is very different from the bound obtained in Theorem 12.2.2. For k -- 3

it turns out that the bound is again 2n-1 + 1 (see Ein, Richmond, Shearer, and Sturtevant 1981). These authors also investigate the problem when all the sets in 4 are required to be distinct. Finally, we note that Li (1978) has extended Exercise 5.1 to multisets: if Si is a collection of divisors of m such that if g.c.d.(a, b) = 1 then

l.c.m.(a, b) = m, then 141 is bounded by precisely the bound of Theorem 10.1.2 due to Erdos and Schonheim (1970) (see Exercise 10.4).

12.3

Using network flow

One of the basic theorems in the study of network flows is the theorem of Ford and Fulkerson (1956) which relates flow through a

Miscellaneous methods 1 195

network to capacities of cuts. If D is a directed graph, suppose that to each arc (x, y) we assign a non-negative real number c(x, y) called its capacity. The capacity can be thought of as the maximum amount of flow which can be sent along the arc. Suppose also that D contains two particular vertices s, t which are the source and the sink of the flow. A flow f of value v from s to t is a non-negative real function f defined on the set of arcs of D such that 0 _--f (x, y) , c(x, y) for each arc (x, y), and, for each vertex x, v

Y, lf(x,y)-Y,2f(y,x)= Y

y

ifx=s

0 ifx=sort 1-v

(12.11)

ifx=t

where E1 is the sum over all arcs (x, y) out of x, and E2 is the sum

over all arcs (y, x) into x. Condition (12.11) just states that the amount of flow leavings is equal to the amount of flow arriving at t, and that, at each vertex other than the source and sink, the amount of flow in is balanced by the amount of flow out. Let X and Y be sets of vertices of D such that s c- X, t E Y and every other vertex in D is in precisely one of X, Y. Let (X, Y) denote the

set of arcs joining a vertex in X to a vertex in Y. Then (X, Y) is called a cut separating s and t, and we define the capacity of the cut (X, Y) to be c(X, Y) = E(X,,,)E(x,Y) c(x, y). Intuitively, a cut is a set of arcs whose removal from the network prevents any flow from s to t. It is clear that since any flow from s to t must pass along the arcs of the

cut (X, Y) the value of any flow cannot exceed the capacity of the

cut. Thus the maximum flow value is less than or equal to the minimum capacity of a cut. The Ford-Fulkerson theorem asserts that in fact we have equality. Theorem 12.3.1 (Ford and Fulkerson 1956) The maximum value of a flow from s to t is equal to the minimum capacity of a cut separating

s and t. Further, if all the capacities are integers, there is a flow of maximum value in which every f(x, y) is an integer. As our first application of the Ford-Fulkerson theorem, we give yet another proof of the marriage theorem.

Proof of Theorem 2.2.1 Suppose that we have r sets A1i ... , A, such that, for each m < r the union of any m of them contains at least m elements. Represent the sets A , ,. .. , A, by vertices x1, ... , xr and the elements a1, ... , a of their union by vertices yl, . . . , y,,, and join x; to y; precisely when a, E A;. Introduce a source s and a sink t, and join s to each x; and each y; to t. Assign capacity I to each of the

196 1 Combinatorics of finite sets

arcs (s, x) and (y,, t), and capacity M to each arc (x,, y,), where M is any integer greater than max(n, r) = n.

We want to show that each x; can be paired with a different y;. Suppose that we cannot find r disjoint arcs from {x,} to {y,}. Then there is no flow of value r from s to t, and so there must be a cut of capacity less than r. Considering the values of the capacities, we see that this cut must consist of, say, u arcs of the form (s, x) and v arcs of the form (y,, t) where u + v < r. There can be no arc from any of the remaining r - u vertices x; to any of the n - v remaining vertices y,, and so the only arcs from the remaining r - u vertices x, go to at most v vertices y,. In terms of the original sets, we therefore have r - u sets which contain in their union at most v, and hence fewer than r - u, elements. But this contradicts the hypotheses, so we must, after all, be able to find r disjoint arcs from {x,} to {y,}. These r arcs yield distinct representatives for the r sets A. As our next example, we show how network ideas can be applied to the theory of ranked posets. We have seen in Theorem 2.3.4 that if .4 is a collection of subsets of an n-set S such that no k + 1 members of 4 form a chain, then Isis is at most the sum of the k largest binomial

coefficients (n). Generalizing the definition of the Sperner property,

we shall say that a ranked poset P has property Sk if the size of the largest set of elements of P, with no k + 1 members forming a chain,

is the sum of the k largest rank numbers. We have thus seen in Section 2.3 that any poset with the LYM property possesses property Sk.

While investigating conditions under which a poset P has property Sk, Griggs (1980a) introduced the further property Tk. We say that P

has property Tk if there exist disjoint chains in P which cover the (k + 1)th-largest rank and each of which meets each of the k largest ranks. For example, any symmetric chain order has property Tk. Griggs showed that if P possesses property Sk for all k then it must possess property Tk for all k also. His proof used a deep result of Greene and Kleitman (1976a) concerning k-saturated chain partitions; this will be described in Chapter 13. However, Griggs,

Sturtevant, and Saks (1980) showed in a later paper that the result can be obtained quite neatly by using a network-flow argument. We now present their proof. Theorem 12.3.2 (Griggs 1980a) If the ranked poset P has property Sk for all k then it has property Tk for all k.

Miscellaneous methods 1 197

Proof (Griggs et al. 1980) Let Ni denote the ith rank number of P, and let n be the largest rank in P. Let Mo, M,, ... , M. be the rank numbers Ni listed in decreasing order. We suppose that P possesses property Sk for each k , n. To prove that P has property Tk for each k we need only show that it has property Tk for all those value k for which Mk > Mk+1 and for k = n. Therefore let k have such a value.

Consider the ranked poset P' obtained from P by taking the elements of the k + 1 ranks of P with largest rank numbers and with the order relation of P. Then P' has property Sk since P has. To show that P' has property Tk we need to show that there are Mk disjoint chains in P' which meet every rank. We now construct a network as follows. Take a source s and a sink t, and vertices ux and vx corresponding to each x E F. Then include the following arcs and capacities (where K is any integer greater than IP'I): (s, ux), with capacity K, for all x E P' of rank 0 in P';

(vx, t), with capacity K, for all x E P' of rank k in P'; (ux, vx), with capacity 1, for all x E P'; (vx, u,,), with capacity K, for all x,y E P' such that y covers x. Note that any flow from s to t will have to travel along arcs as follows: (S, ux(o)),

(ux(o)) Vx(O)),

(Vx(o), Ux(1)),

(Ux(1), 11x(1)),

(11x(1), UX(2)), ... ,

(vx(k), t), where x(i) is of rank i in P' and where x(i) covers x(i-1) for each i = 1, . . . , k. Now the set of arcs (ux, vx) corresponding to elements x E P' of a rank with rank number Mk forms a cut with capacity Mk. We shall show that this is the minimum possible capacity of a cut. It will then follow from the Ford-Fulkerson theorem that there is an integervalued flow from s to t of value Mk. Because of the nature of the network, which permits a flow of at most 1 through each ux and vx, this flow must consist of Mk separate unit flows from s to t. Each of

these follows a succession of arcs as noted above and hence corresponds to a chain in P' which meets every rank. The theorem will then be proved.

Therefore it remains to prove that there is no cut in P' with capacity less than Mk. Let C be any cut with minimum cut capacity. Since its capacity is at most Mk < K, it must consist of arcs (ux,, vx1), ... , (ux,, vx), say, where r is the capacity of C. Now form

a poset P" by removing x1i ... , x, from F. No flow from s to t is possible in P", so no chain in P" meets all k + 1 ranks. Thus P" is a subset of P' containing no k + 1 members forming a chain. Since P' has property Sk, it follows that IP" l is at most the sum of the k largest rank numbers in P', i.e. IP" -- Mo + ... + Mk_l. But IF"! + r = JP' J =

198

1

Combinatorics of finite sets

MO +. . . + Mk, so we must have r a Mk. Thus the capacity of C is at

least Mk as required.

The converse result is not in general true, but can be proved if the rank numbers Ni are unimodal (see Griggs 1980a). Exercises 12 12.1

Exhibit a collection

.

of 2n-1 + 1 subsets of an n-set S

satisfying the conditions of Theorem 12.2.2. 12.2

(i) Show that if 0
f(i) - 1 - u1u2 .. . u, increases with i, and i

g(i) = v1v2 ... Vi - 1 decreases as i increases. (ii) Let tp(n, k, i) be as in the proof of Theorem 12.1.2. Put a, =

(n)/( n ) Verify that a, > a2 > ... > a and that a, > 1

if i , Zn. Verify also that ip(n, k, i) _

k-i akak-1 ... ai+1 - I

if i < k and

i-k Ul(n, k' i)-1-1/a; ... ak+1 if i > k.

(iii) Use (i) to deduce that i is an increasing function of i. 12.3 (Preliminary to 12.4). Let 9 be an antichain on an n-set and let si* = {B : B c A for some A e s1}. Let is. denote the number of i-sets in .s and let x, = u;/(n). If x* denotes the corresponding i

number for sz(! *, show that E, x, -- 1 and x; = E, ; x, and hence isi*I = Ei x,{E;-G;

()}

Miscellaneous methods ( 199

12.4

Prove Kleitman's result that if

is an antichain on an

.s

n-set such that Isil -- (k), k---'2n, then JsV J % E;sk (n) (hints

follow). By 12.3 it suffices to prove that the minimum

value of E, x,{E;;

(n)

subject to x , 0, E, (-x;) , -1 and

1

Ei

x'

(n) , (k) is at least E'. k (n).

(i) Show that the dual problem is to maximize -u + (kn)v subject to u .0, v --- 0, -u + (n)v _ E;s1( ). (ii) Note that if there is a solution for which -u + n (n)v ('f), then E;sk (i )' being a solution of the dual

k=

problem, will be a lower bound for solutions of the original problem.

(iii) Letting Mk = E;=k

show

(n),

that -u + ()v = Mk

requires Mk (kn)

MI `vim m;-Mk (ni)

(n) i

(1--j
(

-\k)

and use the log concavity of binomial coefficients to show that v, u exist. (Odlyzko, quoted in Greene and Kleitman 1978) 12.5 Extend 12.4 to multisets. 12.6 Let s4 be a collection of subsets of an n-set such that there are no three distinct sets A, B, C in s2 for which A U B = C. Show that Is:QI _ ([fl2]){1

-)} as follows.

(i) Recall from Exercise 5.8 that if si contains x, sets of size i then, for each j -- Zn, xk

- (n) ;1. k=;k k 2j

(12.12)

Therefore the problem is to maximize E; xi subject to (12.12)

andx%0.

200 ) Combinatorics of finite sets

(ii) The dual problem is to minimize y,+... +y(,/2, subject to

y,0 and

minn(2D

k

( j1,...n).

(fl)Yk

(12.13)

(iii) The required result will follow if we can exhibit values of the y,

O( n1 )).

satisfying y , 0 and (12.13) for which Ei yi _ (iv) Take n

n

1

n

Yzm2m)-(2m-1)+2m (2m-1) =(2m+1)-(2m)+2m+1 (n (2m)+2m+1Y"" n

Yzm+1

n

1

m

and verify that this choice of y solves (iii). (Kleitman 1976b) 12.7 Use Theorem 12.1.3 to show that C(n, k), the set of all subsets

of an n-set S which intersect a given k-subset T of S, has the normalized matching property. (This gives another proof that C(n, k) has the Sperner property (see Theorem 3.5.1).) (Griggs 1982) 12.8

Let A, c A2 c ... c Ar be a chain of subsets of S, and let a, and b;, i =1, . . . , r, be given non-negative integers such that a; ` b; for each i. lXnA;I -_ b;) for each i.

(i) Show that P is the product of 91(A,) and the poset P' which consists of those subsets of A,. which are in P.

(ii) Prove that P' is a LYM poset with log concave rank numbers.

(iii) Deduce that P is a LYM poset with log concave rank numbers.

(West, Harper, and Daykin 1983)

13 Lattices of antichains and saturated chain partitions 13.1

Antichains

In this final chapter we look at antichains in a general poset. We

begin by showing that a partial ordering can be defined on the set 2t of antichains of a poset P, under which %t is a lattice containing the set of maximum-sized antichains as a sublattice. This is essentially work of Dilworth (1960), although we shall not necessarily follow his development. We shall then study some recent work, initially due to

Greene and Kleitman (1976a), which extends the fundamental

Theorem 3.2.1 of Dilworth on chain partitions of a poset. Let s?t(P) denote the set of antichains on the finite poset P, and, for

4, 2 e 21(P), write

si -,
Then clearly

< si for all Si E 2l(P). (13.1) Next suppose that s8 - R and < s4. If A E Si, there exists B E .

such that A _ B, and, for this B, there exists A 1 E Si such that B _ A1.

We therefore have A _ B _ A1, whence A = A1= B since sat is an antichain. Thus A E 98. Thus every member of 4 is in R, and similarly every member of 98 is in .s4, so that .s4 = @. It follows that

(4<

and 98-< 4)

' 4=@.

(13.2)

and 98 T where sB, 96, ' E W(P). If A E s4 Next suppose that A then there exists B E 96 such that A -_ B, and for this B there exists C E T such that B_ C. Therefore, given A E Si, there exists C E such that A _ C; thus s4 T. We have therefore proved that

(4

96

and 96 ') => s4 S.

(13.3)

The following is now immediate from (13.1), (13.2), and (13.3). Theorem 13.1.1 (W(P), _) is a poset. 201

202 1 Combinatorics of finite sets

We now exhibit a relation between antichains and ideals. This is

really just a reworking of the relation between antichains and increasing functions into {O, 1) discussed in Section 3.4.1. For s4 E ;11(P) define

s2={XEP:X
Lemma 13.1.2 R c A 4 *.4 < 90. Proof Suppose first that sd -- 6B. If X E s?1 then there exists A such that X -- A. Since s4 - 90 it then follows that A ` B for some B E We then have X , B, so that X E A. Therefore 5i

(13.4)

Conversely, suppose that . c A. Then, if A E 4, it follows that A E s¢ and so A E -4, so that there exists B E 1 such that A , B. Thus 4 - 96. We have therefore proved that sat

(13.5)

and the lemma follows from (13.4) and (13.5). Corollary 13.1.3

Proof

4=Aq9_A i,4*

.i _ A q . c y8 and A c

si < a and 90 < si q d = 6A

0

This connection between antichains and ideals sheds light on Theorem 13.1.1, for the partial order < on W(P) reflects the partial order of inclusion on the set of ideals si4. We have

(i) 4 < s4 since si c s (ii)

si4 ' B and B < si

(iii) sd - 99 and 98 < 19

.i c A and .4 c R R = A 4 = s4 c t

4 c A and A c

si 5'C.

We also note that the correspondence between antichains and ideals is in fact bijective, since every ideal in P is an for some antichain 4, namely the set of maximal elements of the ideal. We now use this correspondence to show that (2Y(P), --) is more than a poset-it is a distributive lattice. In what follows we let max(-%) denote the set of maximal elements of the ideal ,%. The fact that max(°,) = i4'' 9 = is proved as Exercise 13.1. .

Lattices of antichains and saturated chain partitions 1 203

Lemma 13.1.4 The set of ideals of a finite poset P, under inclusion, is a distributive lattice.

Proof Let At and X be ideals of P. We first show that At n x is an ideal. Let Y e x fl N and X-,-- Y. Since Y EX, X E .tit also; similarly X EX. Thus X e .tit fl X, and we have shown that .tit f1 x is an ideal. Note that At fl X c At and At fl x !z-: X, so certainly At fl x is a lower

bound for At and X. If

is any ideal such that R c At and R c X

then 1 c At fl x; so At f1X is in fact the greatest lower bound At A X. Similarly we can show that At U X is an ideal which is the least upper bound At v X. Thus the set of ideals is a lattice.

Finally, to show that the distributive law holds we have to show that X V (NA %)=(,R v .N) n (.tit v 9f), i.e. ,tit U (X fl %) = (.tit U X) fl (.tit u 9l). But this is true by the distributive law of set theory.

0 It now follows from the one-one correspondence between antichains and ideals, with the corresponding connection between their partial ordering relations given by Lemma 13.1.2, that (%(P), -,<) is also a distributive lattice. Since the greatest lower bound of .sQ and A is ,sQ fl 66, the greatest lowest bound of si and 94 in 9(P) must be the antichain corresponding to S2' A 68, i.e. .vQ fl 97 = max(saf fl A). A

similar situation holds for least upper bounds, and we have the following theorem.

Theorem 13.1.5 (Dilworth 1960) (91(P), -) is a distributive lattice. If Si, % E 9l(P), Si n 98 = max0i fl A) and si v 98 = max(R U -4). We note that max(R U A) = max(si U 98) (see Exercise 13.2). Thus s:2 v 98 = max(si U 98), a nice result. In contrast, however, Si n 90 is not in general the set of minimal elements of si U 90. 13.2

Maximum-sized antichains

Having shown that the antichains of a poset P form a lattice, we now turn our attention to the maximum-sized antichains of P and show that they form a sublattice. We shall denote by d(P) the size of the

largest antichain(s) in P: by Dilworth's theorem d(P) is also the minimum number of chains into which P can be decomposed. We begin by constructing further antichains from any two given antichains, not necessarily of maximum size. If .s and 90 are

204 f Combinatorics of finite sets

antichains, then 4 U,04 will not in general be an antichain. The members of si U P fall into four categories: (i) members of si n R;

(ii) XESdU03: XcYfor some YES4U04; (iii) X E Si U O8: X Y for some Y E Si U @;

(iv) X E Sd U y8: X4 Y and Y c X for

all

Y(#X) E sd U

We note that a v %, the set of maximal elements of si U @, consists of those elements in categories (i), (iii), and (iv). Now define sd V Oh by

(13.6) si 0 = set of minimal elements of si U y8, and note that si V 99 consists of all the elements of .s4 U R except

those in category (iii). Clearly 4V

is an antichain. Another

antichain is sd A 9 defined by ((sd U

(sd v A)) u (sd n M).

(13.7)

This is the set of all non-maximal elements of sd U 0A together with the elements of Sd n Q; it consists of those elements in categories (i) and (ii). It is clear that

sd0" q; SiV 0.

(13.8)

We can also show easily (see Exercise 13.3) that

sd0a c Si n a,

(13.9)

and it is clear by considering categories (i)-(iv) that

I.iv_o4j+IdAQl=Is4UQl+la n,vl.

(13.10)

Lemma 13.2.1 If si and 90 are maximum-sized antichains, then so

are all of si v R, 4 A R, si A R, and si V R. Further,

SdnR=STAR=S4VP. Proof (Green and Kleitman 1976a). By (13.10),

Isi v A +l4AAl=IsdugoI+I,4 ngel=Isdl+Ige{=2d(P,. It therefore follows from (13.8) and (13.9) that and

l4 v Rl + l4BV 98I.2d(P) IS4 V Rl + Id n 901 a 2d(P).

But none of 14 v 69J, Is4 V 901, and 14 A RI can be larger than d(P), and so they must all be equal to d(P). It follows that sd v !Y3, 4V gD,

Lattices of antichains and saturated chain partitions 1 205

and Si A P are all maximum-sized antichains, and (13.10) then shows that .s4LP is also of maximum size. Finally, (13.8) and (13.9) show

that 4 A R and 4V are both equal to 4A. Corollary 13.2.2 If 4 and -04 are maximum-sized antichains then .4 n is the set of minimal elements of .s4 U&

Proof By Lemma 13.2.1, d n

= d V 0B.

Collecting information together we can now state the following theorem.

Theorem 13.2.3 (Dilworth 1960) The antichains of maximum size d(P) in (s21(P), -) form a sublattice (Ef(P), <). For 4, 90 E Y (P), 4 v 99 is the set of maximal elements in s4 U R and .4 A r' is the set of minimal elements of .4 U A.

We note that the fact that qi A @ = si V for maximum-sized antichains s4, reflects the fact that category (iv) is then empty; anything in (iv) could be added to one of the antichains and increase its size. Since (Y (P), ') is a finite lattice it must have a `greatest' member .s4* such that .4 <.s4* for all antichains of maximum size. We can use the existence of .s4* to obtain yet another proof of Sperner's theorem

(Freese 1974)..4* must be invariant under every automorphism of (PP(S), c). If A E Si *, JAI= k, then A will be mapped to all the

k-subsets of S by permutations of the elements of S, and so all k-subsets of S must belong to .sd*. Clearly 4* cannot then have any other members, so ISO = (k) <

13.3

Saturated chain partitions

Recall that Dilworth's theorem asserts that if dl(P) is the size of the

largest antichain in P then there exists a partition of P into dl(P) chains. It is clear that at least d,(P) chains are needed to cover the whole of P; Dilworth showed that no more than dl(P) are needed. We generalize the concept of an antichain by defining a k-union of

P to be a subset A of P such that no k + 1 members of A form a chain. Thus a 1-union is just an antichain. If A has the maximum size

206 1 Combinatorics of finite sets

of all k-unions in P we say that A is a Sperner k-union. Thus a Sperner 1-union is just a maximum-sized antichain. We denote the size of a Sperner k-union by dk(P).

Lemma 13.3.1 A set 4 of elements of a poset P is a k-union if and only if it is the union of k antichains.

Proofs Any union of k antichains is trivially a k-union. Conversely, suppose that si is a k-union. For A E .s4, let 8(A) denote the size of the largest chain in si whose bottom member is A, and let

a Then the sets Sill ... , .s4k form a partition of 4 into k antichains

0

(some of which may be empty).

There are other ways of obtaining k antichains from a k-union, but

the method used here is particularly natural in that it leads to an ordering on the set of k-unions, as we shall briefly explain later. Note that if P is partitioned into chains then each chain can meet a given k-union in at most k elements (and fewer than k if the size of the chain is less than k). Accordingly, if ' = { C 1 , . . . , C,) is any partition of P into chains, dk(P) -_ IBk(f), where r=t

min(k, JCil ).

Any chain partition 16 for which dk(P) =13kM is called a k-saturated chain partition. Dilworth's theorem therefore asserts that there exists

a 1-saturated chain partition of P. Greene and Kleitman (1976a) proved that k-saturated partitions exist for all k % 1. In fact they proved the stronger result that, for each k -- 1, there exists a chain partition which is simultaneously k-saturated and (k + 1)-saturated. In general, there is no chain partition which is simultaneously k-saturated for all k .1. The example given by Greene and Kleitman

is the poset P with six elements a, ... , f and the partial ordering

a_-b_-c_-d, e--c, b , f. For this poset, dl(P)=2, d2(P)=4,

d3(P)=5, d4(P) = 6. The only partition of P into 2 chains is {a, b, f} U {c, d, e); this is 1-saturated and 2-saturated but not 3-saturated. The partition (a, b, c, d) U {e} U {f} is 2-saturated and 3-saturated but not 1-saturated. However, there do exist posets which have a chain partition which is k-saturated for all k; the set of subsets of a set is one example.

Lattices of antichains and saturated chain partitions 1 207

We are going to give a proof of this result of Greene and Kleitman, using the elegant arguments due to Saks (1979) and Perfect (1984). First we make a formal statement of the theorem.

Theorem 13.3.2 (Greene and Kleitman 1976a) If P is a poset and k % 1 then there exists a chain partition of P which is simultaneously k-saturated and (k + 1)-saturated.

Proof We shall consider product posets such as P x [k] where [k] = {1, 2, ... , k} has the usual ordering < and where P x [k] consists of all pairs (x, i), x E P, i E [k], with (x, i) , (y, #* x , y in P and i j. We shall call any element of the form (x, i) an element at level i.

Now there is an immediate connection between antichains (1unions) in P x [k] and k-unions in P. If si is any antichain in P x [k] we can consider syI U S12 U ... U sik where s 1= {x E P : (x, i) E sf). Each s ii is an antichain in P and, further, the si; are all disjoint. Thus corresponding to each antichain of P x [k] we obtain a k-union of the same size in P. It follows that d1(P x [k]) , dk(P).

(13.11)

The key to the proof is to consider certain chain partitions of P x [k + 1] which we shall call special. For any chain partition 16, call an element x maximal in W if x is the `top' member of the chain in T in which it lies, and let M(') denote the set of all elements which are maximal in W. Note that IM(T)l is therefore the number of chains in 19. If x and y are in the same chain in T, if y <x, and if no element in the chain lies between x and y, we say that x covers y in 16. Also, if T

is a subset of P x [k + 1], the projection of T on P is the set of all x E P such that (x, i) E T for some i , k + 1. If 16 is a chain partition of P, denote by M,(f) the projection on P of the maximal elements of W at level i. Having made all these definitions we now say that a chain partition 16 of P x [k + 11 is special if (i) MI(A) Q M2(f) 2 ... Q Mk(T) (ii) if a E Mk+I('W) and a 0 Mk('), then (a, k + 1) covers (a, k) in 16.

Thus, in a special partition T, if (p, i) is the top element of a chain (i , k), so are (p, i - 1), . . . , (p, 1), and if (p, k + 1) is at the top of a chain but (p, k) is not, then (p, k + 1) covers (p, k) in its chain. Note also that (ii) states that Mk+I(T) = Mk+I(`W) U Mk+I(f)

208 I Combinatorics of finite sets

where Mk+,(w) = Mk+,(T) fl Mk(T) and where Mk+,((C) consists of elements p such that (p, k + 1) covers (p, k) in If. The proof will proceed as follows. We shall show (Theorem 13.3.3)

that there exists a minimum chain partition (i.e. a chain partition containing d,(P x [k + 1]) chains) of P X [k + 11 which is special and which has Mk = d,(P x [k + 1]) - d,(P x [k]) chains consisting entirely of elements at level 1 . Suppose that T = {C1, ... , Cmk, ... , CQ} is such a partition, where q = d,(P x [k + 1]) and where C1, ... , Cmk

are the chains consisting of elements at level 1. Let us now define 19' = {C;, ... , CQ} where C; = {(x, j) : (x, j + 1) e Q. Then C;, . . . , ,,,k are all empty. Let 16" denote the set of non-empty members of Se'; then 19" is a chain partition of P x [k] which, by the definition of mk, is minimum and which is also special because T was special. Returning to 19, we can associate with T a chain decomposition 19* of P which we obtain from by restricting the chains of 16 to level k + 1 and taking projections onto P. The set of maximal elements of C* is just Mk+1(16), so that we obtain IMk+,(19)I chains in V. Now 4)I

N1('6)

k+1

IM((

=

=

i=1

IMr(`')I -k IMk(C)I +IMk+1(flI

k IMk+1('1)I + IMk+1(T)I + _ (k + 1) IMk+1((C)I +IMk+1(`6)I

IMk+1(10I

whereas flk+1(T*) _

min(ICI, k + 1) _- IMk+1(16)I + (k + 1) IMk+1(`W)I

since all the chains in `t* with maximal element in Mk+1(`W) are, by (ii), of size 1. Combining these inequalities with (13.11), we obtain ,ok+1(16*) -- N1(t) = d,(P x [k + 1]) -- dk+1(P)

However, since V is a chain decomposition of P, we trivially must have dk+1(P) -_ Nk+1(`6*), so we must have dk+1(P) = /3k+1('C*) V is therefore (k + 1)-saturated. If we now repeat the argument with W" in place of 16 we obtain the same 16* as before and conclude that T* is

also k-saturated. We therefore obtain a chain partition of P which is

simultaneously k-saturated and (k + 1)-saturated. The proof of Theorem 13.3.2 will therefore be complete as soon as we have established the following theorem.

Lattices of antichains and saturated chain partitions) 209

Theorem 13.3.3 There exists a minimum chain partition of P x [k + 11 which is special and which has mk = d1(P x [k + 1]) - d1(P X [k]) chains consisting only of elements at level 1.

The first step in the proof is to omit the `special' requirement. Lemma 13.3.4 If r > d1(P x [k]) then all antichains of size r in the poset P x [k + 1] contain at least r - d1(P x [k]) elements at level 1.

Proof Let s4 be an antichain of size r in P x [k + 11, and suppose that it contains m elements at level 1. If 4' denotes the set of these elements at level 1, then 4 - sV is an antichain in R = (P x [k + 1]) - (P x [1]). But R is isomorphic to P x [k], so that I4 - s4'l d1(P x [k]). Thus r - m , d1(P x [k]), i.e. m , r - d1(P x [k]) as required. Lemma 13.3.5 There exists a minimum chain partition of P x [k + 1]

which contains at least (and hence exactly) Mk chains consisting entirely of elements at level 1.

Proof Let T be a minimum chain partition of P x [k + 1], and suppose that it contains s chains consisting entirely of elements at level 1. We shall show that if s < Mk we can increase s by 1. Suppose that 1 9 = {C1, ... , Cs, C s , , ,.. . , C9) where q = d1(P x [k + 1]) and where C11... , C, consist entirely of elements at level 1. Let

F= (P x [k + 1]) - (C1 U ... U Q. P' is covered by q - s chains. It cannot be covered by fewer, for otherwise its t < q - s covering chains, together with C1, . , C,., would form a chain covering of P x [k + 1] of s + t < q chains. We can therefore apply Dilworth's .

.

theorem to P' to deduce that any maximum-sized antichain in P' has size q - s. Now by Theorem 13.2.3 the maximum-sized antichains form a lattice, so there is a `greatest' antichain .s4+ such that si sB+ for all maximum-sized antichains s4. By Lemma 13.3.4, si+ contains

at least q - s - d ,(P x [k]) = mk - s elements at level 1. If s < mk, choose any such element a. Without loss of generality we can assume that a e CS+1. Put CS+1 = CS+1 U C +1 where C,+1 consists of all

elements of C,+1 up to a, and put P" = P' - C,+1, noting that C,+1 consists entirely of elements at level 1. We next observe that P" cannot contain an antichain of size q - s. For if s4 were such an antichain then 4 would contain an element b

of C,++ 1 and, since 4 < A+, there would have to be an element

210 1 Combinatorics of finite sets

c E s2+ such that b--c, but then a < b , c where a, cc-,W', contradicting the fact that s4+ is an antichain. Since P" does not accordingly contain an antichain of size q - s, P" can, by Dilworth's theorem, be covered by q - s - 1 chains, say Cs+2, ... , C". Now replace ' by It' = {C1, ... , Cs, CI-11P C"+2, ... , C' }. This also is a minimum chain partition of P x [k + 1], and it contains s + 1 chains, C1, . . . , Cs, C,s+,, which consist entirely of elements at level

1. If $ + 1 < mk, repeat the above procedure; after Mk - s steps a minimum chain partition with mk chains of elements at level 1 will be obtained.

Proof of Theorem 13.3.3 The chain partition whose existence is guaranteed by Lemma 13.3.5 possesses some of the properties required by the theorem, but it may not be special. We are therefore going to alter it by applying two types of transformation which will

eventually change the chain partition into a special one. The two operations involved are those of insertion and transfer. If z covers x in a chain partition and w is in a different chain but satisfies x < w < z, then we can remove w from its chain and insert it between x and z. Again, if x is the maximal element of a chain C1 of the chain partition and x < y where y is in another chain C2, then we can make y cover x by transferring the part of the chain C2 from y upwards to the top of

C1. Note that neither of these operations increases the number of chains in the chain partition.

Let 19 be the minimum chain partition given by Lemma 13.3.5. First of all we perform a sequence of insertions so as to obtain a minimum chain decomposition 19, in which every element (a, j) which

is not a top element of a chain in 16, is covered either by another element at level j or by (a, j + 1). To do this, we take the levels j in turn, in increasing order, looking at each (a, j) which is not at the top

of a chain. If (a, j) is covered by (b, i) where b > a and i > j or by (a, i + 1) where i >j, then insert (a, j + 1), . . . , (a, i) into the chain. This operation does not affect coverings already dealt with since it operates only at levels greater than j. Note also that it does not affect the Mk chains consisting only of elements at level 1.

Having obtained If, we now apply to it a sequence of transfers which lead to a special chain partition. If (a, j) is maximal in %1 with 1 < j -- k + 1, but (a, j - 1) is not maximal, being covered by (b, i) for some i -- k, then (b, i + 1) > (a, j) and we can transfer the part of the

chain containing (b, i + 1) from (b, i + 1) upwards to the chain containing (a, j). Such a transfer does not affect the Mk chains containing only elements at level 1. We continue doing such transfers

Lattices of antichains and saturated chain partitions

1

211

until no further transfers are possible. Assuming for the moment that this process will indeed terminate after a finite number of transfers have taken place, we shall finally obtain a chain partition %2 such that

if (a, j) is maximal in 162 then either (a, j - 1) is maximal also or (a, j - 1) is covered by an element at level k + 1. In the latter case we must, by the property of 191, have j - 1 = k, i.e. j = k + 1, and (a, k) must be covered by (a, k + 1). Thus 192 is special. Why does this transfer process eventually terminate? To see why, define, for a given chain partition ' of P x [k + 11, numbers V 1, ... , vk as follows: let v; denote the number of elements a of P such that (a, i) is covered by some (b, j) and (a, i + 1) is covered by (b, j + 1). Each time a transfer of the type described is used to cover an element at level i, v;_1 increases by 1 and v; is unchanged for each j , i - 1 (although v, may increase). Thus every transfer results in a chain partition for which the vector (v1, ... , vk) is lexicographically

greater than before. Since v, , IPI for each i, such increases in the lexicographic order can occur only a finite number of times. The proof of Theorem 13.3.3, and hence that of Theorem 13.3.2, is now complete.

As mentioned earlier, the proof of Theorem 13.3.2 presented here is the combined work of Saks and Perfect. Saks used these ideas in 1979 to prove that k-saturated chain partitions exist for every k, and

then Perfect exploited his method to prove the stronger result of Theorem 13.3.2 in her 1984 paper. We now give an application of the existence of a k-saturated chain

partition by presenting Griggs' original proof of Theorem 12.3.2.

Recall that if the rank numbers Ni of a ranked poset P are, in decreasing order of magnitude, M0, M1, . . . , M,,, then P is said to have property Tk if there exists a set of Mk disjoint chains in P, each of which intersects each of the k + 1 largest ranks. Theorem 12.3.2 asserted that if, for each k, the size of the largest k-union in P is the sum of the k largest rank numbers, then P possesses property Tk for each k. Proof of Theorem 12.3.2 (Griggs 1980a) As in the first proof, let P' be the ranked poset consisting of the k + 1 largest ranks in P, ordered

as in P. Then P' also has the property that the size of its largest k-union is MO +. . . + Mk-1. Now by Theorem 13.3.2 there exists a k-saturated chain partition T = { C1, C2, ... } of P', for which Z min(lC,l, k) = dk(P') = MO +

... + Mk-1 = IP'I - Mk.

212 1 Combinatorics of finite sets

We therefore have

JP'J - Mk= E JC,j+ku Ic.kkk

where u is the number of chains in cC of size k + 1. Since (P'I = E, IC,I we therefore have u(k + 1) - Mk = ku, whence u = Mk. Thus has

Mk chains of size k + 1, and these chains must intersect each of the 0 k + 1 largest ranks of P.

13.4 The lattice of k-unions Greene and Kleitman (1976a) have extended the ideas of Section 13.1 to the set of k-unions. It turns out that an ordering can be defined on the set of k-unions in such a way that the resulting poset is in fact a lattice. The definition of the ordering depends upon the ideas in the proof of Lemma 13.3.1. There each k-union .s1 is partitioned into k 1-unions (antichains) s$1, ... , dk where ski is the set of all A E .R1 from which the largest chain starting at A has size i. These antichains are in fact totally ordered by the ordering of Theorem 13.1.1: we have

Sik-dk-1 :!

2! d1-

Thus to each k-union 4 we can associate a k-tuple (4l, .

. .

,

4k) of

antichains satisfying

(i) 4,fl4,=Q (i#.l) (ii) 4k - ... - 41.

If now s4 and R are k-unions, we write s4 < R if and only if 4, a for each i =1, . . . , k. It can be shown that 91k, the set of k-unions, is a lattice in which Si v 58 = U (Si, v 9i) i

The formula for 4 A 69 turns out to be much more complicated. The lattice is not in general distributive but, surprisingly, the set of all Sperner k-unions forms a sublattice which is distributive. Exercises 13 13.1

13.2

Show that if s4 is an antichain and 9 is an ideal in P then max(S)=.si<* °S =si. Prove that, if 4 and -64 are antichains then max(.s U y8) _ max(s4 U R).

Lattices of antichains and saturated chain partitions 1 213 13.3 13.4 13.5

Show that, in the notation of Section 13.2, Si A 9J C Si A 9B. Give an example to show that there need not be any inclusion relations between 4 A 99 and s4 V 96. For 4, 98 E Y (P) let

s4qe={Aes4:A;B forsomeBER}, 49° = {A E s4 : A _-B

for some B E 9i3}.

Show that

(i) 4598gd =4g9oj=2 ;

(ii) 4U4=4;

13.6

(iii) d. fl m8, = sim fl R - si fl 94; (Freese 1974) (iv) 4 fl sill _ f,& fl 9$0 = si fl 9o. Let IC be a chain partition of P, let ak denote the number of chains of 19 of size 'k, and let Sk denote the set of elements of P contained in chains of 19 of size less than k. Show that is

13.7

k-saturated if and only if dk(P) = ISkI + kak. Show that dk(P) = mins,_.p {ISI + kd1(P - S)}.

13.8

Show that the antichains of maximum size in P have non-

empty intersection if and only if any two of them have

non-empty intersection. 13.9 Let Ak(P) = dk(P) - dk_I(P). Show that if 'C is a k-saturated chain partition of P with ak chains of size --k then Ak+I(P) ak , Ak(P). Deduce that AI -- A2 % A3 .... 13.10 Prove that if Ak(P) = Ak+I(P) and 19 is k-saturated then T is (Green and Kleitman 1976a) (k + 1)-saturated.

Hints and solutions Exercises 1 1.1

1.2 1.3

Since A U B = S <*A' fl B' = 0, apply Theorem 1.1.1 to the A;. Yes: e.g. if n is odd take all subsets of size > 2(n + 1).

Follow the proof of Sperner's theorem, noting that p; = 0 whenever i > h.

1.4

(a) 6; (b) 20. (Include the empty antichain and the antichaif {0}.)

1.5

For each k-subset Ythere are 2k - 1 choices of X. Use the binomial theorem to evaluate Ik (k) (2k - 1). Alternatively, every, x e S is in 0, 1 or 2 of the sets in a pair X e Y, and not every x belongs to 0 or 2 of them.

1.6

For any B e P there is a subset B, of S which is comparable with B but with no other member of B (otherwise B would not

be needed in PA). Let B* be the larger of B and B, and let * = {B* : B E PA}. Prove that 60 * is an antichain. To eliminate the possibility A * c B * consider four cases: (i) A * = A, B * = B,

(here we would have B, and A comparable); (ii) A* = A, B* = B; (iii) A* = A,, B* = B; (iv) A* = A,, B* = B,. The sets of indices corresponding to sums in 1 must form an antichain. (This is followed up in Chapter 11.) 1.8 Apply Theorem 1.3.1 to A,, . . . , Am, B;, . . . , 1.9 The same as 1.8 with the A. and B; interchanged. 1.10 For each i there is a set T, I T = t, T fl A; = 0' i = j. Apply Theorem 1.3.1.

1.7

1.11 X X means that there exists k such that k E X, k 4 X, i.e. x, E Ak, x; s Ak. 1.12

(i) Every member of si intersects every member of b(si) and so contains a member of b(b(s4)). However, if A e b(b(si)), then A intersects every set in b(sd), so A' contains no member of b(si), so there must be some Al E .St1 such that A' fl A, = 0, i.e. A, cA. Thus every member of b(b(s/)) contains a member 214

Hints and solutions

1

215

of .s4. Thus if X E Si, then X z) Y for some Y E b(b(s4)), and

Y Z for some Z E . i. Since si is an antichain, X = Y = Z; therefore every set in .sf is in b(b(s1)) and vice versa. (ii) Take S = {1, . . . , 4} and let .sf consist of the two sets (1, 2) and (2, 3). Then c(s4) _ {{2, 4), {1, 3, 4}} and c(c(.si)) _ {{1, 2}, {2, 3}, {4}}. 1.13 b(sf) consists of {a, b, c}, {c, d, a}, {b, d, f), {a, e, f}. 1.14

The sets of vertices of minimal paths from x to y form a Menger antichain.

Exercises 2 2.1

Count the pairs (E, B) with E c B, E E AP , B E 90, in two different ways to obtain k 1901, (n - k + 1) X0981. Alternatively, apply the first part of the lemma to the collection

98'={B':BE9i3}. 2.2

Apply the normalized matching property.

2.3

If d contains a,

2.4 2.5

2.6 2.7 2.8 2.9

(n) i

sets of size i then, by repeated use of the

normalized matching property, a; for all i -_ Zn; therefore the average size of the sets of sizes i and n - i is ?Zn. By Corollary 2.1.2 any m sets of size k are contained in at least m sets of size k + 1; therefore apply Theorem 2.2.1. Let Si = { j : A; n B; 0 0}, and use Theorem 2.2.1 to show that the sets Si possess an s.d.r.. Then remove the s.d.r. from the A; and B, to obtain a similar problem with m replaced by m - 1; hence repeat the process. The set of all maximal chains is a regular covering. 1,3,4,3,1. Immediate from Corollary 2.3.3.

(a) If there are sets in si of size h < 2(n - r), consider the smallest such h and replace the sets by an equal number of sets

of size h + r; this is possible by normalized matching since Nh+,. = N,,. In this way push all the sets into the middle r ranks. 2.10

For each X c S take A,(X) = u(X)(IX) where u(X) is the number of collections .c , containing X. Also take R as the union of the . ,. The left-hand side in Theorem 2.3.2 is then E, Idj. On the right, note that, for any chain C, if C contains

216 1 Combinatorics of finite sets

an X occurring in more than one s/1, then A(X) -- t(IX) and X

is the only member of R fl c, whereas if each X in C occurs in at most one .ss, then ExERnc L(X) -- E'7=0

2.12

(a) Induction

step:

= 2n

S(k + 1, 2) = S(k, 1) + 2S(k, 2) = 1 +

2(2k-1-1)=2k-1.

(b) Induction step: S(k + 1, 3) = S(k, 2) + 3S(k, 3) = 2k-1 1 + ?(3k-1 -2 k + 1) = 2(3k -2 k+1 + 1). 2.13 If we require 1, ... , k to be in different blocks, there are then k choices for each of the other n - k elements; thus there are k"-k such partitions. Alternatively, use S(n, k) -- kS(n - 1, k) from Theorem 2.4.1. For the second inequality use induction: S(n + 1, k)

\k - 2/(k n

k-1)k 2.15

-1)n-k+l + k(k

-1)kn-k

n-k+l

Given a permutation of 1, . . . , n take the first kl elements as kl blocks of size 1, the next k2 pairs of elements as k2 blocks of size 2, and so on. This procedure will give all partitions but with duplications. Firstly, there are k,! orderings of the blocks of size i all of which give the same partition; secondly in each

block of size i there are i! orderings of the elements, all of which give the same partition. Therefore we have to divide n by k,! and by (i!)ki for each i. 2.17

2.18

Each surjection f determines a unique partition of X into k blocks, where the ith block consists of those x E X such that f(x) = i. Further, each partition arises from k! surjections. k" = number of mappings from an n-set to a k-set Y = Z IAA! S(n, JAI) (on considering the image A of each f) AcY

k

i!

(')S(n, i

k

S(n, i)[k],.

i=o

For n=3, 1[k]1+3[k]2+1[k]3=k+3k(k-1)+k(k-1)(k-2) = k3.

2.19

If, for each h, any h A,s contain more than h elements in their union, the first part of the proof of Theorem 2.2.1 shows that all the A, have the required property. If IA;, U ... U Ad = h,

Hints and solutions 1 217

where h is minimal, then we claim that all of these Ail have the required property, for if, say, a E A,, is chosen to represent A,,, and t of the remaining A,, had less than t elements other than a

in their union, then they originally had exactly t elements in their union, contradicting the minimality of h. 2.20 Use 2.19 and induction, noting that once an element has been chosen to represent one set, it has to be deleted from the other sets which now have k -1 effective elements. Exercises 3 3.1

The size of a chain beginning with a set of size k is n + 1 - 2k. For the second part, note that n + i is always odd for a chain of subsets of an n-set of size i to exist. The given expression is 0 if

n + i is even. If n + i is odd, put i=n + 1-2k to obtain

(k)-(k-1)-(n-k)-(n-k+1) _ 3.2

n

n

(n + i -1)/2 - (n + 1 + i)/2 = given expression.

The contribution to the sum from each of the chains is a sum of

the form f(1-1+1-...), and so is 1,-1, or 0. 3.3

U sc, can contain at most r members of each chain. Using 3.1,

UA

n+1

1([(n +j)/21) n+1

n

([(n +j)/21

l

- ([(n +j + 1)/2]) min(r, j)

l J{min(r,

j) - min(r, j - 1))

r

3.4

([(n Take B, as the h-subset occurring in the same chain of a given

3.5

symmetric chain decomposition as A,. Then B, c A;. The number of antichains consisting of r subsets of size [n12] is

('n) where m = ([n/2]). Summing over r gives 2(,A)). 3.6

Either there is a chain of size % n 112 or every chain decomposition of {A,} needs more than n'n chains, in which case there is

an antichain of more than n"2 sets. But a chain and an antichain have the required property.

218 1 Combinatorics of finite sets

3.7

(a) For each x E P let f(x) = size of longest chain ending at x. x, y incomparable. Thus if the largest chain size is k, the antichains .s', = {x E P : f(x) = i}, i =1, . . . , k, form a partition of P. This proves that the maximum size of the chain = minimum number of antichains partitioning P. The other way round is trivial. Then f (x) = f (y)

(b) If there is no antichain of size s + 1 then the minimum number of antichains into which P can be partitioned is %(rs + 1)/s > r, so that the maximum chain size is greater than r.

(c) Given a sequence a,, . . . , ars+,, define a; < a; if i <j and a, < a;, and apply (b) to the poset consisting of a,, .... ars+, under <. 3.8

Think how basic elements are paired with previous elements.

There are ([n/2]) sets of basic elements (one for each chain) and each yields a zigzag which does not go below the x-axis from (0, 0) to (n, p), p = n - 2 (number of basic elements). A

chain of length n - 2k + 1 has k basic elements, so that p = n - 2k, i.e. k = 2(n - p), and there are (k)

- (k n 1) such

chains. 3.9

To include the sets of size [n/2] the sequence must be of ([n n

length . /2] 3.10 In a chain of size n + 1- 2k the largest set has size n - k. By 3.1, the sum of sizes of the largest sets is kZ

(n - k){(k) - (k n 1)J +n =()+(1)+...+(A

1)+(n-),)(n)

where A = [in]. Divide by (n/A) and use the facts that

(tz)=r_1{i+o(44}

3.11

and (n/A) (2/2nn)'"22". Thus the average size of each U, and V in the proof of Theorem 3.4.2 is 2k + O(n"2), and this gives a bound asymptotically half that of the previous one. For n = 2, P(S) = {o, {1}} U {{2}, {1, 2}) = {Q, {2}} U {{1},

Hints and solutions 1 219

{1, 2}}. For n = 3, PP(S) = {Q, {1}, {1, 2}} U {{2},

{2, 3},

{1, 2, 3}} U {{3}, {1, 3}) = {{2}, {1, 2}} U {{1}, {1, 3), {1, 2, 3)) U {0, {3), {2, l3)). 3.12 Let f (n) = (m+ 1)

-

(Mn).

Then din+1- d, =f(n) - f (n - k).

If 1, m + 1 _ ?(n - k) then f (n) > f (n - k) is easily verified.

If [J(n -k)]<m+[In], f(n)>0 and f(n-k)<0. If [in] , m -- n, use d,,, = E

((fl m 3.13

-1) - (

'

(12))

to obtain d,

-

n-j )) where each bracket is %0. In

Consider the k. (n - 1)! permutations of {1, . . . , n} beginning with an element of T, I TI = k. If I A n TI = j, the number of such permutations starting with the set A E C(n, k) is

IAnTI (n-1)! (Anj-

-1) Since d is an antichain, such permutations arising from different A e 4 are all different. Therefore IA n TI (n - 1)! AE4

(,n- 1

k(n -1)!

(IAI -1) as required. For C(n, k, h), a ((similar argument gives nhTI)

E (n1 \IAI

3.14 3.15

_

(k).

-h)

Use induction and I L(m, n)I = E7_0 I L(m -1,1)I. (i) Suppose A,, AZ E g(.s4), A2 (-- A1. Then there exist B1, BZ E

d with B;n{1,...,n-1)=A;. But then Ben{1,...,n1) c A, contradicting the definition of g(s4). (ii) (a) If A U {n } e s4, choose D E Si such that D c A U (n)

andIDn{1,...,n-1}Iisminimal.LetB=Dn{1,...,n-

1}. Then B E- A and BE g(s4). Conversely, given BE g(s4),

choose DES4such that Dn{1,...,n-1)=B. (b) If A E .s4, use (a) to obtain B E g(4) with B c A. If C E g(s4), C n A= 0, choose D E s4 with D n (1, ... , n1 ) = C to obtain D n A = 0, which is a contradiction. Con-

220 1 Combinatorics of finite sets

versely, suppose B E g(si), B c A, and there is no C E g(s4) with C fl A = 0. By (a), A U {n) E Si. Suppose A f 4; then

... , n} - A E sat, and by (a) we can choose C E g(sa2) with Cc{1,...,n}-A. But then CflA=0. {1,

Exercises 4 4.1

(a) \k n 1)(k +

(k)

1

2

gk!k!(n-k)!(n-k)! -- (k - 1)! (k + 1)! (n - k - 1)! (n - k + 1)!

qk(n-k)-(k+1)(n-k+1), which is true.

(b) Assume that the S(n, k) are log concave, and consider

n+1.

S(n + 1, k + 1)S(n + 1, k - 1)

= {S(n - k) + (k + 1)S(n, k + 1)){S(n, k - 2)

+ (k -1)S(n, k -1)} = S(n, k - 2)S(n, k) + (k22- 1)S(nk - 1)S(n, k + 1) + (k + 1)S(n, k - 2)S(n, k + 1) + (k -1)S(n, k - 1)S(n, k) {S(n, k -1)}2 + k2(S(n, k))2 + 2kS(n, k -1)S(n, k) _ {S(n, k - 1) + kS(n, k))2 = {S(n + 1, k))2. 4.2

Rank numbers are 1, 3, 6, 9, 11, 12, 12, 12, 11, 9, 6, 3, 1.

L =5; 2k1-K+1=3. 4.3

If d is a collection of divisors of m such that no k + 1 members of d form a chain, then id I is at most the sum of the k largest

rank numbers. The proof is the same; it uses normalized 4.4 4.5

matching which still holds. Identical result.

If Ni_1= N Corollary 4.2.2 gives 10RI - 1,%I, whence the result. If N;-1
N,-1

jA96l/N;_1;1-641/N1 so that

Ni

N,-1

whence Ni_1- IA-641 -- Ni -1.141 as required.

Hints and solutions A-r

4.6

1

221

A-r-1

(a) NA_, 2 (K - 2i)Nj _ NA_, I (K - 2i - 2)N, t=0

i=0

A-r

+ (2r -1)Nx_r + 2NA_r E Ni i=0

where the first term on the right is --N;,_r_1 'i-o(K - 2i 2)Ni+1 = NA-r-1 E o (K - 2j)Nj. Therefore A-r-1

A-r

NA-r E (K - 2i)Nj ; NA_r_1 E (K - 2i)N, t=o

i=o

A-r

+ 2NA_r Y, Ni + (2r -1)Nx_r + (2r + 1)NA_r_1NA_, i=0

whence the result.

A-1

A-r

A-1

(b) NA E (K - 2i)Nj _ 2 Y, Nx_r Z N, + 2 1 rNN_r r=0

t=0

r=0

i=0

A-1

+ 2 1 rNA_r_1NA_r + KN0 r=0 A

2ENj(N) +2NjN,) A

i=0 j=0 A-1 A-1

j=j+1

j=0

+2EN,(EN) i=0 A

j=i A

A-1

A-i

1=0

j=+

2jNj(EN,)+2y, Ni(ENj) j=0

(=o

(A

2

N, )2

+ 2(E Nt) 2 < 4T2(m). i=0

i=0

(c) Since (cf. Exercise 3.1) there are (K) k

size K+ 1-2k, J(m) Since (x 2 1)

K+1) + 2

A

- (Kk-1 ) chains of

K+ 2 2

)(Nk-

N"-1)'

- (x 2 1) =2x-1, we obtain

J(m)=

A-1

(2K - 4k - 1)Nk+ (

k=0 A

2 7, (K - 2k)Nk; k=0

therefore use (b).

K+1 -2A )NA 2

222

1

Combinatorics of finite sets

(d) Consider the case n = 2.1. A-1

(yik)

-1)(n)

k= (2n -1)

J(m) _ Z (2n - 4k k=0

k=O a-'

=

-' n n ()-4n Z ( k-

k=0 k

k=1

(n)j

-1){2n-1-

= (2n

_

n - 4 . k(k)

1) 1

)

- 4n12n-2 -

(11

1)J

2

n

1)-(n = n odd similarly.

4.7

(ii) Multiply out the product of the two expressions, which must be ,0. (iv) The poset of divisors of m = pi' ... pn" is the product of the n chains C; where C; = {1, p,, p?, ... , pk'}. Each C, trivially has log concave rank numbers.

Exercises 5 5.1

Remove the smaller set from each complementary pair of sets, to be left with an intersecting family. By EKR the number of n

sets removed is at most ([n2j -1)' apply Theorem 1.1.1 to 5.2 5.3 5.4

the rest. All [n/2] - subsets of {1, . . . , n} containing 1. If n is even, apply Sperner to the antichain s4 U 4'. If n is odd, apply EKR to the sets of size less than Zn in d U ssf'. If h -- then the result is immediate from 5.1.2. Now suppose h > in.Zn Suppose that jA, I < n - h, 1 i -- m0, and 1A, % n - h,

mo < i , m. Then mo , E? 8 -'n(i-1) by 5.1.2. For sets of size between n - h and h, no set and its complement can both be

A,s, so m-mom2Eh=n_,,(n). Use

n-1 to deduce m - mo -- E;-n-h (i -1)' h

(r)=(r-1)+(n r 1

Hints and solutions 1 223

5.5

Let k be such that f (k) (k

-1)

takes its maximum value. By

Theorem 5.2.2, 1

ll= E

11) 5.6

f(IA,I)

Ef(IA;I).

lf(JA;l)(IAtl?1) f(k)\k-

Let p be the size of the smallest set A;. Replace all sets A, of size p by those sets of size p + 1 containing them. The resulting

collection of sets is also an intersecting antichain, and, by has not decreased. Continue Lemma 5.2.5, E;1/( Ail n -1) -1 5.7

until only sets of size [Zn] are left; the sum is then ;1 by EKR. The first part is clear from the conditions for 2-independence in the text. To prove 5.2.4, replace each set containing a chosen

element of S by its complement to obtain an intersecting antichain A of subsets of an (n -1)-set. By Milner's theorem, ([(nn

-

2)/21)ICI ' ([(n + 1)/2]) (i) A; U A; * C is equivalent to (C - A;) n (C - A;) * 0, so by

ICI

5.8

EKR we have

m=\kkh11) kkh(h)' (ii) Let OD be the set of all h-subsets of C in d. By (i), y8

contains at most k k h (h) sets. Hence the proportion of saturated chains through C from rank h to rank 2h and containing no member of 98 is at least h /k. Hence the proportion of saturated chains through C containing no smaller

member of si is ,h/k; hence their number is

h k! (n - k)! k h! (n - 2h)! Summing over all such C we obtain n!

h!

5.9

ti`

h k' (n - k)!

(n-2h)!J k=h p"`kh!(n-2h)!'

Take saf = 90 U 19 where 90 is the set of all (q - 1)-subsets which

224 1 Combinatorics of finite sets

contain all of 1, ... , k, and where ' is the set of all q-subsets which do not contain all of 1,

. . .

, k.

Exercises 6 6.1

(a) Take "ii = 9'(S) - .4,

= R to obtain (2" - siI) ICI

2"(IRI-IRnII).

(b) Apply (a) to the downsets 4', !' where 4' _ {X : X' E °.

si}. 6.2

As in Exercise 2.3. Alternatively, use Theorem 6.4.1; the

6.3

average size in each pair must be =2n. Take S = {1, . . . , n} and define .sil to consist of the one set

(1), 142 to consist of all 2-sets containing 1 together with {2, ... , n}, S43 to consist of all 3-sets containing 1 together with all (n - 2)-subsets of {2, .

. .

, n},

... , srk to consist of all

sets of size =k containing 1 and all (n - k + 1)-subsets of 6.4

{2, . , n}. W U J is an upset and j U X is a downset, so (I WI + I1I)(I%I +

IfI)>_2"I.0I. But Igei . IXi on substituting for (2 (I

2n'. But 6.5

so that

ILL

Thus I-fI iJO I -- IXI ICI f + I I))2 = (1(2" - I-01 - I1I))2, whence 2".

c J, 2 c J.

Take 91 = .s4, l = si' to obtain Is4I12 + I si "2 , 2r'6.6

(i) Let X E (mid

S4)+. Then X -D Y where Y E mid 1. Y E d+,

so that Y A for some A E sf. So X A and X E si+. The other way round is trivial. (ii) As in (i), saf- _ (mid(sd))-. But 16(si) = sd+ U sbr-.

(iii) Apply Kleitman's lemma to li = si+ and 9 = d-, and use + I.s I) Ijd I)1n 1(I4+I

(ISJ+l

(iv) I.+ U .c I = I.S '1 + Isr I - Isr n d-I, so use (iii). (v) This follows from (iv) since Imid .s % I. , and f (x) _ 2.2'x"' - x is an increasing function of x for 0 < x < 2". 6.7

(i) XE,4 v60X=AUB for some AEsi, BED X= AUB where A U B E s/J n oh

sin

X e s9 n -. Conversely, X e

XEs9andXEQ=>X=XUXE4v1'.

(ii) XES/In911=>X=AnB for some AES4, BER=>X= AnB where A n B e si n 58 X E s4 n c4. Conversely, X e

sin9O'XE.4QandXER=>X=XnXESf A %.

6.8

Put si = °U n o, -64=-9(S), and note that .si v e c'11, d A c-!@.

Hints and solutions 1225 6.9

Obtain

f(i)g(i)t(i) ; µ(1)+i µ(i) r

f(i)i(i) 6.10

;

µ(1)f(1)g(1)

g(1)µ(1) +

g(i)µ(i) >f(j)µ(I).

a(A)f3(B) = µ1(A)µ2(B)h(B) < µ1(A U B)1z2(A n B)h(A U B) = cp(A U B)8(A n B), so by Theorem 6.2.1,

µz(A) AS µ1(A) AcS µ2(A)h(A) AS µl(A)h(A) AS 6.11 Given µ, f, g with f, g increasing, take µl, µ2 as in hint and h =f. Then t

(A)µ2()g(A)µ(A)µ(B) g(A u B)µ(A u B)µ(A fl B) EA g(A)µ(A) `

EA g(A)µ(A)

= µ1(A U B)µ2(A fl B)

so that, by 6.10, A

µ(A)f(A) =

g(A)µ(A)f(A) A EA g(A)µ(A)

and FKG follows in the case E µ(A) =1. The general result follows by scaling.

(i) X E (a4)+qX c B for some B E a4 qaX c aB for some

aB E 4. So I(a4)1= I I. n -I , (ii) 2" i4f , 2"

I(a4)`I by Kleitman. But I.1. I(as4)+I 2" 14+ n (a4)+i by Exercise 6.1(b). Similarly for the last part. 6.13

(i)

n 4'

4' n

=

n

= B;)' _ (A; n B;)

(ii) 4n Q81

A

n

4n

n n ' = O; for A, U B, _ (A; U n

Al c A; where B. Similarly for other intersections.

O A = Al U B1 for some A E s4,

B1 c A, which is a contradiction.

B1 E

226 1 Combinatorics of finite sets

(iii) fill=19011

1191=I{C:C=A;UA,;A1,A;Eg1}l=Is$1-ji11-

(iv) 1-41+1-041=ISiI+I@2I+Al _ISiI+Al +161=IM

2n-' 6.14 If 1 h
I

I

=E

n

1)

(n

Take 98 = si'. The conditions of 6.13 become Al

A2' and Al whenever A1i A2 E .si, i.e. Al fl A2 0 0, Al U A2 * S. The conclusion becomes 21siI -- 2"-' 6.16 Assume that all the A; and B 1 are subsets of {1, ... , r} and use induction on r. Consider the induction step. Put si = 6.15

A2'

{Ai,...,Am}, ={B1i...,B"}, 16 ={A-{r}:A-{r}ESZ2 andAU{r}ESi), ={A-{r}:Aesi}, W={B:BEP,rj B), F={B-{r}:BER}. Show that ITI+I!@l =m.

Apply the induction hypothesis to rg and 9 and to and to obtain I' I _- I W - WI and 121 _- 12 - 91. Then show that if E E T - W then E - {r} and E U {r} are in .sd -.6h. Show also that if E E - 9 then either E - {r} or E U {r} is in d - 90.

Hence Isi

I - IW - WI+Iq-9I so that m=ITI+I.9I_(Ahlswede and Daykin 1979) IT - 91 + I' -,fl -- Id - RI.

6.17

Let 90 be a maximum-sized intersecting family in the ideal si. Extend -64 upwards to an upset T. Then % = T fl si and Kleitman gives 2" 199 1 = 2"1% fl . I , 11. 141 < 2"-' I,s4l by Theorem 1.1.1.

6.18

Suppose si has two bases A1i A2. If Al flA200, the result follows from Theorem 6.4.4. If Al fl A2 = 0, every intersecting family is in !(A1) or P(A2); therefore the maximum size of an intersecting family is max(2IAII-', 2141!_1) But there is a star with this size.

6.19

(ii) EEi?4 EEJf*, EfR*4 E'f.s1*, E'E9Y*aE'E9.

is an upset so !@' is an ideal, so that there is a (iii) permutation a such that D, E Da(j) for each D, E % i.e. DiUD,(;) = S. (iv) By (ii) the sets in W and S are in complementary pairs, the union of each pair being S. Together with the pairs in (iii) they give a pairing for W U 9R*. Finally, each pair contributes

at most one to IqI + I.I.

Hints and solutions 1 227

6.20

(i) By maximality of 1981, A is an upset, so 98' is a downset.

(ii) Extend 4 to an intersecting family 98 of size 2"-1. The condition A U B 0 S is equivalent to A' n B' 0 0, so that 4' is an intersecting family in the ideal 98'. Hence I4l = I.s4'I JgS,j = 19gl =

2"-2

i 6.21 ((AUE)'UE)'=(AUE)nE'=AnE'=A since AnE=Q. z

Therefore B = (A U E)' is also in D. We also have A n B =

An(AUE)'=AnA'nE'=Q, and AUB=AU(AUE)'= AU(A,nE')=S-E.

6.22 Take A E 1.59 and define D1= M1 U (A n M'), D. = Di n E. 6.23 Let 90=4-4. s4n91=o, so that Id U901,2141. No two

sets in s4 U 9$ have union S, so no set and its complement can both be in 4 U 98; therefore Isi U 981, 2"-1 Thus 2 Is4l - 2"-1 Exercises 7 7.1

(a) 23=(4)+(3)+(2)+(1); 7

37 = (4 (36

(b)

27=(4)+(3)+(2)+(1);

2

3

) + (3) + (2)' 4

3

) + (2) + (1) + \01 = 30.

(3) + (2) + (1) so the 25th set is {7, 4, 2}.

7.2

25 =

7.3

{8, 6, 5, 3,1}

is the mth set where m-\Sl+(4)+(3)+

(2)+(1)=32

{8, 7, 5, 4, 3) is the mth set where m = 7.4

(7)

5+

(6)

4+

(5)

3= 46.

(i) A < B 4 largest member of A + B (=A' + B') is in B 4 largest member of A' + B' is not in B''' B' < A'. (ii) and (iii) are now immediate.

7.5

If A is the mth set where m is given by (7.1), then A is given by

(7.3). So a = a, and b = t - 1. The suggested B is therefore {1 + a,,

... , 1 + at;1, ... , t-11 if t> 1; this is the nth set

where

n=(11)+...+(a,)+(r-i)=m+1. If t=1, B= t

{1+a ..... , 1 + a1}. If 1 + a1 < a2 this is the mth set where

228

1

Combinatorics of finite sets

n=(a`)+...+(2)+(1

)=m+1. If 1+a,=a2let s be

1

the smallest i such that 1 +l a;
\s + 1) +

M+1=(k+l)+...+

Also

(s), so that =(k)+.+(SS)+s>m.

IAFk+I(M+1)I

(b) M+m=(k+l)+...+(s+l)+(s)+...+(t+1) so that I,&Fk+I(M+m)I = (1 kak)

+... + (1

s as)

1)=m+IAFk(m)I

1/+...+(t

t Osd = OFk(r)' U Fk_l(s)' U ({1} U OFk_,(s)'). +(s S

7.7

If s , ILFk(r)I then Fk_l(s)' c OFk(r)' so that IA-41 = IAFk(r)'I + 1{1) U z Fk_,(S)'I = ItFk(r)I + IOFk-B(S)I if s > IAFk(r)I, Similarly, I AFk(r + s)I -- IodI.

7.8

IedI = s + IAFk_1(s)I.

First suppose a, > t. Then

r+s=m={(akk 1)+...+(at

1)1

f(k-l)+...+(tt-1)l

Thus

1)+...+(a, I AFk(r) I

s.

But

Hints and solutions 1 229

Next suppose a,= t. Let m = (k) + ... +

(h) + (h -1) +

+ (t) where ah > h. Then

m={(akk

1)+...+(ahh

1))+{(k-1)+...+(h-1)

Ih-2

Thus s<(k-1)+..+(t,-1)

r>(akk 1)+...+(ahh 1)

rakk 1)+...+(ahh Similarly if s =

(k-1) + ...

IAFk(r)I=s.

+(a,- 1). Adapt the argument

if t = 1 or ifm -- k. 7.9

(a) (i) Immediate from Kruskal-Katona on writing v in its (k -1)-binomial representation.

(ii) If m2 = (k) then the right-hand side

is

(k n 1) +

IAFk(m,)I=IAFk(m2)I+IAFk(m1)I. So suppose now 0<m

mi) I > m2 and M2< (k). Use (i) to show that kFk+1((k" + 1) + AFk+i((k

+

1) + m, + m2 - (k)) 1)

whence

(n k 1) +

I+

(n k m2 - (k)) I = I

+ IAFk(mi + m2

1,&Fk+l((k

+

1) + ml + M2)1

+ 1) + mi)

(by Exercise 7.7) = (k) + I AFk(m,)I + AFk(m2)I.

(k))

+ IAFk(m2)I

230 1 Combinatorics of finite sets

(b) Let m, be the number of A E 4 such that A E Si. Then ml + m2 = Is4l and JAsiQl $ IAFk(ml)I + IAFk(m2)I, so use (ii).

7.10 Take k =1; every A is a B, so m --u. If m > (k - 1 ) then u >

n-11) so m < (kn-1 - 1)' which is a contradiction.

(k 7.11

Take m as in (7.1), so that the mth set is as in (7.3). Use the representations of m + 1 as in the solution to Exercise 7.5 and the Kruskal-Katona theorem to obtain expressions for IAFk(m)l and I AFk(m + 1)I. For example, if t>1, the difference is t -1-- k - 1. 7.12 (i) Take si4o to consist of the last m subsets of S of size 2n.

Then by Exercise 7.4 l4 n 4ol = 2 l s$ol - l s$o U -4ol = 2m n

n/2 . But if t(sar) is the number of disjoint pairs of subsets in s$ then t(s$o) = 2 ls4o n s4ol.

(ii) First replace sets of size
2

.i'I=IsiI+Id'I-I.4U4'j 2m-(n/2)' iii Note that 2 ( ) (n2)

n/2

where 19 consists of the first (n - k)-subsets of S. But IAICI = where IV= -64.

7.13

)V41 = IAsi¢'I -- IATI

7.14

IV4I = IAs4'I, so apply Corollary 7.3.6 to the collection 4' of

Isi

l

k)' take u = n -1 (n - k)-subsets. Since Is$'I -- (k -1) ( = \n to obtain

-1 1 n-k-1

Iv,d I=I'&S"I--

(n

IS,I=

k 1-41.

(n-k) 7.15

(n-1

(k-1)

{21 - a;} _ is that of required position The {20, 19, 17, 16, 13, 11, 10, 9, 8, 6, 5, 3} in the antilexicographic ordering, i.e. the rth-last position in the antilexicographic order where r is its position in the squashed ordering. But (19

18

16

(6)+(5)+(4)+(3)+(2)+(1)=96034.

Hints and solutions

So

the

answer is

7.16 Note that

1

231

(2) + 1- r =125971 - 96034 = 29937.

n-3 \+ n-k-1

n-k

n-k-11+

n-2k+1)

l \n - 2k) and deduce that IA(k)dl (n k 1). But m, (k -11' Exercises 8 8.1

(a) S.s4 = ALs4 = si. (b) Ss4 consists of {1, 2, 3), {1, 2, 4}, {3, 4}, {1, S).

8.2

1V = 4 U sV is an antichain with p, =pn_1i so apply 8.2.2 to obtain r(91) + (@) , 2. If n = 2m this gives

AL4 consists of {4, 5}, {3, 5}, {1, 2, 5}, {2, 3, 4}.

z

P,

1
+t
2p_m

--2,

(n-

whence 1sn/2

A ,1. n -1

(i1)

The case n odd is easier. 8.3

Suppose that m , antichain

(note

(nk

i). If M = {B1), then 4' U 90 is an that

A' E- B =>A U B = {1, .

contradicting the given conditions). Thus by 8.1.5, 51+1(51+2(... Sn_k(I4'I )) .

But

. . ,

n},

1.691 +

(l)'

n-1 n-1 ISO-- ( k-1)-(n-k)'

so

(n -

1901'(n)-(n

l 8.4

The following theorem follows from 8.1.5 and 8.3.4. Let P1,

, Pk be given non-negative integers, k, 2n. Then there

exists an intersecting antichain with parameters p, if and only if

P1 + 61(P2 + 62(P3 + ... ))) ' (k -1/'

232 1 Combinatorics of finite sets

8.5

8.6

Apply the ordinary LYM inequality to the corresponding antichain on {1, . . . , n - 1). (i) Each construction puts into 82 pairs of complements in {1,

... , n + 1}. Note that for each i exactly one construction

applies.

(ii) Check all the possibilities. For example, Ai c A; U (n + 1) in B would require A. c A, so that by (a) A; U {n + 1 } is not in £, which is a contradiction. (iii) is obvious. Applying 5.2.3 ngives n

ICI-1I4I-- \[(n+1)/21-1J- \[n/21 +1 8.7

Apply 8.1.5 and divide by (k). This is stronger than the LYM inequality since by the normalized matching property Sk+1(Pk+1)

Pk+1

n

(k)

(k + 1)

8.8

(i) Yes: e.g. 3-sets containing 1 and 4-sets not containing 1. (ii) No: ruled out by LYM. (iii) No: ruled out by 8.1.5.

8.9

(i) By Theorem 8.2.4 we can have equality only when all non-zero pis correspond to the largest denominator, i.e. i = m - 1, m or m + 1. (ii) Considered the squashed antichain @ with non-zero parameters qm =p,,,, qm_i =pm-1 +pm+,. Since s4 is complement2m

free, pm , a ( m) so the sets of size m in 90 are really subsets of

a (2m - 1)-set; thus IA(m)RI ;pm. Thus

(2m) %Pm-1 +

Pm+1 + IA(m)-04I --Pm-1 +Pm+l +pm = \m2m1)'

Let

IA(m)98I =pm.

IA(m),%I . (m

u

so

p.. = (i'), u -- 2m -1; then whence u = 2m - 1.

(

1)'

(i')

Thus pm = 0 or

2m (iii) M

that

{1,...,m-1}E64q{m+2,...,2m}E LI {1,...,m-2; m}

Hints and solutions 1 233

so continue to change one element each time to transform into

0 (Pm , , -Pm *1

any Y other m - I)-set. If tha t I V2(m -1).s1 =

+

(m2_mi), show

so that V2(m -1)d = ((m -1)d)'.

Take 98=(m-1)4: for A,IlA2=0, A,E(m-1)4=>A1c A2'A2' E V2((m - 1) d)

A2 E ((m

A2 e (m - 1)d.

Exercises 9 9.1

d : (1, 0, 2, 1), (0, 1, 1, 2), (2, 0, 2, 0). (0, 0, 0, 4), (0, 0, 1, 3), (0, 0, 2, 2). Ash (0, 0, 2, 1), (1, 0, 1, 1), (1, 0, 2, 0), (0, 0, 1, 2), (0, 1, 0, 2), (0, 1, 1, 1), (2, 0, 1, 0). ACsg : (0, 0, 0, 3), (0, 0, 1, 2), (0, 0, 2, 1). CAd (0, 0, 0, 3), (0, 0, 1, 2), (0, 0, 2, 1), (0, 0, 3, 0), (0, 1, 0, 2), (0, 1, 1, 1), (0, 1, 2, 0). Cs:Q

9.2

Suppose

k = k,.

If 9 6 = {(0, k), (1, k -1), ... , (h, k - h)}

then A= {(O, k -1), (l, k - 2), ... , (h - 1, k - h), (h, k h -1)} if h
are blocks, IedI - Idl = E {IdgO;I - IA,I} % JACs:QI -

1041 since Cs/i is an initial block. Thus I Osf I , I A Cs:Q I . But ACd is compressed, so AC.s/ c CAsl. Next consider the cases k, -_ k , k2 and k > k2 similarly. 9.3

Start with do = F1+1(m). If Aj/I * (l)S then the last member of

(1)S is not in Ado, so remove the last member from do and replace it by the last member of (1)S; call the resulting antichain s:Q,. If A(1 + 1)s:Q, U (1)sei1 * (1)S, remove the last

member from (1 + 1)s/I, and replace it by the second-last member of (1)S to obtain the antichain s42. Eventually we must

obtain A(1 + 1)d; U (1)s4, = (1)S as required; otherwise the process will result in some ,, = (l)S U F,+1(m - N,), which is not an antichain. 9.4

Yes.

9.5

Suppose that y E T' and x is such that x, _- y; for each i. Then (y2, . . . , y,,) is one of the first a,,, members of S(k2s ... , k"). is one of the first a,,, members, and hence Thus (x2, . . . , one of the first ax, members. Thus x E T'.

234 I Combinatorics of finite sets

9.6

Since It s4I % IOCse4I we might as well assume that 4 is

compressed. si is then the first Is4I

(n

k

I)

k-sets, and none

of them contain n. Considering them as k-subsets of an (n - 1)-set; the normalized matching property gives

(IesiI

ICI

`k-1) \nk 1 9.7

(i) The required conclusion will follow if we can show that the shadow of any subcollection {Xi,, . . . , X,} contains at least r

sets. But r--t, so I DF,(r)I %r.

(ii) Let t > M, and take

n {X,, . . . , X,,,+,} = F,(M + 1). . Then there are not enough sets in the shadow of . ' to represent Xi, .. , XM+, since I1 F,(M + 1)I < M + 1. (iii) Let M be the largest integer --I (1)SI such that t --M

implies IA'-"F(t)I > t. Then, if X, ... , X are t -- M I-subsets, there exist t k-subsets Z,, . . . , Z, such that Z; c X for each i. Exercises 10 10.1

(i) f(k...... k.) =

WV).

(ii) Take d to consist of all vectors in S(k,, . .

. ,

with

10.2

If IJI = i, then w(J) = k'.

10.3 10.4

If k; and k; are odd, take all vectors x with x; < 12k; and x; > 2'k1. where g(J)=IxESi :s(x)=JI+IxE Is$I=2E,g(J)

Sd : s(x) = J' I. Suppose there is a J and x, y E Si with s(x) = J

and s(y)=J'. Then we must have x; = k; for i E J, x; = 0 otherwise, and y, = k; for i f J and y, = 0 otherwise. Thus

10.5

g(J) = 2 for such J. Thus I4I _- i r,, max(w(J), w(J'), 2). But k; > 1 for some i, so that w(J)w(J') > 1, whence max(w(J), w(J'), 2) = max(w(J), w(J')). (Gereb-Graus 1983) We can look on 20 as a set of subsets of an (n -1)-set. Thus

I(n-i)-9ol =`ni1 10.6

=(n-11 and

I(i)9oI=(n

q; is isomorphic to S' = S([Z(k;+, - 1)], kj+2, ... , with rank i in S' corresponding to rank i + 2(k, +. . . + k,) in 4.

Hints and solutions 1 235

The rank of S' is K' =

[2(kj+, -

1)] + kj+2 + ...

1(i) fit -- 1([1(K+ q 1(i - Zk, - ... -1kj)S' I -- f ([1(kj+, + ...

10.7

+ k,,. Thus

+ k + 1)])S' l.

Verify that and apply Theorem 3.1.2 to S'. If K is odd, IsiI = 2 Id+j. By Theorem 10.2.6 and Exercise 10.6, E, q,,j = I(?(K + 1)).`DjI for each j, so 1,41-- 2 E;=, i(2 (K + If K is even, Isil = 21st +j or 2 1s4+1 + 1, where I.0+I --E;=o j(1K)!17jj, and 2 jsi+1 = Isi+1 + la i --number of

ZK-vectors other than (Zk...... ZkR). (ii) For K even take all ZK-vectors. For K odd, take Y U Y` where Y = U;=o (2(K + 1))!@,. 10.8

10.9

!2j is isomorphic to S([2(kj+, - 1)], kj+2, the LYM property. The same conclusion holds. We have

I

gI,J.

... ,

which has

II

Zj=i k, I(1)'9ji

for each j, so that E, q,,j- I(K - k)-15djI since, as in Exercise 10.6, j(i)Jjj , (K - k)9j(. Thus E qi,;, Z I(K - k)!@ji = Y, 1(k),04j+11i.j j j 10.10

Clearly 4 will include all divisors of rank --l if

Isis is

maximum. For each divisor of m of rank l + 1, s4 can contain

at most one multiple of it; thus s4 can contain at most N,+, divisors of rank greater than 1. In terms of vectors x= we obtain the following: if for all x, y E 4, (x,, ... , E, min(x,, y,) ,1, then 1.41-- E;±o N,. 10.11

If jyj = Izt then the result is immediate. Suppose if possible that jz j < jy j and z < y. Follow the argument of the last paragraph of the proof of Corollary 9.2.5 with z in place of a and y in place of b to obtain a contradiction. You have to replace ` E 4' by `is a subset of a set in (,sf +) *'.

Exercises 11 11.1

! 91= 21(2(n

2

- 3)) + \?(n -1)I) - 2`2(n -1)).

236 1 Combinatorics of finite sets

11.2 11.3

11.4

Check all possibilities. Note: where does the condition given after Example 11.2.1 fail to hold? When partitioned into chains as in Chapter 3, the rectangle gives min(p, q) chains, each containing one set of size [Zn]. You must verify that w; - x and w1 + x (i 0 j) are at least 2 apart. But their difference in the x,, direction is at least 2. Let S = X, U X2 be the colour decomposition of S. Partition X1, X2 into symmetric chains and combine them to obtain

symmetric rectangles as in the proof of Theorem 11.2.1.

11.5

Partition each rectangle into min(p, q) rows or columns. Each resulting chain is monochromatic and each contains one set of size [in]. Colour all elements of S differently. The set of subsets of S of even size shows that f (n, n) . 2"-'. But if . 4 is a collection of

f(n, n) sets with given property, then, for any x e S, A c 11.6

S - {x}, A e 4 zA U {x} ,$ si. So at most one of A, A U {x} is in d, so that Isil --2--1. Order the elements of the blocks B1, B2 of the first colouring so that in each block the second colouring corresponds to oddness and evenness. Partition B1, B2 into symmetric chains

and form rectangles as in proof of Theorem 11.2.1; then 11.7

follow that proof. (ii) Imitate the growth of block sizes as in proof of Theorem

11.3.2. But this time start with all of 0, x x2, x, +x2 in the one block. Thus at n = 2 we have one block of size 4 instead of blocks,of sizes 1 and 3. The chain construction has blocks of izes 4,2,2 at stage n = 3. Since a 2-block grows into

([2])

blocks after n - 1 steps, i.e. into ([ fin -1)]) blocks

fter n - 2 steps, the 4-block will, in n - 2 steps, give 1 2([? (n -1)]) blocks as required. At most one

(,,(nn

+ 1)]I -

of the sums can be in any given block. Exercises 12 12.1

Take all subsets of S containing a fixed element x, together with the set S - {x}.

Hints and solutions 1 237

12.2

(i) Let u = (u1... u,)'. Then 1 > u 1 . . . u,+1? u`+' so that

f(i+1)%1i ul' (1-u)(1+u+...+u`) c

> (1- u)(1 + ut+ ... +U'-

1

ul -f(i)

Similarly for g. Similarly

(ii) a, =

n-i+1

(iii) Take 12.3

u, =

1 ak+1

and v, = ak+1-i

E, x; ,1 by LYM; x; -- E,.; x; by the normalized matching property. (Compare Exercise 2.2.)

12.4

-u+

WV _

v((k) - (n))

IM

.. .

- If we can chnw that

L

mk-m, `I- (s-i) m,-mk (

(k)-(j) (k)-(kn 1) (k+1)-(k) (i)-(k) then we can choose v in the middle and then u. The middle inequality is precisely the log concavity condition; the first inequality reduces to

(){(k2) +...+(j)l--(kn 1)l(kn

n

1)+...+(j+1)J

which is also true by log concavity. 12.5

Replace ('f) by Ni, and imitate the argument. The LYM and i

normalized matching properties, and log concavity, all hold. Nk, Therefore if szt is an antichain in a multiset with J k -- JK, then E,4k Ni-

238 1 Combinatorics of finite sets

12.6

(iv) y, + ... +(y((,,,21j

-

(l +... + (fl+1\fl(nn))/

ll

(,nn/2)

+ 8\\01 +\1/ +21 + 0((,

n

1

< 6 < 1)

n

([nl2))+n+1+O(\[n/4)l1 on integrating 1 + (1)x + ... + (n)x" = (1 + x)" and putting

x =1. You must verify that the y, satisfy the constraints (12.13). If j

In, E;n-k_,; kyk = j (n) - (1) + E' iyi where I is

the greatest integer less than j and where E' is over all i

12.7

between I and I. This sum is the same form as the first, so repeat the argument. Let Q denote the poset of all non-empty subsets of T and let R denote the poset of all subsets of S - T. Then both Q and R are LYM posets and have log concave rank numbers. Their product is C(n, k), and so the result follows from Theorem 12.1.3.

12.8

(i) If X c A,' and Y c A, and YEP, then X U Y e P since X n A, = O for each i.

(ii) Induction on r and ISj. If A, * S, the result follows by induction. If A, = S, let P" be the subposet of P(A,) obtained by dropping the requirement a, -_ IX n A,I ' b,. By induction

P" is LYM with log concave rank numbers. P is obtained from P" by deleting the bottom a, and top n - b, ranks since X fl A, = X n S = X. But such deletion does not affect LYM or log concavity. (iii) Use Exercise 4.7 and Theorem 12.1.3. Exercises 13 13.1

then X , Y for some Y E Suppose max(S) = s4. If X e max(f) = s4, and so X E s. If X e .s4, X -_ Y for some Y e s4 = max(°.f) and so X -- Y where Y e 9, so that X e 9. Thus S = R. Suppose 9 =,s4. If X e max(.°) then X e 9 so X E .s and so

Hints and solutions 1 239

X _ Y for some Y E sc; but then Y E .s = 9, so X = Y by maximality and so X E &Y. If X E saf then X E R = %, and X E max(°), since if X< Y where Y E 9= si then X< Y_ Z for some Z E Si, contradicting the antichain property; there13.2

fore X E max(g). Thus s&' = max(S). Let X E max(si U 99). Certainly X E R U A. If X < Y where

YEsiUA, then X
,

say, so that X
and so X 0 max(si U A). Therefore X E max(,f U A), and max(,si U Oh) c max(s4 U -4). The converse is similar. 13.3

Consider X in category (i) or (ii) as in the text. You must show that X E max(.s fl A). If X E si fl go then X E ii fl A, and if X _ Y E Q fl -4 then Y A for some A E Si, so that X _ A with A,X E ,sQ and so A = X = Y and X E max(.s fl S3).

If X E Si U A and X c W for some W E si U -64 then, say, X E d, W E - 0 4 ; therefore X E si and X E

so that X E si fl -04.

If X _ V E ' fl -4, proceed as before. 13.4 13.5

Take P = {a, b, c} with c
B A for this B, then B _ A _ B' for some B' E IR since si 90; therefore B = B' and A _ B. Therefore this B is incomparable with every A E a and could therefore be added to .s1, contradicting maximality. Thus 3& = 90. The converse is similar.

(ii) If sego U s1 c si strictly, there is an A E 4 which can be added to 68 to give a larger antichain. (iii) Certainly sie n 3& c si n,130. But if X E s4 fl 6a then X e 34 (since X a X) and X E R,; so X E,4, fl go,,. (iv) A E q, fl d' B' _ A , B for some B, B' B' = A = B A E s4 fl 9d. The converse is easy. 13.6

E; min(ICI, k) = kak + Eici
13.7

For a k-saturated chain partition ', dk(P) = ISkI + kak. But ak ? d,(P - Sk), so that dk(P) l ISkI + kd,(P - Sk). On the other hand, dk(P) _ ISI + kd,(P - S) for all S c P.

13.8

Consider the top and bottom members s4*,sC* of &(P). If they have a member A in common, then A is in every member 64 of .2(P). For .s4* - A, so A _ B for some B E °.4, but then

240 1 Combinatorics of finite sets

B , C for some C E Si * since 58 - .si * ; therefore A -- C where

A,CE9.I*, so A= C=B and A E60. 13.9

dk(P) = ISkI + kak, dk+,(P) -- ISkj + (k + 1)ak, and dk_,(P)

ISkI + (k - 1)ak. The first two give Ak+l(P) -- ak The first and third give Ak -- ak. 13.10

If Ak(P) = Ak+,(P) then, by Exercise 13.9, both are equal to Also dk(P) = ISkI + kak. Thus Ak+](P) = dk(P) + ak = ISkl + (k + 1)ak. ak.

dk+1(P) = dk(P) +

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Index

Ahlswede-Daykin inequality 91 antichain 2, 14, 36, 131, 168, 201 complement-free 138

L(m, n) 46 lattice 97, 203, 212 lexicographic order 36, 112, 114, 129, 145, 211

intersecting 70, 172

squashed 133

antilexicographic order 113, 140 base 105

basic element 33 binomial representation 115 blocker 8 ChvStal conjecture 103 Clements-Lindstrom theorem 145 common system of distinct representatives 48 completely separating system 8 compression 126, 146 crosscut 7

Dilworth's theorem 30, 203, 209 duality 187

Erdos-Ko-Rado theorem (EKR) 70, 79, 124, 142, 173

FKG inequality 95, 97 Ford-Fulkerson theorems 49, 195 generalized Macaulay theorem 145

linear extension 98 Littlewood-Offord problem 176 log concavity 58, 69, 190, 200 log convexity 58, 96 log supermodular 95 Lovdsz's version of KK 123 LYM inequality 3, 15, 63, 73, 131, 138, 143, 171,200 marriage theorem 12, 32, 161, 195 Menger antichain 9 multiset 18, 145 nested chains 44

normalized matching property 12, 15, 51, 60, 138, 159, 190, 200 ordering antilexicographic 113 lexicographic 112 squashed 113 orthogonal chain decomposition 41

parameters of an antichain 131 parentheses 33 partition 19 Perfect, H. 49, 207 poset 14 probability 40, 79, 98 product poset 45, 69, 190

ideal 87, 103, 127, 152, 166, 202 intersecting family 54, 70, 77, 82, 90, 103, 162

k-saturation 206, 213 k-union 205, 212 Kleitman's lemma 87, 108, 110 Kruskal-Katona theorem (KK) 119, 129, 134, 145

rank 14, 27, 56 regular covering 15 regular poset 23 Rota conjecture 21 saturated chain partitions 205

saturated chains 14

249

250 1 Index shade 10, 60, 129, 142 shadow 10, 61, 81, 112, 125, 146 shift operator 81, 120 special chain partition 207 Sperner property 17, 21, 30, 44, 48, 200 Sperner's theorem, proofs of 2, 11, 29, 205 squashed antichain 133, 137, 169 squashed order 113 Stirling numbers 19, 24, 68 subadditivity 125, 152 support 163 symmetric chain order 45 symmetric chains 27, 37, 39, 42, 45, 66, 179, 182

systems of distinct representatives 12, 48, 161, 195

Theorems of Ahlswede-Daykin 91, 110 Anderson 51, 67 Berge 103 Bollobas 4, 73 Bondy 192 Brace-Daykin 75, 79 de Bruijn-Tengbergen-Kruyswzijk 28 Clements 56, 84, 125, 133, 152, 158, 169 Clements-Gronau 138, 169 Clements-Lindstrom 146, 154 Daykin 94 Daykin-Godfrey-Hilton 133, 136 Dilworth 30, 203 Erdos 17, 176, 177 Erdos-Herzog-Schonheim 165

Erdos-Ko-Rado 70 Erdos-Schonheim 163 Ford-Fulkerson 49, 195 Fortuin-Kasteleyn-Ginibre 97 Graham-Yao-Yao 98 Greene-Hilton 136, 140 Greene-Katona-Kleitman 77 Greene-Kleitman 143, 186, 207, 213

Griggs 44, 51, 56, 182, 196 Gronau 171, 173 Hall, M. 26 Hall, P. 13 Hansel 38 Harper 190 Hilton 85, 110, 111

Hilton-Milner 84 Hoggar 69 Holley 110 Hsieh-Kleitman 69, 190 Katona 23, 82, 179, 183 Kleitman 14, 86, 87, 90, 126, 179, 183, 186

Kleitman-Milner 62, 188 Kruskal-Katona 118 Li 193 Liggett 79 Lindsey 156

Lindstrom 47 Lovdsz 123

Marica-Schonheim 94 Mikl6s 106 Milner 81 Purdy 139 Sali 181, 185 Schonheim 79, 105 Seymour-Welsh 96 Shearer-Kleitman 42 Shepp 101 Sperner 2

unimodality 18, 56 upset 87

variance 64

XYZ conjecture 98

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Astronomy BURNHAM'S CELESTIAL HANDBOOK, Robert Burnham, Jr. Thorough guide to the stars beyond our solar system. Exhaustive treatment. Alphabetical by constellation: Andromeda to Cetus in Vol. 1; Chamaeleon to Orion in Vol. 2; and Pavo to Vulpecula in Vol. 3. Hundreds of illustrations. Index in Vol. 3. 2,000pp. 64 x 9/4. 23567-X, 23568-8, 23673-0 Three-vol. set THE EXTRATERRESTRIAL LIFE DEBATE, 1750-1900, Michael J. Crowe. First detailed, scholarly study in English of the many ideas that developed from 1750 to 1900 regarding the existence of intelligent extraterrestrial life. Examines ideas of Kant, Herschel, Voltaire, Percival Lowell, many other scientists and thinkers. 16 illustrations. 704pp. 5% x 8Y.

40675-X

A HISTORY OF ASTRONOMY, A Pannekoek. Well-balanced, carefully reasoned study covers such topics as Ptolemaic theory, work of Copernicus, Kepler, Newton, Eddington's work on stars, much more. Illustrated. References. 521pp. 5'4 x 8'/i. 65994-1

AMATEUR ASTRONOMER'S HANDBOOK, J B. Sidgwick. Timeless, comprehensive coverage of telescopes, mirrors, lenses, mountings, telescope drives, micrometers, spectroscopes, more. 189 illustrations. 576pp. 51 x 8'4. (Available in U.S. only.) 24034-7

STARS AND RELATIVITY, Ya. B Zel'dovich and 1. D. Novikov. Vol.

1 of Relativistic Astrophysics by famed Russian scientists General relativity, properties of

matter under astrophysical conditions, stars, and stellar systems Deep physical insights, clear presentation. 1971 edition. References. 544pp.554 x 84.

69424-0

Chemistry CHEMICAL MAGIC, Leonard A Ford Second Edition, Revised by E. Winston Grundmeier Over 100 unusual stunts demonstrating cold fire, dust explosions, much more Text explains scientific principles and stresses safety precautions 67628-5 128pp 5't x 8' THE DEVELOPMENT OF MODERN CHEMISTRY, Aaron J. Ihde Authoritative history of chemistry from ancient Greek theory to 20th-century innovation. Covers major chemists and their discoveries 209 illustrations. 14 tables.

Bibliographies Indices Appendices. 851pp. 514 x 8'64235-6 CATALYSIS IN CHEMISTRY AND ENZYMOLOGY, William P. Jencks. Exceptionally clear coverage of mechanisms for catalysis, forces in aqueous solution, carbonyl- and acyl-group reactions, practical kinetics, more. 864pp. 51,b x 81A 65460-5

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THE HISTORICAL BACKGROUND OF CHEMISTRY, Henry M. Leicester. Evolution of ideas, not individual biography. Concentrates on formulation of a coherent set of chemical laws. 260pp 5% x 81 61053-5

A SHORT HISTORY OF CHEMISTRY, J. R. Partington Classic exposition explores origins of chemistry, alchemy, early medical chemistry, nature of atmosphere, theory of valency, laws and structure of atomic theory, much more 428pp. 5%. x 84

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GENERAL CHEMISTRY, Linus Pauling. Revised 3rd edition of classic first-year text by Nobel laureate Atomic and molecular structure, quantum mechanics, statistical mechanics, thermodynamics correlated with descriptive chemistry. Problems 992pp. 5% x 81/2

65622-5

Engineering DE RE METALLICA, Georgius Agricola The famous Hoover translation of greatest treatise on technological chemistry, engineering, geology, mining of early mod60006-8 em times (1556). All 289 original woodcuts. 638pp. 6/. x H.

FUNDAMENTALS OF ASTRODYNAMICS, Roger Bate et al. Modem approach developed by U.S Air Force Academy Designed as a first course. Problems, exercises. Numerous illustrations. 455pp. 5% x 8'4

60061-0

DYNAMICS OF FLUIDS IN POROUS MEDIA, Jacob Bear. For advanced students of ground water hydrology, soil mechanics and physics, drainage and irrigation engineering and more 335 illustrations Exercises, with answers. 784pp 6'. x 9'/,. 65675-6

ANALYTICAL MECHANICS OF GEARS, Earle Buckingham Indispensable reference for modern gear manufacture covers conjugate gear-tooth action, gear-tooth profiles of various gears, many other topics. 263 figures. 102 tables. 546pp 5% x 8Y. 65712-4

MECHANICS,]. P Den Hartog. A classic introductory text or refresher. Hundreds of applications and design problems illuminate fundamentals of trusses, loaded 60754-2 beams and cables, etc 334 answered problems. 462pp. 5% x 8'/!.

MECHANICAL VIBRATIONS, J. P Den Hartog. Classic textbook offers lucid explanations and illustrative models, applying theories of vibrations to a variety of practical industrial engineering problems. Numerous figures. 233 problems, solu64785-4 tions Appendix. Index Preface. 436pp 5% x 8'/. STRENGTH OF MATERIALS, J. P. Den Hartog. Full, clear treatment of basic material (tension, torsion, bending, etc.) plus advanced material on engineering methods, applications. 350 answered problems. 323pp 5% x 814

60755-0

A HISTORY OF MECHANICS, Rene Dugas. Monumental study of mechanical principles from antiquity to quantum mechanics. Contributions of ancient Greeks, Galileo, Leonardo, Kepler, Lagrange, many others. 671 pp. 5% x 81/.

65632-2

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Physics OPTICAL RESONANCE AND TWO-LEVEL A'l'OMS, L Allen and,J H. Eberly Clear, comprehensive introduction to basic principles behind all quantum optical resonance phenomena 53 illustrations Preface. Index 256pp. 5% x 8%. 65533-4

ULTRASONIC ABSORPTION. An Introduction to the Theory of Sound Absorption and Dispersion in Gases, Liquids and Solids, A B. Bhatia. Standard reference in the field provides a clear, systematically organized introductory review of fundamental concepts for advanced graduate students, research workers Numerous diagrams. Bibliography. 440pp. 5% x 81h

64917-2

QUANTUM THEORY, David Bohm This advanced undergraduate-level text presents the quantum theory in terms of qualitative and imaginative concepts, followed by specific applications worked out in mathematical detail. Preface. Index. 655pp. 5% x 8'h

65969-0

ATOMIC PHYSICS (8th edition), Max Born. Nobel laureate's lucid treatment of kinetic theory of gases, elementary particles, nuclear atom, wave-corpuscles, atomic structure and spectral lines, much more Over 40 appendices, bibliography 495pp. 5% x 8'h.

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AN INTRODUCTION TO HAMILTONIAN OPTICS, H. A. Buchdahl Detailed account of the Hamiltonian treatment of aberration theory in geometrical optics. Many classes of optical systems defined in terms of the symmetries they possess Problems with detailed solutions 1970 edition xv + 360pp 5% x 81h.

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THIRTY YEARS THAT SHOOK PHYSICS- The Story of Quantum Theory, George Gamow Lucid, accessible introduction to influential theory of energy and matter Careful explanations of Dirac's anti-particles, Bohr's model of the atom, much more. 12 plates. Numerous drawings 240pp. 5% x 8%.

24895-X

ELECTRONIC STRUCTURE AND THE PROPERTIES OF SOLIDS The Physics of the Chemical Bond, Walter A. Harrison. Innovative text offers basic understanding of the electronic structure of covalent and ionic solids, simple metals, transition metals and their compounds. Problems. 1980 edition 582pp. 6A x 9'/4. 66021-4

HYDRODYNAMIC AND HYDROMAGNETIC STABILITY, S. Chandrasekhar. Lucid examination of the Rayleigh-Benard problem; clear coverage of the theory of instabilities causing convection 704pp. 5`h x 8'%4

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INVESTIGATIONS ON THE THEORY OF THE BROWNIAN MOVEMENT, Albert Einstein Five papers (1905-8) investigating dynamics of Brownian motion 60304-0 and evolving elementary theory Notes by R. Furth 122pp 5% x 8Y,.

THE PHYSICS OF WAVES, William C. Elmore and Mark A Heald. Unique overview of classical wave theory. Acoustics, optics, electromagnetic radiation, more. 64926-1 Ideal as classroom text or for self-study. Problems 477pp. 5% x 8%.

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METHODS OF THERMODYNAMICS, Howard Reiss. Outstanding text focuses on physical technique of thermodynamics, typical problem areas of understanding, and significance and use of thermodynamic potential 1965 edition. 238pp 5% x 81A 69445-3

TENSOR ANALYSIS FOR PHYSICISTS, J. A Schouten Concise exposition of the mathematical basis of tensor analysis, integrated with well-chosen physical examples of the theory. Exercises. Index Bibliography 289pp 514 x 814 65582-2

RELATIVITY IN ILLUSTRATIONS, Jacob T. Schwartz Clear nontechnical treatment makes relativity more accessible than ever before Over 60 drawings illustrate

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THE ELECTROMAGNETIC FIELD, Albert Shadowitz. Comprehensive undergraduate text covers basics of electric and magnetic fields, builds up to electromagnetic theory Also related topics, including relativity Over 900 problems 768pp. 5k x 8'/,

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GREAT EXPERIMENTS IN PHYSICS. Firsthand Accounts from Galileo to Einstein, edited by Morris H. Shamos 25 crucial discoveries: Newton's laws of motion, Chadwick's study of the neutron, Hertz on electromagnetic waves, more Original accounts clearly annotated 370pp. 5% x 81k.

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RELATIVITY, THERMODYNAMICS AND COSMOLOGY, Richard C. Tolman. Landmark study extends thermodynamics to special, general relativity; also applications of relativistic mechanics, thermodynamics to cosmological models. 65383-8 501 pp. 5% x 81h.

LIGHT SCATTERING BY SMALL PARTICLES, H C. van de Hulst. Compre-

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STATISTICAL PHYSICS, Gregory H. Wannier Classic text combines thermodynamics, statistical mechanics and kinetic theory in one unified presentation of thermal physics. Problems with solutions. Bibliography. 532pp. 5% x 86 65401-X

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