Transversal Theory (Mathematics in Science & Engineering, Vol. 75)

TRANSVERSAL THEORY

This is Volume 75 in MATHEMATICS IN SCIENCE AND ENGINEERING A series of monographs and textbooks Editor RICHARD RELLMAN, University qf Southern California A complete list of the books in this series appears at the end of this volume.

TRANSVERSAL THEORY

An account of some aspects of combinatorial mathematics L. Mirsky

1971

A C A D E M I C P R E S S New York and London

COPYRIGHT 0 1971, BY ACADEMIC PRESS. INC. ALL RIGHTS RESERVED NO PART OF THIS BOOK MAY BE REPRODUCED I N ANY FORM,

BY PHOTOSTAT, MICROFILM, RETRIEVAL SYSTEM, OR ANY OTHER MEANS, WITHOUT WRITTEN PERMISSION FROM THE PUBLISHERS.

ACADEMIC PRESS, INC.

1 1 1 Fifth Avenue, New York, N e w Y o r k 10003

United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. Berkeley Square House, London, W l X 6BA

LIBRARY OF CONGRESS CATALOG CARDNUMBER: 71-142083 AMS 1970 SUBJECT CLASSIFICATIONS 05-00,OSAOS

PRINTED I N THE UNITED STATES OF AMERICA

PREFACE Transversal theory, the study of combinatorial questions of which Philip Hall’s classical theorem on ‘distinct representatives’ is the fount and origin, has only recently emerged as a coherent body of knowledge. The pages that follow represent a first attempt to provide a codification of this new subject and, in particular, to place it firmly in the context of the theory of abstract independence. I have sought to make the exposition leisurely, systematic, and as nearly self-contained as possible; but since the length of the book had to be kept within conventional bounds, it has been necessary to exclude certain topics even though they impinge on my central theme. Thus I say nothing about the subject of ‘flows in networks’ initiated by Ford and Fulkerson; I pass in silence over the exciting possibilities of establishing combinatorial theorems by the method of linear programming; and I refer only occasionally to the theory of graphs. I hope that as a result my presentation has gained in care and clarity what it has undoubtedly lost in breadth of treatment. The account offered here is intended primarily for three classes of readers. It aims to serve as a detailed introduction to the methods of transversal theory for postgraduate students who wish to specialize in combinatorial mathematics. It will, perhaps, provide a convenient work of reference for experts in the field. And finally, it is a repository of combinatorial results which those engaged in the application of mathematical techniques to practical problems may find occasion to invoke. The stock of knowledge requisite for the study of the book is modest, although a few of the arguments presuppose some degree of mathematical sophistication. The reader needs to be conversant with a small number of results from the theory of sets, including Zorn’s lemma, and with some concepts in general topology: as much of this as is necessary is summarized in the first chapter. Beyond this, T assume some familiarity with the theory of vector spaces and, here and there, a nodding acquaintance with other basic structures of elementary algebra. An early ancestor of the book is the survey article ‘Systems of Representatives’ contributed by Dr Hazel Perfect and myself to Volume 15 of the Journal of Mathematical Analysis and Applications. I have made entirely free with material from this source and 1 have to thank the editor and publishers (Academic Press, Inc.) for permission to do so. V

vi

PREFACE

1 am eager to record my very deep sense of gratitude to a number of friends. My indebtedness to Dr Perfect will be plain to anyone who compares the survey with the present account. Indeed, D r Perfect’s influence has been pervasive, for she and I discussed at length almost every topic treated here, and she has put me under yet a further obligation by scrutinizing the entire manuscript. Dr J. S. Pym has read and commented on several chapters and has saved me from many blunders. Further, he extracted from his own investigations the proof of the difficult Theorem 10.4.4 presented below. I must add that it was only his and Dr Perfect’s active encouragement which enabled me to complete the project. I have had many discussions with Professor R. A. Brualdi and with Dr D. J. A. Welsh, and I have benefitted greatly from their insight into combinatorial problems. My debt to Professor Richard Rado is very extensive. I owe to him the pleasure and stimulus of countless mathematical conversations and the use of much unpublished material that he most generously placed at my disposal. Above all, his contributions to transversal theory have had a decisive influence on the growth of the subject and, consequently, on the shape of this book. I am grateful to the editor, Professor R. Bellman, for inviting me to write a volume for his series ‘Mathematics in Science and Engineering.’ Finally, I should like to express my appreciation of the helpfulness and impressive efficiency of Academic Press, Inc. and of the excellence of their printing.

University of Shefield August 1970

L. Mirsky

CONTENTS 1 Sets, Topological Spaces, Graphs 1.1 Sets and mappings 1.2 Families 1.3 Mapping theorems and cardinal numbers 1.4 Boolean atoms 1.5 The lemmas of Zorn and Tukey 1.6 Tychonoff‘s theorem 1.7 Graphs Notes on Chapter 1

1

5

9 14 16 20 21 23

2 Hall’s Theorem and the Notion of Duality 2.1 Transversals, representatives, and representing sets 2.2 Proofs of the fundamental theorem for finite families 2.3 Duality Notes on Chapter 2

24 27 32 38

3 The Method of ‘Elementary Constructions’ 3.1 ‘Elementary constructions’ 3.2 Transversal index 3.3 Further extensions of Hall’s theorem 3.4 A self-dual variant of Hall’s theorem Notes on Chapter 3

39 40 44

48 50

4 Rado’s Selection Principle 4.1 Proofs of the selection principle 4.2 Transfinite form of Hall’s theorem 4.3 A theorem of Rado and Jung 4.4 Dilworth’s decomposition theorem 4.5 Miscellaneous applications of the selection principle Notes on Chapter 4 vii

52 55 59

61

64 71

...

CONTENTS

Vlll

5

Variants, Refinements, and Applications of Hall’s Theorem 5.1 Disjoint partial transversals 5.2 Strict systems of distinct representatives 5.3 Latin rectangles 5.4 Subsets with a prescribed pattern of overlaps Notes o n Chapter 5

74 78 81 84 88

6 Independent Transversals 6. I Pre-independence and independence 6.2 Rado’s theorem o n independent transversals 6.3 A characteristic property of independence structures 6.4 Finite independent partial transversals 6.5 Transversal structures and independence structures 6.6 Marginal elements 6.7 Axiomatic treatment of the rank function Notes on Chapter 6

90 93 99

loo 101

105

107

110

7 Independence Structures and Linear Structures 7.1 A hierarchy of structures 7.2 Bases of independence spaces 7.3 Totally admissible sets 7.4 Set-theoretic models of independence structures Notes on Chapter 7

112

119

124 125 128

8 The Rank Formula of Nash-Williams 8.1 Sums of independence structures 8.2 Disjoint independent sets 8.3 A characterization of transversal structures 8.4 Symmetrized form of Rado’s theorem on independent transversals Notes on Chapter 8

130 134 138 140 145

9 Links of Two Finite Families 9.1 The notion of a link 9.2 Common representatives 9.3 The criterion of Ford and Fulkerson 9.4 Common representatives with restricted frequencies 9.5 An insertion theorem for common transversals 9.6 Harder results for a single family Notes o n Chapter 9

147 148 150 154 158

161 167

CONTENTS

ix

10 Links of Two Arbitrary Families 10.1 The theorem of Mendelsohn and Dulmage and its interpretations 10.2 Systems of representatives with repetition 10.3 Common systems of representatives with defect 10.4 Common transversals of two families 10.5 Common transversals of maximal subfamilies Notes on Chapter 10

i69 173 175 176 181

182

11 Combinatorial Properties of Matrices The language of matrix theory 11.2 Theorems of Konig, Frobenius, and Rado 11.3 Diagonals of doubly-stochastic matrices 11.4 Doubly-stochastic patterns 11.5 Existence theorems for integral matrices Notes on Chapter 1 1 11.1

183

187 192 199 204 21 1

12 Conclusion 12.1 Current trends in transversal theory 12.2 Future research and open questions

Miscellaneous Exercises

214 218 229

Bibliography

236

Index of Symbols

247

Index of Authors

24 9

General Index

252

This page intentionally left blank

T R A N S V E R S A L THEORY

This page intentionally left blank

1 Sets, Topological Spaces, Graphs In this introductory chapter, we shall pass in brief and somewhat informal review a series of definitions and results from set theory, general topology, and the theory of graphs. This preliminary discussion will provide the necessary background for the study of transversal theory with which we shall come to grips in Chapter 2. 1.1 Sets and mappings We shall assume that the reader is familiar with the elementary algebra of sets. Consequently, much of what is said here is intended to serve as no more than a reminder. We regard the notion of a set as primitive and shall not attempt to subject it to further analysis. A set is specified by the elements (or members) which belong to or are contained in it. Two sets are therefore called equal if they contain the same elements. A particular set we often have to consider is the empty set, denoted by the symbol 0, which contains no elements at all. If a n element x belongs to a set X, we write x E X ; in the contrary case, we write .x$ X. We shall, whenever possible, use lower case italic letters for elements and capital Roman letters for sets. Complete notational consistency is, however, impossible since the terms ‘element’ and ‘set’ are only relative: thus given sets can themselves be elements of other sets. To avoid verbal contortions such as ‘set of sets’ we shall occasionally use the term collection as a synonym for a set. A set is calledznite if it contains only a finite number of elements; otherwise it is called inznite. We can often specify a set by listing its elements. Thus { 1, 2, ..., k } denotes the set whose elements are the first k natural numbers. It is an immediate consequence of the notion of a set that if, in such a catalogue, an element is mentioned more than once, then all but one of its occurrences can be ignored. Again, if the order in which the elements are listed is changed, the set is not affected. Thus, for example,

{ I , 1,2)

= {1,2,

I}

=

(2, I , I }

=

(1,2)

= (2, l};

(1)

and each of the above expressions stands for the set whose elements are the integers 1 and 2. Order and repetition of elements are, then, irrelevant in a set. I

2

SETS, TOPOLOGICAL SPACES, GRAPHS

1,s 1 . 1

We shall use the symbol {x,,..., x,}, to denote the set consisting of the elements xI, .. ., x k and at the same time express the fact that these elements are distinct. If the suffix ‘ # ’ is not appended to the curly bracket, then no assumption is made about the distinctness of the elements listed. If x is an element of some set, then, by the convention just laid down, {x} denotes the set containing x as its only element. A set of this type is called a singleton. The objects x and {x} are logically quite distinct. Thus, for example, ( 0 )is a set containing one element, namely 0. When the elements present in a set cannot be catalogued, we may still be able to give a ‘descriptive’ definition of the set by specifying some distinguishing property of its elements. Let X be a given set and let G ( x ) be a statement about the element x of the set X. Then either of the expressions

{x E x: G ( x ) } ,

{x:x E x, G ( x ) }

denotes the set of all those elements x in X for which the statement G ( x ) is valid. More concisely, if less explicitly, we can also write {x: G(x)} for this set if it is clear from the context that we are concerned with the elements of X. Next, let X, Y be sets. We write X c Y (or, equivalently, Y z X) to indicate that every element of X is also an element of Y. We then say that X is contained in, or is a subset of, Y. (This relation does not preclude the possibility of X and Y being equal, i.e. X = Y). The empty set is consequently a subset of every set. If X is not contained in Y, we write X $ Y. If X c Y but X # Y, we write X c Y (or Y I> X) and we say that X is properly or strictly contained in Y, or that it is a proper subset of Y. The relation c is called the relation of inclusion, c that of strict (or proper) inclusion. We shall write X cc Y to indicate that X is a finite subset of Y. The basic operations by means of which sets can be combined are the formations of unions, intersections, and differences. The union of X and Y, denoted by X u Y, is the set consisting of all elements which belong to at least one of X, Y. The intersection of X and Y, denoted by X n Y, is the set consisting of all elements which belong to both X and Y. Analogous definitions and obvious notational modifications apply in the case of more than two sets. The difference of X and Y, denoted by X \ Y, is the set of all elements which belong to X but not to Y. We say that X and Y are disjoint if they have an empty intersection, i.e. X n Y = 0. We say that X and Y intersect if they have a non-empty intersection, i.e. X n Y # 0. Let X be a subset of a set E. Then by the complement of X (relative to E) we mean the set E \ X, and we denote this set by gE(X)or, more briefly, by U(X) or %‘X. The collection of all subsets of X, including the empty set and X itself, is

8 1.1

3

SETS AND MAPPINGS

called the power set of X and is denoted by S(X). If X is a (finite) set of n elements, then S ( X ) has 2” elements. Our next step is to introduce the fundamental notion of a mapping. Let X, Y be sets and suppose that, with each element x of X, is associated a definite element of Y, which we denote by #(x). We then say that # is a mapping of X into Y, and we express the situation symbolically by writing 4: X Y. The element #(x) is called the image of x (under 4). The sets X and Y need not be different: if Y = X, then # is called a mapping of X into itself. The set X on which # is defined is called its domain. When A c X, the subset #(A) of Y is defined by the equation --f

#(A) = (#<x>:x E A)

(so that, in particular, #(0) = 0).The set #(X) ( G Y) is called the range of #. In other words, the range of # is the subset of Y consisting of those elements which are images, under #, of elements in X. We shall normally use Greek letters for mappings. Let #: X -+ Y be a mapping, suppose that Y G Y’, and let the mapping 4’ :X -+ Y’ be defined by the equation #’(x) = #(x) (x EX). Strictly speaking, 4 and #’ are different objects but we shall, in practice, not distinguish between them and shall even designate them by the same symbol. Again, let X’ c X and let the mapping $ : X ’ - + Y be defined by the equation $(x) = $(x) (x E X’). We then call $ the restriction of # to X‘, and write )I

= #lX’.

It is useful to specify certain types of mappings. The mapping #: X Y is called injective (or an injection) if distinct elements have distinct images, i.e. if 4(x1) # # ( x 2 ) whenever xl, x2 E X and x1 # x2.It is called surjective (or a surjection) if every element of Y is the image of at least one element in X, i.e. if the range of # is Y. A mapping which is both injective and surjective is called bijective (or a bijection). Thus, a bijection is what, in traditional mathematical language, is known as a ‘one-one correspondence.’? A bijection of a set X into itself is called a permutation of X. Further, if #: X + Y is an injection, then 4: X #(X) is a bijection. In some mathematical literature, the terms ‘oneone’ and ‘onto’ are used where we employ ‘injective’ and ‘surjective’. Happily, this base coinage is likely before long to be withdrawn from currency. Let 4 :X -+ Y be a mapping. It is then easily shown that --f

---f

4 ( A \ B)

= #(A)

\ 4(B)

for all A, B c X if and only if # is an injection.

t More precisely, a one-one correspondence is a bijection together with its ‘inverse’.This last term is defined a few lines below.

4

SETS, TOPOLOGICAL SPACES, GRAPHS

1, !j 1.1

L e t 4 : X 4 Y b e a bijection. Wethendefineitsinverse(mupping)4-': Y-+X by the specification that, for each y E Y, 4 - ' ( y ) is the unique element x of X such that 4(x) = y. Clearly 4-l is again a bijection and (4-')-' = 4. If 4: X + Y is merely an injection, we can still define its inverse since 4: X+4(X) is a bijection. In that case, 4 - l is a mapping of 4(X) into X. Next, let 4 : X -+ Y, $: 2 + W be two mappings such that the range of the first is contained in the domain of the second. Then the product (or composition) $4 is defined as the mapping of X into W given by the equation

($4)(XI = ${4(x>l (x E XI. The mappings $4 and 4$ are, in general, quite distinct:

indeed, it may well happen that one of them i s defined while the other is not. However, if 4,$, 0 are mappings, then 4($@= (4$)0 provided all the products are defined. It is therefore permissible t o omit brackets and to write 4$6J without risk of ambiguity-a remark which applies equally to the product of any number of mappings. If 4 is a mapping of a set into itself, we define its powers inductively by the relations

4'

=

4,

(y = & j f - 1

(n = 2, 3, . ..).

x,

Let X,, ..., X, be pairwise disjoint sets and let 0,: + Y ( 1 < k d n) be mappings. Let the mapping CT: X I u ... u X, + Y be specified by the requirement that C T ( X ) = ak(x)when X E X , . We then call CT the direct sum of the mappings crl, .. ., on.

Exercises 1.1 1. Show that two subsets A, B of a set E are unequal if and only if

( A n 'CB) u (%?An B) # 0.

2. Let A, B, C be sets. Show that C = A u B if and only if both the following conditions are satisfied. ( i ) A G C , B G C. (ii) The relations A c D, B G D imply C c D. 3. Let

4:X + Y

be a mapping. Show that, for any subsets A and B of X, &A n B) E &A) n 409. Establish the equivalence of the following statements. (i) 4 is injective. (ii) 4 ( A n B) = $(A) n $(B) whenever A, B _C X. (iii) &A \, B) = di(A) \ &(B) whenever A, B G X. 4. Show that the number of surjective mappings of a set of rn elements into a set of n elements is equal to

(-I>"

f (-1)'(i)

/l=O

k".

p

FAMILIES

1.2

5. Let A

A

5

B, the symmetric difierence of A and B, be defined by the equation A A B = (A u B) \ (A n B).

Prove that

A

is an associative operation and verify that, for n 3 2, A,

A

A,

A

... A A,,

is the set of all those elements which belong to precisely an odd number of A’s. 6. Let S be arbitrary set consisting of at least 2 elements and let P(S)be the power set of S. Show that, with respect to the operation A as addition and n as multiplication, Y(S)is a commutative ring with identity. Determine all divisors of zero in this ring.

7. A is a set and 8 is a collection of subsets of A with A € 3 which is closed under arbitrary intersections. Prove that, for each subset X of A, there exists one and only one subset X* of A such that (a) X X I ; (b) X* E 5 ; (c) if X s Y and Y € 5 ,then X* G Y. Also show that (X*)* = X.

1.2 Families In a set, the order in which the elements occur and the frequency of their occurrence play no part. Since we need, at times, to consider totalities in which these features are present, we shall now introduce the notion of a ‘family’. Strictly speaking, this is not a new object at all: we have at hand just what we need in the concept of a mapping. Let, then, E and I be sets; let 4 : I + E be a mapping, and write 4(i) = x i for all i E 1. We shall often find it useful to denote the mapping 4 by the symbol (xi:ie I) and to call it a family (or system) of elements of E indexed by I (or with index set 1). A family is thus a mapping and not a set.? This distinction must be firmly maintained throughout our discussion : in the theory we shall develop, failure to d o so is visited by calamity. To emphasize the distinction, we shall adopt the convention of using curly brackets for sets and round brackets for families. We note, in particular, that the domain of the family ( x i :i E I) is 1 and that its range is { x i :i E I}. We shall normally use capital German letters for families. Let X = ( x i :i E I) be a family of elements of E. Unless the contrary is expressly stated we do not, of course, require that xi,xi should be distinct whenever i # j . Indeed, a given element x of E may occur infinitely often among the x,. Thus the phenomenon of repetition is allowed for in the notion of a family. And equally, if the index set I possesses a n ‘ordinal structure’, say if it is the set of real numbers, then The distinction between sets and mappings is not as absolute as may appear from our remarks. If we were offering a systematic treatment of the theory of sets, we should take care to define mapoings as certain kinds of sets (of ‘ordered pairs’).This is, however, irrelevant in the present context, for we are now merely concerned to avoid confusion between ( x i : i E 1) and { x ~i:E 1). $ Cf. 5 1.5 below.

6

SETS, TOPOLOGICAL SPACES, GRAPHS

1,51.2

an ‘order of precedence’ is set up among the elements of the family. Thus, if i, j E 1 and i < j , we may say that x iprecedes xi. The family ( x i : 1 < i < n) will often be written in the alternative form (xl,x2, ..., x,J. This notation conveys clearly that the index set (or domain) is { 1,2, .. ., n } and that the family is the mapping which, for 1 < i < n, carries i into x i . On the other hand, the meaning of a succession of n unindexed symbols such as

is less immediately obvious. However, we shall agree t o regard (1) as a legitimate notation for the family in which (unless the contrary is stated) the index set is understood t o be { 1,2, ..., n } and its elements 1,2, ..., n are mapped into a, b, ..., Y respectively. We often call an object such as ( I ) an ‘ordered set’ or, more explicitly, an ‘ordered n-tuple.’ In particular, when n = 2, we speak of an ‘ordered pair.’ With the conventions just laid down, (1, 1,2) for example denotes the family (xi:1 < i < 3), where x1 = 1, x2 = I , x3 = 2. We note that in contrast to the relation (1) in 91.1, the families

are all different (and that they d o not even have the same index set) although all have the range { 1,2}. Again, a sequence ( x n :n = 1,2, ...) is simply a mapping of the set of positive integers into, say, the set of real numbers: here we have a further instance of a family. We shall frequently denote this family by the symbol (xl, x2,x3, ...). For us the distinction between sets and families is fundamental. Nevertheless, it is often convenient to allow ourselves a certain latitude of language by using expressions about families which are, strictly speaking, only appropriate to sets. Indeed, we have already anticipated this convention by speaking about the ‘elements of a family’. To be more explicit, we shall say that an element x of E belongs to, or is contained in, or is an element of the family S = ( x i :iE 1) if x = x, for some i E I . Again, let I ‘ c I. Then the family (xL: i E 1’) will be said t o be a subfamily of X; we shall denote it by X(l’), and we shall write X(I’) c .X = X(1). Further, when we speak of ‘kx,’s’ in the family S = ( x I: i E l), we simply mean a subfamily X(1’) in which the index set I ’ consists of k elements. (The k x’s i n question need not, of course, be distinct.) A family will be called finite or infinite according as its index set is finite or infinite. If 91 = (a,: i~ I), 23 = ( 6 , : j E J) are two families with disjoint index sets, then the family 9Z + 23 is defined as ( c k :k E 1 u J), where ck = ak or b k according a s k E 1 or k E J. The notions of union and intersection are easily extended t o families of sets.

4 1.2

FAMILIES

7

Thus, let ( A i: i E I) be a family of subsets of E. Then the union

U Ai

icl

denotes the set consisting of all those elements of E which belong to at least one A,; and the intersection iel

Ai

(3)

is the set of those elements which belong to all Ai. It follows that, for I = 0, the union (2) is 0 while the intersection (3) is E. A family ( A i :iE I) of subsets of E is called a partition of E if the Ai are pairwise disjoint and their union is E. We shall sometimes describe the same situation by saying that E = IJ (Ai: iE I) is a partition. Again, let as before (Ai: iE I) be a family of subsets of E ; let F be a set; and consider any mapping Q, : E + F. Then

On the other hand, we have

but if Q, is injective, then the relation of inclusion can be replaced by that of equality. We next turn to the subject of Cartesian products. Let (A, B) be a family of two sets (or, more precisely, a family of sets whose index set consists of two elements). We define the Cartesian product A x B of this family as the set of all ordered pairs (x,y ) with x E A, y E B. More generally, the Cartesian product of the family (Al, ..., A,) is defined as the set of all ordered n-tuples

...)xnl

(XI,

(4)

with x1 € A I , ..., x, E A,. This product is denoted by A, x ... x A, or, alternatively, by

i = 1

Ai.

It is not immediately obvious how this definition is to be extended to the case of infinite families. To meet the difficulty, we shall make a fresh start. Let

8 A

SETS, TOPOLOGICAL SPACES, GRAPHS

1,

p 1.2

.. ., A,, be subsets of E, and consider a mapping

4 : (1, ...,n>-+

E

with the property that

~ ( I ) E A... ~ ,4 ( n ) ~ A , , 9

Thus 4 is simply a mapping which ‘chooses’ one element from each of the sets A , , ..., A,,; and we shall accordingly call it a ‘choice function’ of the family (Al, ..., A,,). We now recognize that the n-tuple (4) is just such a choice function and that the Cartesian product A , x ... x A,, is simply the set of all choice functions of the family (Al, .. ., A,,). When the definition of a Cartesian product is framed in this manner, it can easily be extended to families with arbitrary index sets. Let, then, E, I be sets and let BI = (Ai: i E 1) be a family of subsets of E. Any mapping 4 : I + E such that +(i) E A; for all i E I is called a choicefunction of ‘ill;and the set of all such choice functions is known as the Cartesian product of YI and is denoted by

X Ai.

i s 1

If A; is empty for some i E I, then the Cartesian product is evidently also empty. Now there are many mathematical situations which call for the converse inference; and in the usual axiomatic presentations of the theory of sets (such as that associated with the names of Zermelo and Fraenkel) the legitimacy of this inference is guaranteed not by a theorem but by a separate axiom. This axiom, which is known as the axiom of choice, asserts that every (nonempty).faanlily ofnon-empty sets possrsses at least one choicefunction. Its status in the theory of sets is accounted for by the fact that it cannot be derived from the other axioms. The importance of the axiom of choice can hardly be overstated: modern mathematics would be a different and a poorer thing if a selfdenying ordinance compelled us to relinquish its use.

Exercises 1.2

1. Let E, F be sets; let subsets of E. Show that

4 : E + F be mapping;

and let (Ai: i E I) be a family of

and that the relation of inclusion can be replaced by that of equality when 4 is injective. 2. Let X, Y be sets and let $: .9(X) + Y ( Y ) be a mapping. Show that, if $(A u B) = $(A) u $(B) for all A, B G X, then $(A n B) G $(A) n $(B) for all A, B G X.

8 i.3

9

MAPPING THEOREMS A N D CARDINAL NUMBERS

3. Let (Ai: i E I) be a family of subsets of E. Establish the 'De Morgan identities':

4. Let (Xi: i E I) and (Yj: j relation

E

J) be two families of subsets of a set E. Verify the

u X i n u Yj u =

isI

jsJ

(i,j)slx J

(XinYj).

5. If (Ai: i E I), (Bi: i E I) are two families of subsets of E with the same index set I, show that

X

is1

Ai n

X Bi = X

ic I

is I

(Ai n Bi).

6. Let (Ai: i E I) be an infinite family of sets of which at least one is finite. Show that, if (-)(Ai: i E J) # 0 for every J c c I, then n ( A i : i E 1) # 0. Show also that the qualification 'of which at least one is finite' cannot be omitted. 7. Let (Ai: i E I) be a family of sets. Can every subset of the Cartesian product X(Ai: i E I) be expressed in the form X ( B i : i E I), where Bi E Ai (i E I)?

1.3 Mapping theorems and cardinal numbers We shall next develop a number of results on pairs of mappings. These will

be crucial to many of the arguments in combinatorial theory. Our first result is concerned with mappings of power sets. A mapping 0:P ( X ) P ( Y ) (which maps every subset of X into some subset of Y) is said to be isotone if the relation X , E X, implies @(XI) G @(X,). --f

LEMMA 1.3.1. Let X , Y be sets a n d @ :9 ( X ) + 9 ( Y ) ,Y : 9 ( Y ) 9 ( X ) isotone mappings. Then there exist partitions X = XI* u X2*, Y = Y,* u Y,* such thatO(X,*) = Y,*,'€'(Y,*) = X,*. --f

To prove this, we observe that the collection

9 = {F E X : X \ Y ( Y \ @ ( F ) ) (of subsets of X) is non-empty since X E 9. Let

xl*=

n F,

F € 9

so that

X I * E F foreach F E F .

c F}

10

SETS, TOPOLOGICAL SPACES, GRAPHS

Let A c B

I,

0 1.3

c X. Then, since 0 and Y a r e isotone, we have X \ Y ( Y \@(A))

c X\Y(Y\O(B)).

(2)

In view of (I), this implies that, for each F E F,

X\Y(Y\O(X,*))

c F.

c X\Y(Y\@(F))

Hence

Denoting the expression on the left-hand side by XI**, we thus have XI** c

X I * . Hence, by (2),

X\Y(Y\O(X,**))

E X\Y(Y\O(X,*))

i.e. X1** ~ P a n so, d by (I), X , *

5

=

X \ X I * ,Y,*

Y(Y,*)

=

xl**,

X,**. Tt follows that X,**

X \ Y ( Y \O(X,*)) Finally, put X,*

=

=

= O(X,*),

Y(Y\O(X,*))

=

=

XI*, i.e.

XI*.

Y2*= Y \ Y1*. Then X\XI*

=

x* 2

3

as required.

THEOREM 1.3.2. (Perfect & Pym) Let X, Y, X', Y' be sets with X' E X, Y' E Y, and let 0 X' Y, $ : Y' X be mappings. Then there exist sets X,, Yo with X' c X, c X, Y' E Yo L Y and partitions X, = X I u X,, Yo = Y, u Y , s u c h f h a f X , c X',Y, E Y',tl(X,) = Y,,$(Y,) = X,. -+

--f

We define isotone mappings O:Y(X) means of the equations

+g(Y)

and Y : P ( Y ) -+ P ( X ) by

n X')

(A

c X),

$(B n Y')

(B

E

@(A)

= 8(A

Y(B)

=

Y).

Let X,*, X,*, Y,*, Y,* be sets with the properties specified in Lemma 1.3.1. Write

X,

=

X,* n X',

X,

=

X,*,

x,

=

x, x,,

Y,

=

Y,*,

Y,

=

Y,* n Y',

Yo

=

Y, v Y,.

LJ

It is clear that these equations define partitions of X, and Yo, and that

Q 1.3

X,

G

11

MAPPING THEOREMS A N D CARDINAL NUMBERS

X’, Y, E: Y’. Furthermore X,

=

(X,* n X’) u X,*

=

X n (X’ u X2*) = X’ u X,*

=

(X,* u X,*) n (X’ u X,*) 2

X‘,

and similarly Yo z Y‘. Finally

e(x,)= qxl* xi)= o(x,*)= Y,* = Y $(Y,)

= $(Y2*

l?

n Y’) = V(Y2*) = X,* = X 2 ,

and the assertion is therefore established. The case X’ = X, Y’ = Y of the resukjust proved is of sufficient interest to be worth stating separately. COROLLARY 1.3.3. (Banach) Let X, Y be sets and let 0: X + Y, $: Y X be mappings. Then there exist partitions X = X I u X,, Y = Y v Y, such that W,) = YI,$(Y,) = x,. --f

Another consequence of Theorem I .3.2 runs as follows. THEOREM 1.3.4. (0.Ore) Let X, Y, X , Y’ be sets with X’ c X , Y‘ c Y, and let A be a subset of the Cartesian product X x Y . Further, let 0: X’ Y, $: Y’ + X be injective mappings and suppose that (x, O(x)) E A for all x E X‘ and ($(y ) , y ) E A for ally E Y’. Then there exist sets X,, Yo with X’ c X , G X, Y‘ 5 Yo G Y anda bijection 0:X , -+ Yo such that (x,~ ( x )E) A,for all x E X,. --f

Let X,, X I , X,, Yo, Y,, Y, be sets with the properties specified in Theorem 1.3.2. Then (x, e(x)) E A for all

xE X,,

($(Y), Y ) E I4 for all Y E y,.

(3)

NOW$ is an injective mapping and $(Y,) = X,. Hence the restriction of $ to Y, is a bijection of Y, into X,. Thus I,- exists, and (3) can be written in the form

’

(x, $-‘(x)) E A for all x E X,.

Defining the mapping

6 :X,

-+

Yo by the equations

we arrive at the desired conclusion.

12

SETS, TOPOLOGICAL SPACES, GRAPHS

I,

As for Theorem I .3.2, we again formulate the special case X‘ = X, Y’

5 1.3

=

Y.

COROLLARY 1.3.5. Let X, Y be sets and let A be a subset of X x Y. Further, let 0: X + Y, $ = Y -+ X be injective mappings and suppose that ( x , 0(x)) E A (x E X) and ( $ ( y ) ,y ) E A ( y E Y) .Then there exists a bijection 0 :X + Y such that (x, ~ ( x )E) A (x E X). Much of the subsequent discussion of combinatorial theory is concerned with problems involving the ‘size’ of sets. To be able to handle such questions, we need to introduce some measure of size; and this is provided by the notion of a ‘cardinal number’. With each set X, there is associated a certain uniquely defined object called its cardinal number (or simply cardinal) and denoted by 1x1. When X is finite, 1x1 is defined as the number of elements in X. In the case of infinite sets, the definition of cardinal number presupposes an elaborate technical discussion, but fortunately we can dispense with it for our limited purpose. We shall not, in fact, need to define cardinal numbers: it will be sufficient to explain what constitutes equality and inequality between them. Let, then X and Y be arbitrary sets. We say that their cardinal numbers are = lYl, if and only if there exists a bijective mapping equal, and we write of X into Y (or, equivalently, of Y into X). It is plain that, for finite sets, this definition is in conformity with the definition laid down a few lines earlier. Inequalities between cardinal numbers can also be defined without difficulty. If X, Y are sets and there exists an injective mapping of X into Y, then we say that the cardinal of X is not greater than the cardinal of Y, and we write 1x1 < IYI. If, in addition, 1x1 # lYl, we write 1x1 < IYI. We leave it to the reader to verify that this notation, too, is consistent with the definition of cardinal numbers for finite sets. If a, b are real numbers, then the relations a < band b < a imply the equality a = b. The analogous inference for cardinal numbers is not at all obvious. That it is nevertheless valid is guaranteed by the following important theorem.

1x1

THEOREM 1.3.6. (Schroder-Bernstein) If X and Y are any sets, then the relations 1x1 < IYI and IYI < 1x1imply that 1x1 = IYI. By hypothesis, there exist injective mappings 0: X + Y, $: Y -+ X. It follows by Corollary 1.3.5 (with A = X x Y) that there exists a bijection of X into Y, so that (XI = iY/. Another result of a similar kind will be needed subsequently. LEMMA 1.3.7. I f X’ E X, Y’ L Y, IX’J < IY’l, set X* such rhat X‘ E X* c X andIX*( = IY’I.

1x1 = lYl, then there exists a

8 1.3

MAPPING THEOREMS A N D CARDINAL NUMBERS

13

Since IX’I < IY’I, there exists a n injection 8 : X‘ + Y’. Moreover, since IY‘I < [YI = 1x1,there exists a n injection $: Y’ + X. Hence, by Theorem I .3.4 (with A = X x Y’),there exists a set X* with X’ c X* G X and a bijection 6 :X* 4 Y‘. Hence IX*I = IY’l. In the discussion of cardinal numbers, the set (say N ) of natural numbers plays a special role since [ N J(universally denoted by No) is the smallest infinite cardinal. This follows from the fact, the proof of which requires the axiom of choice, that every infinite set has a subset of cardinal KO. Any set which can be mapped bijectively into N i s called an (infinite) denumerable (or countable) set. The elements of such a set can be ‘numbered off, i.e. the set can be written in the form {xl,x2,xj, ...}. We recall a basic result proved in almost all books of analysis, namely that while the set of all rational numbers is denumerable, the set of all real numbers (or, indeed, the set of all real numbers in any non-degenerate interval) is not denumerable. We sometimes speak of the cardinal number of a family, say ’LL = ( A i :i E I). By this we mean the cardinal number of its index set I , and we denote this occasionally by 1911.

Exercises 1.3 1. Show that, for any finite sets XI, ..., X,,

Also show that this identity remains valid if the symbols u and n are interchanged. 2. Establish the following implications. (i) If X c Y, then 1x1 < IYI. (ii) If 1x1 6 IYI and lYI < 121, then (XI < 121. 3. Prove Cantor’s theorem that, for any set X, 1x1 x) and we shall call ‘6’a rdation defined on X. (It need not, of course, be connected in a n y way with the special relation of numerical inequality.) Suppose, further, that this relation satisfies the following axioms.

< x for all x E X. (ii) If x < y and y < x, then x = y . (iii) If x < y and y 6 z, then x < z. Then ‘),

we have 1111 +

1121

3 IS11 + IS21 = IS1 u S2I + IS, n s21 3 IA(I1) u MI,) Bl + M I , ) n W 2 ) I 3 I N 1 1 u 12) u BI + IA(I1 n 1211 3 11, u 121 1 + 11, n 121 = 11,l + 1121

”

+

+ 1.

We arrive, then, at a contradiction. Hence at least one of the two elements x l , x2 has the required property. LEMMA 2.2.3. If B is ajinite set and the,faamily ( A i :i E I) + (B) satisfies Hall’s condition, then so does the.farnily (Ai: i e I ) + ({x}),forsome x E B.

To establish this result, we simply apply the preceding lemma IBJ - 1 times.

t T o be entirely precise, we ought t o add that the index set, which is naturally a singleton, of the family (B) is assumed to be disjoint from I.

30

HALL’S THEOREM A N D THE NOTION OF DUALITY

2, 5 2.2

We now come t o our next proof of Hall’s theorem. Suppose that the family = (A,, ..., A,,) satisfies X’. Define Ai’ as A i if Ai is finite and as any subset of cardinal n of A i if A i is infinite. Then (Al’, ..., A,,’) again satisfies 2‘.We now apply Lemma 2.2.3 repeatedly, taking B in turn as A1’, ..., A,,‘. We then eventually obtain a family ([xl},..., {x,,}) of singletons which satisfies X‘ and the relations x i € A i ’ c Ai ( 1 < i < n). Hence {x,, ..., x,}+ is a transversal of ?I. ?[

It is also worth noting that Hall’s theorem is an easy consequence of Konig’s theorem 1.7.1, i.e. of Menger’s theorem 1.7.2 for the case of bipartite graphs. Write the given family i n the form 41 = ( A i :i e l ) , where I = ( I , ..., n } and the Ai are subsets of E ; assume, as may be done without loss of generality, that I n E = 0; and consider the bipartite graph G = (N, F), where N = I u Eand

Let I ‘ G I , E’ G E and suppose that S = I’ u E’ is a set separating I and E in C. Then A(I \ 1’) n (E \ E’) = 0 and so A(I \ 1’) G E’. Hence, by %,

I S I=

11’1

+ IE’I3 11’1 + J A ( I \ ri)i

3

1 ’1

+ IT\ rq = n.

T h u s no set separating I and E contains fewer than n nodes of G. Hence there exist n disjoint paths linking 1 and E, i.e. ?I possesses a transversal. We shall conclude the present section by deriving a result which stands in a ‘dual’ relation to Hall’s theorem. The precise meaning of this term will be elucidated in 5 2.3. The theorem in question is as follows.

THEOREM 2.2.4. Let ?1 = (A,, ..., A,,) be a family of subsets of E. Then E is a partial transversal of ?I if‘and only ifevery subset F of E intersects at least IF1 A’s, i.e. I{i: 1

< i < n,

A i n F # 0)l 2 IF1 foreach F

5

E.

(3)

The necessity of condition (3) is obvious and it only remains to establish its sufficiency. If (3) holds, then E is finite (in fact, ]El < n) and we shall write E = ( x l , ..., x,}+. Wedefine

B,

=

(i: 1 6 i < n, x j € A i }

(I

< j < m),

so that 23 = ( B l , ..., B,) is a family of subsets of { I , ..., n}. Let 1 1 < .jl < ... < j , < m. Then

B ~ u, ... u B .

Jk

=

( i :1

I

< i < n, {xj,, ..., xi*)n

z 0 \J

< k < m,

0 2.2

PROOFS OF THE FUNDAMENTAL THEOREM FOR FINITE FAMILIES

31

and so, by (3), IBj, u ... u Bjkl 3 k . therefore satisfies condition 2” and so, by Theorem 2.2.1, The family possesses a transversal, say i , E B , , ..., im€B,,,, where { i l ,..., im}+E { 1, ..., n}. Thus

€ A i l , ...>x

m

Ai,,, ~

i.e. {x,, ...,x,~)= E is a PT of (21. It is worth observing that Hall’s theorem can, in turn, be deduced from Theorem 2.2.4. For let BI = ( A l , ..., A,,) satisfy .%?.If F = {i,, ..., ik}+ c { 1, .. ., n } , and the B’s are defined as above, then

{ j : 1 d j d m, B j n F # 0}= { j : 1 < j d r n , x j ~ A iu , ... u A i k } and hence l{,j: I

< j < pi, Bj n F # 011 = \Ai, u ... u Aikl 3 k

=

IF].

Therefore, by Theorem 2.2.4 (applied to %), (1, ..., n } is a P T of ( B , , ..., B,,,); and this means that YI possesses a transversal. Finally, we mention a slight generalization of the preceding theorem.

COROLLARY 2.2.5. Let ‘LI = (Al, ..., A,) be a family ofsubsets of E, and let E* 5 E. Then E* is a partial transversal of ?I if and only if every subset F of E* intersects at least IF1 A’s. It is clear that E* is a PT of 21 if and only if it is a PT of ( A , n E*, A, n E*), and the assertion now follows at once by Theorem 2.2.4.

...,

Exercises 2.2 1. Use Theorem 2.2.1 to show that the family (Al, ..., A,,) of subsets of a finite set E possesses a transversal if and only if is 1

whenever I

E {I,

..., n} and F E E.

2. Let ‘Ql= (Al, ..., A,,) be a family of sets which satisfies Hall’s condition. Show that the additional requirement [ A , u ... u A,,I = n is necessary and sufficient for the transversal of rU to be unique.

32

HALL'S THEOREM A N D THE NOTION OF DUALITY

2, 5 2.3

3. Show that the family 91 = (A,, .._,A,) of sets satisfies Hall's condition if and and only if, for each I c { I , ..., 171,

c

N o l f 0

where A"],

0cN

c

lA"lI

3 111,

[ I , ..., n>, denote the boolean atoms generated by 91.

4. Let A , , ..., A ,,,, B,. ..., B, be the nodes of a bipartite graph (in which no two A's and equally no two B's are linked by an edge.) Let the degrees of Ai, Bj be the positive numbers ai,h j respectively. (The 'degree' of a node is the number of edges incident with it.) Show that, if min ai3 max bj, then m n and among the edges of the graph there are rn edges linking the A's to m of the B s .


, then the relation {x} ++ { y ) is written as x - y . This is consistent with the usage of the symbol n, then k = 0 or k > 0. For k as before. F o r k > 0, the requirement is that IA(1) u (E\ M)I 2 III whenever1

c (1, ..., n>, 0 < k < IEl

=

0, we obtain (i)

(all I).

(5)

+k

- n, i.e.

IA(1) u (E \ M)I 2 111 + IEl - r

+

But the left-hand side is equal to IEl - [MI IA(1) n MI, and so (5) is equivalent to (ii). We have thus shown that 2I* has a transversal if and only if both (i) and (ii) are satisfied. The proof is therefore complete; and the reader may well feel that, though brief, it is not strikingly illuminating. We shall, however, gain further insight in 9 6.6, where the discussion of marginal elements will be resumed in the context of the theory of abstract independence. The conditions in Theorem 3.3.6 admit of an alternative formulation.

COROLLARY 3.3.7. Let 2I = (Al, ..., A,,) be a family of subsets of aJinite set E, and let M G E. Then 2I possesses a transversal which contains M if and only if both the following conditions are satisfied: (i) each F E E contains at most IF1 A's; (ii) each N E M intersects at least IN1 A's.

48

THE METHOD OF ‘ELEMENTARY CONSTRUCTIONS

3,

5 3.4

By Theorem 2.2.4, applied t o the family (A, n M: 1 < i < n), condition (ii) is satisfied if and only if M is a PT of 41. If we can show that (i) is equivalent to Hall’s condition, then the assertion will follow by Theorem 3.3.6. Suppose Hall’s condition is satisfied. Let F E E, and put

I Then

IF( 3

j ,!

A,\

=

=

( i : 1 < id n, A ,

F).

5

(A(I)I 3 \ I ( = \{i : 1 < i

< n , A, c F](.

Conversely, let condition (i) be satisfied. Then, for each 1 E (1, ..., n}, we have IA(1)l 3 I(i : 1

< i < n, Ai E A(I))( 3 111,

and the proof is complete.

Exercises 3.3 I . Show that Theorem 3.3.6 remains valid for an infinite ground set E. 2. Let ‘?[= ( A , , ..., A,) be a family of subsets of E; let M E, E; and let [MI < p < n. Show that ‘?t possesses a PT of cardinal p and containing M if and only if, for all I 5 ( I , ..., n}, 111

< min (IA(1)l + n - p,

n - IM(

+ IA(1) n MI).

3. Let 41 be a finite family of sets, YI’ a subfamily of 91, and k an integer such k < I%I. Establish the equivalence of the following statements: that IYZ’I (i) YI possesses a PT of cardinal k which contains a transversal of%’; (ii) ’21 possesses a PT of cardinal k , and ??I’ possesses a transversal.


(i). Write 2I = (Al, ..., A,), 2I’ = (Al, ..., A,), IEl = m. Denote by D any set such that D n E = 0, (Dl = n, and by 2I* the family consisting of the sets A,, ..., A,, A,, u D, ..., A, u D and m copies of (E \ E’) u D. Assume, for the moment, that 2I* possesses a transversal. Since PI* consists of m n subsets of E u D and since J Eu DI = m n, it follows that the transversal of 2I* is E u D. Hence E is a transversal of a family consisting of the sets A,, ..., A,, certain of the sets A,+1, ..., A,, and a number (say k ) of copies of E \ E’. We now remove from E the elements representing these k copies of E \ E’. Denoting the resulting set by E,, we see that E‘ G E, s E and that E, is a transversal of the family (11, consisting of A,, ..., A, and certain among the sets A,+1, ,..,A,. Thus 2l’ E 21, E 21, and (i) is valid. It remains to show that (ii) implies the existence of a transversal of 21*. Now, by (a) and Hall’s theorem, we have

+

+

IA(I)I > 111 Further, by (b) and Corollary 3.2.2, IA(1) n E’I Z 111

(I

E

(1, ..., v)).

+ IE’I - n

(I

L

(1)

(1, ..., n ) ) .

(2)

Again, by Hall’s theorem, 2I* has a transversal precisely if

i ( y,

Ai) u

u (Ai u D) u p

isL

j ~ E’) \ u D > /3

IKI + ILI

+ 1(

(3)

+

whenever K c { l , ..., v}, L E {v 1, ...,n } , 0 d p d m. We shall now verify that this set of conditions is fulfilled. If L = 0, p = 0, then ( 3 ) reduces to (1). If L # 0, p = 0, then ( 3 ) reduces to the statement that (A(K) u A(L) u DI 3 IKI + ILI

for all K E { I , ..., v}, L E {v + 1, ..., n } ; and this holds trivially since ID1 = n > IK1 IL1. Finally, if p > 0, then ( 3 ) is the condition that

+

(A(K) u A(L) u (E\E’) u DI 3

IKI + ILI

whenever K E (1, ..., v>, L c {v + 1, ..., n } , 0 < p equivalent to the requirement that the inequality IA(I) u ( E \ E)I 2 (11

< m.

+m -n

+p

This, in turn, is

50

THE METHOD OF ‘ELEMENTARY CONSTRUCTIONS

3, g 3.4

should hold for all I E (1, ..., n>. But IA(1) u ( E \ E’)I = m - IE’I

+ IA(1) n E’I,

and hence the condition in question is simply (2). Thus PI* possesses a transversal, and the proof is complete. We note that the case 91’ = ‘I1 of Theorem 3.4.1 is simply Theorem 3.3.6. Again, the dual result of Theorem 3.3.6 is obtained if we take E’ = E in Theorem 3.4.1. The detailed statement is as follows.

COROLLARY 3.4.2. Let 9I be afinitejamily of subsets of afinite set E, and let ‘21‘ be a subfamily of %. Then there exists afamily 91, such that ‘? EI21, ‘ c 2I which has E as a transversal if and only if 21’ possesses a transversal and E is a partial transversal of 91. If we write down the deltoid form of Theorem 3.4.1, we shall recognize at once that it is identical with Theorem 2.3.1 for the case of finite deltoids. Thus the somewhat technical argument in the discussion above could have been by-passed. It is, nevertheless, of interest to push as far as possible the method depending on elementary constructions. Subsequently, when the transfinite form of Hall’s theorem becomes available, we shall be able to formulate the transfinite analogue of Theorem 3.4.1 (cf. 510.1).

Exercises 3.4 1. State the deltoid version of Theorem 3.4.1. 2. By specializingthe proof of Theorem 3.4.1, frame a new proof of Theorem 3.3.6.

3. Let (X, A, Y ) be a finite deltoid; let X c X, Y Y ; and let r , s be positive integers. Establish the equivalence of the following statements. (i) There exist linked sets X o ( & X ) , Y , ( s Y ) such that /Xo n XI 3 r , ( Y on Ul 3 s. (ii) (a) IA(X*)] + IX \ X*] 3 r forall X* G X. (b) lA(Y*)I + lY \ Y*l 3 s for all Y* s Y. Give an interpretation of this result in terms of sets andelements.

Notes on Chapter 3 4 3. I . Nearly everyone who has worked in transversal theory must have made use of one or other of the elementary constructions. The method is described and illustrated in Mirsky’s papers (6)and (7).

9 3.2. The defect form of Hall’s theorem (Theorem 3.2.1) is contained in a much more general result of 0. Ore (I). For a generalization in a different direction, see Theorem 5.1.1 below. Theorem 3.2.4 was noted by Kuhn & Tucker (1, Preface).

NOTES ON CHAPTER 3

51

Q 3.3. Theorem 3.3.1 is due to Halmos & Vaughan (I). Theorem 3.3.2 was shown to me by Professor R. Rado in 1965. Corollary 3.3.4 will be seen later to be a special case of a much more comprehensive result (Corollary 8.2.2). The problem of marginal elements was first raised by Mann & Ryser (1) and was settled definitively by Hoffman & Kuhn (1) with the aid of linear programming. The proof given here was devised by Hazel Perfect (2). Q 3.4. Theorem 3.4.1 is, in essence, due to Mendelsohn & Dulmage (l),though these writers stated their result in terms of abstract binary relations rather than sets and elements. Their proof, if lengthy, was entirely constructive. Hazel Perfect (1) gave a matrix formulation of the theorem (cf. Theorem 11.1.4 below) and based her short and elegant argument on ideas drawn from linear algebra. The proof offered above is due to Mirsky (7).

4 Rado’s Selection Principle We shall now discuss a very general principle which enables us, in particular, to prove a transfinite analogue of Hall’s theorem. However, as we shall see, its effective range is not limited to this application, nor indeed to transversal theory.

4.1 Proofs of the selection principle Let ’9t = (A,: iE I) be a (generally infinite) family of subsets of E. Denote by 2 the collection of all finite subsets of I and, for J ~ f let, 0, be a choice function of the subfamily %(J) of 9I (cf. $1.2). The functions S,, J €2,will be called the ‘local choice functions’ (on a).We should like to be able to assert the existence of a ‘global’ choice function 0 of the entire family ‘21 which should, in some sense, reflect the behaviour of the local functions. Now we cannot demand that, for each i E I, 0 should agree on i with every local choice function which is defined on i, for two local functions need not agree with each other on i. On the other hand, it would be pointless merely to seek a global choice function 6 such that, for each i E I, 8 should agree on i with some local function defined on i ; for such a function 0 is given by 6 ( i ) = O f i , ( i ) (i E I). We shall, in fact, establish the existence of a global choice function which satisfies much more stringent requirements. THEOREM 4.1.1. (Rado’s selection principle) Let 9t = ( A i :i E I) be a family of finite subsets of a set E. Let f denote the collection of allJinite subsets of the index set I and, for each J E f , let 6, be a choice function of the subfamily (Ai: ig J).t Then there exists a choice,function 6 of ‘21 with the property that, for each J E f , there is a K with J G K E 9and 01J = OK[J. We shall refer to 0 as a ‘Rado choice function’ corresponding to the given of local functions. system O,, J E 9, The most succinct proof of Theorem 4.1.1 depends on properties of topological spaces. Write X = X ( A i :iE I) and let each A i be endowed with the discrete topology. Since the A i are finite, they are compact topological spaces. Let X be endowed with the resulting product topology. Let J E f and denote by F, the (obviously non-empty) set of all choice

7 The existence of the choice functions implies that all A i are non-empty. 52

$ 4.1

PROOFS OF THE SELECTION PRINCIPLE

53

functions 6 on such that OIJ = e,lJ for some K with J E K E Y . Then X \ FJ is the set of all O E X such that OIJ # O,IJ whenever J 5 K €9. For OEX, JEf,wewrite

Then clearly sJ(e)= verification that

EX: 4IJ = OlJ}, X\Fj

=

u

and it is a matter of immediate

6€X\Fj

sJ(6).

It follows that, for each J €3,X \ F, is open in the product topology. Now assume that

U (X\

Jsf

Fj)

= X.

By Tychonoffs theorem 1.6.1, X is a compact topological space and the open covering (1) possesses therefore a finite subcovering, say

( X \ F j , ) u ... u ( X \ F J , ) = X , where J,, ..., J,

E$.

Taking complements, we infer that

Fj, n ... n F,,

= 0.

But FJ,

...

FJ,,,

F J ~ u... wl,

#

IZI,

and the contradiction shows that (1) is false. Hence

u (X\FJ)

JEJ

and, taking complements, we see that

n Fj # 0.

JEB

Any choice function in the above intersection satisfies the requirements of the theorem. The proof just concluded is extremely short, but there may be some advantage in supplementing it with an alternative treatment which relies solely on set-theoretic ideas. As before, let O,, J €9, be local choice functions on 'II.Denote by 0 the collection of all families (Bi: i E I) with B, c Ai (i E 1) and such that, given any

54

RADO'S SELECTION PRINCIPLE

J E I, there exists a K with J c K E Y and O,(i) order R by declaring that (Bi: L E I )

< (B,':

E

4,

3 4.1

Bi (i E J). We shall partially

iEl)

if and only if Bi E Bi' ( i E I). Next, we consider chains in R. Let A be an ordered set and, for each I* E A, let %(A) = (B,(A): i E I ) E R. Moreover, suppose that %(A), A E A, is a chain, i.e. for each i E I , Bi(2) c B,(n') whenever I., A' E A, I < 2'. Write Bi* =

n Bi(A)

( i s I).

A€ A

Let i E I. Assume that, given any A E A, there exists some A' < A such that Bi(i') c Bi(I-). Then there is an infinite sequence _..c Bi(i") c Bi(I') c Bi(A),

(2)

where ... < A" < 3,' < A ; and this is impossible since (in view of the finiteness of Ai) the cardinals of ail terms in (2) are positive integers. Hence there exists a Aisuch that Bi(A') = Bi(Ai) whenever I' < Ii.It follows that

Bi* = Bi(Ai). Now let J

€9.Writing&

=

rnin(A,:

i E

(3)

J), we infer from (3) that

Bi(An) E Bi(Ai) = B,* E Bi(I,)

(iE

J).

Hence

Bi*

=

Bi(In)

( i cJ).

Now (Bi(ILn):i E I) E R and so, corresponding to the chosen set J €2, we have a K with J c K E Y and O , ( i ) eBi(An) = B,* ( i J).~ Moreover, we see by (3) that, for i E I, Bi* = Bi(Ai) c Ai. Thus (Bi* : i E I) E R. In other words, every chain in R possesses a lower bound. Hence, by Zorn's lemma, R has a minimal element, say 911 = (M ;: i E 1). Let i, E 1. Clearly M i , # 0. Suppose that x,, x2 E Mia. We shall show that, in fact, x , = x2 -an inference which implies that Mi, is a singleton. We define M ;' as M i or M ;,\ (x,} according as i c 1 \ {i,} or i = in. Since 9Jl E R. we know that, given any J €2, there exists some K with J c K E Y and O,(i) E M , ( i E J). Any such K will be called an 'associate' of J. We now assert that there exists a set J I with inE J, E $ such rhat OK(&) = x, for every associate K o f J , . For assume that this assertion is false, i.e. assume that each J with inE J €9possesses some associate K with O K ( i 0 ) # xl. For any J €2, write J ' = J u {in}. By hypothesis, J' possesses an associate K such

5 4.2

TRANSFINITE FORM OF HALL‘S THEOREM

55

that 8,(i,) # x l . Hence 0,(i) E Mi’ ( i E J’) and, a fortiori, O,(i) E Mi’ (iE J). We have therefore shown that (Mi’: i E I) E Q, contrary to the minimality of ‘Jn. This establishes the italicized statement. In the same way it follows that there exists a set J, with i, E J, €9such that OK(io)= x, for every associate K of J,. Next, let K* be an associate -of J , u J,, and so of J , and of J,. Then e,,(io) = xl, 8,*(io) = x 2 and therefore x, = x,. Thus Mi,is a singleton and, since i, is an arbitrary element of 1, every M iis a singleton. We shall write ’nZ = ({zi> : i E I). Finally, let the mapping 8: I -+ E be specified by Q(i) = zi ( ~ E I ) Since . ~JIEESZ,we know that, given any J E ~ there , exists K such that J 5 K E Y and O,(i) E {zi}(iE J), i.e. O,(i) = zi= O ( i ) (i E J). The proof is now complete.

COROLLARY 4.1.2. Let the assumptions and notation be as in Theorem 4.1.1. I f all local choice functions O,, J €2, are injective, then so is the global choice function 8 of 41. Let i,, i, €1, i , # i,, and take J = {i,, i,}. a set K with { i l ,i,} c K E 9such that

8(i1) = Q,(il), But 8, is injective, so that OK&) jective.

O(i,)

By Theorem 4.1.1, there exists =

OK(i2).

# OK(i2). Hence 0(iJ # O(i2), i.e. 0 is in-

4.2 Transfinite form of Hall’s theorem We shall now investigate whether Hall’s theorem can be extended to families with infinite index sets. It will be recalled that the family CU = (Ai: i~ I) is said to satisfy ‘Hall’s condition’ ( 2 )if the inequality IA(J)I 3 IJI is valid whenever J cc I (i.e. J is a finite subset of I). A family which possesses a transversal certainly satisfies Hall’s condition. However, the converse inference is false as is shown, for example, by the family consisting of the sets

Thus the obvious generalization of Hall’s theorem fails. We shall show that its validity can be restored if we confine our attention to families of finite sets. The case of a family with a denumerable index set is particularly easy, and we consider it first. Let, then, the family 41 = (Al, A,, A,, .. .) consist of finite sets and let it satisfy Hall’s condition. For each Y 1, there exist (by the finite case of Hall’s theorem) r distinct elements x,, , ...,x,,such that x , i ~ A 1 , xr2€A2, ...,x,,EA,-

56

RADO’S SELECTION PRINCIPLE

4,

0 4.2

N ow the elements x,, ( r = 1,2, ...) all belong to the finite set A,. Hence there exists an infinite subsequence N of natural numbers such that the elements xrl ( r E N ,) are all equal, say to y , . Similar reasoning establishes the existence of a subsequence N , of N ,such that the elements x,, (r EN,) are all equal, say to y 2 . Evidently y , € A , , y 2 € A 2 ,y1 # y z . Repetition of this process yields a sequence of distinct representatives y, E A , (k = 1,2, ...). The case of a denumerable index set is rather special and our next theorem yields more comprehensive information. THEOREM 4.2.1. (Transfinite form of Hall’s theorem) Let CU = ( A i :irz I) be a family of finite subsets of a set E. The following statements are then equivalent. (a) ‘Lt saiisjies Half’scondition. (b) Everyfinite subfamily of 2I has a transversal. (c) 21 has a transversal.

The implication (a) * (b) holds by the finite form of Hall’s theorem (Theorem 2.2. l), and the implication (c) (a) is trivial. To establish (b) * (c), let J c c I. The subfamily YI(J) possesses a transversal by (b), i.e. there exists an injective mapping 0,: J + E such that O , ( i ) E Ai (i E J). The functions 0, are injective local choice functions and it follows by Theorem 4.1.1 and Corollary 4.1.2 that there exists an injective choice function 8 of 21. Thus 8 : T + E is an injective mapping such that 0 ( i ) E A i (i E I), and therefore 2I possesses a transversal. We have now shown that the step from the finite to the transfinite form of Hall’s theorem can be readily taken with the aid of Rado’s selection principle. At the same time it may be of some interest to sketch a direct proof of Theorem 4.2.1 : this will follow in essence the set-theoretic proof of the selection principle but will naturally be somewhat simpler. We shall establish the implication (a) * (c). Denote by R the collection of a11 families ( B i :ie I) of subsets of E which satisfy 2 as well as the inclusion relations Bi E Ai (i E 1). We define a partial order in R by declaring that

if and only if Bi c Bi’ ( i E I ) . It is easily verified that every chain in (0,(i E I), we see that x i E Ai ( i I)~ and that all x i , i E I, are distinct. Thus ’L[ has a transversal. It is worth observing that in this argument we made no use of the finite version of Hall’s theorem.

As an illustration of the use of Theorem 4.2.1, we shall show that any two bases in a vector space have the same cardinal number. (The standard proof of this result operates directly with Zorn’s, or Tukey’s, lemma and with certain properties of cardinal numbers.) Let { x i :i E I} #, { y j:j E J} # be any two bases of a given vector space. For each i E I, let Ai denote the (necessarily finite) subset of J such that j~ A i if and only if y j occurs with a non-zero coefficient in the linear expression for x i in terms of y’s. Assume that the family (Ai: i E I) fails to satisfy condition A!. Then, for some integer k and certain distinct elements i,, ..., ik of I, we have JAi,u

... u Ai,l < k .

Write Ail u ... u A i k = { j , , ...,J m } + , so that m < k. We then have relations of the form Xi,

= P I IY j ,

+ ... + P l m Y j ,

xil,

= Pkl

+ ... + P k m Y j , ,

.................................... vjl

where the p’s are scalars. Hence xii,..., xikare linearly dependent, contrary to hypothesis. It follows, therefore, that ( A i :iE I) satisfies 3;hence, by Theorem 4.2.1, it possesses a transversal. In particular, there exists an injective mapping 4 : I + J (such that 4(i)E A i for all iE 1). Hence (11 < IJ( and, by symmetry, we also have J J (< 111. It follows, by the Schroder-Bernstein theorem 1.3.6, that 11) = J J / ;and this is the required conclusion. The proof just given will be subsequently adapted to the more general case of ‘independence spaces’ (cf. $7.2). Next, we give the deltoid version of Hall’s theorem. (For the relevant notation and terminology, see $ 2.3).

58

4, 0 4.2

RADO’S SELECTION PRINCIPLE

THEOREM 4.2.2. Let (X, A, Y) be a locally right-finite deltoid. The following stateniewts are then equivalent. (a) IA(A)I 3 / A / foreuery A cc X.

(b) EveryJinite subset of X is admissible. (c) X is admissible.

A proof is hardly necessary, but if a formal derivation of Theorem 4.2.2 from Theorem 4.2.1 is desired, then we consider the family 8 = (A(x):x E X) of (finite) subsets of Y. With this definition, statements (a), (b), (c) above reduce to the corresponding statements (for the family 5) in Theorem 4.2.1. Conversely, Theorem 4.2.1 follows if we apply Theorem 4.2.2 to the deltoid ( I , A, E), where (i,e ) E A if and only if e E A,. If, on the other hand, we take the deltoid (E, A, I ) where (e, i )E A if and only if e E A,, then we obtain the dual of Hall’s theorem; this reads as follows. THEOREM 4.2.3. Let Y I = ( A i :i~ I) be a restricted family of subsets of E. The,following statements are then equivalent. (a) Every E* c c E intersects at least IE*I A’s. (b) Eueryjinite subset of E is apartial transversal of ?[.

( c ) E is a partial transversal of ‘u. We note that this result is the transfinite analogue of Theorem 2.2.4.

COROLLARY 4.2.4. Let ‘21 = (Ai: i E I) be a restricted ,family of subsets

of E. Then the set ojallpartial transversals of Y l hasfinite character.

Let X E E and write 2t* = (Ai n X : i E I). Then ‘u* is a restricted family of subsets of X. By Theorem 4.2.3, X is a PT of and so of if and only if every finite subset of X is a PT of 9I*, and so of ?IThe . assertion therefore follows.

a,

a*,

Exercises 4.2 1. Let Y [ = (Ai: i E I) be a family of subsets of E, let E* 2 E, and suppose that no element of E* belongs to infinitely many A’s. Show that the following statements are equivalent. (a) Every F* c c E* intersects at least IF*[ A’s. (b) Every finite subset of E* is a PT of 91. (c) E* is a PT of 2[.

2. Show that Corollary 4.2.4 ceases to be valid if the qualification ‘restricted’ is dispensed with. 3. Deduce Theorem 4.2.3 from Theorem 4.2.1.

8 4.3

A THEOREM OF RADO AND JUNG

59

4. Let 21 = (Ai: i E I) be an infinite family of finite sets, and let d be a natural number. Show that 21 possesses a transversal with defect dif and only if the union of any k (>d) A’s contains at least k - d elements. Also state the dual of this result. 5 . Let d be a natural number, 3 = ( x i :i E I) an infinite family of elements of a set E, and suppose that, for each J c c I,

I{x~:

i~ J}I 3 IJI - d.

Show (i) by Ex. 4.2.4, (ii) by Zorn’s lemma, (iii) directly (and without invoking the axiom of choice) that there exists a set I, 5 I with 11 \ I,] 6 d such that all elements in the family 3E(I,) are distinct. 6. Let % = (Al, A,, ...) be a denumerable family of non-empty subsets of E. Show that % possesses a system of representatives in which no element of E occurs infinitely often if and only if, for every infinite set S of natural numbers, U(Ai: i E S ) is an infinite set. (This question does not depend on the use of Hall’s theorem.) [R. Rado]

7. Let % = (Ai: i E I), B = (Bj:j E J) be two families of subsets of E. Suppose that each A is finite and that it intersects only a finite number of B’s. Use Rado’s selection principle (Theorem 4.1 .l) and the result of Ex. 4.2.1 to show that the collection of subsets I* of I for which %(I*) has a transversal which is a PT of 23 is of finite character.

8. Deduce Hall’s theorem (Theorem 4.2.1) from the following proposition. ‘Let % = (Ai: i E I), B = ( B j : jE J) be families of subsets of E. Suppose that each A intersects only a finite number of B s and that, for each natural number k < (11, the union of any k A’s intersects at least k B’s. Then there exists an injective mapping 8:I -+ J such that Ai n Bs(i) # 0 for i E I. ’Also obtain this proposition from Theorem 4.2.1. 4.3 A theorem of Rado and Jung In the transfinite form of Hall’s theorem (Theorem 4.2. l), we operate with families o f j n i t e sets. This restriction is extremely irksome as it greatly narrows the field of possible applications of Hall’s theorem, but it is not at all easy to see how it might be relaxed. The next theorem records a slight progress in this direction: here we permit just one of the sets to be infinite. Let % = (Ai: i E I) be a family of sets. We shall call a subset J of I critical if it is finite and IA(J)I = IJI.

THEOREM 4.3.1. The infinite family ‘21 = (Ai: i E I) of sets among which exactly one (say A,) is in$nite possesses a transversal if and only if it satisfies Hall’s condition and the relation Aio

$ A(I*),

(1)

where I* denotes the union of all critical subsets of I.

Let % possess a transversal, say {xi:i E I ) + , such that xi E Ai ( iE I). Hall’s

60

4,0 4.3

RADO'S SELECTION PRINCIPLE

condition is then, of course, satisfied. Moreover, for any critical set J,

l{xi:i~ J}I

< IA(J)I = IJI

=

l{xi:iE J}I

and therefore A(J) = {xi: i E J}. Hence

A(I*)

=

{xi: i E I*>.

(2)

Now i, does not belong to any critical set, and so i, $ I*. Thus I* and consequently

x i o ~ A i o \ { x i : i ~ l \ { i O }E } A,\

E

1 \ {i,}

{ x i :i E I * } .

By (2) we now infer that xioE Aio\ A(I*), and this establishes (1). Next, let Hall's condition and relation (1) be satisfied. Choose an element xioE Aio\ A(L*), and consider the family

9I*

=

( A , : i ~ l *+ ) (Ai\{~io}:i~I\Il),

whereI, = I * u { i , } . F o r K c c I*, L c c I \ I , , w e h a v e

3 If L # 0, then K u L

~

u

isKuL

Ail - 1 . I

Q I* and so K u L is not a critical set. Hence

and therefore

This inequality is clearly still valid for L = 0. Thus 'u* is a family of finite sets which satisfies Hall's condition. By Theorem 4.2.1, it possesses a transversal, say {xi:i E 1 \ {i,}}, such that xi E Ai

( i I*), ~

Thus xio # xi for all i E 1 \ I

xi € A i \ {xio}

,. Moreover

(iE

I \ I1).

g 4.4

DILWORTH’S DECOMPOSITION THEOREM

61

and so xi,, # xi for all i E I*. Hence { x i :i E I}+ is a transversal of ‘ill(with xi E Ai for all i E I).

4.4 Dilworth’s decomposition theorem In this section, we shall establish a fundamental result on partially ordered sets. The transition from the finite to the infinite case will be effected by the use of the selection principle. THEOREM 4.4.1. (Dilworth’s decomposition theorem) Let S be an arbitrary partially ordered set and let m be a natural number. If S contains no antichain of cardinal m + 1, then it is the union of m (pairwise disjoint) chains.

For the case of a finite set S , we argue by induction with respect to ISI. If

I SI = 1, the assertion is clearly true. Let ISI > 1, and let C be a maximal chain

in S. If no antichain in S \ C has m elements, then, by the induction hypothesis, SC \ is the union of m - 1 chains and therefore S the union of m chains. On the other hand, let S \ C possess an antichain of m elements, say A = {a,, ..., a,,,}+. Put

L R

< ak forsome k},

={x~S:x =

{x~S:x 2 ak forsome k } .

Let z be the maximal element of C . If zE L, then z < ak for some k , in contradiction to the maximality of C. Hence z $ L, ILI < ISI, and by the induction hypothesis

L

=

L, u ... u L,,

where L,, ..., L, are chains and ak E L k (1 < k < m). Next, let x € L k so that x < a j for some j . Hence ak < x would imply ak < aj, which is false. This means that ak is the maximal element of L k . By symmetry, we infer a relation of the form

R

=

R, u ... u R,,

wheie R, is a chain with minimal element ak. Now, by hypothesis, S has no antichain of cardinal m + 1. Hence every element in S is comparable with some ak,and therefore S

=

R u L = (R, u L,) u ... u (R, u L,).

Thus S is the union of m chains, which can be taken as pairwise disjoint.?

t It will be recalled that a chain can be empty.

62

RADO’S SELECTION PRINCIPLE

4, 5 4.4

We have now established the assertion for finite sets. When S is infinite, we consider the family 91 = (Ax: x E S), where A, = (1, ..., m } for all x E S. Let T cc S. Then, by the result already proved, there exists a partition T = T, u . .. u T, of T into m chains. For each x E T, there is thus a unique integer k in the range 1 d k < m such that x E T,. Writing t,bT (x) = k , we specify a choice function t,bT of the subfamily (A,: x E T). If x,x’E T and t,bT (x)=t,bT(x‘), then x,x’ belong to the same chain and so are comparable. Denote by t,b a Rado choice function of 2I corresponding to the local functions t,bT, T c c S. For x,x’E S, there exists a set K such that

{x,x’}E K c c S and = $lC

$(x’)

= 1//K

(x’>.

If $(x) = t,b(x’),then t,bK(x)= t,bK(x’)and so x, x’ are comparable. Writing, for 1 < k < rn,

s, = {XE s : $(x)

=.k),

we see that S ..., S , are pairwise disjoint chains whose union is S. We record the following neater formulation of the theorem just proved.

COROLLARY 4.4.2. lf the maximum number of elements in an antichain of a partially ordered set S is finite, then it is equal to the minimum number of pairwise disjoint chains into which S can be decomposed. An obvious deduction is as follows.

COROLLARY 4.4.3. A partially ordered set of rs + 1 elements possesses a chain of cardinal r + 1 or an antichain of cardinal s + 1. If there is no antichain of cardinal s + 1, then the given set, say X, can be expressed as the union of s pairwise disjoint chains, say X = C, u ... u C,. Hence rs 1 = IC,I ... ICJ

+

+ +

and therefore max lCil > r + 1, as required. We note, in particular, that a partially ordered set of n > r 2 + 1 elements possesses a chain or an antichain of cardinal r 1. An application of this result is given in the next theorem.

+

THEOREM 4.4.4. A sequence of n 3 r2 + 1 real terms possesses a monotonic subsequence of r + 1 terms.

5 4.4

DILWORTH’S DECOMPOSITION THEOREM

63

< k < n). Put x = { ( k ,ak) : 1 < k < n }

Let the given sequence be (ak:1

and let X be partially ordered by the requirement that (k, ak) < ( j , a j ) if and only if k < j and ak < aj. Suppose that X has an antichain of cardinal r + I , say

(kl,akl),

...2

(kr+l,akr+I)>

(1)

where 1 < k , < ... < k , + , < n. Then ak, > ... > ak,+, and we have a (strictly) decreasing subsequence of r + I terms. If, on the other hand, X has no antichain of cardinal r + 1, then, by Corollary 4.4.3, it has a chain of cardinal r + 1, say (l), again with 1 < k , < ... < k , + , < n. In that case, we have a,, < ... < a,,,, and there is an increasing subsequence of r + 1 terms. We shall conclude the present section by pointing out that the Hall-Ore theorem 3.2.1 (for finite sets) is an easy consequence of Dilworth’s theorem. Other deductions from Dilworth’s theorem will be found in Chapter 11. Let rU = (Al, ..., An)be a family of subsets of ( x , , ..., x,>+ ;let 1 < r < n ; and suppose that

IA(1)I 2 111

+r -n

forall I

s (1, ..., n}.

(2)

In the set P of ‘objects’ XI,

..., X m , A,, ...>An,

we introduce a partial order by declaring that x i < Aj if and only if x i E Aj. Lets denote the maximum number of objects in an antichain, and let

A,,

.-.,xk,

be an antichain of s objects (so that k

A,

V

...?Ah

+ h = s). Then

... V A, C

{Xk+I,

..., X m Ij

and therefore, by (2), h

+ r - n < ( A , v _..uAhI < rn - k .

Hencem + n - s 2 r. Now, by Dilworth’s theorem, P can be decomposed into s pairwise disjoint chains, say {XI,

Al},

‘..7

{xi, Ai},

{xi+,},

...) ( x m } , (Ai+1}, ..-){An}

64

4, 0 4.5

RADO’S SELECTION PRINCIPLE

+

(with the A’s and x’s suitably renumbered). Then s = m n - i and we see that the transversal index t* of ’21 satisfies the relation t* > i = m + n - s 2 r. Exercises 4.4 1. Let S be an arbitrary partially ordered set and let rn be a natural number. Show (e.g. by induction with respect to rn) that, if S has no chain of cardinal m 1, then it can be expressed as the union of m antichains. Also verify that Corollary 4.4.3 is a consequence of this result.

+

2. (i) Show that a sequence of rs + 1 real terms either contains an increasing 1 terms or a decreasing subsequence of s 1 terms (or both). subsequence of r (ii) Give an example of a sequence of rs real terms which possesses neither an increasing subsequence of r + 1 terms nor a decreasing subsequence of s 1 terms.

+

+

+

3. Let (d,, ..., d,) be a sequence of positive integers. We say that Y is an ‘independence number’ if it is possible to select r d’s such that none of them divides any other. Further, we say that s is a ‘decomposition number’ if it is possible to partition (d,, ..., d,) into s subsequences such that, of any two d’s in the same subsequence, one divides the other. Show that the greatest independence number is equal to the least decomposition number.

4.5

Miscellaneous applications of the selection principle

The wide range of problems on which the selection principle can be brought to bear has already been hinted at. Here we shall consider several applications in different branches of (not necessarily combinatorial) mathematics.

Chromatic number The notion of the chromatic number of a graph was introduced in $1.7.

THEOREM 4.5.1. Let G be an infinite graph and k a natural number. Then the chromatic number of G does not exceed k if and only if every finite subgraph of G has thisproperty. Suppose that the chromatic number of every finite subgraph of G is at most k . Denote the set of nodes of G by I and consider the family (Ai: iE I), where Ai = { 1,2, ..., k } for every i~ I. Let J cc I. By hypothesis, a colour (i.e. one of the integers 1,2, ..., k ) can be assigned to each node in J such that, if i, i’ are any two nodes in J which are linked by an edge of G, then different colours are assigned to them. This means that the subfamily (AL:iE J) possesses a choice function 8,such that O,(i) # 8,(i’) whenever i, i‘ E J and i, i‘ are linked by an edge. Let 8 denote a Rado choice function of the entire family ( A i :i E I). If i, it are any two nodes which are linked by an edge, then there exists a set K such that ( i , i’} G K c c I and O ( i ) = O,(i), O(i’) = &(if).But eK(i)# O,(i’)

5 4.5 MISCELLANEOUS APPLICATIONS OF THE SELECTION PRINCIPLE

65

and so e ( i ) # O(i’). Thus we can paint the nodes with a stock of k colours such that no two nodes linked by an edge are painted alike. It is, of course, possible to interchange the roles of nodes and edges and, by an argument almost identical with that given above, prove the following proposition. The edges of a graph can be so painted with k colours that no two concurrent edges are painted alike provided this is true of every jinite subgraph.? Problems on partitions Let 8 be a non-empty collection of subsets of a set E, and let k be a natural number. If a subset F of E can be expressed as a union of k pairwise disjoint sets all of which are members of d,we shall say that F is (a, k)-divisible.

THEOREM 4.5.2. Suppose that a non-empty collection d of subsets of E has jinite character. Then the collection of all (&, k)-divisible subsets of E also has finite character. Let E* G E and suppose that every finite subset of E* is (8,k)-divisible. We shall verify that E* is itself (8,k)-divisible. For each x E E*, let A, = { 1,2, ..., k } . For each F cc E*, there exists a partition F = F, u ... u F,, where F,, ..., F, E 8.Hence, for each x E F, there exists a unique integer i such that 1 < i < k and x E Fi. Writing &(x) = i, we define a choice function & of the family (Ax: x E F) with the property that, for 1 < i < k , (x E F : &(x)

=

i}

(= Fi) E &.

(1)

Denote by 6 a choice function of the family (Ax: x E E*) of finite sets whose existence is guaranteed by Rado’s selection principle. Write Ei* = { x E E * : 4(x)

=

i}

(1

< i < k).

(2)

Then E* = El* u ... u E,* is a partition. It remains to show that El*, ..., E,* are members of &. For each F cc E*, there exists G with F G G cc E* and 41F = &IF. It follows that, for 1 < i < k , { ~ E FCp(x) : =i} Now, by (I),

{XE

G : 4c(x)

=

=

{ x E F : &(x) = i }

c

{XE

G : &(x)

=

if.

i} € 8 .But B has finite character and so every

t While this result is correct as stated it is, perhaps, more convenient in the present instance t o define a subgraph of G = (N, E) as a graph G’ = (N’, E’) such that E’ C E and N’ is the set of those nodes in N which are incident with edges in E’.

66

RADO’S SELECTION PRINCIPLE

4, 5 4.5

subset of a member of 8 is again a member of 6. Hence (x E F: 4(x) = i } E 8. In view of (2) this implies that every finite subset of Ei* is in 6. Hence Ei* is in G,and the proof is complete. Next, we consider an application of Theorem 4.5.2 to transversal theory.

THEOREM 4.5.3. Let 21 = (Ai: i E I) be a restricted family of subsets of an arbiirury set E. Thefollowing statements are then equivalent. (a) E can be partitioned into k partial transversals of ‘21. (b) For eaclzjniie subset F of E,

IF( d k l ( i € I : A in F # 011. (c) For eachjnite subset F of E, we have IF( < k . p(F), where p(F) denotes ihe maximum cardinal ofpartial transversals of 2l contained in F. Let & denote the set of all PTs of 21. Then, by Corollary 4.2.4, d has finite character. Hence, by Theorem 4.5.2, statement (a) is valid if and only if every finite subset E* of E can be partitioned into k PTs of 2l. In view of Corollary 3.3.3, this is the case precisely if, for each F G E*, the inequality in (b) is valid. In view of Corollary 3.3.4, it is also the case precisely if, for each F E E*, the inequality in (c)is valid. The desired conclusion therefore follows. A result much more general than Theorem 4.5.3 will be discussed subsequently (cf. § 8.2). Ordered groups

Let G be a group (and denote by the same symbol the set of its elements). We shall say that a relation ‘ (‘! t !/ ( t - n ) !

(t (t

< n) > n).

(1)

Although this result constitutes a quantitative refinement of Hall’s theorem, this last theorem is not invoked in the proof. However, the argument is based on the same case distinction that we used in the first proof of Hall’s theorem. We shall denote by $(t, n) the expression on the right-hand side of (1). It is plain that 4 ( t , n) increases with t. For n = 1, the assertion is clearly valid. Let n > 1 and assume that the assertion holds for all families consisting of fewer than n sets. Case 1 . Suppose that IAil u ... u A,,! > k whenever 1 < k < n and 1 < ii < ... < ik < n. Since (Al, ..., A,) satisfies 9, we have Al # 0. Let x1E A,, and put

(2 < i

Bi = Ai \ (x,) Then, for 1 < k < n, 2

< i,

< n).

< ... < ik < n,

and so (B2, ..., B,) satisfies 3.Moreover, if (xz,..., x,) is any SSDR of (B2, ..., B,), then (xl, x 2 , ..., xn)is a SSDR of 81. Hence N(%)

>

N(B,, ..., BJ.

XIEA~

Now we have IBil 2 IAJ-1 2 t - 1

(2 < i < n).

Write u = min (IB21, ..., IB,I). Then, for t > 1, we have u 2 t - 1. On the other hand, if t = 1, then since the B s are non-empty, u > 1. By (2) and the induction hypothesis, we have

80

VARIANTS A N D APPLICATIONS OF HALL'S THEOREM

5 , § 5.2

If t > 1. then N('11) 3

1 4(t - 1, n

-

xi E A L

z t 4(t Iff

=

1, n

1) = IA1I 4(f - 1, n - 1)

1) = 4(t,n).

-

1, then

NWLI) 2

4(l,n

xi E A T

-

>, 4(1, n - I)

1)

=

=

lAl14(1,n - 1)

1 = 4(1,n).

Cuse 2. Suppose that, for certain values of k, i,, ..., ik with 1 1 6 i, < ... < ik < n, we have



and y , €A1, ..,,~ , E A , .Adjoining the column ( y l , ..., y,) to R , we obtain a Latin rectangle R' of type (r, s + 1, n).

84

VARIANTS A N D APPLICATIONS OF HALL'S THEOREM

5 , s 5.4

Denote by N ' ( i ) the number of elements of R' equal to i, and by Y ( i ) the number of integers among y , , .. ., y , equal to i. Then = N(i)

"(i) If i E P, then N ( i ) = r

+s

-

n and Y ( i ) = 1, so that

N'(i) I f i $ P, then N ( i ) > r

+s

-

+ Y(i).

=

r

+ (s + I ) - n.

n and so

3 N(i) > r

"(i) Hence, for all i,

"(i) 3 r

+ s - n.

+ (s + 1) - n.

We can now repeat the process described above until we obtain a Latin rectangle R* of type (r,n,n). For this rectangle, N ( i ) = r and so ( I ) is satisfied. Hence R* can be extended to a Latin square of order n (a fact which also follows by virtue of Theorem 5.3.1). 5.4 Subsets with a prescribed pattern of overlaps When we say that the family (Ai: 1 < i < n ) possesses a transversal, we assert the existence of subsets of A,, ..., A, which exhibit a particular settheoretic structure : they are pairwise disjoint singletons. It is therefore natural to extend the discussion by seeking to determine conditions which ensure that A , , ..., A, should possess subsets the pattern of whose overlaps is prescribed in advance. To formulate our question with greater precision, let us say that the families ( A i :1 < i < n ) and (Bi: 1 < i < n) are combinatorially equivalent if there exists a bijection n:

6 A i + 6 Bi

i= 1

i= 1

such that a(A,) = Bi (1 < i d n). We shall first prove a preliminary result which will help us to visualize the meaning of combinatorial equivalence. By a boolean polynomial in A,, . .., A,, we understand a finite expression involving the A's and formed by means of unions, intersections, and differences. If unions and intersections only are admitted, then we speak of a restricted boolean polynomial. We also recall, from $1.4, the notion of boolean atoms A"] generated by (Al, ..., A,). We shall write No = (1, ..., n } .

THEOREM 5.4.1. Let 91 = (Al, ..., A,), 23 = ( B l , ..., B,) be two families of sets. Each of the following three statements then implies the other two.

p 5.4

SUBSETS WITH A PRESCRIBED PATTERN OF OVERLAPS

85

(i) 9L and B are combinatorially equivalent. (ii) ID(A1,..., A,)I = Ip(B,, ..., BJ for every booleanpolynomialp. (iii) lA[N]I = lB[N]I whenever0 c N E No. Further, if 3,B are families of finite sets, then the phrase 'boolean polynomial' in (ii) can be replaced by 'restrictedboolean polynomial'. Let (i) be satisfied, and let a be an associated bijection, as in (1). Since, in particular, B is injective, we have (cf. Exs. 1.1.3 and 1.2.1) B(B,,

...?

Bn)

= B(a(A,), ...>a(An)) =

o(B(A,, ...)An))

for every boolean polynomial p. Hence (i) implies (ii). Moreover, (ii) implies (iii) trivially since A"] is a boolean polynomial in A,, ..., A,,. Next, let (iii) be given. Then there exist bijections cN:A"] -+ BCN] (0 c N c No). Now the atoms A"], 0 c N E No, are pairwise disjoint; and we shall denote by B the direct sum (cf. $1.1) of the 2"- 1 bijections cN. By Lemma 1.4.1, the union of the A"] is A, LJ ... LJ A,; and an analogous statement holds for the family 8.Hence a is a bijection of type (1). Moreover, by Lemma 1.4.2,we have

=

u

keNCNo

B[N]

=

Bk;

and so (i) is valid. This completes the proof of the first part of the theorem. Suppose, next, that 2I and 8 are families of finite sets, and substitute 'restricted boolean polynomial' for 'boolean polynomial' in (ii). The proofs of the implications (i) =- (ii) and (iii) 3 (i) remain precisely the same as before. To establish (ii) => (iii), we note that, for 0 c N E No,

A"] Hence

and so, by (ii),

i.e. (iii) is valid.

=

n A,\

i EN

u

i9N

Air\

n Ail.

i EN

86

VARIANTS A N D APPLICATIONS OF HALL'S THEOREM

5 , Q 5.4

We are now able to answer the question raised at the beginning of the present section.

THEOREM 5.4.2. (R. Rado) Let ( A i : 1 < i < n), (Bi: 1 < i < n) be two families of5nite sets. Then there exist sets X i E A i (1 < i < n) such that the family (Xi: 1 < i < n ) is combinatorially equivalent to ( B i : I < i < n) if and only if Ip(A,, ..., A")I

z

Ip(B1, ..', B,)I

(2)

for every restricted boolean polynomial p . The stated condition is certainly necessary. For suppose that the sets Xi with the stated properties exist. Since X i G A i , we have Ip(A1,

..., A,)I 3 IP(X1, ..., X,)l

for every restricted boolean polynomial p. Further, since ( X , , ..., X,) and ( B l , ..., BJ are combinatorially equivalent, we have, by Theorem 5.4.1, (i) and (ii), Ip(X1, ..., Xn)l = Ip(Bi,

..., BJI;

and so (2) holds for every p . Conversely, suppose that (2) holds for every p . For 0 c N c No, write A{NJ =

B{N)=

A,, keN

n Bk.

ktN

,,

Let 1 < k < 2" - 1 and denote by N ..., N, any k different non-empty subsets of No. Then, in view of our hypothesis, I A ( N , } u ... u A{Nk)I 3 IB{N,} u ... u B(Nk}I

... U B[Nk]I = lBIN1]l + ... + IBINk]l. 3 lB[Nll

U

Hence, by Theorem 3.3.1, there exist 2"- 1 pairwise disjoint sets, say X { N } (0 c N G No) such that

Xk=

u

(1 < k = ~ ( A u ) )+ IDI

+ d = 111.

It follows by Theorem 6.2.1 that ( A i u D: 1 < i < n) possesses a transversal which is a member of €*, and so (I1 possesses a PT with defect d which is a member of 8. This establishes the sufficiency of condition (2): we leave it to the reader to verify its necessity. The next result is a simple application of Theorem 6.2.2.

COROLLARY 6.2.3. Let 0 < m < n and let A,, ..., A,, be subsets of a vector space V. Thefollowing statements are then equivalent. (i) Whenever x iE A i (1 d i < n), the vectors x , , ..., x, span a subspace of V ofdinzension not exceeding m.

96

INDEPENDENT TRANSVERSALS

(ii) There exists an integer fi with 0 < h < m and a collection of h A’s contained in a subspace of V ojdimension fi.

6, 0 6.2

+n

-

m

Let G be the independence structure consisting of all linearly independent subsetc of V, and denote its rank function by p. Statement (i) then means that the family 91 = ( A , , ..., A,) does not possess an independent PT of cardinal m I . i.e. of defect n - ni - 1. In view of Theorem 6.2.2, this is the case precisely if, for some 1 5 { 1, ..., n } ,

+

p(A(I))
IDI. By the replacement axiom, there exists an element X E C \ D such that {x}u D E G . Now x # B and so X E C \ B c A \ B . Hence x E Ai \ B for some i . In particular, x 4 B, and therefore

{XI

u Bi 6 &.

(4)

Since B , E & , we have lBil = p(Bi) d p(B) = ID\. Hence, by Lemma 6.1.1, there exists a set Fi such that Bi 5 Fi

E

B,

Fie&,

IFi[ = ID1 = p(B).

Hence, by (4),

(x} u Fi $8.

(5)

Finally, applying the replacement axiom to the independent sets Fi and {x) u F i € G , which contradicts (5); or else { y > u Fi E B for some y E D \ Fi c B, which contradicts the definition of p(B). We conclude, then, that p(A) = p(B). {x>u D, we see that either

We are now able to formulate a refinement of Theorem 6.2.4.

COROLLARY 6.2.6. Theorem 6.2.4 r( mains valid if the phrase tfinite subsets’ is replaced by ‘rank-Jnite subsets’.

-

The implications (c) 3 (b) and (b) (a) are trivial. and it suffices to establish (a) * (c). For each i~ 1, let Bi be any (necessarily finite) subset of

98

INDEPENDENT TRANSVERSALS

6, 5 6.2

A i such t h a t B i € & , IBiJ = p(A,). Let 23 be t h e family ( B i : i E I ) . Now, by Lemma 6.2.5, p(B(J)) = p(A(J)) whenever J cc I, a n d so P(B(J)) 3

IJI

(J cc

1).

Hence, by Theorem 6.2.4, ’H possesses a n independent transversal which is, of course, also a n independent transversal of 41.

Exercises 6.2


>

for all t E T, then

1, then taking t E T and using R(3) for the sets A = S u (T\ { t } ) ,

we obtain P ( S u T)

B

=

S u(t},

+ A S ) < P ( S u (T\ W)) + P(S"

(t>)

d P ( S u (T\ ( t } , ) + P ( S ) .

The lemma now follows by induction with respect to [TI. Our next theorem furnishes an axiomatic characterization of those mappings which are rank functions of independence structures.

108

6, 0 6.1

INDEPENDENT TRANSVERSALS

THEOREM 6.7.2. Let E be a,finite set and p a mappitig o/Y(E) into the set of nori-negative integers. Then p is the rank function of sotwe independence structure OM E if and only if it satisjies the conditions R( I), R(2), R(3). That every rank function satisfies the stated conditions has already been noted. Suppose, next, that p satisfies R(I), R(2), and R(3), and define

d

=

{X c E : p ( X ) =

1x1).

Since p(A) is non-negative for all A G E, it follows by R(1) that p ( 0 ) = 0. Hence 0 E 6. We note further that, in view of R(3), p(A u B) d p(A) Now let X E 6 ,Y

+ p(B)

(A, B

G

E).

(1)

c X. Then p ( X ) = 1x1and so, by (1) and R( I),

1x1 = p ( X ) d

p(Y)

+ p ( X \ Y) d IYI + IX \ YI = 1x1.

Hence p(Y) = IYJ,i.e. Y E 8. Again, let X, Y € 6 , IYI = Assume that

1x1 + I .

Then p ( X )

P(X u iY1)

P(X)

for all y E Y \ X. Then, by Lemma 6.7.1 (with S that p ( X u Y) < p(X). Hence, by R(2),

=

=

1x1, p ( Y ) = 1x1 + 1.

X, T

=

Y \ X), we infer

and we arrive at a contradiction. It follows that

for some y o E Y \ X. But, by (1) and R( I),

+

and so p(X u ( y o ) )= 1x1 1 = IX w fy,)l, i.e. X u [ y o )€8. We have now proved that R' is an independence structure. Denoting its rank function by a, we have a(A)

=

rnax

(1x1:X c A, p ( X ) = 1x1)

(A 5 E).

0 6.7

AXIOMATIC TREATMENT OF THE RANK FUNCTION

109

Let A E E and write a(A) = IBI, where B is a suitable subset of A with p(B) = (BI.Then 4 A ) = IBI

=

p(B)

and so, by R(2), 4A)G

Furthermore, no subset of A of cardinal IBI P ( B u {XI) # IB

+1

is a member of 8,so that

” {x}l

for all x E A \ B. Hence, by R( I), p(B u

(4) G p(B)

(X

f

A \ B)

and therefore, by Lemma 6.7.1,

Thus p =

0,i.e.

p is the rank function of the independence structure 8.

COROLLARY 6.7.3, Let 8 be a collection ofsubsets of a$nite set E such that 0 € 8 . Then F is an independence structure if and only if the mapping p, defined by the equation p(A)

=

max {[XI:X

5

A, X E F }

(A

G

E),

satisfies the ‘modular inequality’.

I f & is a n independence structure, then its rank function p certainly satisfies R(3). Suppose, on the other hand, that p satisfies R(3). Trivially, it also satisfies R(1) and R(2). Moreover, p(A) = IAJif and only if A E 8 ; and so W

=

{A G E: p ( A ) = [ A ] } .

It follows, by the proof of Theorem 6.7.2, that & is an independence structure. Exercises 6.7 1 . Let E be a finite set and p a mapping of Y(E) into the set of non-negative integers which satisfies the ‘modular inequality’. Show that, for A, B E,

2. Show that the conditions R(I), R(2), R(3) are independent.

110

4 6.1.

INDEPENDENT TRANSVERSALS

6

Notes on Chapter 6

The study of independence structures is the axiomatic investigation of linear independence in vector spaces. This study was initiated some thirty-five years ago. We should mention H . Whitney’s pioneering paper (1) in which finite independence structures (there called ‘matroids’) were subjected to a searching analysis. Again, in the second edition of van der Waerden’s Moderne Algebra (3), the theory of linear dependence and the theory of algebraic dependence were derived from a common axiomatic source. Abstract independence has by now acquired a sizeable literature and, in particular, there exist numerous studies of the relation between different sets of axioms. For example, we may refer to the work of Birkhoff (l), MacLane (2), Lazarson (I), Bleicher & Preston (l),Dlab (l),Ashe (l), and Brualdi (6). Very substantial contributions to the theory of independence structures have been made by W. T. Tutte (2, 3, 4). Further information will be found in Rado’s survey article (9). In addition to structural questions, some interesting quantitative problems arise in the study of independence. Thus D. J. A. Welsh (3) obtained a result which implies, in particular, that there are at least 2” ‘non-isomorphic’ independence structures on a set of n elements. It should, perhaps, be pointed out that the terminology of the subject is as yet far from standardized. In addition to ‘independence structure’ (or ‘space’) and ‘matroid’, terms such as ‘incidence geometry’, ‘matroid lattice’, ‘combinatorial geometry’, and ‘geometric lattice’ have been o r are being used in the same, o r in a very similar, sense.

3 6.2. The credit for recognizing the significance for transversal theory of the study of independence structures belongs undoubtedly to R. Rado who discovered both Theorem 6.2.1 (3) and Theorem 6.2.4 ( 5 ) . A good case can be made out for regarding Theorem 6.2.1 as the central result in tranversal theory; in this connection, see Mirsky’s expository paper (6). The proof of Theorem 6.2.1 given above is based on Dr Perfect’s observation that Rado’s proof of Hall’s theorem (i.e. the second proof of theorem 2.2.1) can be adapted to the more general case of independent transversals. P. Scherk (1) gave an a d hoc proof of Corollary 6.2.3; the combinatorial treatment offered here was suggested to me by Professor H. Tverberg. The very interesting observation embodied in Corollary 6.2.6 is due to J. H . Mason (1). Rado’s theorem 6.2.1 passed almost unnoticed for many years, and its dominant position i n tranversal theory has emerged only very recently. There is now n o shortage either of its applications o r extensions. We mention, in particular, the work of R. A. Brualdi (2, 7, 8, 9). Hazel Perfect ( 5 , 7), and D. J. A. Welsh (1, 4, 5 ) . An exceptionally interesting generalization of Rado’s theorem, due to Brualdi, will be discussed in 4 8.4. 4 6.3. The result in this section is based on Rado’s paper (3). For further investigations, see D. J. A. Welsh (9).

4 6.4.

Theorem 6.4. I is due to Hazel Perfect ( 5 ) .

9: 6.5. Matrices whose elements are indeterminates and zeros were used in combinatorial investigations by W. T. Tutte (1) as early as 1947. The idea of a formal incidence matrix associated with a family of sets appears in Hazel Perfect’s paper ( I ) . Theorem 6.5.2 was discovered, in essence, by Edmonds & Fulkerson (1) and,

NOTES ON CHAPTER 6

111

independently, by Mirsky & Perfect (2), to whom the proofs discussed here are due. The very elegant demonstration of Hall’s theorem given at the end of the section is taken from the work of Edmonds (3).

5 6.6. The history of the problem of marginal elements for the finite case is referred to in the Notes on $ 3.3. The transfinite version contained in Theorem 6.6.3 was communicated to me by Professor Rado in 1965. Here I follow, more or less, the treatment of Mirsky & Perfect (2). 5 6.7.

Theorem 6.7.2 i s implicit in the work of lngleton (I).

7 Independence Structures and Linear Structures Below we continue the study of independence structures begun in the preceding chapter. I n particular, we shall analyse the relation between independence structures, transversal structures, and ‘linear’ structures.

7.1 A hierarchy of structures We recall that the collection of all PTs of a family % of subsets of E is called the ‘transversal structure of PI’. Let, now, d be a collection of subsets of E; we then call 8‘ a transversal structure if there exists a family 91 of subsets of E such that b is the transversal structure of !!I. (This family need not, of course, be unique.) Again, let x‘ be a non-empty collection of subsets of E ; write

E*

=

{ x E E : {x}E&};

and let 1) be a division ring, We call 8 a linear structure Over D (or say that it is linear ouer D ) if there exists a vector space V over D and an injective mapping I/I:E* + V such that a subset X of E* belongs to d precisely if $(X) is a linearly independent subset of V.7 It can be shown that, if D , D‘ are division rings such that D E D’ and if 8 is linear over D , then it is also linear over D’. This result is of particular interest for us when D, D’ are fields. In that case the assertion is not difficult to prove, and we shall leave the details to the reader. Let 8 be a collection of sets. If there exists some division ring over which 8 is linear, we say that d is a linear structure (or simply that it is linear). A number of alternative expressions are to be found i n the literature. Thus, it is said that 8 is ‘linearly representable’ (over D). To put the matter more loosely, x‘ is a linear structure if its members can be ‘identified’ with certain linearly independent subsets of a vector space. It is, of course, trivial that every linear structure is an independence structure. Our first object is to clarify the relation between transversal structures and linear structures, and we begin by considering the case of finite transversal The reason for considering a mapping on E* rather than on E is plain. For suppose that x , y E E \ E*, x # y . Then each of v/ ({x)), v/ ( { y } ) is a linearly dependent subset of V. As they are singletons, both must be equal to the zero element of V ; and this is incompatible with the injective character of y . 112

0 7.1

A HIERARCHY OF STRUCTURES

113

structures. We recall from the discussion in 5 6.5 that every such structure is an independence structure; but on re-examining the arguments leading to Lemma 6.5.1 and Theorem 6.5.2, we realize that more has, in fact, been proved. Denote by Q the field of rational numbers. Let & be the transversal structure of the family 2I = (Al, ..., A,) of subsets of a finite set E. We may, of course, regard 2l as a family of subsets of E* = A, u ... u A,. Let M be the formal incidence matrix in which the rows correspond to elements of E* and the columns to sets of 2I;and denote by z l , ..., z, the indeterminates appearing in M . Let F E E*; then, by Lemma 6.5.1, the rows of M corresponding to F are linearly independent over the field Q(z,, ..., z,) (of rational functions, with rational coefficients, in the z’s) if and only if F is a PT of 9I. Hence d is not merely an independence structure: it is actually a linear structure over the function field Q(zl, ..., z,). This conclusion admits of yet a further refinement, which is contained in the next theorem. THEOREM 7.1.1. Every finite transversal structure is linear over the field of rational numbers.

We continue to use the notation introduced above. The product of the determinants of all non-singular square submatrices of M is a polynomial p ( z , , ..., z,), with coefficients in Q , which is different from thezero polynomial. Hence there exist distinct, non-zero numbers t , , ..., t, in Q such that dtl,

.-.,t,> # 0.

Denote by fi the matrix obtained when the indeterminates z,,...,z, in M are replaced by f l , ...,t, respectively while the zeros are left unchanged. Then a n y non-singular submatrix of M is transformed into a non-singular submatrix of fi and, trivially, any singular submatrix of M is transformed into a singular submatrix of A. Now denote by I/ the vector space consisting of all n-tuples with entries in Q . Write E* = {x,,..., x,}+ and let the mapping $: E* -+ V take x, into the k-th row of fi. Since t , , ..., t, are distinct, $ is plainly injective. If F E E*, then F is a PT of 2I precisely if the rows of M corresponding to F are linearly independent over Q ( z , ,...,z,), and this in turn is the case if and only if the rows of fi corresponding to F are linearly independent over Q . The mapping $ has therefore the required properties, and the assertion is proved. We shall next widen the scope of our inquiry by considering transversal structures not necessarily finite. We first need a preliminary result. THEOREM 7.1.2. L e t & be a transversal structure offinite character on a set E.

114

INDEPENDENCE STRUCTURES AND LINEAR STRUCTURES

7 , s 7.1

Then there exists a restricted,fumily ‘21 of subsets of E such that & is the collection ojall partial transversals of CU. Let 8 = ( B i : i E 1) be any family of subsets of E whose transversal structure is identical with 8. Denote by E, the set of all elements in E which belong t o infinitely many B’s, and write

9I’= ( B i n ( E \ E o ) : i c I ) ,

P I ” = ({e}:eEE,).

We suppose, as may be done without loss of generality, that I n E, = 0;and we put 9I = PI’ + 91”. It is plain that PI is a restricted family. Further, let X be any PT of 8 and write X = Y u Z, where Y c E \ E,, Z G E,. Then Y is a PT of ’2” while Z is a PT of 41”. Hence X is a PT of CU. Conversely, let X be a PT of 91. Let X* c c X, and write X* = Y* u Z*, where Y* E E \ E,, Z* E: E,. Now Y* is a PT of 8. Also, each element in Z* belongs to infinitely many B’s, and we can therefore avoid those B’s which are represented by elements of Y*. Hence Y* u Z* = X* is a PT of 8. Thus every finite subset of X is a PT of 8.But the set & of PTs of 23 has finite character and so X itself is a PT of 8. We have thus shown that 9I and 8 have exactly the same PTs. Consequently, W is the set of PTs of the restricted family 91. We now come to the main result of the present section.

THEOREM 7.1.3. A transversalstructure ofjinite character is a linear structure. Let d be the given transversal structure on a set E. In view of Theorem 7.1.2, there exists a restricted family PI = ( A i :i c I) which has 8 as its set of PTs. We shall find it convenient to frame the argument in terms of this family. The method t o be described constitutes a natural extension of the proof of Theorem 6.5.2, which is based on the notion of a ‘formal incidence matrix’. Write Z = { z e i :e E E, i E I , e E A,}, where the z’s are independent indeterminates over the field of rational numbers. Denote by K the field of rational functions, with rational coefficients, in the z’s (each function involving only a finite number of indeterminates). Write E* = { X E E : { x }E 8). For each e E E*, let the mapping $ e : I K be defined by the equations --f

For I Y ~x 2, E K and e l , e2 E E*, let the mapping a1 $el defined by the equation (

~ $1c ,

+ a2 $ e 2 ) ( i ) = a1 $ e l ( i )+ ~2 $e2(i)

+ a2 (iF

tje2:

I).

1 + K be

5 7.1

A HIERARCHY OF STRUCTURES

115

We denote by V the set of all finite linear combinations, with coefficients in K , of the mappings $,, e E E*. Then V is a vector space over K ; and the mapping $ : E* + V defined by the equation $(e) = $ e (e E E*) is injective. Now let F E &, i.e. let F be a PT of CU. We shall show that $( F) is a linearly independent subset of V, i.e. that for each F* cc F the set $(F*) is linearly independent. Write F* = {el, ..., ek}?. Then there exists a set {i,, ..., i k ) + c I such that e l E A i l , ..., ekE Aik.

(1)

Let M denote the k x k matrix ~ ~ $ e r ( i(1s )< ~ ~r, s < k). Each element of M is either zero or an indeterminate in 2, and the indeterminates are independent. Moreover, in view of (I), all places on the main diagonal of M are occupied by indeterminates. Hence M is non-singular. Now assume that $(F*) = {$,,, ..., $e,> is a linearly dependent set, i.e. $q

+ ... +

c(k$ek

=

0

for certain elements ctlr .. ., ak,not all 0, of K . Then

+ ... + ak$ek(is) = 0

a1 $,,(is)

(1

< s < k),

and the rows of M are therefore linearly dependent (over K ) . We thus arrive at a contradiction and conclude that $(F*) is a linearly independent set. Consequently, $( F) is also linearly independent. Next, let G G E*, G $ &. Since 8 has finite character, there exists a set G* cc G, say G* = { e , , ..., e,},, such that G* $€. Now the family Y I is restricted and there exists, therefore, a finite subset J = {i,, ..., i,>+ of I such that

(1

e,#Ai

IYI. Then X possesses a subset X’ such that IX’I = IYI + 1 and also, of course, X’IY. Therefore, by the replacement axiom, there exists some element xo EX’ c X such that xo $ Y and {xo}u Y E 8. Hence xorY, and this contradicts the hypothesis XIY. THEOREM 7.2.7. Let (E,6) be an independence space, and let A be an independent subset of E. Then there exists a basis B of E such that A E B. This result is simply a restatement of Theorem 6.1.2. COROLLARY 7.2.8. Let (E, F ) be an independence space, and let X c E. Then X contains a maximal independent subset. We consider the independence structure 6’ consisting of all independent subsets of X. By Theorem 7.2.7, the independence space (X, 8’)possesses a basis.

THEOREM 7.2.9. Any two bases in an independence space have the same cardinal number. Let B, B‘ be two bases in an independence space (E, &). In view of the maximality of B’, it is obvious that xlB’ for each x E E. Hence, by Lemma 7.2.3, for each b E B there exists a finite subset S, of B’ such that 61 S,. Write 6 = (Sb:bEB).LetB* c c Bandput

In view of Lemma 7.2.2, we have B*IS. Hence, by Lemma 7.2.6, IB*I < ISI. Thus the family 6 (of finite subsets of B’) satisfies Hall’s condition and so possesses a transversal. In other words, there exists an injection 8: B + B’ (such that 8(b)E S, for all b E B). Therefore I BI d IB’I. By symmetry, we have IB’I < IBI; and the assertion now follows by the Schroder-Bernstein theorem 1.3.6. COROLLARY 7.2.10. Let (E, &) be an independence space and A c E. Then any two maximal independent subsets of A have the same cardinal number. The collection of independent subsets of A is an independence structure, and the Corollary therefore follows by Theorem 7.2.9. Corollary 7.2.10 enables us to frame a new definition of rank which, for independence structures, supersedes the cruder definition given in 5 6.1 for the

I22

INDEPENDENCE STRUCTURES A N D LINEAR STRUCTURES

7 , 5 7.2

wider class of pre-independence structures. If (E, 6) is an independence space and A c E, then the rank of A can now be defined as the common cardinal of all maximal independent subsets of A. (For a finite set A, the new definition of rank coincides, of course, with that introduced earlier.) In particular, if € is the universal structure on E, then. for any subset (finite or infinite) A of E, the rank of A is simply IAl.

COROLLARY 7.2.1 I. Let (E, &) be an independence space. I f IYI. pendent szrhset andY a basis ofE, then

1x1


min ( p i ( N i \ r)

r G A

+ Pz(Nr2)).

(7)

Let x1E E,, x2 E E,, (xl, x2)E A, I- c A. Then either (xl,x2) E A \ I-, in which case x1E N i \ r ; or else (xl, x 2 )E r, in which case x2 EN,’. It follows that Hence, for each

(Ni \ r, Nr2)E 3E

A,

~ i ( N i \ r+ ) Pz(Nr2) and so, by (7), m < M . The proof is now complete but it is worth observing that, in applications of the theorem, an expression formally different from (6), namely

min ( P l ( E I \ X I

XCEi

+ PZ(A(X>)>?

(8)

IS sometimes more convenient. T o show that the two expressions, which we shall denote by rn and m‘ respectively, are equal, we first note that, for any X G E l , (El \, X, A(X)) is a disconnecting pair. Hence

mG

\XI

+ PZ(A(X))

(X

c El)

< i d . Furthermore m‘ < min ( p l ( E l \ X , ) + p 2 ( X 2 ) : X I c E,, X, c

and so m

=

min ( p l ( X l )

=

m.

+ pz(X2) : X ,

E

E,, X,

E

E,, A(Xl>G Xz>

E,, A(El \ X I ) E X,)

This establishes our assertion. We shall conclude by mentioning briefly three specializations of Theorem 8.4.3. ( I ) The proof of Theorem 8.4.3 just exhibited is based on Theorem 8.4.2, and it is of interest to note that we can recover Theorem 8.4.2 by applying Theorem 8.4.3 to the deltoid (E, A, E), where A = {(x, x ) : x ~ E } and , the given independence structures Q,, Q, on E. (ii) Next, let 91 = (Al, ..., A,) be a family of subsets of a finite set E, and let & be an independence structure, with rank function p , defined on E. Taking E l = {1,2, ..., n ) , E, = E , 6 , = . q ( E , ) , B , = 6,

A

=

{ ( k , e ) :1 < k

< n, e E E , e E A k }

NOTES ON CHAPTER 8

145

in Theorem 8.4.3, we are led to the conclusion that the maximum cardinal of independent partial transversals of % (i.e. partial transversals of B[ which are members of &) is equal t o

n

+

min

IG(1,

..., fl)

{p(A(I)) - 111).

This is, in essence, Theorem 6.2.2, i.e. the defect form of Rado’s theorem on independent transversals. (iii) Let G be a finite bipartite graph and denote by El, E, the two (disjoint) sets of its nodes such that every edge links a node in E, and a node in E,. Further, let A be the set of all pairs (el, e,) such that el E El, e2 E E,, and { e , , e , } is an edge. Taking &,, &, as the universal structures on E l , E, respectively and using Theorem 8.4.3, we infer at once Konig’s theorem 1.7.1 (for the case of finite graphs). An alternative proof will be offered in $1 1.2.

Exercises 8.4 1. Write out the proof of Konig’s theorem 1.7.1 (for finite graphs) by specializing the proof of Theorem 8.4.3. 2. Supply the full details of the applications of Theorem 8.4.3 indicated at the end of the section. 3. Let &,, &, 8,be independence structures on a finite set E, and let p , , p 2 , p3 denote their rank functions. Show that max { 1x1 : X E Q, n Q, n &,} < min {pl(x,) p2(x2)

+

+ ps(x3): xl u x2 u x3= E},

and that the sign of inequality cannot be replaced by that of equality.

4. Let &,, &, be independence structures on a finite set E. Using asterisks to denote complementary structures, show that

+ &,*)* c Q, n &,.

(&,*

Can the sign of inclusion be replaced by that of equality? 5. Let the notation be as in the preceding question, and denote by p l , p,, p the rank functions of &,, &, ,(dl* b2*)*respectively. Show that, for A E E,

+

p(A) =

min f ( X ) - minf(X),

AGXGE

where f ( X ) = pl(W + p2W) -

X C E

1x1.

Notes on Chapter 8 $8.1. Theorems 8.1.1 and 8.1.2 both occur in the work of C . St. J. A . NashWilliams (l),although the basic idea seems to be due to J. Edmonds. The proof of Theorem 8.1.1 given above was devised by A. P. Heron, that of Theorem 8.1.2 by

146

THE RANK FORMULA OF NASH-WILLIAMS

8

D. J. A. Welsh (5). The rank formula contained in Theorem 8.1.2 is a result of crucial importance in the theory of abstract independence since it enables us to treat without difficulty problems which had previously been accessible only to extremely complex arguments. The relation between the Nash-Williams rank formula and Brualdi’s symmetrization of Rado’s theorem on independent transversals (Theorem 8.4.3) has been investigated by Welsh (5). Transfinite analogues of Theorems 8.1.I and 8.1.2 were discussed by J. S. Pym & Hazel Perfect (1).

$ 8.2. All results in this section other than Corollary 8.2.3 are due to Edmonds (1) or Edmonds & Fulkerson (1). The latter paper contains a wealth of further results of the same general character as those discussed here. For the special case of linear independence, Corollary 8.2.3 was originally proved by A. Horn (1) and also by R. Rado (8). The work of all these authors is difficult: I owe the very transparent treatment offered here to a communication of Dr D. J. A. Welsh; see also Harary & Welsh (1) Far-reaching extensions of the findings of this section will be found in Brualdi’s paper (9). Some of the results are discussed by C. Berge (2) in the more general context of ‘graphoid’ theory.

5 8.3. The treatment in this section is based largely on Dr Welsh’s ideas; in particular, Theorem 8.3.1 is due to him (Welsh ( 6 ) ;cf. also (8)).A characterization, different from that contained in Theorem 8.3.1, of independence structures which are transversal structures has been given by J. H. Mason (1). Theorem 8.3.3 was noted by Professor Brualdi. $ 8.4. The notion of a complementry structure derives from Whitney’s fundamental work (1). Theorems 8.4.2 and 8.4.3 made their first appearance in Brualdi’s unpublished manuscript (2) (see also Aigner & Dowling (l)),but the proofs given here are due to D. J. A. Welsh (5).

9 Links of Two Finite Families All the investigations sp far have been centred on the existence of transversal-like objects associated with a single family. We shall now extend the scope of the discussion by considering more than one family. In practice, this will amount to a study of pairs of families, since the difficulties of dealing with more than two families have not yet been surmounted. 9.1 The notion of a link Let 2l, 23, ... be families of subsets of E. If a family X of elements of E is a system of representatives of each of these families, then it is called a common system of representatives (CSR) of 2l, 23, ... . In particular, then, X = (xk:kEK) is a CSR of 2l = ( A i : i e I ) and 23 = ( B j : j e J ) precisely if there exist bijections 4 :K + I, t+b: K + J such that

Hence 2l and 8 possess a CSR if and only if there exists a bijection 8: I such that Ai n Be(i) # 0

-+

J

( i I).~

Again, let a set X be a transversal resp. partial transversal of each of the families (It, 8,... . Then X is called a common transversal (CT) resp. common partial transversal (CPT) of these families. It is clear that X is a CT of 2l = (Ai: ~ E I )and 23 = ( B j : j € J ) if and’ only if there exist bijections p : X + I, 6 :X -+ J such that

This statement remains valid if the term ‘bijection’ is replaced by ‘injection’ and ‘common transversal’ is replaced by ‘common partial transversal’. We note that, if several families of sets possess a CSR or a CT, then their cardinals must be equal. The term ‘link’ will be used as a joint designation for a common system of representatives or a common (partial) transversal. In the present chapter our discussion will be confined to finite families. 147

148

9, 5 9.2

LINKS OF TWO FINITE FAMILIES

9.2 Common representatives We begin our examination of pairs of families with a result closely related to Hall’s criterion.

THEOREM 9.2.1. The families ‘u = (A,, ..., A,,) and b = (B,, ..., B,) of sets possess a common system of representatives if and only if, for each k with 1 < k < n, the union of any k A’s intersects at least k B’s. The argument here is very similar to that used in the deduction of Theorem 2.2.4 from Theorem 2.2.1 (Hall’s criterion). We write

Ci = ( j : 1 d j d n, Ai n Bj # 0} so that j € C i if and only if A, n Bj # 0. For < i, < n, we therefore have

Ci, u ... u C i R= { j : 1 < j = { j :1 < j

< i < n), 1 d k < n and (1

1 d i , < ...

< n, A i , n B j # 0 or ... or A i k n B j# 0) < n, (Ai, u ... u Aik)n B, # 0}.

Thus, if 2I and 23 satisfy the intersection condition stated in the theorem, then 6 = (Cl, ..., C,,) satisfies Hall’s condition and so (by Theorem 2.2.1) possesses a transversal, say j l EC1,

...,J,,EC,,,

wherej,, ... , j nare the numbers 1, ..., n taken in a suitable order. Hence A, n Bj, # 0,..., A, n Bin # 0

and so 2l and 23 possess a CSR. This establishes the sufficiency of the intersection condition ; its necessity holds trivially. The intersection condition in Theorem 9.2.1 is, on the face of it, asymmetric with respect to 2l and %3 (so that it can be restated with the roles of 2l and 23 interchanged). The next result gives a formally symmetric criterion for the existence of a CSR. In formulating it, we shall make use of the function 1: Z -+ (0,I ) (where Z is the set of integers) which is defined by the equations X(X) = 1 (x > O), x(x) = 0 (x < 0). COROLLARY 9.2.2. The families ‘u = (A,, ..., A,) and b = (Bl, ..., B,) of sets possess a common system of representatives if and only if, for all pairs I, J of subsets of { 1, ..., n } ,

1Am n B(J)I 2 x(lII

+ IJI - n).

(1)

§ 9.2

149

COMMON REPRESENTATIVES

Suppose, in the first place, that 2l and 23 have a CSR so that, without loss of generality, A, n B, # 0, ..., A, n B,,# 0 . Let I, J E (1, ..., n > . If I n J If I n J # 0,then

= 0, then

(I(

+ (J(- n d 0 and (1) is satisfied.

IA(1) n B(J)I 2 I 3 x(lII

+ IJI

- n).

Suppose, next, that (1) holds for all I, J. Take any subset I of { 1, ..., n} and let {I, ..., n } = J, u J, be a partition such that A(I)nBj#O

(jeJ1),

A(I)nBj=O

(jeJz).

+

Then A(1) n B(J,) = 0 so that, by (l), 111 IJzI - n d 0, i.e. III d IJ,I. Thus, for any I, A(1) intersects at least 111 B’s. Hence, by Theorem 9.2.1, 2l and 23 have a CSR.

,

THEOREM 9.2.3. Let A u ... u A,, = B u ... u B, where the A’s, andequally the B’s, are pairwise disjoint and where each A and each B has the same, $finite and non-zero, cardinal. Then the families (A,, ...,A,,) and (B,, ..., B,) possess a common system ofrepresentatives (and also a common transversal). Let the cardinal number of each A and each B be s. Assume that the union of a certain set of k A’s (say A, u ... u Ak) intersects fewer than k B’s. Then A, u ... u A, fails to intersect the union of a set of n - k + 1 B’s, say B, u ... uB,-,+,.Itfollowsthat

ns = IA, u ... u A,I = IA, u ._.u A, u B, u ... u B,I 2 (A, u ... u A, u B, u ... u B,-,+,( = (A, U ... u AkI + (B, u ... U B,-k+,I = ks + ( n - k = (n + 1)s.

+ 1)s

We thus arrive at a contradiction and conclude that, for each k with I < k < n, the union of any k A’s intersects at least k B’s. Hence, by Theorem 9.2.1, the two families possess a CSR. But any two A’s (or B’s) are disjoint. Hence the range of the CSR is a CT of the two families. An immediate consequence of Theorem 9.2.3 relating to groups is as fOIl0ws . COROLLARY 9.2.4. Let H be a subgroup of aJinite group G . Then the.fami1y of left cosets of H and that of right cosets of H possess a common transversal.

150

9, Q 9.3

LINKS OF TWO FINITE FAMILIES

Exercises 9.2 1 . Let ( A l , ..., A,,), (Bl, .._,B,,) be two families of sets. Show that they possess a CSR if and only if 111 IJI n whenever I, J 5 (1, ..., n } and A(1) n B(J) = 0.

+


, II*l -

IJ*I.

(1)

Since 23 is restricted, the set of all its PTs is, by Theorem 6.5.3, anindependence structure on E. Thus (i) means that '.It has an independent transversal. Now all A i are finite and so, by Theorem 6.2.4, this is the case if and only if every finite subfamily of 2I has an independent transversal. This establishes the equivalence of (i) and (ii). Again, by Theorem 10.4.1, the finite subfamily %(I) of 21 has a CT with some subfamily of 23 if and only if, for all I* c I and all cofinite J* G J, we have (A(I*) n B(J*)( 3 Ill - I T \ I*l - IJ \ J*), i.e. IA(l*) n B(J*)l 3 11*1 - IJ \ J*I.

178

LINKS OF TWO ARBITRARY FAMILIES

10, 3 10.4

Thus (ii) holds if and only if (2) is satisfied for all I c c I, all I* c 1, and all cofinite J* L J. This is plainly equivalent to the requirement that ( 2 ) should be satisfied for all I* cc I and all cofinite J* c J ; and this, in turn, means that (1) is satisfied for all I* cc I and all J* c c J. Thus (ii) and (iii) are equivalent. Next, we recall the insertion theorem 9.5.2. It can be shown that this theorem remains valid for arbitrary families but the proof falls outside the scope of the present treatment, and we shall therefore content ourselves with establishing the transfinite analogue of the special case 2l’ = %, 23‘ = ‘23. It will be useful to record first an almost obvious preliminary result.

LEMMA10.4.3. Let H, K, E be sets. Let ( c h : h E H) be a family ofsubsets of E, and let p : K -+ H, o: K E be mappings. Suppose that (i) f o r any k , k’ E K, the relations p ( k ) = p ( k ’ ) and o(k) = a(k’) imply each other; (ii) o ( k )E C p ( k ) for all k E K. Then o(K) is a transversal of (Ch:h E p(K)). --f

For h E p(K), we shall write p*(h) = ( k E K : p ( k ) = h )

( # 0).

If k , k‘ E p*(h), then p ( k ) = h = p ( k ’ ) and so, by (i), o ( k ) = o(k’). It follows that o(p*(h)) is a single element in o(K). We shall denote it by r(h), so that z is a mapping of p(K) into o(K). It is a matter of immediate verification that z is bijective. Moreover, let h E p ( K ) and k E p*(h), so that p ( k ) = h. Then r(h) = o ( k ) and, by (ii), we infer that 7 ( h )E ch for all h E p(K).

THEOREM 10.4.4. Let 9I and 23 be two arbitrary families of subsets of an arbitrary set E. I f 91 resp. 23 has a common transversal with a subfamily of 23 resp. ?I, then 9I and 23 have a common transversal. We shall write 9I = ( A i :i c I), 23 = ( B j : j € J ) and shall assume, as may be done without loss of generality, that I n J = 0. By hypothesis, P[ and a subfamily of % have a CT, i.e. there exists a family (x,: i E I) of distinct elements J such that x, E A i n B+ci,( i s I). Further, 23 and of E and an injection 4: I a subfamily of ?I have a CT, i.e. there exists a family ( y j : j €J) of distinct elements of E and an injection $: J 4 I such that y j E A,,j, n B j ( j J).~ We define --f

-

Ai=

[

(xi,yjj

if i = $ ( j )

{xi}

if i $ $(J)

(i E I),

Q 10.4

COMMON TRANSVERSALS OF TWO FAMILIES

[

-

Bj=

if i

{Yj}

if i 6 441)

-

=

(iE J),

B = ( B , : j e J).

% = (A,: i E I ) , It is plain that

4(i)

{ y j , xi}

-

(~EJ).

G

Ai ( ~ E I ) ,

=

{(i,eij)EI x E x J : e E A j n B j ) ,

Ai

179

Bj

G

Bj

Further, we define F

Fio = {(i,e , j ) E F: i = i,} (i, E I), Fjo= { ( i , e , j ) E F : j = j , ] ( ~ , E J ) ,

We note that every set Fk is finite. For let i E 1. Then e E Ai for only finitely many e E E. Further, if e E E, then e E Bj for only finitely manyjE J. Therefore e E Ai n B j for only finitely many pairs ( e , j ) ,and this means that Fi is finite. Similarly, Fj is finite for eachjE J. Let I‘ E I, J’ c J, and denote by a choice function of the subfamily %(I’ u J’) of 8.We shall then write V(k) = (V1(k)>V2(k), V 3 ( k ) ) E Fk

( k €1’ U J’).

The choice function q will be said to be ‘coherent’ if, whenever h, k E I’ u J’ and q‘(h) = f ( k ) for some r with 1 < r < 3, then q‘(h) = f ( k ) for all r with 1 0, a 1 + cz2 = 1 . Hence, by obvious induction.

t See e.g. Titchmarsh (1, 172).

I96

COMBINATORIAL PROPERTIES OF MATRICES

11,§11.3

whenever B , , ..., B, E 9,,+ and z,, ...,a, > 0, L Y ~+ ... + a,.= 1. we have 4 ( P B ) = Now for any permutation matrix P and any B E gin+, 4 ( B ) . Hence, taking 0

I

0 0

P=

0 0

... 0

1 0 ... 0 ... ... ...

... ... ...

0 0 0 0

... I

I

...

0 0 0

0

and using (9,we obtain

=

1

4(JnB ) = 4 ( J n ) = n log-.n

Thus

4 ( B ) 3 nlog-

1

(BE9,,+).

n

(6)

Next, let A E 9,, and put B = ( A + EJ,)/(I + E ) , where E > 0. Then B E 9,,+. Denote by 7-r a permutation of ( I , ..., n ) which satisfies

Then

n

n

where P = /Ipkj/ldenotes a typical permutation matrix. I n view of Corollary I I .3.2 and inequality (6), we infer that max P

n

1

k.j=l

p k j l o g h k j= max

n

1

D E ~k , ,j =~ 1

dkjIogbkj

DIAGONALS OF DOUBLY-STOCHASTIC MATRICES

Q 11.3

197

Thus

i.e.

Letting E

+ 0, we

obtain the desired result.

We next turn to a somewhat different kind of statement. THEOREM 11.3.5. Let 1

< k < n and write if n

kink =

\(n +4kk,,_l

+k

iseven,

if n + k isodd.

Then every doubly-stochastic n x n matrix possesses a diagonal on which at least n - k + 1 elements are greater than or equal to Lfnk. Moreover, this result is best possible in the sense that the phrase ‘or equal to’ cannot he omitted. Assume that every diagonal of the d.s. n x n matrix A has fewer than n - k 1 elements 3 P,k. Thus, every diagonal contains at least k elements < pnk. Hence, by Corollary 11.2.5,t A possesses a submatrix B of width n k in which all elements are < pnk.Without loss of generality, we can write

+

+

+

where B is of type p x q and p q = n + k. Let b, c, d, e denote the sum of all elements in B , C, D, E respectively. Then b + c = p , b d = q and so

2b Butb

+

+ c + d = p + q = n + k.

+ c + d + e = n a n d s o b - e = k. Hence

In applying Corollary 1 1.2.5, we distinguish not between zero and non-zero elements but between those 3.pnkand < pnr.

198

COMBINATORIAL PROPERTJES OF MATRICES

11,

0 11.3

say (where x , y take positive integral values). Now it is an easy matter to verify that 111 = k / p n k .and we thus arrive a t a contradiction. It follows that some diagonal has at least 11 - k I elements > p n k . T o show that this result is best possible, we shall denote by Up, the p x q matrix all of whose elements are equal t o 1. Let n + k be even, and write ( n + k ) / 2 = r, ( n - k ) / 2 = s. Then

+

+

is a d.s. n x n matrix. Its submatrix kr-’ U,, is of width n k and all its elements are equal to p n k . Hence, by Corollary 11.2.5, every diagonal of A , contains at least k elements equal to p n kand so no diagonal contains n - k 1 elements greater than / i n k . The case when ti + k is odd is dealt with similarly. We write

+

r = (n

+k

+1)/2, p = ( n

+k

- l)/2, s = (n

-

k +1)/2,

G =

(n

-

k -l)/2.

The ti x n matrix

+

is then d.s. Its submatrix k ( r p ) - ’ U,,, i s of width n k and has all its elements equal to / i n k . Hence no diagonal of A contains n - k 1 elements greater than / i n k . The case k = I of the theorem just proved is worth stating separately.

+

COROLLARY 1 1.3.6. Let p n be defined as 4/n(n + 2 ) or 4/(n + I)’ according us i i is m e n or odd. Then every doubly-stochastic n x n matrix possesses a diagonal each of whose elements is greater than or equal to p,,. Moreover, the phrase ’ o r equal to’ cantiof be omitted. Exercises 11.3 I . Deduce Theorem I I . I .3 from Frobenius’s theorem (Corollary 11.2.6). .._,.Y,} and [ y , , ..., y,) be orthonormal sets of vectors in unitary 2. Let n-dimensional space with inner product ( , ). Writing aij = I(xi,yj)12, show that llajill is a. d.s. matrix. Deduce that. for each k with 1 < k < n, the x’s and y’s can be renumbered such that [ ( . x i ,yi)l 3 j ~ , , ~ ;

(I d i d n - k

+ I ).

3. Determine the largest number I., such that at least half the elements on some diagonal of every d.s. N x n matrix are greater than or equal to A,,. In particular, verify that I,, > 8/9/7 ( n 3 I ) .

$1 1.4

DOUBLY-STOCHASTIC PATTERNS

199

11.4 Doubly-stochastic patterns The result of Theorem 11.1.3 prompts a more comprehensive inquiry into the distribution of positive elements in d.s. matrices. More precisely, we ask whether it is possible to prescribe in advance the position of positive elements in a d.s. matrix. This question is now to be discussed both for finite and for infinite matrices. Two matrices of the same type are said to have the same pattern if their non-zero elements occupy the same places. A square matrix i s said to have a doubly-stochasticpattern if it has the same pattern as some d.s. matrix. Again, if A , B are matrices of the same type, then A is said to be contained in B if a i j # Oimpliesbij # 0. The solution of the problem for the finite case i s entirely straightforward. THEOREM 1I .4.1. Let M be a finite, non-zero, square matrix. The ,following statements are then equivalent. (i) M has a doubly-stochastic pattern. (ii) M cannot be reduced by means of permutations o j rows and of columns to the form

where X is a square matrix (of order less than that of M ) and Y # 0 . (iii) Every non-zero element of M belongs to a non-zero diagonal. Suppose, in the first place, that (i) holds and let D be a d.s. matrix with the same pattern as M . If (ii) were false, then D (as well as M ) could be reduced to the form (1). In that case, denoting the type of X by k x k , we see that the sum of all elements in X would be k when computed by rows and less than k when computed by columns. Thus (i) implies (ii). Next, suppose that (iii) does not hold. Then there is a non-zero element x of M which does not belong to any non-zero diagonal. By permutations of rows and columns, we can ensure that M assumes the form

where the matrices of M , p , q, N are of type n x n, ( n - 1) x 1, 1 x (n - l ) , (n - 1) x (n - 1) respectively and N has no non-zero diagonal. Hence, by Corollary 11.1.2, there exists an integer k in the range I < k < n - 1 and a set of k rows in N whose non-zero elements lie in (at most) k - 1 columns. If these k rows and k - 1 columns are moved into initial positions in N, then M assumes the form (1) with Y # 0 since x is an element of Y. Thus (ii) is violated, and we conclude that (ii) implies (iii).

200

COMBINATORIAL PROPERTIES OF MATRICES

1 1, 9 11.4

Finally, let mij be any non-zero element of M . If (iii) holds, then there exists a permutation matrix P c i i )which is contained in M and has a 1 in the (i,j)-th place. Denoting by t the number of non-zero elements in M , we see that t-1

C mij #

p(ij) 0

is a d.s. matrix with the same pattern as M . Thus (iii) implies (i), and the proof is complete. Next, we turn to the case o f infinite matrices. An infinite matrix is said to be lineTfiniteif none of its Iines contains infinitely many non-zero elements. The following criterion is analogous to Corollary 1 1.1.2. THEOREM 1 1.4.2. An injinite, line-finite matrixpossesses a non-zero diagonal if and only iL for each natural number k , any k rows resp. columns contain between them non-zero elementsfrom at least k columns resp. rows. The necessity of the stated condition i s obvious. To prove its sufficiency, we denote by I the set of all natural numbers and by M = l\mijll the given matrix. We now define

Since M is line-finite, it follows that each A and each B i s finite. Since, moreover, the A’s and also the B’s are pairwise disjoint, it is clear that the families (Ai: i E I), ( B i : iE I) are relatively finite. Lastly, by our hypothesis, for every natural number k , the union of any k A’s resp. B’s intersects at least k B’s resp. A’s. Hence, by Theorem 10.1.5, the two families possess a CSR, and the desired conclusion follows.

I n our next result, the assumption of line-finiteness is dropped.

LEMMA11.4.3. Let M be an infinite non-zero matrix. If every non-zero element of M belongs to a non-zero diagonal, then M has a doubly-stochastic pattern. Let m i j be a non-zero element of M . Then, by hypothesis, M contains at least one permutation matrix which has a 1 in the (i,j)-th place. Choose one such permutation matrix and denote it by P ( ; j ) .Since the elements of M are denumerable, so are the permutation matrices P( ii) associated with the non-

0 11.4

DOUBLY-STOCHASTICPATTERNS

20 1

zero elements of M . Arranging them in a sequence P , ( k = 1,2, ...), we see that the matrix

is d.s. and has the same pattern as M . We are now in a position to establish the analogue of Theorem 1 I .4.1 for line-finite matrices.

THEOREM 11.4.4. The following statements relating to an infinite, line-finite, non-zero matrix M are equivalent. (i) M has a doubly-stochastic pattern. (ii) M cannot be reduced by permutations ojrows and columns to either o j the twoforms

where X,, X , are finite square matrices and Y, # 0 , Yz # 0. (iii) Every non-zero element o f M belongs to some non-zero diagonal. It should be noted that (in contrast to the finite case) the introduction of both forms (a) and (b) is essential. Thus, for example, the matrix

cannot be reduced to the form (a); for every set of k rows in this matrix has non-zero elements belonging to exactly k columns, and the non-zero elements in these k columns lie entirely in the k rows in question. But plainly the matrix does not have a d.s. pattern and, of course, it is of the form (b). We now come to the proof of the theorem. The fact that (i) implies (ii) is demonstrated in precisely the same way as the corresponding step in the proof of Theorem 11.4.1, except that now both the forms (a) and (b) need to be considered. Again, Lemma 11.4.3 shows that (iii) implies (i). It remains to show that (ii) implies (iii).

202

COMBINATORIAL PROPERTIES OF MATRICES

11,s 11.4

Assume that (iii) does not hold. Then there is a non-zero element x in M which does not belong to any non-zero diagonal. Hence the matrix N , obtained from M by the deletion of the row and column through x, has no non-zero diagonal. Hence, by Theorem 11.4.2, there is a natural number k and a set of k rows (or columns) in N whose non-zero elements lie in at most k - 1 columns (or rows). T o fix our ideas, consider the first alternative and take the k rows and k - I columns in question as occupying initial positions i n N , so that

where Q is of type k x ( k - 1). By permutations of rows and columns, we can move x into the ( k + 1 , k)-th place in M . Then M assumes the form

Q P 01 R where X ,

=

s

S

IlQ p(j is of type k x k , while the submatrix

is non-zero since x is one of its elements. Thus M has been reduced t o the form (a). Similarly, by considering the second alternative, we find that M can be reduced to the form (b). In either case, then, condition (ii) is violated; and we conclude that (ii) implies (iii).

Let us next consider the infinite matrix

! I

1 1 1 1 M = i 1 0 I 1 ~

1 I 0 1

1 1 0 0

1 ...I 1 ... 0 ... 0 ...'

Now any k rows resp. columns of M contain non-zero elements which belong to at least k 1 columns resp. rows. and therefore M satisfies statement (ii) of Theorem 11.4.4. On the other hand, the matrix obtained by the deletion

+

8 11.4

DOUBLY-STOCHASTIC PATTERNS

203

of the first row and first column in M has no non-zero diagonal, and so M fails to satisfy (iii) in Theorem 11.4.4. It follows therefore that Theorem 11.4.4 ceases to be valid for unrestricted infinite matrices. Nevertheless, something can be salvaged: our next theorem shows that (i) and (iii) remain equivalent.

THEOREM 11.4.5. An infinite non-zero matrix M has a doubly-stochastic pattern if and only if every non-zero element of M belongs to some non-zero diagonal. In view of Lemma 11.4!3, it suffices to show that, if M has a d.s. pattern, then every one of its non-zero elements belongs to some non-zero diagonal. We shall prove this by reducing the case of a general infinite matrix to that of a line-finite matrix. Let, then, M have a d.s. pattern and denote by D a d.s. matrix with the same pattern as M . Assume that some positive element d of D (which we may take as occupying the leading position) does not belong to a positive diagonal of D. Then, writing

we see that E has no positive diagonal. It is plain that the sum of all elements in any k rows or any k columns of E is greater than or equal to k - 1 + d. Denote by p m the sum of all elements in the m-th row of E. We now replace by zeros all but a finite number of positive elements in the m-th row of E , in such a way that the new row-sum exceeds pm - 3Y"d; and we perform this operation for every natural number m. We then obtain a row-finite matrix F ; and since the sum of all elements in E which have been replaced by zeros is less than

C 3Y"d

m= 1

=

+d,.

it follows that the sum of elements i n any k rows or k columns of F exceeds k -1 td. Next, denote by gm the sum of all elements in the m-th column of F . We replace by zeros all but a finite number of positive elements in the m-th column of F , in such a way that the new column-sum exceeds B, - 3-"d; and we perform this operation for every natural number m. As a result, we obtain a line-finite matrix G in which the sum of elements in any k rows or k columns exceeds k - 1.

+

204

COMBINATORIAL PROPERTIES OF MATRICES

l l , $ 11.5

N o w the positive elements of G in a n y k rows (columns) c a n n o t be contained in fewer than k columns (rows) since t h e s u m of elements in a n y k - 1 columns o r k - I rows of G is a t most k - 1. Hence, by Theorem 11.4.2, G possesses a positive diagonal a n d so, therefore, does E . We t h u s arrive at a contradiction a n d conclude t h a t the element d is, in fact, p a r t of a positive diagonal of D.T h e proof is therefore complete. Finally, we note t h e following immediate consequence of Theorem 11.4.5.

COROLLARY1 1.4.6. posi f ive diagonal.

Every infinite doubly-stochastic matrix possesses a

T h e infinite analogue of Theorem 11.1.3 is therefore valid.

Exercises 11.4 1. Show that the omission of the phrase ‘line-finite’ invalidates Theorem 11.4.2.

2. An infinite, line-finite matrix is such that, for each natural number k, any k rows contain between them non-zero elements from at least k columns. Show that the matrix need not possess a non-zero diagonal. 3 . Theorem 1 1.4.5 guarantees the existence of an infinite d s . matrix all of whose elements are positive and also of one whose only zero elements are precisely the elements on the main diagonal. Give actual examples of such matrices.

4. Show that there exists an infinite d.s. matrix lldk,ll such that, for every permutation 7-c of { I , 2, 3 , ...},

5. (i) Let M be infinite matrix; let p, r s be permutations of the set of natural numbers; and suppose that, for every n 3 1 , the n-th row of M contains at least p(n) non-zero elements and the n-th column at least o(n) non-zero elements. By means of Corollary 1.3.5, show that M possesses a non-zero diagonal; and also verify that it need not possess more than one non-zero diagonal. (ii) Let M be an infinite matrix and suppose that every line of Mcontains infinitely many non-zero elements. Prove that M has a non-zero diagonal. 6. Let A be an n x n matrix and, for 1 < k < n, denote by rk resp. s, the number of non-zero elements in the k-th row resp. k-th column of A. Show that the sequences of integers (rlr ..., r,,), (s,, ..., s,,) d o not determine whether A possesses a d.s. pattern.

11.5 Existence theorems for integral matrices For a (rectangular) matrix Q, we shall denote by R,(Q) a n d C,(Q) t h e s u m of its elements in t h e i-th r o w a n d .j-th column respectively. W e shall determine necessary a n d sufficient conditions f o r t h e existence o f a n integral matrix

5 11.5

205

EXISTENCE THEOREMS FOR INTEGRAL MATRICES

of given type whose elements, row-sums, and column-sums all lie between prescribed bounds. THEOREM 11.5.1. Let 0 < ri' < r i , 0 < sI' < s j , cij > 0 ( I < i < rn, 1 < j < n) be integers. Then there exists an rn x n matrix Q = llqijll with integral elements such that

< Ri(Q) < ri (1 < i < rn), < C j ( Q ) d ~j (1 < j < n), (1 < i < rn, 1 < j < n ) 0 < qij < cij ri'

(1)

sj'

(2)

ifand only $ f o r all I c { 1,

(3)

..., m>,J c { 1 , ..., n } ,

To prove this result, we put

< rn, 1 < j < n, 1 < k < c i j ) , Ei = { ( i , j , k ): 1 < .j < n, 1 < k < c i j } ( 1 < i < rn), F j = { ( i , . j , k ) :1 < i d rn, 1 < k < cij} (1 < . j < n). E

=

{ ( i , j ,k ) : 1 < i

Then (El, ..., E,) and ( F l , ..., F,) are partitions of E. Given a set X define them x n matrix Q = 11 qijll by the formula

G

E, we

q i j = IX n Ei n Fjl.

Then (3) is satisfied. Again, given Q by the formula

X

=

{(i,j , k ) : 1

=

IIqij/lsubject to (3), we define X

E

E

< i < m, 1 < ,j < n, 1 < k < q i j ) .

In either case, we have

&(Q)

=

IX n Eil (1 d i d m),

C j ( Q ) = IX n Fj( (1

< j < n).

Hence there exists a matrix Q with the requisite properties if and only if there exists a set X E E such that ri'

< IX n Eil < ri

(1

< i < m),

sj'

< IX n Fjl < sj

Now, by Theorem 9.6.5, this is the case if and only if

(I

<j

I? 49. Let condition (4) in Theorem 11.5.1 be satisfied. What additional requirements must be met if the matrix Q , subject t o (l), (2), and (3), is to be unique.

50. What can be said about the number, say P(n), of n x n incidence matrices which possess a positive diagonal? In particular, is the relation P ( n ) = o(2”’) valid? We can also seek to estimate the number of n x n incidence matrices with a d.s. pattern.

Miscellaneous Exercises 1. Let (sk:1 < k with 0 c K E { I ,

< n) be a family of n integers. Show that there exists a set K

..., n} such that

is divisible by n. 2. Deduce Theorem 3.3.1 (for a finite E) and Theorem 5.1.1 (for finite E and I) from Theorem 8.2.4.

3. Let E be a finite set, a finite family of subsets of E, B the transversal structure of 91, and k a natural number. Show that the collection of sets {X E 8: 1x1 k } is not necessarily a transversal structure.


lI*l,

+

I*] IJ \ J*( 3 k , lB(J*)l 2 IJ*l

233

MISCELLANEOUS EXERCISES

(ii) Show that % and 23 possess transversals the cardinal of whose intersection is at most k if and only if

IA(I*) u B(J*)l IA(I*)I 3

whenever I*

C_

> II*l + IJ*l

P*L

- k,

IB(J*)l 3 IJ*I

1, J* E J.

[H. Perfect]

29. Let 9" denote the set of all d.s. n x n matrices and, for A = Ilukjll~ g , , write

7c E B,,

4JA) = ~ I , , ( I ) + ... + a n . n ( n ) . Show (without invoking the deeper Theorem 11.3.3) that, given A ~!2?.,, there exists n E 6, such that d,(A) 3 1 . Deduce that, for all n 3 1 ,

min max d,(A)

= 1.

A E % nsB,

30. Show that the sum of two linear structures need not be a linear structure. 31 Let 91 = ( A l , ..., A,, be a family of non-empty subsets of a set of cardinal n. Show that there exist non-empty, disjoint subsets I, J of ( I , 2, ..., ?I 1 1 such that A(1) = A(J). [B. Lindstrom]

+

32. Let E be a n arbitrary set and let 8 denote the collection of all subsets of E whose cardinal does not exceed the natural number m. Show that & is linear over any field F such that \El I F / .




235

MISCELLANEOUS EXERCISES

(iii) Let & be an independence structure on a finite set E, let (1l be a family of subsets of E which possesses an independent transversal, and let M E 8. Show that the set of integers

{IT n MI : T an independent transversal of %} need not be an interval. 45. Show that the complementary structure of an independence structure on an infinite set need not have finite character.

46. Let E = { I , 2, ..., 6 ) and denote by & the collection of all subsets X of E such that 1x1 < 3 with the exception of ( 1 , 2, 6}, { I , 4, 5 } , (2, 3, 5 } , (3, 4, 6}. Show that & is linear over every field. 47. Let E be a finite set; (11 = ( A l , ..., A,,) a family of subsets of E; M E; and 1 k n. Use the theorem of Hoffman & Kuhn (Corollary 9.6.2) to establish necessary and sufficient conditions for ‘91 to possess a transversal X such that k ; (iii) IX n MI = k . (Cf. Ex. 6.2.8.) (i) IX n MI 3 k ; (ii) IX n MI Suppose that 21 possesses transversals X I , X, with IX, n MI 2 k , IX, n MI < k . Show that (11 possesses a transversal X with IX n MI = k . (Cf. No. 44.)

<