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Dissertation UMI Information Service University Microfilms International A Bell & Howell Information Com pany 300 N. Zeeb Road, Ann Arbor, Michigan 48106 8625259
Miklos, Dezso
SOME RESULTS RELATED TO A CONJECTURE OF CHVATAL
The Ohio State University Ph.D. 1986
University Microfilms
International300 N. Zeeb Road, Ann Arbor, Ml 48106 SOME RESULTS RELATED TO A CONJECTURE OF CHVATAL
DISSERTATION
Presented in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the Graduate School of the Ohio State University
By
Dezsti Miklos, M.S.
The Ohio State University 1986
Dissertation Committee: Approved by Professor D. K. Ray-Chaudhuri Professor T. Dowling Professor G. N. Robertson Adviser Department of Mathematics ACKNOWLEDGMENTS
M y research on the topic of this dissertation began in Hungary, while I was a student, and later with a colleague of Professor G. Katona and was finished (as far as research can ever be finished) at the Ohio State University under the supervision of Professor D. K. Ray-Chaudhuri. I wish to express my deepest gratitude for their valuable assistance and patience. I wish also to thank the members of my dissertation committee, Professors T. Dowling and G. N. Robertson, as well as the entire combinatorial community of the Department of Mathematics for maintaining a pleasant working environment. Among them, I would like to mention Professor E. Bannai, from whom I learnt about a new topic of combinatorics. I offer my sincere thanks to my fellow student, Ms. Gail Gill, who read my dissertation and corrected several English (and sometimes not only English) mistakes. VITA
October 28, 1957 ...... Bom - Budapest, Hungary
1982 ...... M. S., Etitvos University, Budapest 1982 - 1984 ...... Junior Fellow, Hungarian Academy of Sciences, Budapest, Hungary 1984 - ...... Junior Research Fellow, Mathematical Institute of the Hungarian Academy of Sciences (on leave of absence) 1984 - 1986 ...... University Fellow and Teaching Assistant (one year), The Ohio State University, Columbus, Ohio
PUBLICATIONS
[1] D. E. Daykin, A. J. W. Hilton and D. Miklos, Pairings from down-sets and
U p :SStS in distributive lattices. J. Combin. Theory (A), 34(1983), pp. 215-230.
[2] D. Miklos, Great intersecting families of edges in hereditary hypergraphs. Discr. Math., 48(1984), pp.95-99.
[3] D. Miklos, Linear binary codes with intersection properties. Discr. Appl. Math., 9(1984), pp.187-196.
FIELDS OF STUDY
Major field: Mathematics
Studies in Combinatorics: Professor D. K. Ray-Chaudhuri iii TABLE OF CONTEST
ACKNOWLEDGMENTS...... ii V ITA ...... iii LIST OF TABLES ...... vi LIST OF FIGURES ...... vii INTRODUCTION ...... 1 Chapter Page 1. PRELIMINARIES...... 4 1.1 Conventions 4 1.2 Definitions, notations 4 1.3 Early results 7 1.4 Chvatal's theorem 10 1.5 Berge's theorem 14 1.6 Sterboul’s theorem 22 1.7 Some other results 25 2. ANOTHER PROOF AND COMBINATORIAL GENERALIZATIONS OF BERGE'S THEOREM ...... 28 2.1 Introduction 28
2.2 Another proof of Berge's theorem 31
2.3 Some combinatorial generalizations of Berge's theorem 33 2.4 Some other generalizations for hereditary multi- hypergraphs 38 3. A GENERALIZATION OF BERGE’S THEOREM FOR DISTRIBUTIVE LATTICES ...... 41 3.1 Statement of the main result 41 3.2 An algebraic lemma 45 3.3 Proof of the main result 47 3.4 Generalization of Chvatal's conjecture for distributive lattices 60 Chapter Page 4. MAXIMUM INTERSECTING FAMILIES IN A HEREDITARY HYPERGRAPH IN THE EXTREME CASE ...... 63 4.1 Introduction 63 4.2 Description of the maximum intersecting families in H 64 4.3 Proofs of Theorems 4.2 and 4.3 66 4.4 The generalizations for hereditary multi-hypergraphs 73 5. A MORE GENERAL CONJECTURE...... 77 6. THE INFINITE CASE ...... 81 6.1 Introduction 81 6.2 If there is an intersecting subfamily of H of infinite cardinality 84 6.3 The general infinite case 86 LIST OF REFERENCES...... 90
v LIST OF TABLES
Table Page
1. Pairings from D and the bijection 9 ...... 29
2. Pairings from D' and the bijection 3. The bijections f\ and f z ...... 43 LIST OF FIGURES Figure Page 1. Pairings from D ...... 29 2. Pairings from D '...... 30 3. For the proof of Lemma 2.2 32 4. For the proof of Theorem 3.1 51 5. The lattice K ...... 62 6. For the proof of Theorem 4.2 67 7. T, M and M' in S ...... 68 8. For the proof of A2cS\T ...... 69 9. The choice of A and B ...... 70 INTRODUCTION Let S be a finite set and H cP(S) be a family of subsets of S satisfying the hereditary property, i.e., if A cB e// , then A tH as well. H is then called a hereditary hypergraph. We denote the maximum cardinality of the pairwise intersecting subfamilies of H by u>(H ) and the maximum degree of the elements of S in H by d (H ). Chvatal in 1972 conjectured that for a hereditary hypergraph H on a finite set w(LT) = d (H ) holds. This conjecture is the main core of the dissertation. In Chapter 1, we give a list of the earlier results related to the conjecture. There will be given detailed proofs for Berge's, Chvatal's, and Sterboul's results ([2], [5], and [23], respectively). Among them, the following theorem of Berge, which is given in a different form here to reduce the number of the necessaiy definitions, will play an important role throughout Chapters 2, 3, and 4. Theorem 1.12 (Berge [2]) If H is a hereditary hypergraph, then the sets from H (if |H | is even) or from H \{ 0 } (if |H | is odd) can be partitioned into pairs, such that the intersection of the two sets of a pair is empty. Chapter 2 is devoted to several generalizations of Berge’s theorem. The whole chapter, together with Chapters 3 and 4, is based on a different proof of the above theorem, which is also given here. We give the generalization of Berge's theorem, as well as some other results, for hereditary multi-hypergraphs. These results encourage us to generalize Chvatal's conjecture for hereditary multi-hypergraphs. The main result of Chapter 3, Theorem 3.1, generalizes Berge's theorem for distributive lattices with polarity. An algebraic lemma, Lemma 3.5, helps us to use combinatorial rather than algebraic methods to prove this theorem. At the end of the chapter, we give the following conjecture, which generalizes Chvatal's conjecture to distributive lattices. 1 Conjecture 3 If L is a finite distributive lattice, D is a down-set of L and w(D) = max {|M|: M cD and V a,beM => aAb^O}, then there is a star CcD in L such that |C|= From Berge's theorem, it follows that for a hereditary hypergraph H to (ff) cannot be larger than g |H |. With the help of this result, it is easy to see that if d (H ) = £|H |, then Chvatal's conjecture holds. Chapter 4 handles the case when only u >(H) = i\H | is known by proving the following result. Theorem 4.1 If H cP(S) is a hereditary hypergraph on a finite set S and w ( //) = iz\H J], then u>(//) = d (H ), where [.] denotes the integral part of a number. In fact, we give a more general result, describing, from a certain point of view, all of the intersecting subfamilies with maximum cardinality of a hereditary hypergraph satisfying the assumption of the above theorem. Also,we give the generalizations for hereditary multi-hypergraphs. Chapter 5 investigates the possible counterexamples to Chvatal's conjecture. Also, the following, more general conjecture is given. Conjecture 5 If H is a hereditary hypergraph and M is an intersecting subfamily of H , then every edge of M contains a point whose degree in H is at least \M |. We can prove a weaker form of this conjecture, which at least assures us that as to (H) goes to infinity, d (H ) goes to infinity as well. (A similar result is given for hereditary multi-hypergraphs.) Theorem 5.3 If H is a hereditary hypergraph and M is an intersecting subfamily of H with |M |= m, then every edge of M contains a point of degree at least m/log 2 m. Finally, Chapter 6 proves the infinite version of Chvatal’s conjecture. Of course, in this case u>(H ) denotes the supremum of the cardinalities of the pairwise intersecting subfamilies of a hereditary hypergraph and d (H) denotes the supremum of the degrees of the points in the hypergraph. We will also give, and prove the infinite version of Conjecture 5: Theorem 6.2 If H is an infinite hereditary hypergraph and M is an intersecting subfamily of H of infinite cardinality, then for every MeA/ the supremum of the degrees of the points of M, counted in H , is at least as large as \M |. 3 This result, together with Theorem 5.3, helps us to prove Chvatal's conjecture, whenever we have a really infinite hypergraph. Theorem 6.6 If H is an infinite hereditary hypergraph and w (H )> ^ 0, then w (H ) = d(H ). Again, these results are generalized for infinite hereditary multi-hypergraphs. The proofs of these generalizations are easy consequences of the proofs given for infinite hereditary hypergraphs. CHAPTER 1 PRELIMINARIES 1.1 Conventions In the first five chapters we will assume that all of the sets and objects are finite, while in Chapter 6 we will work with infinite sets, in general. The symbol 0 will denote the empty set, i.e., the set which has no elements. If S is a finite set and A is any subset of it, then the complement of A, which is denoted by S\A=A, is the subset of S which contains exactly those elements which are not in A. For a finite set S, P(S) will denote the set of all of the subsets of S. Two sets, A and B are said to intersect if they have nonempty intersection. We will use the usual notations (e.g. u,c,n ,\, etc.) for the operations and relations of sets. Throughout this dissertation, lemmas, propositions, corollaries, and theorems are numbered sequentially by chapter. For example, Proposition 1.5 is the fifth statement of Chapter 1 which is designated as a proposition, lemma, corollary, or theorem. The symbol □ is used to indicate the end of a proof. If □ stands immediately after a statement, it indicates that the proof is omitted, either because the statement is trivial, or because the proof can be found in the references and it is not necessary to follow the flow of the dissertation. If a statement is not finished by the symbol □ and is not followed by a proof immediately, it indicates that it will be proven somewhere else in the dissertation. 1.2 Definitions, notations I f S is a finite set and A|,A2,...,Ai is a collection of distinct subsets of S (and thus l 4 5 vertices of the hypergraph, while the sets A|, A 2,..., A f are called the edges of the hypergraph. We will refer to the edges as the elements of H , as well. Usually, we denote the points by lower case and the edges by upper case letters, though in some part of Chapter 2 and and in Chapter 3 the edges will also be denoted by lower case letters. If the edges of a hypergraph have one or two vertices, we call it a graph. The edges of cardinality one in a graph are called loops . If a graph does not have loops, it is called simple. The above definition does not allow an empty hypergraph, i.e., a hypergraph without edges, and in general, we will suppose that the hypergraphs are nonempty. However, in a few exceptional cases, we will need the notation of the empty hypergraph for induction proof. In those cases the induction hypotheses will be trivial for the empty hypergraph. If the sets A 1,A 2,...,Ai are not all distinct in the above definition of the hypergraph, i.e., if we allow repetition among the edges, we call H = {A ,,A 2,...,A j} a multi-hypergraph .An empty multi-hypergraph is simply an empty hypergraph. With the help of this notation, a multi-graph is a multi-hypergraph consisting of edges of one or two vertices. Two or more edges which appear as the same set are called parallel. For any hypergraph or multi-hypergraph H and xeS, where S is the underlying set of H , we define the degree of x in H as the number (or cardinality) of edges of H containing x. It is denoted by (x), or, if it is obvious what hypergraph we mean, simply by d(x). d (H ) will denote the maximum degree of the points of the (multi-)hypergraph H , i.e., d (H )=max{d^ (x ): xeS}, and will be called the degree or the maximum degree of the (multi-)hypergraph. Sometimes we will call a collection of edges containing a given point, i.e., a subset R of H such that for some xeS R = (H e// , xeH}, a star .The star R = {H eH : x«H} is centered at x, and x is called the center of the star. In general, if M is a subset of the hypergraph H, and thus is a hypergraph itself, we call M a subhypergraph ox family . Two special types of hypergraphs play important role in this dissertation. If M is any hypergraph such that any two edges of it have nonempty intersection, then M is called an intersecting or sometimes more precisely, a pairwise intersecting hypergraph. (We can define similarly intersecting multi-hypergraphs.) In particular, an intersecting hypergraph may not contain the empty set and thus, the hypergraph {0 } is not intersecting. This is the only hypergraph of one edge which is not intersecting. On the other hand, the empty 6 hypergraph is supposed to be intersecting. If H is a hypergraph satisfying the so called hereditary property , i.e., for any AzH and BcA, B is also an edge of H , then H is called a hereditary hypergraph . A hereditary hypergraph, unless it is empty, always contains the empty set as one of its elements. The empty hypergraph is supposed to be a hereditary hypergraph, for induction purposes. UH is a hypergraph on a finite set S, then the hereditary closure of H is defined as the collection of all of the subsets of S which are contained by any element of H. In other words, the hereditary closure of H is the hypergraph {AcS: 3 BzH such that AcB}. Thus, the hereditary closure of H always contains H as a subhypergraph. For a hereditary hypergraph H , we call those elements of it which are not contained by any other element maximal elements , or maximal sets , or sometimes bases of the hereditary hypergraph. Thus a hereditary hypergraph H is the hereditary closure of its bases. The collection of the bases is the smallest possible collection of edges which has H as its hereditary closure. We will define two other parameters of hypergraphs. For any hypergraph H,u>(H) will denote the maximum number of pairwise intersecting edges of H , that is w(H )= max {|M |: M is an intersecting subfamily of H }. The chromatic index of the hypergraph H , denoted by q (H ), is the smallest possible number of colors needed to color the nonempty edges of H such that two edges with the same color have empty intersection. The concept of hereditary multi-hypergraphs is rather complicated, and it will be discussed in Chapter 2, where we will need it. We will also give there the precise definition of multi-hypergraphs and the generalization of the definitions given here for hypergraphs only. The next proposition is valid for multi-hypergraphs as well, but here we give it for hypergraphs only, because of the lack of the precise definitions. Proposition 1.1 For any hypergraph H q(H ) > w(H ) > d(H ) . Proof Consider a coloring of the edges of H with q(H ) colors such that no two intersecting edges have the same color. Now any pairwise intersecting subfamily of H may contain at most one edge with a particular color, and so q (H )>u(H). On the other hand, for any element of S the edges of H containing this element form an intersecting family, and so every star in H is also an intersecting subfamily of H . This gives that 7 v(H )i d(H ). □ When q (H )=d(H ) for a hypergraph H , we say that H has the edge-coloring property . Hence, every hypergraph H which has the edge-coloring property also has the property u (H )=d(H ). On the other hand, we can find examples of hypergraphs H for any of the following combinations of the properties: q(H) > u>(H) = d(H), q (H ) = u>(H ) > d(H ), q(jH ) > u (H ) > d(H ). For example, the hypergraph H ={{1,2}, {2,3}, {3,4}, {4,5}, {5,1}} has parameters q (H )=3, )=d(H )=2, and so satisfies the first combination. If H ={{1,2}, {2,3}, {3,1}}, then q(H )=w(H )=3, while d(H )=2, so this hypergraph gives an example for the second combination. The third combination requires the most complicated example. Replace the edge {3,4} by {3,4,6} in the first example and add the edges {2,5,6} and {1,3,4.6}, resulting the hypergraph H = {{1,2}, {2,3}, {3,4,6}, {4,5}, {5,1}, {2,5,6}, {1,3,4,6}}. Not too complicated counting shows that q(H )=5, u>(// )=4, and d(H )=3 for this hypergraph. The above examples are, of course, examples of multi-hypergraphs as well satisfying the given combinations of equalities and inequalities among these parameters. Chvatal [4] in 1972 conjectured the following: Con jecture 1 For any hereditary hypergraph H on a finite set S, ) = d (H ). In other words, among the maximal intersecting subfamilies of a hereditary hypergraph, there is a star always. It means, that if we want to maximize the number cf edges in an intersecting subfamily of a hereditary hypergraph H , it is enough to pick the elements of H containing a fixed point of S and find the maximum one among these families. 1.3 Early results Since 1972 many mathematicians contributed to this conjecture, but the conjecture itself is still unproven. In this chapter we will list these results, including the proofs of the most important ones. First, we mention that, as opposed to Chvatal's conjecture, the edge-coloring property is not true even for hereditary hypergraphs, as the following example of Berge [2] shows. Let n and i be two integers with n > i > 0 and S be a set of n elements. Let K" denote the i-uniform complete hypergraph on S, i.e., the hypergraph 8 whose edges are all the subsets of S having exactly i elements. Then the hereditary closure of K 73 , which will be denoted by J here, does not have the edge-coloring property. To see this we first mention th at/ consists of all the subsets of a 7-element set having 0,1,2, or 3 elements. Thus, the degree of any element of S is 15+6+1=22, and thus, d (/ )=22. On the other hand, simple, but long calculations show that / can not be colored by fewer than 23 colors such that the intersecting edges each get different colors. Though written much earlier than when Chvatal formulated his conjecture, the first results about this conjecture are found in the famous paper of Erdos, Ko, and Rado [11]. They have two different, but corresponding results about extremal hypergraphs. The first of them essentially handles the case of Chvatal's conjecture when H is the complete hereditary hypergraph which contains all of the subsets of its underlying set S. Proposition 1.2 [11] IfH =P(S) for a finite set S of n elements, then Chvatal's conjecture holds for it. Moreover, every intersecting subfamily can be completed to a maximal intersecting subfamily having 2n_1 edges. Proof It is an easy counting proof to see that in P(S) every point has degree 2n_1, and thus the degree of this hypergraph is 2"'1. On the other hand, we may form pairs {A ,A } from the elements of P(S) for every AcS. The number of these pairs is trivially 2n_1, and since from every pair we may take only one element into an intersecting subfamily of P(S), w(P(S)) <, 2"'1. Now, let M be any intersecting family of P(S). I f M has 2n_1 elements, then it is maximal. If it has less than 2"'1 elements, then there must be a pair {A ,A } of subsets of S such thatM does not contain any of these subsets. We will show that either every element of M intersects A, or every element of M intersects A. Suppose that there is an edge B in M which has empty intersection with A. Then Be A, and since every edge of M intersects B, every edge of M must intersect A as well. Thus, we can complete M with A or A, resulting in a larger intersecting subfamily. □ We note that the second part of this result implies that in the case of the hypergraph P(S) not all maximal intersecting subfamilies are stars. Moreover, we have the following immediate corollary of this proposition, which we will need in Chapter 4. Corollary 1.3 An intersecting subfamily M of P(S) is a maximal intersecting subfamily (of cardinality 21s!'1) if and only if for every AcS, M contains exactly one of A 9 and A. □ The following case will give us an example of the situation where all the maximal intersecting subhypergraphs are stars in a hereditary hypergraph. This comes from the next, rather famous result from the above mentioned paper. Theorem 1.4 (Erdos, Ko, and Rado [11]) If M is an intersecting subfamily of the i-uniform complete hypergraph K" on n vertices and i most the degree of the hypergraph K". □ Later Hilton and Milner [17] proved for this hypergraph that we not only cannot have any larger intersecting subfamily of K? than the stars, but that here the only intersecting subfamilies of this maximum cardinality are the stars. These theorems together give us the following result. Proposition 1.5 In the hereditary closure of the i-uniform complete hypergraph K" on n vertices with i Proof The hereditary closure of K? is the union of the j-uniform complete hypergraphs with 0 < j < i. Theorem 1.4 holds for all of them, so we cannot have a larger intersecting subfamily than the (common) cardinality of the stars. On the other hand, the Hilton-Milner theorem implies that a maximum intersecting subfamily must contain exactly one star from each of the j-uniform hypergraphs. But these stars cannot be centered at different points, since every star contains the one-element set consisting only of the center of the star, and for different stars, these sets do not intersect. □ We can say even more about this case. The following result is found in the paper of Berge [2] and based on a theorem of Baranyai [1] about the factorization of the complete uniform hypergraph. Theorem 1.6 [2] The hereditary closure of the i-uniform complete hypergraph K" has the edge-coloring property if and only if there are integers rjk > 0 for j=l,2,...,i and k=l,2,...,l+(n ;')+ ... +(":!) such that: I j J rjk = n v k and 2 k rjk = (?) v J- n 10 1,4 Chvatal's theorem Chvatal's conjecture was suggested by his next theorem [5], Here he supposes that the hereditary hypergraph has an even stronger property. Suppose that now we have the set S={l,2,...,n}, and H is a hypergraph on S such that if AsH , if BcS and if there exists a one-to-one mapping f from B to A such that f(x) > x for every element x of B, then B e// also. A hypergraph satisfying this property called a strong hereditary hypergraph , or a hypergraph having the strong hereditary property . If X and Y are two finite subsets of the positive integers and there is a one-to-one mapping f: X —» Y such that x < f(x) for every xeX, then we write X < Y. In his paper Chvatal used a similar technique to the one that ErdBs, Ko, and Rado used for the proof of Theorem 1.4 (and which technique later was used by several authors to prove various results on extremal set theory) to prove the following result. Theorem 1.7 (Chvatal [5]) L e t // be a hypergraph on S={ 1,2,...,n} having the strong hereditary property and let M be an arbitrary intersecting subfamily of H . Then |M | £ |{H€/ / :1<=H}|. Proof The proof is by induction on n. The statement is trivial if n=l. Let now n be larger than 1 and suppose that the statement is true for every strong hereditary hypergraph having an underlying set of cardinality less than n. For every hypergraph H * c P(S) we assign a weight w( // *) = 2 H€// * 2k€H k-Since we are interested only in the cardinality of M , we may assume, without loss of generality, that M is one of the intersecting subfamilies of H having the minimum possible weight among all the intersecting subfamilies of cardinality |M |. We will prove, using the technique developed in [11], thatAf is "compressed to the left". That is to say, if M e M , t and s are two distinct elements of S (in other words l (M \{t})u {s} is also an edge of A / . Suppose the contrary, i.e., let s and t be two integers such that l< s < t< n and we have an M te M with te M ), st}M (, and (M|\{t})u{s}^M. We define M * , F *, and F in the following way. M * = {MeM : teM, s F * = {(M\{t})u{s} : MeM *}, and F = F *u (M \M *). 11 Here, we have that M|€Af *, thus M| (F\{s})u{t) is an element of M *, and since both (F\{s})u{t> and M are elements of the intersecting family M , then they must have a nonempty intersection. This, from the fact that F n M = 0, cannot be anything else but {t}. We have that M<=M , teM, s M (F \{s })u {t) and (M \{t})u{s} are elements of the intersecting family M , that is, 0 * [(M\{t})u{s}] n [(F\{s})u{t)] = (FnM)\{s,t>. This contradicts the fact that F and M are disjoint. Thus, we really have proved our statement, i.e., if M e M , t and s are two elements of S such that Is set sn and teM, s |X n {k,k+l, . . . ,n}|. This implies that the statements Y < X and S\X < S\Y are equivalent. Let H ,={XeH : S\Xe// }. Now, it is clear that if Xe// \H , and Y < X, then Yeff \H ,as well, that is the hypergraph H \H , also has the strong hereditaiy property. Indeed, if XeH \H (and Y YiH \HU that is, if Y is also an element of H u then, by definition S\Y <=H. But from 12 the above remark, S\X < S\Y and so H , being a strong hereditary hypergraph, contains S\X as well. This implies that XeH ,, so Xi\H \H ,, a contradiction. Define H 2, H 3, and H 3* as follows: H z = {XsH \// ( : n H 3={Xe// \H ,: neX}, and H 3* = {X\{n} : XeH 3}. If X«H 2 and Y < X, then, since X does not contain n, Y may not contain it either, and so Y zH 2 as well. Similarly, H \H | being strong hereditary implies that if X « // 3* and Y < X, then Y e // 3*. In other words, we divided our original hypergraph H into three parts, H ,, H 2, and H 3, such that H 2 has the strong hereditary property and it is defined on the smaller underlying set {l,2,...,n-l}. While for H z we defined H 3* such that it also has the underlying set {l,2,...,n -i} and has the strong hereditary property as well. Divide M similarly, that is, define M , = M r\H j for i=l,2,3, and let M 3* be {X\{n} : XsM 3}. Finally define F = {XzH : UX], jF j = F o f f , for i=l,2,3, and F 3* = {X\{n} : XeF 3}. Since M is a pairwise intersecting subfamily of H and A/t and M z are subsets of M, we have that both of them are pairwise intersecting themselves. We will prove that M 3* is also pairwise intersecting. I f X and Y are two elements of M 3*, then XuYc{l,2,...,n-l}, but XuY ^ {l,2,...,n-l}, since otherwise (Xu{n})u(Yu{n}) = {1,2,...,n} and so Xu{n} and Yu{n} would be elements of //, rather than elements of /Z 3. Thus, we have a k€{l,2,...,n-l} such that kcjX, k proven fact that since ncXu{n} but k is also an element of M . It follows then from the facts that k<}Y and n$X that XnY = (Xu{k})n(Yu{n}). But Xu{k} and Yu{n} are elements of M , and so X n Y = (X u {k})n (Y u {n » * 0, that is ,M 3* is also an intersecting family. We will now use the fact that H 2 and H 3* are strong hereditary hypergraphs and that they are defined on the smaller underlying set {1,2,..,n-l}. It is easy to see that {XeHz : leX} = F 2 and {XeH 3* : 1«X} = F 3*. Since M z and M 3* are intersecting subfamilies of H z and H 3*, respectively, then by induction, we know that 13 |M z\ <, |F 2| and |M 3| = \M 3*| ^ |F 3*| = |F 3|. At the same time, if XsH ,, then, by definition, S \X e H ,, which implies that M |, being pairwise intersecting, has cardinality at most {\H ,|. On the other hand, F , contains every set from H , containing 1, thus exactly one of X and S\X is in F ,, and so \F 11= ±\H , |. These imply that \M 1 1 < \F , {• Putting these inequalities together we obtain that \M | = \M , |+|M z\+\M 3| < |F , |+|F 2|+|F 3| = |F |, which is what we aimed to prove. □ We note that this theorem is really a special case of Chvatal's conjecture. It is obvious that a strong hereditary hypergraph has the hereditary property as well. Chvatal's theorem says that for every intersecting family M in a strong hereditary hypergraph H , defined on the set S={l,2,...,n}, \M (1) holds, and thus|M | < d(H ). In particular, this result also gives that in the strong hereditary hypergraphs 1 is one of the points of the underlying set having the maximum degree. Beyond Chvatal's conjecture, there could be another generalization of this theorem: Let H be a strong hereditary hypergraph on the set S={l,2,...,n} and let M be a subfamily of H such that \M \ > k and M does not contain k+1 pairwise disjoint sets. Then |M |< \{X {l,2 ,...,2 k + l}. Then \{XeH : {1, 2,. .., k} n X * 0}| = 22k+1 - 2k+1, while the family M = { X c { l, 2 , , 2k+l} : |X| > 2} obviously does not contain k+1 paiwise disjoint sets, but its cardinality is 22k+1 - (2k+2) > 22k+1 - 2k+1. Later on, D. L. Wang and P. Wang [24] found a more general result. In order to state this we need some definitions. Let T be a tree , i.e., a finite connected simple graph without cycles. Let r be a specified vertex of T, called a root, and x,, x2, ..., xn the list of the other vertices of this tree, called sometimes nodes . We will call the vertices of degree one leaves in this tree, with the exception of r, which is not a leaf, even if it has degree one. If v is any node of T, then there is a unique path from r to v, which will be denoted by [r,v]. Here, we represent a path by the sequence of the vertices in the path, thus [r,v] is an ordered subset of the set V(T), the set of the vertices of T. The path [r,r] is defined to be {r}. We define a partial ordering on the set V(T) in the following way. If x 14 and y are any two vertices, then x < y if and only if [r,x] c [r,y]. In particular, r is the only minimal element of this partial ordering. A hypergraph H will be said to have a hereditary tree structure if and only if H is a hereditary hypergraph and there exists a tree T with root r such that H is defined on the vertex set of T and the following conditions are satisfied: (i) Every edge of H is contained in [r,aj] for some leaf a| of T. (ii) If {xb ..., xm} is an edge of H , and there are yf's such that yf < x( for 1< i in the above defined partial ordering, then {y ,,..., ym} is also an edge of H. It is easy to see that a strong hereditary hypergraph also has the hereditary tree structure, where the tree is a path, and thus the following theorem is really a generalization of Chvatal's result. Theorem 1.8 (D.L. Wang and P.Wang [24]) If the hypergraph H has a hereditary tree structure, then Chvatal's conjecture is true for it. □ The proof of this theorem is very similar to the one used by Chvatal to prove Theorem 1.6. In particular, it also uses a "compression" technique, and then proves the statement by induction on the number of nodes, leaving out a leaf each time. .1,5 Berge's theorem In a hereditary hypergraph there is an almost trivial upper bound for the degrees of the points of the hypergraph and so for the maximum degree of the hypergraph. Proposition 1.9 If H is a hereditary hypergraph defined on the underlying set S, then for every xeS d// (x) Proof For every xeS and H eH such that xcH, H , being hereditary, also contains H \{x}. Since for different H, and H 2, such that xeH, and xeH2, H ,\{x) and H 2\{x } are also different, and obviously H \{x) is not counted in the degree of x, then it immediately follows that the degree of x in H cannot be larger than {\H |. The second part of the statement is obvious now. □ If Chvatal's conjecture is true, then a similar statement has to be valid for the maximum cardinality of the pairwise intersecting subfamilies of a hereditary hypergraph. 15 This was shown independently by Berge [2] and Schbnheim [19]. Theorem 1.10 If H is a hereditary hypergraph and M is an intersecting subfamily of H , then |M \ < i\H |. Schonheim's proof uses the following result of Erdos et al. [10] and, independently, Marica [15]. Lemma 1.11 If H = {A (, A2, ..., An} is a hereditary hypergraph, then there exists a permutation a of {1, 2 ,..., n} such that A, c A a(i) for every i€ {l, 2, ... , n}. □ We do not prove this result here, since, as we will see in Chapter 2, Berge's result is an immediate generalization of this one. However, with the help of this lemma, it is easy to see that Theorem 1.10 holds, since a pairwise intersecting subfamily of H cannot contain both A f and Ac^ . Berge proved a stronger result. We will discuss the relation of these results in Chapter 2, together with another proof of Berge's theorem. Here, we give the original form and proof of it, which requires some definitions. With every hypergraph H we will associate a graph G(H ). The vertices of G (H) represent the edges of H , and two vertices V) and v2 are joined in G(H ) if and only if they represent two disjoint edges H, and H 2 of H . In other words, G(H ) is the complementary graph of the usual line-graph of the hypergraph H . We will use the usual concept of matching of a simple graph G. That is, a subset of pairwise disjoint edges of a simple graph is called a matching. The vertices which are the endpoints of the edges of a matching are said to be saturated by the matching. A matching which saturates all the vertices of a graph (when the cardinality of the vertices is even) is called a perfect matching . With the help of these concepts we can state Berge's theorem now. Theorem 1.12 [2] If H is a hereditary hypergraph, then G(H ) has a perfect matching, whenever |H | is even. If \H | is odd, then G(H ) has a matching which saturates all the vertices, except the one corresponding to the empty set 0 . Proof In general, we want to prove the existence of a matching which saturates all the vertices of G(H ) with the only possible exception being 0 . Indeed, if |jfiT | is even, then a matching is either perfect, or misses at least two vertices of G(H ), which cannot be the 16 case. We will say that such a matching is almost perfect We will prove the theorem by induction on the number of edges of the hypergraph. If the hereditary hypergraph is empty or has one edge (the empty set), the statement of the theorem is trivially true. While if it has two edges, i.e., G(H ) has two vertices, 0 and the set, let us say, {1}, the required matching is [0 ,{ 1}]. Similarly, for the hereditary hypergraph of three edges 0 , {1}, and {2}, the required matching is [{1},{2}]. It is easy to see that this is the only hereditary hypergraph (up to isomorphism) of three edges, since if one contains an edge of more than one point, then its cardinality is at least 4. Assume that the statement of the theorem is true for every hereditary hypergraph with less than n edges, and let H be a hereditaiy hypergraph, defined on the underlying set S, of n edges. Let x be any point of S such that {x} e H . (In general, we may suppose that for every xeS, {x}eH as well, since otherwise we could delete x from S resulting in a smaller underlying set of the same hypergraph.) We will divide the hereditary hypergraph H into three parts. Let {x}=K (, K 2, ..., Km be those edges of H which contain x. Let M ( = Ki\{x} for every 1 < i < m, then the sets M, for 1 < i < m are also elements of H . Finally, let N (, N 2, ..., N t be those elements of H which do not contain x and for which N)u{x} is not an element of H . Write H in the form of K uM uN , with K ={K,, K 2, ... , Km}, M = {M „ M 2, ... , M m}, and N ={ N„ N 2, ... , N t }. H , = M uN is the collection of those edges of H which do not contain x, and thus, it can easily be seen, that it is a hereditary hypergraph as well. We also know that |AT | > 1, and so, by the induction hypothesis, G (H ,) has an almost perfect matching, which we will denote by F (. The hypergraph H 2 =M is also hereditary, for if M 'cM eA f, then M u { x } « fl , and thus , which implies that M '<=M. By the induction hypothesis, H 2 also has an almost perfect matching, which will be denoted by F 2. Note that the set M ,= 0 belongs to M , and thus, obviously, to both of the hypergraphs H , and H 2. It may be saturated or unsaturated by both F , and F 2. Since the graphs G(H 0 and G (H 2) are defined on subsets of the vertices of the graph G(H ), the subsets of the edges of G (H ,) and G(H 2), F , and F 2, respectively, define subgraphs of G (H ). Consider the partial graph of G(H ) (i.e., a multi-graph, 17 defined on the vertex-set of G(H ) whose edges are edges of G (H ) ) with the edge-set of F ivF 2. Note that an edge of G (H ) may appear in both F | and F 2, and then it is represented twice in this partial graph. Since every vertex of the graph G(H ) is covered by at most one element of F t and at most one element of F 2, the maximum degree of the partial graph generated by F ,uF 2 is at most two. Thus, each connected component of this partial graph is either an isolated vertex, or an edge, or a path of length at least two, or a cycle, including the cycles of length two, i.e., two parallel edges. If the connected component is a path or a cycle, then its edges must belong alternately to F | and F 2. We will now define the almost perfect matching F of the graph G(H ). With every connected component of the partial graph generated by F |uF 2 we will associate some edges to be included in this required almost perfect matching. We will consider different cases, according to the nature of these connected components. Case 1 The connected component is a single vertex . This single vertex can only be associated with the edge M (= 0 of the hypergraph, and only in the case when it is unsaturated in both F t and F 2. We will take the edge [0,{x}] into the almost perfect matching F. Case 2 The connected component consists of a single edge IN-.N^ where Nj and N. are elements of N and so this edge belongs to F ,. (From now on we will identify an edge of H with the associated vertex of G( H )). In this case we will take the edge [N-,Nj] into F. Case 3 The connected component is a path 7t. which mav consist of only one edge. but not like Case 2. Its edges belong alternately to F ( and F 2. In this case we know that at least one of the endpoints of the path is in N , fo r, if M, and M 2 are the endpoints, then both of them are in the hereditary hypergraphs H , and H 2, and at most one of them is the empty set. Then the other is saturated both by F , and F 2, and the degree of it is two in the graph generated by F 2, a contradiction. On the other hand, the vertices of the path having degree two may only belong to M , since the vertices belonging to N have degree only at most one in the partial graph generated by F tuF 2. Thus, we know that the path is either 18 it = {N,, M jj, M,2, ... , M ik, Nj} or 7t = (N,, M,2, ... , M|k = 0}. In the first case, we will take the edges [Nt , K fl], [M fl, K )2], ... , [M ik j, K ik], [Mjk, Nj] into the required almost perfect matching F , where K, = M ju{x}. In the second case, F will not include the last edge, and thus the vertex M ik=0 will remain unsaturated in the almost perfect matching F . Case 4 The connected component consists of a cvcle of length two, that is an edge which belongs to both F , and F ?. In this case the endpoints of this edge have to belong to the hereditary hypergraph H 2, and thus the edge is of the type [M j, Mj]. Now, since the edge [M j, Mj] is in the matchings F ( and F 2, M ;nMj = 0 , when we also know that none of M j and Mj contain x. So if we denote M ju{x} and M ju{x} by Kj and Kj, respectively, then M ;nKj = 0 and MjnKj = 0 hold as well. Thus, [Mj ,Kj] and [Mj ,Kj] are both edges of the graph G (H) and we will include them among the edges of the almost perfect matching F . Case 5 The connected component is a cycle whose edges belong alternately to F , and F 7. Here, the vertices of the cycle are of degree two in the partial graph generared by F , and F 2, and thus they must belong to the hereditary hypergraph H 2 = A f. Thus the cycle is in the form {M ^ , M i2 , ... , M f)c, M fl}. We will include the edges [M fl, K i2] , [M j2 , K l3] , .. ., [M(k , K,j] among the edges of the almost perfect matching F , where again K fj denotes M (jU{x}. We have now to prove that the resulting set of edges F of the graph G(H ) is an almost perfect matching, i.e. every vertex of G(H ) is an endpoint of exactly one edge of F, with the possible exception of the set 0 , which is the endpoint of at most one edge. Since H was decomposed into K , M , and N , it is enough to show that the above statement holds for the sets belonging to K , M , or N . Let KjeK andM f=Kj\{x}. Then either M j=0, and then, by Case 1, M ( is in one of the edges of F, or M ^ 0 , and then M , belongs to a connected component of one of the Cases 3, 4, or 5. In all of these cases, for every Mj the corresponding Kj=MjU{x} was defined to be in an edge of F . On the other hand, since every M j is in exactly one connected component belonging to one of the Cases 1, 3, 4, or 5, we defined at most one edge of F going through K t. Similarly, if Mj is an element of M , then one of the Cases 1, 3,4, or 5 holds for it. In Case 1, though M ,=0, it is defined to be the endpoint of an edge of F . In Cases 4 and 5, no matter what is the cardinality of M (, we defined one edge of F going through it. Finally, in Case 3, unless M t=0, we also found an edge from F having as one of its endpoints M (. Again, exactly one of these cases holds for M f, thus, we did not define more than one edge going through it. Finally, if Nj is an edge of the hereditary hypergraph from, the set N , then exactly one of the Cases 2 or 3 holds. In Case 2, we did not even change the edge from F ,, thus F will also contain this edge going through N f. In Case 3, we also preserved N ( as one of the endpoints of the resulting edges in F . Thus, for every edge of the hereditary hypergraph H we have at most one edge of F going through it (and for the non-empty edges at least as well), and thus the statement of theorem is proven. □ In other words, Berge's result states that for every hereditary hypergraph H we can partition the edges of H into pairs (with the exception of the set 0 , if |H | is odd), such that the intersection of the sets belonging to the same pair is empty. Now, we may include at most one set of every pair in a pairwise intersecting subfamily of H , hence its cardinality cannot be larger than the half of the cardinality of H . Thus, we have another proof of Theorem 1.10. With the help of this theorem we can prove a special case of Chvatal's conjecture, which is a generalization of the following, earlier results. Theorem 1.13 (Sterboul [23]) Let H be a hereditary hypergraph such that the number of maximal sets in H is at most 3. Moreover, suppose that the intersection of these three bases is not empty. Then Chvatal's conjecture holds for H , i.e., the cardinality of every 20 pairwise intersecting subfamily of H is at most d(H ). Theorem 1.14 (Sterboul [23]) Let H be a hereditary hypergraph defined on the underlying set S such that the bases of H form a simple star , i.e., there exists a KcS, such that if M ,, M 2, ..., M n are the bases of H , then MjnMj=K for every 1 < i < j < n. Then the cardinality of every intersecting subfamily of H is at most d(H ). In the generalization of these two results, we do not suppose any restriction on the number of the bases of H , and we only suppose that the intersection of these edges of H is not empty. Theorem 1.15 (Schonheim [20] and D.L. Wang and P. Wang [24]) If the intersection of the bases of a hereditary hypergraph H is not empty, then Chvatal's conjecture holds for H. In fact, we can say even more about this case, but the result in its full strength will be stated in Chapter 4, where we will use it, and give a generalization of it as well. Here, we give Wangs' proof, which uses the following lemma of Kleitman and Magnanti [14]. Lemma 1.16 [14] L etF ) and F 2 be two hypergraphs defined on the underlying set S such that each set in F , has a non-empty intersection with every set of F 2. Let F Lj ={ AcS: AcB for some B«F {and A$F .} for i=l,2, and let ft, f2, f1^, and fLz denote the cardinality of F ,, F 2, F L,, and F L2, respectively. Then the following inequality is satisfied: fS fS > f| f2. □ Proof of Theorem 1.15 Let M be a maximal intersecting family in H . Since we suppose that B |n B 2n ... n B n ^ 0 , where B,, B2, ... , B n are the bases of the hereditary hypergraph H , we have an x€B,nB2n ... nBn. Define M 0 = {HeM : x M | = {HeAf : xsH} and M 2 = [H\{x}:H€M (}. If M is a maximal intersecting family, then M, c M ze H , M ,eM implies that M zsM as well, as it can be easily seen. Thus, in this case,M 0cM 2, which implies that (*) fo + fLo c f, + fL, where f, and fL, denote the cardinality of M 0 and M L0, respectively, while ^ denotes the 21 common cardinality of M , and M 2, and denotes the common cardinality of M L( and M L2. Here, by Lemma 1.16, we have that f^o f^( > f0 f,, which implies that either fL0 > f| or f^ > f 0. But if fL0 £ f/, then from (*) we have again that f0 < f1^, that is, we have f0 ^ f^| in any case. Define G in the following way. G = {Hu{x}: HeAf , or HeM L,} = {Hu{x}: HsMeM Then G is a star inH and |G |= f, + ^ f( + f0 = |M |. This is exactly the required result. □ It is immediately obvious that Theorem 1.13 and Theorem 1.14 are really special cases of Theorem 1.15, if the kernel K of the simple star is not empty in the latter case. On the other hand, if every two bases of a hereditary hypergraph H have empty intersection, then every intersecting family of H may contain subsets only one of the bases, which easily implies the statement of Theorem 1.14 in this case too, by Proposition 1.2. To finish this section, we mention another corollary of Theorem 1.10. Daykin asked the following question: If M sP(S) is a hypergraph defined on the underlying set S of n elements such that for every two edges M t and M 2 of M M)nM2 ^ 0 and M,uM2 * S hold, then what is the maximum of \M |? If we fix two elements x and y of S and consider the hypergraph M ={HsS: xeH, y<]H}, then M satisfies the above properties and \M \ = 2n'2, which shows that max \M \ > 2n'2. Daykin and Lovasz [7], Kleitman [12], Seymour [17], and Schonheim [18] independently showed that max |M \ = 2n"2 by proving the following result. Theorem 1.17 If S is a finite set of cardinality n and M AP(S) such that for every two edges M , and M 2 of M , M tnM 2 * 0 and M ,uM 2 * S hold, then |M | < 2n"2. Proof The proof we will give here is essentially SchOnheim's proof in [18]. Let M be any hypergraph with the required properties, and consider the hereditary closure of M , which we will denote by H . Then, obviously, for any two edges H| and H 2 of H we still have that H (uH2 ^ S. It implies, by a very similar argument to the proof of Proposition 1.2, that |H | < 211'1. But now M is an intersecting subfamily of the hereditary hypergraph H , and thus \M\ < i\H \ = i 2n'1 = 2n'2. □ 22 1,6 Sterboul's theorem A finite projective plane is a hypergraph P defined on the finite set S such that the following properties are satisfied: (i) Every PeP has the same, fixed cardinality. (ii) The intersection of every two distinct edges o fP has cardinality one. (iii) For every two elements x and y of S, there is exactly one element of P containing both x and y. (iv) There are four elements of S such that no three of them are contained in any edge of P . The edges of a finite projective plane are called lines. We note here that there are known finite projective planes only if |S| is of the form q2+q+l, where q is a prime power, and the number of lines of a finite projective plane is always equal to |S|. One of the first questions raised about Chvatal's conjecture was whether one can prove it for the hereditary hypergraph which is obtained as the hereditary closure of the lines of a finite projective plane. Sterboul's following theorem [23] answers this question. Theorem 1.18 (Sterboul [23]) Let H be a hereditary hypergraph defined on the finite set S satisfying the following properties: (i) Every base of H has the same fixed (b > 3) cardinality. (ii) The intersection of every two bases of H has cardinality at most 1. (iii) There are two bases of H with intersection of cardinality one. Then Chvatal's conjecture holds for H . Moreover, in this case every maximal intersecting family of H is a star. Proof Let M be an intersecting family of H which is not a star. We have to prove that there is a star in H which has cardinality larger than |M |, and thus we may suppose that for every H e // \M , M u{H} is not intersecting any more, i.e., M is not contained by any other intersecting family of H . It implies that for any M e M , M cN, N e // , NeM as well, i.e. every maximal set of M is also a maximal set, a base of H , and thus, has cardinality b. Let n denote the number of the maximal sets in M and let M b M 2, ..., M n denote these maximal sets. In this section, we will denote (x) by d(x), since it is clear which is the hypergraph we mean. Let f be the function defined on S such that for an xeS f(x) = |{M eM : xeM and M is a maximal set of M }| and let k(i) denote the number 23 KxeMj: f(x)>2}| for every maximal set o fM . Let F denote the maximum of the function f on S. For every maximal set M ( of M let x,, x2, ..., xk^ be those points of M f where the function f takes value at least two. Here, M is an intersecting family, thus, in particular, every maximal edge of M intersects M (. From the assumptions of the theorem, it follows that these intersections have cardinality one. Hence, we can count the number of the maximal edges of M , which was denoted by n, in the following way, for every M f. One of them is M f itself. All of the others intersect M ( in exactly one element This element is contained in M ( and in the other maximal set of M , therefore, the function f takes value at least two on it. On the other hand, for every xeM t satisfying f(x) > 2 there are exactly f(x)-l maximal sets of M , excluding M (, going through this point. Collecting these facts together, and using that f(x) < F for every xeS, and thus, in particular, for every xsM, we get that n = 1 + {f(*0-l} + {f(x2)-l} + ■ • • + {f(xk(i) )-l} < 1 + k(i) (F - 1). This implies that (1) k(i) > (n - 1)/(F - 1). Let M| be any maximal edge of M and N be a subset of M ,. If N«M as well, then N has to have non-empty intersections with the other maximal edges of M . For every xmcM |, 1 < m < k(i), such that f(xm) > 2, there is another maximal edge of M intersecting M ( exactly in this xm, thus, N must contain xm. On the other hand, if N contains all the elements Xj, x2, ..., xk^ , then, since every edge of M intersects Mj in one of the points xm, 1 < m < k(i), every edge of M intersects N as well. The uniform cardinality of Mj's is denoted by b, and so, using the previous fact, we have that there are exactly 2b'kW subsets of M (, including M ( itself, in the intersecting family M . We assumed that M is not a star, i.e., it does not contain any edge of cardinality one, and the intersection of every two bases of H has cardinality at most one. Thus, every edge from M is in exactly one of the maximal sets of M. We have that (2) |M | = 2?=. 2b-k. 24 Let x be a point of S where the function f takes its maximum, that is, f(x)=F. Then x is in F maximal sets of M , each having cardinality b. Thus, all of these maximal sets of M have 2b_1 subsets containing x, and so 2b' 1-l subsets of cardinality at least two which contain x. The pairwise intersection of the above maximal sets of M is {x}, which implies that no set of cardinality larger than one can be contained by two or more of them. This gives 1+F(2b_1-1) edges of H containing x. Thus, we get that d (H ) > d(x) > 1+F(2b'1-1), and so to prove the theorem it is enough to prove that \M | < F(2b_1-1). We assumed that M is not a star which implies, in particular, that for every maximal set M j of M k(i) > 2. We have that | M \ = 2"_t 2b'kW, where b-k(i) > b-2 for every 1 < i £ n, and so |Af | ^ n 2b'2. If n < F(2-22'b), then \M \ <> n 2b’2 < F (2-22’b) 2b'2 = F (2b'1-l), and so we have the required result. Suppose that n > F (2-22"b), which implies that there is a positive integer m such that (3) F 2m (l-21_b) < n < F 2m+1 (l-21"b). Using (1) and the left hand side of (3) we get that (4) k(i) £ (n-l)/(F-l) > { F2m (l^ 1’1’) -1 }/(F-l) = (F 2m-F 2 m21'b-1 )/(F-l) = { (F-l) 2m + 2m - F 2m 21_b -1 }/(F-l) = 2m + (F -l)’1 (2m - F 2m 21_b -1). If 2m-F 2m 21_b-l is non-negative, then k(i) > 2m +1 > m+2 for every 1 < i < n, since m is a positive integer. Now by (2) we have that \M \ < n 2b‘(m+2), which together with the right hand side of (3) imply that \M | < F 2m+1 (l-2 1‘b) 2b-(m+2> = F (2b_1-l), which is the required result. We will prove that if M is not a star, then F < b. Suppose otherwise. This means that there is a point x of S such that there are F, i.e., more than b, maximal edges of M containing this point. Let these maximal edges be denoted by M|, M 2 M p. From the assumptions of the theorem, it follows that the pairwise intersection of these maximal edges is exactly {x}. M is not a star, thus there is another maximal edge of it, M ' which does not contain {x}, but intersects all of the sets M |} M 2, ... , M p. Here M* cannot intersect two of these M j’s in the same point, and so M' contains at least F>b elements, a contradiction. 25 Let 2m - F 2m 21'b - 1 be denoted by D. As we have seen, D > 0 implies the aimed result. If b > 4, then b 21'b < £, thus, 2m- F 2 m 21_b-1 > 2 m- b 2 m 21_b -1 > 2 m _ 2m‘ l - l ^ 2m_1 -1 > 0 , we are done. We assumed that b ^ 3, and so the only remaining case is if b=3. M is not a star, thus F > 2, and now F < b implies that F= 2 or 3. If b=3 and F=2, then D = 2m - F 2m 21_b -1 = 2m -2 2m 2‘2 -1 = 2m_1 -1 ^ 0 , again. The only case we still have to handle is when b = F = 3. I f n < F (2-22‘b) =4.5, then, as we previously saw, we are done again. Since b=3, k(i) < 3 for every 1< i < n, which, together with (1), imply that n < 7. We want to show that \M | <, F (2b' 1-l) = 9. Suppose otherwise, that is, that |M | S 10. We know that M contains 5, 6, or 7 maximal edges, i.e. sets of three elements, and thus at least 3 sets of two elements. An easy, but long investigation of the cases shows the impossibility of this situation. n The last condition of the theorem, i.e., the existence of two bases of H with non-empty intersection can be omitted, since otherwise the same conclusion follows by Theorem 1.14. This theorem obviously answers the question about the finite projective plane, since by definition, the lines of of a finite projective plane satisfy the condition of the theorem. 1.7 Some other results In this section, we will list the other existing important results concerning Chvatal's conjecture without proofs. First, we mention two other results of Sterboul [23]. We will say that the rank of a hypergraph H is the maximum cardinality of the edges of the hypergraph, and we will denote it by r (H ). Theorem 1,19 (Sterboul [23]) If H is a hereditary hypergraph of rank 2 or 3, then Chvatal's conjecture holds for it □ In fact, Sterboul proved something more. We say that an intersecting family M of a hypergraph H is a clique if there is no other intersecting family of H which strictly containsM . The cliques of a hereditary hypergraph of rank two (i.e., of a graph) are only the stars and the triangles, i.e., three edges of the type {a,b}, {b,c}, and {c,a} (which result is well-known and very easy to prove). Sterboul [23] lists the maximum cliques (i.e., cliques of maximum cardinality, which are exactly the maximal intersecting subfamilies) of the hereditary hypergraphs of rank 3, proving the above theorem for 26 hypergraphs of rank 3. Here the maximum cliques are the stars, or one of the following types: - {{1,2}, {1,3}, {2,3}, {1,2,3}} - {{l,2,x,}, ... , {l,2,xn}, {1,3.x,}, ... , {l,3,xn}, {2,3,x,}, ... , {2,3,xn}, {1,2} {2,3}, {1,3}, and eventually {1,2,3}} - {{1,2}, {1,3}, {1,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}}. The proof of the previous Theorem 1.18 can be finished now with the help of this list, since the last, exceptional cases were the cases where b=3, i.e., the rank of the hereditary hypergraph is 3. Schanheim and Stein [22] proved a generalization of the previous result of Sterboul, of Theorem 1.13. Theorem 1.20 If H is a hereditary hypergraph such that the number of bases is three, then Chvatal's conjecture holds for it. □ In the same paper they proved the next result, which was also proven independently by D. L. Wang and P. Wang. Theorem 1.21 ( Stein and Schanheim [22], Wang and Wang [24]) If H is a hereditary hypergraph and any three bases of H have empty intersection, then Chvatal's conjecture holds for it. □ A similar result, which is another generalization of Theorem 1.14, can be found in a paper of Stein [21]. Theorem 1.22 If H is a hereditary hypergraph, the number of bases of H is n, but n-1 bases ofH form a simple star, then Chvatal's conjecture is true for H . □ The proofs of the previous two theorems are very technical, involving lots of counting, and are quite similar to one another. A similar technique was used by Gritsak to prove an important special case of the conjecture. We say that a hereditary hypergraph H is of matroid type , if for every H t, H zsH there is an H 3zH such that H ,n H j* 0 and H 2u H 3* 0 . Theorem 1.23 (Gritsak [12]) If H is a hereditary hypergraph of matroid type, then Chvatal's conjecture holds for it. □ A hereditary hypergraph H is called matroid, if the cardinality of the bases of H is uniform and for every two edges H| and H 2 of H with |H, | < |H2|, there is an X€H2 27 such that as well. We mention, without proof, that this theorem answers the question of several mathematicians, whether Chvatal's conjecture holds for matroids, since a matroid is of matroid type. Finally, we mention a result of Kleitman and Magnanti, for its similarity to the previous results. Theorem 1.24 (Kleitman and Magnanti [14]) Let X! and x2 be two elements of the finite set S and let M cP(S) be a pairwise intersecting hypergraph on S such that every edge of M contains at least one of x, and x2. Then there is another pairwise intersecting hypergraph M 'cP(S) such that the followings are satisfied: (i) M eM ' 3 M ,eM such that M s M t (ii) \M'\2\M\ (iii) Either x, is contained by every edge of M ', or x2 is contained by every edge of M '. □ We note, that a similar statement for three different elements x,, x2, and x3 of S is not true. Let S = {xt, x2, x3, x4, x5, x6} and M = {{Xj,X 4 ,x5}, {xbx4,x6}, {x,,x5,x6} : i = 1,2,3}, then \M |= 9. On the other hand, as it can be checked, every intersecting hypergraph M' on S satisfying properties (i) and (iii) above has cardinality at most 7. (But there is an M ' of 12 elements which satisfies (i) and every edge of M ' contains x4.) Theorem 1.24 gives Chvatal's conjecture in a rather abstract special case. Namely, if H is a hereditary hypergraph defined on the finite set S such that for every intersecting subfamily M cH there exist two points x, and x2 of S satisfying that every edge of M contains either x, or x2 (or, eventually, both), then Chvatal's conjecture is true for this H , as it immediately follows from the above theorem. CHAPTER 2 ANOTHER PROOF AND COMBINATORIAL GENERALIZATIONS OF BERGE'S THEOREM 2.1 Introduction The main result of the next chapter, Theorem 3.1, is a generalization of Berge's theorem for distributive lattices. We will need some terminology, which partly will be used in this chapter too, in order to be able to state that result. If P is any partially ordered set and D is a set of points of P such that a Similarly, if U is a set of points of P such that a ordered by inclusion, and acS then the complement S\a of a is denoted by a here, and if AcP(S), then A denotes the set {a:ae A}. We restate Berge's theorem here using this new terminology. Theorem 2.1 Any down-set D in P(S) is the disjoint union of pairs of disjoint subsets of S, together with the set { 0 } if |D| is odd. In other words, if |D| is even, D can be partitioned into pairs of subsets of S, such that for every pair {a,b}, anb = 0 holds. If |D| is odd, we have the similar partition of D \{0 }. This theorem, as it was mentioned earlier, can be considered as a generalization of the result due to Erdbs, et al. [10] and, independently, Marica [15], that for any down-set D of P(S) there is a bijection 9:D —> D such that dc9(d) for all de D. Berge's result is equivalent to demanding that the bijection 9 has the additional property that 9(9dj=d 28 29 (V de D). The following two examples illustrate this, giving the pairings of D mentioned in Theorem 2.1 as well as the bijection D -» D. First let D = {0 , {1}, {2}, {3}, {1,2}, {1,3}} defined on the set S={1,2,3}. Then D is the union of the pairs of sets (which are all disjoint from each other) shown on Table 1, and at the same time we have the correspomding bijection f as well. Figure 1 shows the down-set D and the pairings from its elements. Table 1 Pairings from D and the bijection f pairs deD f(d )e D {{2}, {1,3}} {2} {2}={T;37 {1,3} {1,3}={27 {{3}, {1,2}} {3} {3 }={T,27 {1,2} {1»2}={37 {0 , { ! } } 0 {2,3}={T7 {1} {1>2,3}=0 (1,3} {1,2} {1} {2} {3} 0 Figure 1 Pairings from D Table 2 Pairings from D' and the bijection f ' pairs deD' {>(d)e D 1 {{!}, {4,5}} { 1} {1,2,3}={^“3T {4,5} {2,3,4,5}={T7 {{4}, {1,2,3}} {4} {4,5}={r,2;37 {1,2,3} {1,2,3,5}={47 {{1,2},{3,5}} { 1,2} {1,2,4}={3}57 {3,5} {3,4,5}={I}27 {{2},{1,3,5}} {2} {2,4MT-,^57 {1,3,5} {1,3,4,5}={27 {{5},{1,4}} {5} {2,3,5}={Tj47 {1,4} {1,2,3,4}={57 {{1,5}, {2,3}} {1,5} {1,4,5}={T,?7 {2,3} {2,3,4}={T}57 {{3}, {1,2,4}} {3} {3,5}={T,T,*1 {1,2,4} {1,2,4,5}={37 {{1,3}, {2,4}} {1,3} {l,3,5}={i;47 {2,4} {2,4,5}={T;37 {0} 0 {1,2,3,4,5}=0 {1,3,5} {4 {2,3} { 1, {3} {5} 0 Figure 2 Pairings from D' 31 The next, more complicated example illustrates the case when |D| is odd and thus the empty set is not included in any pair. Let D'= {0 , {1}, {2}, {3}, {4}, {5}, {1,2}, {1,3}, {2,3}, {1,4}, {2,4}, {1,5}, {3,5}, {4,5}, {1,2,3}, {1,2,4}, {1,3,5}}, defined on the set S={1,2,3,4,5}. Similarly to the previous example, the pairs from D ’ (in this case together with the singleton {0 }) and the corresponding mapping 9’: D' —»IT is shown on Table 2 on the previous page. Figure 2, on the same page, illustrates this down-set D' and shows how we form the pairs from its elements. Note again that, since |D| is odd, the empty set stands alone, it is not paired. 2.2 Another proof of Berge's theorem Here we prove Theorem 2.1 (Berge's theorem) in another way, which is just the specialization and simplification of the proof of Theorem 3.1 to P(S) and so it may help to understand the main ideas of the second proof. This is the main reason why we use the terminology of Chapter 3 here, though otherwise it would not be necessary. This proof is a common generalization of Berge's result [2] (Theorem 1.12) and the result of Erdos, et al. [10] and, independently, Marica [15] (Lemma 1.11). The method of the proof seems to be simpler than the one Berge used. The induction step used here will appear in a more complicated form in Chapter 4 as well. This simpler version of it makes easier to understand the main proof of that chapter. We mention that the forthcoming proof can be found in [7]. Proof of Theorem 2.1 We will prove that there is a bijection 9: D -» D such that d c fd and 9(93)=d (V deD ) using induction on the size of D. Once we have this bijection 9 we will form the pairs {d, 93}. The union of these pairs is really D. They are also disjoint, since if d2 is the pair of d(, i.e. d2=9 3 ], then by the property of 9 d|=9(93])=9cT^ and so d| is the pair of d2. The only possibility to have a singleton, i.e. a set d such that d=9cl is if d=0, since in this case both dc9d and dc93 have to hold. Thus the existence of this bijection implies the theorem. Let e be a subset of S of minimum size such that duee B for some de D. We have such an e, since S satisfies 0u S = S = 0eB . Let A ={deD :dueeB }. By the choice of e, A is a non-empty set. We proceed via some lemmas. 32 Figure 3 For the proof of Lemma 2.2 Lemma 2.2 D\A is a down-set of P(S). Proof Let dcd'e D\A, then d' and d are elements of D and d'g A. We need to prove that d£ A. If de A, then by definition duee D , and since here D is an up-set and d'ueDdue, d'uee D as well. This implies that d'e A, which is a contradiction. Thus, dg A, and so D\A is a down-set. o We note the following lemma which points out why this induction proof works and will be used in Chapter 4 as well. Lemma 2.3 If de A then dne=0 and (S\e)\d is also an element of A. Thus, A can be partitioned into pairs of subsets of S such that the sets from a pair form a partition of S\e into two parts. Proof If for a de A d n e *0 holds, then d and e\d satisfy du(e\d)=duee D and e\d is a proper subset of e, which contradicts the minimality of e. Hence, for every de A d n e = 0 holds, de A implies, by definition, that d u e e D , i.e., Da H ue = S\(due) = (S\e)\d. We also have that {(S\e)\d}ue = S\d = 3 e D, which together imply that (S\e)\d is 33 really an element of A. Having these results we can write A in the form {{d, (S\e)\d}: de A}. The pairs (d, (S\e)\d} and {(S\e)\d, (S\e)\[(S\e)\d)]} are the same, since dne=0 implies that (S\e)\[(S\e)\d)]=d. A is partitioned into the pairs {d,(S\e)\d}, and now dn(S\e)\d=0 and du(S\e)\d)=S\e imply the last statement of this lemma. □ We finish the proof of Theorem 2.1 now. Since D\A is a proper subset of D and D\A is also a down-set by Lemma 2.2, we may suppose by induction that there is a map 9 from D\A to I5\A such that if de D\A, then dc9d and 9(93)=d. Define 9 by 9d=due, if de A, and 9d=9d, if deD\A. Then 9 easily seen to satisfy dcfd. If de A, then f 3=3uee A, since (3ue)ue=3eB and thus 9(93")=93ue= (3uejue=3. IfdeD\A, then 9 satisfies 9(931=3 by induction, and so 9 satisfies the required properties. □ 2.3 Some combinatorial generalizations of Berge's theorem In this section we prove three generalizations of Berge's theorem. These generalizations turn out to be no more than corollaries of the theorem. In the first one we require that the two sets of each pair in the partition of the down-set intersect in precisely k elements. Theorem 2.4 [7] If k>0 is an integer and D is a down-set of P(S), then D is the disjoint union of two sets E,F such that (i) E is a down-set and ae E => |a| such that |anb|=k. Proof We iterate the following procedure. Let xcS and |x|=k and suppose 3 ae D, xca, x^a. (If we do not have an xcS and aeD with these properties then |a| < k for every ae D, and so D=E.) Let G={a:aeD, x Then G is a down-set. For, if ycae D, then xcty, thus it follows that yeG. 34 Let T=S\x and K={a\x: ae H}. Then K is a down-set in P(T). For, if yca\xe K, then xn y = 0 and xuye D, therefore ye K. By Berge's theorem, K is the disjoint union of pairs {u,v} with unv=0, with the possible exception of the set {0 }. Therefore, H is the disjoint union of pairs {a,b} with |anb|=|x|=k, with the possible exeption of the set {x}. Put the pairs of H into F. If {x} is an exceptional set, put it into G. Then G remains a down-set. If G contains no set of size greater than k, put E=G. Otherwise repeat the operation letting G replace D. The theorem now follows by induction on |D|, since |G|<|D|. □ The second generalization concerns hereditary multi-hypergraphs We. have different possibilities to define a hereditary multi-hypergraph. For example, we may call a multi-hypergraph H hereditary if the underlying hypergraph of H, the hypergraph which contains exactly one copy of the edges of H, is hereditary. (In the remainder of this chapter we resume the terminology of Chapter 1.) Thus, the multi-hypergraph H ={{1,2}, {1,2}, {1,3}, {1,3}, {2,3}, {1}, {2}, {3}, 0 } would be called hereditary, since its underlying hypergraph, {{1,2}, {1,3}, {2,3}, {1}, {2}, {3}, 0 } , is a hereditary hypergraph. On the other hand, the subfamily M ={{1,2}, {1,2}, {1,3}, {1,3}, {1}} of H is an intersecting multi-hypergraph and \M |= 5 > 4.5 = i\H |, which means that we have no hope to prove the generalization of Berge's theorem for this type of multi-hypergraphs. Thus, we will define the hereditary multi-hypergraphs on another way. A multi-hypergraph H is called hereditary if for every A e H and BcA we have at least as many copies of B among the edges of H as the number of the copies of A in H. A more rigorous definition is as follows. A multi-hypergraph defined on the finite set S is a function f: P(S) -> N, where N denotes the set of the non-negative integers. O f course, for an AcS, f(A) is the multiplicity of the set A, the number of the copies of A in the multi-hypergraph. f(A)=0 means that A is not in the multi-hypergraph. A multi-hypergraph f is called hereditary if and only if for every AcBcS f(A)>f(B) holds. It is easy to see that this definition coincides with the previous definition. In the future we will denote a (hereditary) multi-hypergraph by H , but we will use the function f as well, whenever it is convinient. If it happens, f denotes the same multi-hypergraph as H, and f will be called the multiplicity function of H sometimes. We will say that an AcS is an edge (element) of 35 the multi-hypergraph H (and we will denote this fact by A e H ) if and only if f(A)>0. The cardinality of a multi-hypergraph H on the underlying set S is defined to be 2 AeS and is denoted by \H |. We give now the precise definitions of some other concepts and parameters for multi-hypergraphs. If H and H ' are multi-hypergraphs with multiplicity functions f and f , respectively, given on the same underlying set S, then H ' is said to be a sub-multi-hypergraph , or subfamily of H if and only if f (A) AcS. A multi-hypergraph M said to be intersecting if every two elements of it have non-empty intersection. If H is a multi-hypergraph on the underlying set S with multiplicity function f, then the underlying hypergraph of H is defined to be the hypergraph (AcS: f(A)>0}. We mention here that the underlying hypergraph of a hereditary multi-hypergraph is hereditary as well. For a multi-hypergraph H on the set S with multiplicity function f the degree of an element x of S is defined as dH (x)= 2xe AeS and the degree of the multi-hypergraph is defined to be d (H )= max{d^(x): xe S}. u>(H) is the maximal cardinality of the intersecting subfamilies of H, as it was in the case of the hypergraphs. Finally, q (H ), the chromatic index of the multi-hypergraph H, is the smallest possible number of colors needed to color the non-empty edges of H (where we have to color every copy of an edge) such that two edges with the same color have empty intersection. Since the different copies of the same edge have to get different colors, q (H ) is at least as large as the maximal multiplicity of the non-empty edges of H. We mention again that Proposition 1.1 is valid in the same form and with the same proof for multi-hypergraphs as well. For the above definition of the hereditary multi-hypergraph we are able to prove the generalization of Berge's theorem. First we mention that the generalization of Proposition I.9 is valid. Proposition 2.5 IfH is a hereditary multi-hypergraph defined on the underlying set S, then for every xe S d^(x) < ||H |, and so the maximum degree of the hypergraph is at most i\H\. Proof For every xe S and He H such that xe H, H , being hereditary, also contains H \{x}. Moreover, f(H \{x}) > f(H), where f is the multiplicity function of H . Since for 36 different H| and H 2, such that xeH , and x g H 2, H (\{x } and H 2\{x} are also different, and obviously the copies of H \{x} in H are not counted in the degree of x, then it immediately follows that the degree of x in H cannot be larger than £| H |. □ We w ill give now the generalization of Berge's theorem for hereditary multi-hypergraphs and two proofs of it. Theorem 2.6 Any hereditary multi-hypergraph H defined on the finite set S is the disjoint union of pairs of disjoint subsets of S, together with the set { 0 } if \H | is odd. (Here a subset A of S appears in as many pairs as the number of the copies of A in H , while the set { 0 } contains exactly one copy of 0. A pair may contain two copies of the same set now, but from the other assumptions it follows that this set has to be the empty set.) First proof We will prove the theorem by induction on |H |. If |H |=0, i.e., H = 0 , the statement is trivial. If H * 0 , we will use the following induction step. Let f:P(S) -» N be the multiplicity function of the hereditary multi-hypergraph H . Consider the hypergraph H ,={AcS: f(A) >1}. If BcAe H ,, then f(B)>f(A)>l, and so B e //, as well. Thus, H , is a hereditary hypergraph and so we may use Berge's theorem for it. Also, H , is trivially non-empty. Let the multi-hypergraph H Z=H \ H , be defined by the multiplicity function f, where f,(A) = max {0, f(A )-l} for every AcS. If BcAcS, then f(B )-l > f(A )-l, and so we can easily see that f,(B) > f,(A). These imply that iy2 is also a hereditary multi-hypergraph and its cardinality is less than |jF7 |. Thus, we may suppose that both H , and H 2 are disjoint union of pairs of disjoint subsets of S, with the possible exception of 0 in both cases. If the set 0 is not left out in any of these cases, H is the union of the pairs which occur in the partitions of H , and H 2. If the set 0 is left out in exactly one of the cases, then H is the union of the pairs which occur in the partitions of H , and H z and we have to include the set { 0 } as well. If the set 0 is left out in both of the cases, then H is the union of the pairs which occur in the partitions of H , and H 2 together with the pair {0 ,0 } . □ Second proof A little modification of the proof of Theorem 2.1 (the second proof of Berge's theorem) works as well. We will give a sketch of this proof here, leaving out 37 those details which have exactly the same proofs which were given in the proof of Theorem 2.1. Again, the proof goes on by induction on |H |. Let T be a subset of S of minimum size such that S\(HuT)e H for some He H . Define the hypergraph A ={He H: S\(H uT)e H }. The multi-hypergraph A , will have the underlying hypergraph A and the following multiplicity function f: for AcS f (A)=0 if Ae A and f(A )= min{f(A), f(S\(A uT))} if AeA, where f is the multiplicity function of H . By Lemma 2.3 we have that for an A eA , S\(AuT) is also an element of A , and thus f(A )=f(S \(A uT )). Let the multi-hypergraph H |-H \A , be defined by the multiplicity function f ((A )=f(A )-f(A ) (V AcS). Here (A ^O, and thus |H ||< |// |. We will prove the following lemma, similarly to the case of Theorem 2.1. Lemma 2.7 H t is a hereditary multi-hypergraph. Proof Let B cA cS . We have to prove that f((B) > f/(A). If Bg A , then f,(B)=f(B)>f(A)>f(A)-f(A)=f,(A). If Be A , then, by Lemma 2.3, S\(BuT) is also an element of A and S\(BuT) d S\(AuT), since AdB, This implies that f(S\(BuT)) < f(S\(AuT)). If Ae H , then f(A)=0, and so f(B)>f(A) is satisfied. If Ae H , then, by Lemma 2.2, A eA as well. We have two possibilities. If f(A)>f(S\(AuT)), then we have that f(B)>f(A)>f(S\(AuT))>f(S\(B%T)). These imply that f (A)=min{f(A), f(S\(AuT))}= f(S\(AuT)) and f (B)=min{f(B), f(S\(BuT))} = f(S\(BuT)), and so f,(B) = f(B)-f(B) = f(B)-f(S\(BuT)) > f(A)-f(S\(AuT))=f(A)-f(A)=f,(A). If f(A) Now, by induction, we may suppose that the statement of Theorem 2.6 is true for H ,. On the other hand, A , can be partitioned into pairs such that the sets from each pair form a partition of S\T, and so, in particular, are disjoint. This fact comes from Lemma 2.3 and from the fact that the multiplicities of an A eA and S\(AuT) are the same in A,. This concludes the second proof. □ 38 We finish this section with a combination of the previous two generalizations, Theorem 2.4 and Theorem 2.6. Theorem 2.8 If k>0 is an integer and H is a hereditary multi-hypergraph on the underlying set S, then H is the disjoint union of two multi- hypergraphs E , F such that (i) E is a hereditary multi-hypergraph and E e F =>|E| such that |AnB|=k. Proof We can give two different proofs again. One, using the technique of the first proof of the previous theorem, and another, generalizing the proof of Theorem 2.4. We will use the first possibility here, proving the statement by induction on \H | again. Consider the hereditary hypergraph H t and the hereditary multi-hypergraph H 2 defined in the first proof of Theorem 2.6. H x and H z can be partitioned into E\ and F, and into E z and F 2 on the desired way by Theorem 2.4 and by the induction hypothesis, respectively. F =F ,uF 2 will satisfy the assumption about F given in the theorem. The elements of E =E (uF 2 will satisfy the assumption that E e F => |E| thing we have to prove is that E is a hereditary multi-hypergraph. Let g2 be the multiplicity function of the hereditary multi-hypergraph E 2 and g| be the multiplicity function of the hereditary hypergraph H ,, i.e., gt(H )=l if H e //, and gt(H)=0 otherwise. Then the multiplicity function of E , g=gi+g2, satisfies that for Be AcS, g(B)>g(A), since g,(B)>g,(A) and g2(B)>g2(A), and so E is really a hereditary multi- hypergraph. □ 2.4_Some.other_generalizations for hereditary multi-hvpergraphs. First of all, we mention the generalization of Theorem 1.10, which will be an immediate corollary of Theorem 2.6. Corollary 2.9 If H is a hereditary multi-hypergraph and M is an intersecting subfamily of H, then \M | < i\H \. Proof By Theorem 2.6, H can be partitioned into pairs of disjoint subsets, together with the set { 0 } if \H | is odd. M may contain at most one set from every pair and obviously may not contain the empty set, which imply the result. a 39 We can also prove a result similar to Theorem 1.15. If H is a hereditary multi-hypergraph with multiplicity function f and He J/ such that for every H b H , f(H') H is not empty, then )=d (H ) holds for H . Proof Let x be any element of the non-empty intersection of the semi-bases in H . We will prove, by induction on | H |, that H can be partitioned into pairs of edges such that for every pair exactly one member of the pair contains x, proving that (x)= ||H |. Let H \ be the underlying hypergraph of H and H z the the multi-hypergraph defined by the multiplicity function f(A ) = max{0,f(A)-l} for every every subset A of the underlying set of H. Then H \ is a hereditary hypergraph and H z is a hereditary multi-hypergraph. The semi-bases of H are exactly the bases of H) and the semi-bases of H z, and thus the intersection of the bases of H ( and the intersection of the semi-bases of H z contain x. Now, H\ can be partitioned into pairs of the required type by the proof of the forthcoming Lemma 4.4 and H z can be partitioned into pairs of the required type by induction. Putting these pairs together we get the desired partition of H . Thus, we really have that dH (x)= i\H |, which implies that d(H ) < i\H |. On the other hand, for every intersecting subfamily M of H we have that \M | < ±\H | by Corollary 2.9, which implies that (H )< i\H |. Proposition 1.1 gives us that u>(f/ ) < d(H ). Putting these facts together we get that i\H | > w (//) > d(H ) > £|H |, which proves the theorem. □ We prove one more result for hereditary multi-hypergraphs in this chapter, which is the generalization of Chvatal's theorem, Theorem 1.7. We define the strong hereditary multi-hypergraphs . Let H be a multi- hypergraph defined on the set S={l,2,...,n} for some positive integer n. We recall that for two subsets X and Y of S we say that X < Y if and only if there is a function f: X -» Y such that x < f(x) for eveiy xeX. In particular, if XcY, we have that X < Y. We say that H is a strong hereditary multi-hypergraph if for 40 every X < Y c S, f(X) > f(Y), where f is the multiplicity function of H . Theorem 2.11 Let H be a strong hereditary multi-hypergraph on the set S={l,2,...,n} and M be an arbitrary intersecting subfamily of H . Then |M l< d „ ( l) . Proof We will prove the theorem by induction on |/sT |, using a similar technique to the one used in the first proof of Theorem 2.6. If \H |=0, i.e., H = 0, both |M | and dw(l) are 0. Suppose that H &0. Let f:P(S) —> N be the multiplicity function of the strong hereditary multi- hypergraph H . Consider the hypergraph H |={AcS: f(A) >1}. If X < Y e H u then f(X )>f(Y )>l, and so X e //, as well. Thus, H , is a strong hereditary hypergraph and so we may use Chvatal's theorem for it. Also, H , is trivially non-empty. Let the multi-hypergraph H 2=H \H ( be defined by the multiplicity function ft where f,(A) = max {0, f(A )-l} for every AcS. If X easily see that f|(X ) > f t(Y). These imply that H z is also a strong hereditary multi-hypergraph and its cardinality is less than |H |. Let M t be the underlying hypergraph of M, and M 2 be M \M U i.e., let M 2 be defined by the multiplicity function Sz- g2 (X)=g(X) if X g M , and g2(X)=g(X )-l if X e M ,, where g is the multiplicity function of M. Now M , is an intersecting subhypergraph of H ,, and thus by Chvatal's theorem \M || < d^ (l). On the other hand, M 2 is an intersecting subfamily of H 2, and thus by the induction hypothesis |M 2 I - dH . ;(1). Thus, we have that |M |=|M ,|+|M 2| < djy,(l)+dfl2(l)=djir(l). □ O f course, this theorem implies that u>(H )=d(H ) for strong hereditary multi-hypergraphs, since we have that u>(H ) u>(H )> d (H ) by Proposition 1.1. We also have that in a strong hereditary multi-hypergraph on the set S={ 1,2,...n}, 1 is one of the points of the largest degree. The previous results suggest that the generalization of Chvatal's conjecture may be also true for hereditaiy multi-hypergraphs. We conclude the chapter with this conjecture. Conjecture 2 For any hereditaiy multi-hypergraph H on a finite set S, u >(H )=d(H ). C H A PTE R 3 A GENERALIZATION OF BERGE'vS THEOREM FOR DISTRIBUTIVE LATTICES 3.1 Statement of the main result The main result of this chapter, Theorem 3.1 is a generalization of Berge's theorem for finite distributive lattices. In order to state the theorem we need, beyond the previously defined notations, the notion of a polarity, which will take the role of the set-theoretical complement in Berge's theorem. If P is any partially ordered set, then a bijection co: P —> P such that to(cop) = p (V pe P), and p /=/(D ,U ): D -» U such that 41 42 (i) d < /d (V de D) (ii) U \/D is an up-set, (iii) if D|, D 2 are down-sets of L and U t, U 2 are up-sets of L such that D2 cD,, |D,|£|U,| and U,cU2, and if /|=/(D |,U ,) and / 2 = /(D 2 ,U2), then U,\/,(D,)cU 2 \/( D 2). If L has a polarity n, then we also have: (iv) deD, 7tdeU, fn fd is defined, n f'lnde U and nd (v) (7t/)* d is defined for i= l,2 .... => d=ufnfd. Note (a) To say that fn fd is defined for some de D is another way of saying that 7t/de D. (b) In (v), (n f)1 d is to be defined according to the following inductive definition: for de D, (7t/)°d=d; for i=l,2,..., if ( n f y ^ d is defined and lies in D, then (nf)ld=nf((nf)‘‘1 d), otherwise ( 7t/)'d is not defined. (c) In the case when L = P(S), and the polarity n is complementation, then d^rc/d (= Jd) except possibly when d=0. In general, there may be more than one d in L satisfying the equation d=rc/d. Remark (a) We can not say anything about the uniqueness of the injection /= /(D ,U ). In fact, if L = P(S) where S={1,2,3}, D = {0 , {1}, {2 }, {3 }, {1,2}, {1,3}} and U = D ={{1,2,3}, {2,3}, {1,3}, {1,2}, {3}, {2 }} (and so the injection f has to be a bijection in this case), then we have two bijections / , and / 2, shown on Figure 3, from D to U satisfying (i), (iv), and/or (v), with the usual polarity on P(S), i.e., with the set theoretical complement. On the other hand, if S is any finite set and D=U=P(S), then it can easily be seen that there is only one bijection, namely the identity, which satisfies Theorem 3.1. During the proof of Theorem 3.1 we will inductively construct (again, not necessarily uniquely) such injection(s), and (iii) of the theorem refers to this or these injection(s). (b) One can formulate the dual of Theorem 3.1, which also holds. That is, if D is a down-set and U is an up-set of a finite distributive lattice L with |U| < |D|, then there exists an injection / = /(D,U): U —> D satisfying the duals of the statements (i) - (v). 43 Table 3 The bijections / ( and f z deD /,(d )e D / 2 (d)eD 0 {2,3} { 2 } { 1 } {1,2,3} {1,3} { 2 }{ 2 } {2,3} {3} {3} {3} { 1 ,2 } { 1 ,2 }{ 1 ,2 } {1,3} {1,3} {1,2,3} There could be a more natural generalization of Berge's theorem replacing the conditions (iv) and (v) by (*) nfde D, Ttde U => 7t / 7t/d=d (V de D). In other words, (*) states that fnfd=nd whenever it is possible. Of course, in order to get this equality fn fd has to be defined, that is, 7c/d has to be an element of D, and since by definition fn fd is an element of U, xcd has to be an element of U as well. It is easy to see, however, that there is no injection /: D —> U satisfying (i), (ii), and (*) in the case when L=P(S), S={0,1,2}, D={0, {0}, {1}, {0,1}}, U={{2}, {0,1}, {0,2}, {1,2}, S}. This example does not show that (iv) and (v) could not be replaced by (*) in the case when |U|=|D|. This is shown, however, in the following, more complicated example. In this example, there is no bijection /:D —» U satisfying (i) and (*). Let L=P(S), S={0,1,2,3}, D={0,{O}, {1}, {2}, {3}, {0,1}, {0,2}, {1,2}, {0,1,2}}, and U={{2,3}, {0,1}, {0,3}, {1,3}, {0,1,2}, {0,1,3}, {0,2,3}, {1,2,3}, S}. To see that there is no such / , suppose the contraiy. Observe that apart from {0,1}, each set in D has its complement in U. The set in U with no complement in D is {0,1,2}. By (i), /{0,1,2}={0,1,2} or S. Suppose first that /{0,1,2}={0,1,2}. Denote /{ 3 } by u, then ueU. By (i), {3}cu, so u^{0,l}. Therefore, ueD. Since {TJeU and7T37=ueD, by (*), it follows that {3}=/(7T^7), so {0,1,2}={37= / u - Since / is a bijection and 44 /{0,1,2}={0,1,2}, it follows that u={0,l,2}, so u={3}. But {3}«sU, a contradiction; therefore /{0,1,2}=S. Now let/(0)=ueU. If ueD, then, since 0=SeU by (*), we have J I 0 1 = / '1 S={O,1,2}. Therefore, /(0 } = { 3 ). But {3 }e U , a contradiction. Therefore, u eD , in other words, J ^ 0 )e D . Since /(0 )e U , it follows that /(0)={O ,1}. Now let D '= D \{0 , {0,1,2}} and U'=U\{S, {0,1}}. Then B ^ U ’. If there is a bijection / satisfying (i) and (*), then / | D. :D’ -» U ’=IT is a bijection, so /(J3 )= 3 for all deD'. Let Since |D| is odd (=7), there exists d|eD' such that so d|=0. But 0 e D 1, which is a contradiction. Therefore, there is no bijection /:D —» U satisfying (i) and (*). Here, of course, Berge's theorem is a specialization of Theorem 3.1 with the choice of L=P(S) for a finite set S, and U=B. Now (i), and by Note (b), (v) of Theorem 3.1 apply, and they yield the existence of the bijection / used in Section 2.2. In fact, we can state an intermediate result: Theorem 3.2 Let L be a finite distributive lattice with a polarity k . If D is a down-set of L, then D is the disjoint union of sets {a,b} where a=nfb and b=7i/a and where /: D —> nD has the property that d c / d for every de D. Proof A simple induction shows that in this case, the assumption of (v) holds for every deD. By Note (b), it is enough to show that (7c/)°d=d (which is trivial) and that if ( 7t/)i_1deD then (rc/yde D. But ( 7t / ) id = ( 3t / ) ( 7t:/)i' 1 d= 7c ( /( 7c /)i' 1 d) and since /(tc/ ) 1' 1 de 7tD, 7c(/(7t /) 1' 1d)eD holds. Thus, in this case,nfnfd=d (V deD) and the sets {a, b=7t/a } really satisfy the property nfb=a. O f course, as it was mentioned in Note (c), the sets {a,b} do not have necessarily two distinct elements. □ Another nice corollary of Theorem 3.1 is given by parts (i) and (ii). Theorem 3.3 [7] Let L be a finite distributive lattice, let D be a down-set of L, let U and V be up-sets of L and let |D|<|U|. Then |DnV|<|UnV|. 45 Proof By Theorem 3.1, there is an injection /: D —> U such that dc/d for all deD. Since V is an up-set, if de V, then /d e V also. Therefore, /(D n V )c U n V which implies the required result. □ We note the following result, similar to parts (i) and (ii) in Theorem 3.1, due to Dowling and Wilson [9]. Theorem 3.4 Let G be a finite geometric lattice of rank r and let 0 3.2 An algebraic lemma Daykin [ 6 ] has shown that if a distributive lattice L has a polarity n, then L may be represented by means of a partially ordered set P with a polarity co as follows: If a is a lattice element then a may be thought of as a down-set of P; coa is the image in P of this down-set under the polarity co and will therefore be an up-set in P; coa=P\coa is the set theoretical complement of coa in P and so is another down-set, and thus may be thought of as another member of L. Then the representation of na. is given by the formula 7ta=coi. Furthermore, as it was mentioned earlier, the meet and join of two elements of L may be thought as the intersection and union, respectively, of the corresponding down-sets of P. Here we state a slightly weaker result, together with its straightforward proof, which is sufficient for Theorem 3.1. Lemma 3.5 Let L be a finite distributive lattice with a polarity n. Then there is a finite partially ordered set P with a polarity co, and an injection from L into the down-sets of P, such that if the image of ae L under this injection is denoted by a’, the followings are satisfied: a < b o a 's b1 V a,b e L (aAb)' = a 'n b ' V a,b e L 46 (avb)' = a' u b' V a,b e L (Tea)' = P\(oa' V a e L. Proof Consider the distributive lattice K on the set {0,1,2,3}, where 0 is the smallest, 3 is the largest element, and 1 and 2 are non-comparable. Let co be the following polarity on it: coO=3, co3=0, col=2, co2=l. It is well-known that in a distributive lattice the complement of every prime ideal is a dual prime ideal. For every (P,Q) prime ideal - dual prime ideal decomposition of L consider the following homomorphism of L into K, which keeps the polarity: the image of the elements from QnrcP is 3, from PnrcP is 1, from Qo jiQ is 2, and from PnrcQ is 0. It is easily seen that in a lattice the image of the meet (join) of two elements under a polarity is the join (meet) of the images of elements. It follows immediately that if P (Q) is a prime ideal (dual prime ideal), then rcP (tcQ) is a dual prime ideal (prime ideal), and so the above described map is really a lattice homomorphism. We also can see easily that this homomorphism keeps the polarity. For example, if aeQnrcP, then obviously rcae TtQnP, since 7t7cP=P. In this way, of course, for the pairs (P,Q) and (7tQ,7cP) (where nQ is a prime ideal and 7tP is a dual prime ideal trivially) we would define the same homomorphism twice, unnecessarily. Thus we will take every homomorphism only once. It is a well-known fact from lattice theory that for every two elements a,b of a distributive lattice L with bj«a there exists a (P,Q) prime ideal- dual prime ideal pair such that ae P and be Q. If for some two elements a and b every above described homomorphism takes them into the same element of K, then for every pair (P,Q) either both of them are in P or both of them are in Q, thus neither bga nor a*b hold, so a must be equal to b. So if we take as many copies of K as the number of these homomorphisms ( or equivalently as the number of the prime ideal - dual prime ideal decompositions of L ) then we can map L into the product of these copies of K on an injective, operation and polarity preserving way. Let these copies of K be denoted by K, (l In X?=|Kj (the algebraic direct product of the lattices K () we define a polarity by co(a1,a2 ,...,an)=(to(a)),co(a2 ),...,co(an)). Consider Uj = IK, (the disjoint union of the sets 47 Kj,) and the following partial ordering on it: elements from different Kj's are independent, while in every K| we have the same ordering as in K. For every a(e K ( define a subset of K| in the following way: 3l,= {0i,li,2i}, lj'={0|,l|}, 2jI={0,,2j>, and consider the following polarity on Uj,_,Ki: co'3|=0j, co'Oj=3|, co'2j=lj, co’l f= 2 F in a lly define the map %: L -» P(U"_,Kf) in the following way: if the image of aeL in the direct product of Kj's is (a,,a2 ,...,an) then X(a)= Uf., aj1. Since all a,1^ are down-sets in K s, and the elements from different K f's are independent, %(a) is a down-set in U"=, K (. For every a^b we have (al,...,an)*(b 1,...,bn) and thus % (a)*x(b). Since the map a —» (a,,...,an) is operation-preserving in the lattices, then the map a -4 x(a) is also operation-preserving in the sense of the lemma. Finally, we have to prove that for every aeL, (7ta)'=U"=,Kj\co,a’ holds. We know that (7ua)'=%(7ca), and since the image of 7tae L in the direct product of the Kj's is (coa,,...,coan), we have j^Tta^UjLjCcoaj)1. We also have co’a'=co’(U ? =1 aj1). It has to be proven that U^jCcoai) 1 n co’(U" =1 ai')=0 and U^jCcoaj)' u to'(U? =1 ai,)=U? = )K f. Since we know how to and to' act, it is enough to prove this componentwise, i.e. that for every i (toaj)' n to'(aj') = 0 and (xaj ) 1 u toXaj1) = K (. If aj=3f, then (toai),=(Oi),={Oj} and co'(ai,) = to'({O j,lj,2,}) = {3,,2j,lj}. If af= 1,, then (toaj)'= {0 f, 1 f} and to'(ai,)=to'({Oj,lj})={3j,2I}. If aj=2,, then (toaj)'={Oj,2j} and to’(aj')=to,({0j,2,})= {3j,lf}. Finally, if aj=0j, then (toajJ^-COj, 1 j,2j} and to'(ai,)=to'({0,})={3j}. □ 3.3 Proof of the main result Let L be a partially ordered set such that the lattice L consists of some down-sets of it, and if L has a polarity ti, suppose that P has a polarity to and that7 ta=toa (V aeL). It turns out to be veiy convenient to think of the order relation in the lattice L as set inclusion and to denote it by c. 48 Let (f,,f2,...,f2|P|) be a sequence of subsets of P, let each subset occur exactly once and let |fj < |fi+1| (1 < i < 2|P|-1). Note that the f,'s are not necessarily all the elements of L. For given D,UcL with |D|<|U|, if L does not have a polarity n, let e be the first fi such that the set A is nonempty, where A = {a: aeD, ane=0, aueeU } If L does have a polarity 7t, let L be the first f, such that at least one of the sets A, C is nonempty, where A is as above and where C = {a: ae D, ang=0, auge U} and g=coe. Note that |g|=|e|. In the remainder of the proof, if we refer implicitly or explicitly to g or C, then we will, at those points, be assuming that L has a polarity jt, even though we may not expressly say so. Throughout, we do not in general assume that L has a polarity, and if it does not have it, then much of the discussion may be ignored. We show first that D\A is a down-set. Suppose otherwise. Then there are xeD \A , ye L, ys D\A with xoy. Since xe D and D is a down-set of L, it follows that ye D, and consequently ye A. Since both yue and x are down-sets of P, so also is xue=(yue)ux; and since both yue and x are elements of L, so also is xue=(yue)ux. (In the remainder we will use the fact, even if we do not expressly mention it, that the join or meet of any two down-sets of P, which are themselves in L, is also in L.) Because yuecxue and yuee U, it now follows that xuee U. If xne=0, it would follow that xe A. Since this is not the case, we have xne^0. Let z=xn(yue); since both x and yue are down-sets of P, so is z, and since z c x e D , it follows that ze D. Let f= (y u e )\z . Then f=(yue)\((xny)u(xne))=(yue)\(yu(xne))=e\x, so |f|<|e|. z and y satisfy zeD, znf = 0 and zuf=yuee U, since ycz. This contradicts the minimal property of e. Consequently, D\A is a down-set of L. 49 If L has polarity 7t, then, by symmetry, D\C is a down-set of L. Consequently (D\A)n(D\C)=D\(AuC) is also a down-set of L. (In fact, to say that D\C is a down-set of L "by symmetry" might, at first, be thought to be slightly misleading, since e comes before g in the sequence of fj's. However, |e|=|g|, and so the sequence could be rearranged so that g comes before e. The same use of the phrase "by symmetry" occurs elsewhere.) Let A+e denote the set {aue:ae A}. We show next that U\(A+e) is an up-set of L. Suppose otherwise. Then there are xeU\(A+e), yeL, ygU\(A+e) with xcy. Since xeU and U is an up-set of L, it follows that ye U, and consequently ye A+e. Let s=e\x and suppose that s*0. Because y\s=y\(e\x)=(y\e)ux and both y\e and x are down-sets of P, so is y\s. Since xcy\s and x e U , it follows that y\se U. Therefore, we have y\eeD , (y\e)n(e\s)=0, (y\e)u(e\s)=y\seU and |e\s|<|e|. But this contradicts our choice of e. Therefore, s=e\x=0, and so ecx. Therefore, x\e=(y\e)nx. Since both y\e and x are down-sets of P, so also is x\e. Because xcy and y\ee D, it follows that x\ee D. We have now proved that x\ee D, (x\e)ne=0 and(x\e)ue=xe U. Therefore, xe A+e, which is a contradiction. Therefore U\(A+e) is an up-set of L. By symmetry, U\(C+g) is an up-set of L, and consequently, (U\(A+e))n(U\(C+g))=U\{(A+e)u(C+g)} is also an up-set of L. We split now the proof Theorem 3.1 into four main parts. Parti. A nroof of (\) and fiil For this we consider two cases. Case 1 (e=g). (If L=P(S), then we may take to to be the identity map, and then e=g. Thus, the argument here reduces to the argument already given for Berge’s theorem. By contrast, the argument of Case 2 is not needed for a proof of Berge's theorem.) If D =0, there is nothing to prove. So by induction on |D|, we may assume that a map In this case, if L has a polarity it, then A=C and A+e=C=g. Let A -» A+e be defined by £a = aue (V ae A). 50 Now define /: D -» U by /x = /x = £x, if xe A. Then f has properties (i) and (ii). This completes the induction step in Case 1. When the corresponding step in Case 2 is completed, the existence of f satisfying (i) and (ii) will be established. Case 2 (e*g). Throughout the discussion of this, the most difficult part of the proof, we assume, of course, that L has polarity 7t. Let H={a: ae A, aueeC+g}, J=(D\(AuC))uH K={x: xg (Ae C)+g}, N=(U\((A+e)u(C+g)))uK. Also, let H * = {a: aG A\H, 3 hG H such that ach}, K * = {x: xg (C+g)\K, 3 Icg K such that xsk}. Similarly, let H' = {a: ae C, angG A+e}, K' = {x: xg (AnC)+e}, H'* = {a: aGC\h', 3 Ii 'g H' such that ach1}, K'* = {x: xg (A+e)\K', 3 Ic'g K' such that xsk'}. Facts concerning H', K', H'*, and K '* will follow by symmetry from similar facts about H, K, H *, and K * for which detailed proofs are given. To follow this case, it may be helpful to keep Figure 4 in mind. First, we show that H *= 0 . Suppose otherwise. Then there are hG H, h*e H * such that h3 h*. By the definition of H, hue'(C+g)n(A+e), and by the definition of H and H*, h*ue'(A+e)\(C+g), and futrhermore, hue:>h*ue. Since hueG C+g, there is an x g C such that hue=xug, and since hueoh^ue, it follows that there are sets y and g 1 such that h*ue=yug’, where g'cg and ycx. Since xng=0, it follows that xng'=0. Since x and 51 C+g Figure 4 For the proof of Theorem 3.1 52 yog' are points of L, they are down-sets of P, and so, since xng'=0 and ycx, it follows that y=xn(yvjg') also is a down-set of P and a point of L. Since ycxeD , and D is a down-set of L, it follows that ye D. Now if g=g', then since yug'=h*uee U, it follows that ye C, so h*ueeC+g, which is a contradiction. Therefore ggg'. Then we have |g'|<|e|, ye D, yug'e U which contradicts the definition of e. It follows, therefore, that H *= 0, as required. Since e=cog, it follows, by symmetry, that H'*=0. Next, we show that K *= 0. Suppose otherwise. Then there are ke K, k*e K * such that k*ok. Let z=k\g and z*=k*\g. Then, by the definitions of C,K, and K *, we have ze A nC and z*e Q A . We also have z*dz. Since ze AnC, we have ze A so zg D\A. Since z*e C\A, we have z*e D and z*g A, so z*eD \A . Since D\A is a down-set and z* o z, it follows that ze D\A. This contradicts the preceding remark and it follows, therefore, that K *= 0 , as required. By symmetry, we have K '*=0. From the definitions of H and K we have g\e c h for all he H, kne = gne for all ke K. Consequently, we have HnC = 0, Kn(A+e) = 0. We now show that J=(D\(A uC))uH is a down-set of L. Suppose otherwise. Since D\(AuC)=J\H is a down-set of L, there are he H and xch, xe L but xg J. Since he HcD, and D is a down-set, we know that xeD. From the definition of H*, if xe A, then it would follow that x e H *. But since H *= 0 , it follows that xg A. Therefore, xe C \A . Consequently, xn g =0 and xuge C+g, so xuge U. We may assume, without loss of generality, that x and h differ by only one element, say y. Thus h=xu{y}. Since g\ech and xng=0, it follows that |g\e|=l and ye g\e, i.e., {y}=g\e. We have that 53 U\((C+g)n(A+e))=(U\(C+g))u(U\(A+e)) is an up-set of L. Since hue=xu{y}ue= xu(g\e)ue=xugu(e\g), we have hueoxug. Since hues (C+g)n(A+e), it follows that hue£ U\((C+g)n(A+e)). It also follows that xuge U\((C+g)n(A+e)). Since we know that xugeU , it follows that xe ge A+e, and so ecxug. Since e\g*0, it follows that x r w 0 . But xch and hne=0, so xne=0. This contradiction proves that J is a down-set of L, as required. Next, we show that N=(U\((A+e)u(C+g)))uK is an up-set of L. Suppose otherwise. Since U\((A+e)u(C+g))=N\K is an up-set of L, there are ke K and xok, xe L but xeN. Since ke KcU and U is an up-set of L, we know that xe U. From the definition of K*, if xe C+g, then it would follow that xe K *, but since K *= 0 , it follows that xe C+g. Therefore, xe (A+e)\(C+g). We may assume that x and k differ by only one element, say y, so x=ku{y}. Let x\e=x' and k\g=k'. Then k'e AnC, so k'ne=0 and k'ng=0, whereas x'e A, so x'ne=0. Since ecx, we have e=xne=(ku{y})ne=(kne)u({y}ne)= (gne)n({y}ne). But since gne^e, it follows that {y}=e\g. Now x=eux', where e and x' are disjoint, and also x=ku{y}=k'ugu(e\g)=k'u(g\e)u(eng)u(e\g), where k', g\e, eug, and e\g are all disjoint. Therefore x'=k'u(g\e). Now D \(A n C )= (D \A )u (D \C ) is a down-set of L. But k'e AnC, so k'e D\(AnC). Since xbk', it follows that x'e D\(AnC). But x'eD, so x’e AnC, and thus x’ng=0. But since g\ecx, and g\e^0, it follows that x'ng?*0. This contradiction shows that N is an up-set of L, as required. We now show that J is a proper subset of D. By definition, |AnC|=|K|. Since K n (A + e )= 0 we have |C|=|C+g|>|K|+|(A+e)n(C+g)|=|K|+|H|. Therefore, |D|-|J|= |(AuC)\H|=|A|+|C|-|AnC|-|H|>|C|. Similarly, |D|-|J|>|A|+|C|-|K|-|H|>|A|. Thus |D|-|J|> max(|A|,|C|)>0, by the definition of e. Therefore, J is a proper subset of D. 54 Now let A\H - » (A\H)+e be defined by ^a=aue (V ae A\H), and let cp:C\(AnC) —> (C\(AnC))+g be defined by cpx=xug (V xe C\(AnC)). Note that (C\(AnC))+g=(C+g)\K. Since |J|<|D|, we may assume by induction that a map \j/=/(J,N): J -» N exists satisfying properties (i) and (ii). Let /: D —> U be defined by /x = £x, for xe A\H, = 9 X, for xe C\(AnC), = \|tx, for xe (D \(A uC ))uH . It is easy to check that / is well defined and satisfies (i) and (ii). Part II. A oroofof ('iii) To prove (iii), let D,, D 2, U,, U 2 be as specified. Let /, and / 2 be the corresponding maps / , defined inductively as in Case 1 or 2 of Part I. We have to show that Ui\/|(D|)cU 2\/ 2 (D2). Let e* be the first f( such that either {d: deD,, dnf,= 0 , duf,eU,} * 0 or {d: deD,, dntofj=0, dU(of,eU,} * 0 (of course, the second possibility arises only if L has a polarity). For i=l,2 let A, = {a: aeDj, ane*=0, aue*eUj}, C, = {a: aeDj, ang*=0, aug*eUj}, where g*=(oe*. We again consider two cases. Case 3 (Either e* = g* or L does not have a polarity n). Let ^f:A f —> A (+e* be defined by ^a = aue* (VaeAf). Then, as shown above, Dj\Aj is a down-set and Uj\£j(Aj) is an up-set, and if we put Vi = /(D|\Aj, Uj\^|Aj), then 55 / iX = £jX, if xe Aj, = \jrtx, if xeDj\A|. We now show that D 2 \A 2 c D,\A,, |D(\A,| < |U|\(A,+e*)|, U,\(A,+e*) c U 2\(A 2 +e*). To see the first of these, suppose otherwise. Then there is x e D 2 \A 2 such that xeD|\A,. Since D 2 cD (, it follows that xe D (, so xe A j. Therefore, xne*=0 and x u e *e A |+ e*cU tcU 2. But if xu e *e U 2, xn e *= 0 , and xe D 2, then xe A 2, which is a contradiction. This proves the first of these remarks. The last is proved similarly, for suppose it is not true. Then there is yeU|\(A,+e*), where y<£U 2\(A 2 +e*). Since U jcU 2, it follows that y e U 2, so ye A 2 +e*. Therefore, e*cy, y\e*e A 2 cD 2 cD,. But if y\e*eD,, e*cy and yeUj, then y\e*e A,, so ye A|+e*, a contradiction again. This proves the final remark. The middle one is obvious. If |D,|+|D 2 |=0, then there is nothing to prove. We may assume by induction on |D||+|D2| that (Uj\£|A|)\\jri(Dj\A|) c (U 2\£ 2A 2)\\|/2(D 2\A 2). Since /(D|)=£|AjU\j/j(D,YAj) (i=l,2), (iii) follows by induction in Case 3. Case 4 (e^^g*). We assume in this case that L does have a polarity jc. Define the sets Hj, K|, Jf, N ( (i=l,2) in the obvious way. Then J, and J2 are down-sets and N| and N 2 are up-sets. For i=l,2, let A (\H j —> (Aj\Hj)+e* be defined by £ja=aue* (V ae Aj\Hj) and let Cj\(AjnCj) —> (Cj\(AjnCj))+g* be defined by (V xeCj\(AjnCj)). If we put Yi=/i(Ji»Nj): Jj—> Nj, then /(X = ^x, if x e AjYHj, = = \|tjX, if x e Jj. 56 Now we show that J2 c J„ |J,| < |N,|, N, c N2. To prove that J2 cJ,, suppose otherwise. Then there is xe J2, such that xg J(. Since D 2 c D f, it follows that xgD|, s o xe (A^GOVHi. Therefore, since H ,n C |= 0 , we have either: (I) xe A,\H, and xue*e (Al+e*)\(Cl+g*), or (II) xg AI? xe Ct, xug*e Ct+g*. Suppose (I); then x u e * e U ,c U 2. Thus we have x e D 2, x n e *= 0 and x u e * e U 2. Therefore, xe A 2. Since xe J2, it follows that x g A 2\H 2, and consequently, that x e H 2. Thus x e A ,n H 2. Now if y e C 2 and ygC |, then y u g * e U 2 and, since C 2 c D 2 cD,, yO g*g UThus (C 2 \C l)+ g *c U 2 \U|. Therefore, since x e H 2 n A (, then x e A ^ A g , xne*= 0 , and xue* e (C 2 +g*)n((A,nA 2 )+e*) = (C 2 +g*)n((A,nA 2 )+e*)nU , = (CI+ g *)n ((A ,n A 2 )+ e *)n U 1, since (C 2 \C ))+g*cU 2 \U 1. Therefore, xue* e (Cl+g*)n(Al+e*)cU 1 = (C!+g*)n(A,+e*). Thus xeH,. This contradiction proves that (I) does not happen. Suppose, therefore, that (II) occurs. Then xng*=0, xug*eC,+g*cU|cU 2 and x e D 2 as well. Therefore, xeC 2. Since xe J2, it follows that xg (A 2 u C 2 )\H 2. But since C 2 n H 2 =0, we have xg C 2. This contradiction shows that (II) also does not occur, and therefore, that J2 cJ(, as required. The inequality |J,| < |N,| follows from the fact that |D,| < |U||, together with the fact that |(A 1u C 1)\H 1|=|Al\H,|+|CI\A l|= |(A,+e*)\(C,+g*)|+ |(C,+g*)\K,|=|((A 1+e*)u(C 1+g*))\K1|. 57 To prove that N ,cN 2, suppose otherwise. Then there is an xeN , such that x g N 2. Since x g N 1cU 1cU2, it follows that x g U 2 \N 2 = ((A 2 + e *)u (C 2 + g *))\K 2. Therefore either: (III) x g (C 2 +g*)\K 2 and x\g*GC2, or (IV) x g (A 2 +e*)\(C 2 +g*), x\e*GA 2\H 2. Suppose (III) occurs. Then x\g*GC 2 cD 2 cD,. Thus we have x\g*GD,, (x\g*)ng*=0, x g U ) . Therefore, XGCt+g*. But since x g N,=(U|\((A,+e*)u(CI+g*)))uK 1 and since K,cC|+g*, it follows that x g K(=(A,nC|)+g*. Thus, x g K ,n (C 2 +g*). Now if e*cy for some y c L such that y\e*eA , and y\eg A 2, then y\e*GD, and, since A ,+ e*cU |cU 2, y\e*g D 2. Thus, A,\A 2 cD|\D2. Since x g K tn (C 2 +g*), we therefore have x\g* g A, n C (n C 2 = A ,nC iO C 2 n D 2 c A 2 n C ,n C 2 n D 2 since A ,\A 2 c D ,\D 2. Therefore, x\g* g A 2 n C 2 n D 2 = A 2 n C2. Thus, xg (A 2 n C 2 )+ g *= K 2. This contradiction proves that (III) does not happen. Suppose, therefore, that (IV ) occurs. Then x\e*G A 2 \H 2 cD 2 cD,, (x\e*)ne*=0, and xgN,cU,. Therefore, xGA,+e*. Since xgN,, x«5 [(A,uC,)+g*]\K|. But since (A |+g*)nK )=0, it follows that xg A (+e*. This contradiction shows that (IV ) also does not occur, and therefore, that N,cN2, as required. If |D(|+|D2 |=0, there is nothing to prove. Therefore, suppose that |D,|+|D 2 |>0. By the first part of the proof, at least one of J,, J2 is a proper subset of D t, D 2, respectively, so |J,|+|J2 |< |D ,|+|D 2|. Therefore, we may assume by induction on |D ,|+|D 2|, that N,\\|/,(J|)cN2 \\j/2 (J2). But since, for i = 1,2, N i\x|/i(Ji)=[U,\(^(Ai\H i)ucp,(Ci\(A(n C j)))]\Vi(Ji)=Ui\/(D i). Thus, (iii) follows by induction in Case 4 also. 58 Part ITT. A proof of (iv’) We assume that / is defined inductively as described in Part I. Let deD satisfy the conditions of (iv). We consider three cases. Recall that the order of /d\d and co(/d\d) depends on their places in the sequence (f|,f2,...,f2|P|). Case 5 (/d\d = co(/d\d)). Let /d\d=e and let A be the set corresponding to e described in Part I. Then e=/d\d=co(/d\d)=(co/d)\(cod)=£o 3 \d5 7 3 = 7td\7c/d c /7c/d\7t/d. Also, (7t/d)ue = [ 7t(due)]ue = w(3ue)ue = (co3nwe)ue = (co 3 Ue)ue=(o 3 =7td. Thus, (u/d)uee U, it/de D, ec/ 7t/d\jt/d. Therefore, Kfde A and c=f%fd\ufd, so fKfd=nd, and so d-Kfnfd, as required. Case 6 (fd\d < to(/d\d)). Let e and g denote /d\d and co(/d\d), respectively. Let A, C, etc., describe the sets corresponding to e and g of the construction in Part I. Then g=co(/d\d)=(co/d)\(tod)=co3 \co7 3 =Jtd\7t /d c /7t/d \ 7t/d. Also (7c/d)ug=[u(due)]ug= to(3 u e )u g = (d j 3 n(oe)ug=(co 3 og))ug=co3 = 7ud. Thus we have that (jt/d )u g e U , Kfde D, g c /7t/d \ 7t/d. Therefore, Kfde C. If Kfde A as well, then it would follow that c - / 7t/d \ 7t/d so that gee. But since |g|=|e|, it would follow that g=e, which is excluded from this case. Therefore, 7t/d g A and thus Kfde C\A. Therefore g = /jt/d \ 7t/d, so / 7t/d= 7td, and d=7t / 7t/d, as required. Case 1 (f d \d > part of the argument in Case 6 , we have that e= 7td\7t/d c /7i/d \ 7t/d and (7i/d)ue= 7td. Thus we have (Trfd)ueeU, Kfde D, ec/ 7t/d\ 7t/d. Therefore, Kfde A. Since /d\d>co(/d\d), we also have that de QA. If Kfde H, then it would follow that 7td=(7t/d)uee (C+g)n(A+e). Therefore, there would be d*eC \A such that /d*=d *u g= 7td.Then K(due)= 7t((7t/d*)ue)=to(^ 7t7 3 3,r5 ue)= 6 ^ 7 3 ^ ncoe = ( 7t7t/d*)ng = /d*ng = (d*ug)ng' = d*, since gnd * = 0 and therefore, 59 clue=Jid*. Since, by assumption, Tt/^rcdeU and since Jtd*=7t/" * 7rd, we have rcd*eU, so it would follow that d*£ C\A, which is a contradiction. Therefore, nfds H, and thus, nfds A\H. Therefore, e= fnfd\nfd, so fnfd=(nfd)'ue=nd and so %fnfd=d, as required. Part IV. A proof of (VI As well as the definition of /(D ,U ), let the definitions of e, g, C, H and K be as in the proof of Part I. Let \|/=/(D\A,U\(A+e)) if e=g and \|/ = /(D \((A uC )\H ), U\(((A+e)u(C+g))\K)) if e^g. Then the map \|/ has properties (i) and (ii). As an induction hypothesis, suppose that \|/ has the property (v) as well. Clearly, if D =0, there is nothing to prove. Suppose D?*0. We proceed via a number of lemmas. Lemma 3.6 (i) If e=g, ae A, a= 7t/d for some de D and fn fa is defined, then de A. (ii) If e*g, ae (A u C )\H , a=nfd for some d u D and f n fa is defined, then de (AuC )\H. (In fact, if ae AMI (or C\A) and the other conditions are satisfied, then de C\A (or AMI, respectively).) Proof (i) As in Case 5, (nfa)ue=na=fd. Since n f as D, it follows that de A. (ii) If ae AMI, then, as in Case 6 , (nfa)ug=na, so (jt/a)ug=7ia=/d. Since n f as D, it follows that de C\A. The case when ae C\A is similar. □ Lemma 3.7 (i) If e=g, de A, a= 7t/d and ae A, then d= 7c/rc/d. (ii) If e*g, de (AuC)\H, a=nfd and ae (AuC)\H, then d=nfnfd. P roo f (i) n f n f d = n f a = 7t(aue) = 7t{(7t/d)ue} = 7t{(7t(due))ue} = 7t{((7td)n(7re))ue} = 7t7td = d. (ii) Suppose that ae AMI. Then, as in Case 6 , (7t/a )u g = 7ca, so (7t/a )u g = /d . Therefore, de C\A. Since dng=0, it follows that (cod)ne=(cod)n(Sg)=0, so that ecco 3 = 7td. Therefore, nfnfd = nfa = it(aue) = 7c{(7t/d )u e } = 7t{ ( 7t(dug))ue} = 60 7u{co^30g)ue} = 7i{(co3ncdg)ue} = 7t{((7td)ne)ue} = 7t7td = d, as required. The proof for ae G A is similar. □ Lemma 3.8 (i) If e=g, ae A, a= 7t/d for some de D and fn fa is defined, then d=7t/jt/d. (ii) If e*g, ae (AuC)\H, a=nfd for some de D and fn fa is defined, then d=nfnfd. Proof (i) By Lemma 3.6, de A, and so, by Lemma 3.7, d= 7t/jt/d . (ii) This proof is similar to that of (i). □ Lemma 3.9 (i) If e=g, de D\A and fn fn fd is defined, then nfde D\A. (ii) If e*g, deD\((AuC)\H) and fn fn fd is defined, then nfde D\((AuC)\H). Proof (i) Let d,= 7t/d and d 2 = 7t /d (. Then d,e D and d2e D. If d,e A, then, by Lemma 3.6, de A. Since d£ A, it therefore follows that d,eD\A, as required. (ii) This proof is similar to that of (i). □ We now prove (v). Suppose first that e=g. Let de D. If de D\A, then, by repeated application of Lemma 3.9, (n fd )1 de D\A, i=0, 1, 2,... . Therefore, by induction, d=nfnfd. Now suppose that de A. If nfde D\A, then, by repeated application of Lemma 3.9, (nf)htfde D\A for i=0, 1, 2,.... By the induction hypothesis, nfd=nfnfnfd. Since n and / are injective, it follows that d=nfnfd, which is the required result. We might well note, however, that de A, whereas nfnfde D\A, which is a contradiction. This shows, in fact, that if de A, then nfd£ D\A. Finally, suppose that de A and nfde A. Then, by Lemma 3.8, d=nfnfd, as required. The case when e^g is proven similarly. Part (v) now follows by induction. This completes the proof of the main theorem. □ 3.4 Generalization of Chvatal's conjecture for distributive lattices As it was mentioned earlier in Note (c) of Theorem 3.1, we do not have, in general, the same result about pairings from a down-set D of a distributive lattice L as we have in the case if L is P(S), the set of the subsets of a finite set S. However, we still can 61 formulate a similar corollary of Theorem 3.1, which is an immediate consequnce of Theorem 3.2. Corollary 3.10 If L is a finite distributive lattice and tc is a polarity on L, then eveiy down-set D of L is a disjoint union of pairs {a,b} and singletons {c}, where a<7tb (and thus b<7ta) and c<7tc. Proof Take the partition of D into the pairs {a,b}, where a=Jt/b and so b= 7t/a, as in Theorem 3.2. If a^b, then we have that nb=nnfa=fa and so by Theorem 3.1 a (0,S), ({1 },{2,3}>, ({2},{1,2}), ({3},{1,3}). (It can be easily shown that 7t is a polarity.) Now, for example, the subset {{2},{2,3},{1 ,2},{ 1,2,3}} of L (which is obviously a down-set of itself) is pairwise intersecting in the sense of the first definition. However, in the sense of the second one it is not, because the set {2} does not intersect itself, since { 2 } < { l , 2 }= 7t({ 2 }). On the other hand, the subset {{3},{1,2},{2,3}, {1,2,3}} of L is not intersecting in the sense of the first definition, since {3 }n {l,2 }= 0 . However, it is in the sense of the second one, as it can be easily seen. Another difficulty arises from the fact that not every distributive lattice allows a polarity. For example, if we consider the distributive lattice K on the set {0,1,2,3,4} as it is shown on Figure 5, i.e., where 0 is the smallest, 4 is the largest element, 1<2 and 1<3, but 2 and 3 are non-comparable, then we can not define a polarity on this distributive lattice. (This can be easily verified.) If we want to generalize Chvatal’s conjecture for distributive lattices we have to define d(D), i.e., the degree of a down-set We call a subset C of L a star in the obvious case, 62 4 2 0 Figure 5 The lattice K i.e., if there is an atom aeL (that is an element a of L such that 0 element b of L such that 0 the lattice are intersecting if aAb? 0 , and thus, we may state the following generalization of Chvatal's conjecture. Conjecture 3 If L is a finite distributive lattice, D is a down-set of L and w(D)= max {|M|: M cD and V a,beM => aAb^O}, then there is a star CcD in L such that |C|=w(D). The generalization of Berge's theorem, Theorem 3.1, for distributive lattices, however, does not give any meaningful result for this conjecture. Moreover, we can not even state a similar result to Berge's theorem, since, for example, in the case of the lattice K shown on Figure 5, w(K)=4 and |K|=5, i.e. o>(K)> 2 |K| (K obviously can be considered as a down-set of itself.) CHAPTER 4 MAXIMUM INTERSECTING FAMILIES IN A HEREDITARY HYPERGRAPH IN THE EXTREME CASE 4.1 Introduction In this chapter we will prove a special case of Chvatal's conjecture, when the cardinalities of the maximum intersecting subfamilies of edges in the hereditary hypergraph are as large as possible. From Berge's theorem, it follows that they can not be larger than half of the number of the edges of the given hereditary hypergraph. So it is easily seen that, using again the notation of Chapter 1, if d (H ) = £|H | for a hereditary hypergraph, then Chvatal's conjecture is true. Our theorem basically says that it is enough to suppose that u ( H ) = %\H \ in order to be able to prove Chvatal's conjecture in this special case. Moreover, we also include the case when |H | is odd, and so the maximum of the cardinalities of the intersecting subfamilies is at most ?(iH |-1 ). Theorem 4.1 [16] If H cP(S) is a hereditary hypergraph on a finite set S and w (//)= [{\H |], then w (H ) = d(H ), where [.] denotes the integral part of a number. Moreover, during the proof of this theorem we will describe the structure of all of these maximum intersecting subfamilies from a certain point of view. It is easy to see that the following conjecture is equivalent to Chvatal's conjecture: Conjecture 4 If H cP(S) is a hereditary hypergraph on a finite set and all intersecting families M c H of maximum cardinality contain all maximal sets of H, then the intersection of all maximal sets of H is non-empty (Ne/7 is a maximal set of H , if there is not He H such that H2 N.) Actually, if Chvatal's conjecture is true, then there has to be a point p of S such that the sets of H containing p form an intersecting family of maximum cardinality in H. Thus, they include all the maximal sets of H, which in particular means that the intersection of these sets contains p. Therefore, it is not empty. On the other hand, if Conjecture 4 is true, 63 64 it gives a special case of Chvatal's conjecture. This leads to an induction proof of it. If there is an intersecting subfamily M of ff which does not contain all the maximal sets of f f , then deleting a maximal set from f f which is not in M, we get a smaller hereditary hypergraph. If Chvatal's conjecture is true for this one, then it is also true for the original one. Now, with the help of Conjecture 4, Chvatal's conjecture would follow by induction on I f f |. For, if we can not use the previous induction step, then all intersecting subfamilies of f f of maximum cardinality contain all maximal sets of f f , and we can use Conjecture 4. Thus, in the sense of Theorem 4.1 if we would like to prove Chvatal's conjecture, then it is enough to prove the following statement. If f f is a hereditary hypergraph and every intersecting family M c ff of maximum cardinality contains all maximal sets of ff, then u>(ff )={\H |. 4.2 Description of the maximum intersecting families in f f . Theorem 4.1 follows from the stronger results below by Lemma 4.4. Theorems 4.2 and 4.3 also give descriptions of the maximum intersecting families in these extreme cases. Theorem 4.2 [16] Let f f cP(S) be a hereditary hypergraph on a finite set S, satisfying w(ff )= {\H | (and thus |ff | is even ) and let N,, N 2,..., N n be all the maximal sets of ff. Then D"., N f = M * 0 and every intersecting family M c ff of maximum cardinality arises in the following way. There is a maximum intersecting family M ' in P(M) such that Af = {A e ff : 3 B e A f', BcA}. Conversely, every intersecting family M given in the previous way has cardinality ||ff | and thus Chvatal's conjecture holds for ff. Theorem 4,3 [16] If ff cP(S) is a hereditary hypergraph on the finite set S, |ff j is odd, and M c ff is an intersecting family of cardinality gflff |- 1 ), then there exists either a maximal set N, of f f such that M c ff \{N ,} or an N n + 1 c S, N n+1«?ff such that A / u { N n+1} is an intersecting family and f f u { N n+1} is a hereditary hypergraph (i.e., either \M |= ^|ff \{ N f}| or |M u {N n+1}|= i\H u { N n+1}|). Thus, Chvatal's conjecture holds for f f . We will give three preliminary lemmas for the proofs of these rather complicated statements. These lemmas are interesting in themselves as well, since they give some 65 special cases of Chvatal's conjecture. The first one is a more detailed version of Theorem 1.15. Here we give it with Schtinheim's proof. Lemma 4.4 (Schtfnheim [20]). If N„ N 2,..., N n are all the maximal sets in a hereditary hypergraph 7/cP(S) and fl " =1 Nj = M * 0, then u (H )= d(H )=||H |. Proof It follows from Berge's theorem that d (H )< u (H )< i\H \. Take now an element x of M and form the pairs from all the sets of H : {A,AA{x}}, where AAB denotes the set theoretical symmetric difference of the two sets A and B, i.e. A A B = { x g A u B: x g A n B }. Observe that if A^B are two edges of the hereditary hypergraph H, then the pairs {A ,A A {x}} and {B, B A {x}} are either disjoint (if A ^ B A {x }) or the same (if A=B A{x}). We need to prove that {A ,A A {x }}= {AA{x},(AA{x})A{x}} and that A e H implies A A {x}eH . However, (AA{x})A{x}=A. Also, A e H implies that there exists an N f such that A c N ,e H . Consequently A u {x }c N ie flr holds, since xeN ,. But A A {x}cA u {x}, and thus H, being hereditary hypergraph, contains AA{x} also. Hence, it follows that there is exactly one member of each pair containing x, i.e. d(x)= £|H | and then d (H ) > d(x) = £|H |> Lemma 4.5 If H cP(S) is a hereditary hypergraph and xe S, then d(x)= {\H | holds if and only if xe n?=, N f, where N,, N2,..., N n are all the maximal sets in H . Proof If d(x)= i\H | and x g N ( hold for some l Lemma 4.6 If H cP(S) is a hereditary hypergraph of odd degree and for an xe S d(x)= i(\H |-1 ) holds, then there exists exactly one maximal set of H not containing x. Proof If d(x)= U\H |-1) and xe n?=l N {, then d(x)= i\H | by Lemma 4.5, which is a contradiction. On the other hand, if xe N ;, Nj holds for some 1 < i,j < n then the 66 degree of x in the smaller hereditary hypergraph H \{Nj,N.} is stilli(\H |-1), which is greater than [{\H -{Ni(Nj}|], a contradiction again. □ 4.3 Proofs of Theorems 4.2 and 4.3 Proof of Theorem 4.2 First we prove that the families M c H described in the theorem are really intersecting families of maximum cardinality. It is easy to see that M is an intersecting family. If A and B are elements of M, then by definition there are two elements A ’ and B' of M 'cP(M) such that A ’cA and B'cB. ButM ' is an intersecting family, so A 'n B V 0 which implies that A n B *0 . We also prove that | M |= {\H |. By the argument used in Lemma 4.4, we can form disjoint pairs from the sets of H: {A,AAM}. Since M ' is a maximum intersecting family in P(M), we know, by Corollary 1.3, that |M j= £ 2lM| and for every subset N of M, M ' contains exactly one of N or M \N. Thus, there exists a B e M ' such that either BcA or BcAAM, so either A e M or A A M e M . Then \M |> £|H | holds and this implies, by Berge's theorem, that | M |= z\H |. We will prove the other part of the theorem by induction on |H |. The following induction step is based on the same idea which was used in the proof of Theorem 3.1, more exactly on the simplified version of it used in the second proof of Berge’s theorem. Let T again be a subset of S of minimum cardinality for which there exists an He H such that S\(HuT)eJT. As it was mentioned earlier, there exists such a T, possibly empty. Let A ={Ae H:S\(AuT)e H). By Lemmas 2.2 and 2.3 we have that H\A is also a hereditary hypergraph and A is the disjoint union of pairs {A,B} of disjoint subsets of S. Also, A=S\(BuT) and A eA implies Be A (see Figure 6 ). So the union of our pairs from A is S\T. I f M cH is an intersecting family of maximum cardinality, that is \M |= 2 | H |, then \M r\A |= 2 IA |. Here \M r\A \ < {\A | is trivial, since M is a family of pairwise intersecting sets and A consists of pairs of disjoint sets, and Mif | nA |< 2 IA |, then M\A is an intersecting family in the hereditary hypergraph H\A such that \M\A \= \M |-|MnA|> 67 S Figure 6 For the proof of Theorem 4.2 i\H |- {\A |= {\H \A |, which is a contradiction. Hence, we know that \M \A|= i\H \A|, u (H \A )= m \A| and \H \A |<|tf |. Therefore, we may suppose that the statement of the theorem is true for H\A (if H M ). Let M ’ be the non-empty intersection of the maximal sets of H \A . We will use some lemmas during the proof of the theorem. Lemma 4.7 M' and S\T are non-disjoint sets. Proof Every maximum intersecting subfamily of P(M') contains M' itself, since it contains one of M' and NT=0, but it obviously cannot contain 0. Then, by induction, all intersecting families M *c H \A of maximum cardinality contain M'. Consequently, M 'eA f \A . If M 'cT holds, then A n M '= 0 holds for all A eA , since every set from A is completely inside S/T. But \Mc\A |= A |S1 and thus it would give two sets, M ' and an A gM &A , from M such that M 'uA =0. This is a contradiction, sinceM is a family of pairwise intersecting sets. This implies then that M'n(S\T)?t0. □ Lemma 4.8 The intersection of all the maximal sets of H is exactly M=M'n(S\T)?t0. Proof If xe M, then x is in exactly half of the sets from A, since A consists of pairs of disjoint sets whose union is S\TdM. On the other hand, x is in every set from M\A , i.e. in half of the sets of H\A. These together give us that the degree of x in H is exactly 68 2 1H |. Using Lemma 4.5, we obtain that M is contained by the intersection of all the maximal sets of H. If xg M , then x is either not in M' or not in (S\T). In the first case x is in at most half of the sets from A, but in less than half of the sets from H \A . Thus, the degree of x in H is less than £\H |. In the second case x is in at most half of the sets from H V I, but it is not in any set from A, since all of the sets of A are subsets of S\T by Lemma 2.3. But \A |>2, which implies that the degree of x in H is less than i\H\ again. Then, by Lemma 4.5, x can not be in the intersection of all the maximal sets of H. Now we know that the intersection of all maximal sets in H is exactly M (see Figure 7).n Since M\A is an intersecting family in H\A of maximum cardinality, it follows that M V I is like we described in the theorem. Also M V l can be completed to an intersecting family of cardinality z\H | from the pairs of A (since M d M V I and \M \=i\H |). Lemma 4.9 If M'cS\T, i.e. M=M', then M satisfies the description of Theorem 4.2. Proof M contains M VI , and M V I contains a maximal intersecting subfamily M' of P(M')=P(M). Thus M, being maximal intersecting subfamily, contains {Ae H : 3 B e M ', Be A}. The cardinality of this set is ?\H |, as it was shown in the beginning of the proof of Theorem 4.2. Since |M | can not exceed it, M = {A e H : 3 BeM ', BcA }. □ S T * N. "V* / Figure 7 T, M and M' in S 69 Lemma 4.10 If M'nT?t0, and M ( and M 2 are two subsets of M such that M tu M 2 =M, M ,n M 2 = 0, then M VI contains exactly one of M| and M 2. Proof If M\A does not contain M , and M 2, then M ,u (M ’n T ) and M 2 u(M'nT) are elements of M\A , since M\A contains exactly one set from the every two complementary sets of M', by the induction assumption and Corollary 1.3. We will prove that there exists a pair {A,B} of sets from the above described A such that one member of this pair, say A, intersects M in M lf so it does not intersect M 2 u (M 'n T ), and at the same time the other member of the pair, B intersects M in M 2. Consequently, it does not intersect M)U(M'nT). Let A), B,eA such that A|nB|=0, A,uB)=S\T. Choose two maximal sets, A2 and B 2 from H such that A2d A ( and B 2 oBt. Now A 2 cS\T, since otherwise S\(A 2 u (T \A 2 ))cB, and thus S\(A2 u (T \A 2 ))e //. But now |T\A 2 |<|T|, which contradicts the minimality of T (see Figure 8 ). Similarly, B 2 cS\T, and so S\TcA|UB,c A 2u B2cS\T. Here A 2 and B 2 are chosen to be maximal sets in H , and so, as was proven previously, both contain M. Let A=A 2 \M 2 and B=(B 2 \M /)\(A n(B 2 \M ,))=B 2\A, as shown on Figure 9. Then, since A cA 2 andBcB2, A and B are elements of H , and T\A A2 Figure 8 For the proof of A 2 cS\T 70 A2 b 2 Figure 9 The choice of A and B A n B = 0 , AuB=S\T, which implies that A,Be A . Moreover, they really satisfy the conditions A n M = M , and B n M = M 2. Now it follows thatM does not contain any member of the pair {A,B} and this contradicts | M r A |= A |. This contradiction implies that M \A contains exactly one of M, and M 2 (it obviously does not contain both of them). □ We got that M \A, and thus M also contain a maximum intersecting family of sets of P(M). Let us denote it by M '. Similarly to the proof of Lemma 4.9, we have, by the maximality of M, that M o{Ae H : S B eM ', BcA). Here the cardinality of the right hand side is {\H\, and since \M\ can not exceed it, M = {A e H : 3 B e M ', BcA}. Finally, we have to consider the case H =A. But then, since every element of A is a subset of S\T, every edge of H is a subset of S\T as well, and thus S\T=S. It means that SgH and so H contains P(S). On the other hand, H cannot contain any superset of S, 71 since S is the underlying set of H, and thus H =P(S). The statement of the theorem becomes trivial, by Corollary 1.3. We have shown that our assumption that ti>(H )= ||H | implies that the intersection of the bases of H is not empty. By Lemma 4.4, d (H )= %\H \ as well, and thus Chvatal's conjecture is true for H . □ Proof of Theorem 4.3 Now we have that |H | is odd and u>(// )=? j\H |- 5 . Suppose that M is an intersecting family in H of maximum cardinality and let TcS and A cH be the same as in the case i>>(H ) = %\H \. Now, by the argument used in the previous proof, we can see again that | M n A |= ||A | and u(H \A )=|fl/\A|= 5 |fAA|- £. We may suppose that the statement of the theorem is true for H \A and M \A because the different parity of \H | and \A | implies that H\A contains at least one element. In the case when H\A = {0 }, the statement of the theorem is trivial. Lemma 4. 11 Chvatal's conjecture is true for the hereditary hypergraph H, i.e., there exists an xe S such that d# (x)= £|H |- £. Proof We know that there exists either a maximal set of H\A (let us denote it by N *) such thatM\A is an intersecting family in ( H \A )\{N *} or there exists an N**cS, N **&H\A such that (M \A ) u { N * * } is an intersecting family in the hereditary hypergraph ( H \A )u {N **}. In both cases we can find an xe S\T such that the degree of x in the hereditary hypergraph H\A is ||//\A |- 5 . The proof of this is very similar to the one used in the case )= £\H |. First we prove that there is an element of S\T wh^se degree in H\A is £|H\A|. IfM\A is an intersecting family in (H\A)\{N *}, then \M\A |= g|(//\A)\{N*}| and so we may use the result already proven for w (// )= ||H |. The intersection of all the maximal sets in (H \A )\{N *} is not empty. If we denote it by M ' then we know that M'<=M\A , again using the result for the case when | H \ is even. But half of the sets from A (i.e., at least one set) is in the intersecting family M and thus have non-empty intersection with M'. On the other hand, the sets from A are completely inside S\T, and thus M ’n(S\T>*0. Let x be any element from M'n(S\T), then its degree in H\A is 72 \M \A |= g|(H \A )\{N *}|= ?\H \A |- The case when we have the intersecting family (M \A )u { N * * } in the hereditary hypergraph (H \A ) u { N * * } is similar, though we have to be more careful. It could happen that M', the non-empty intersection of the maximal sets of (H\A )u{N **} which is in (M \A )u {N **}, is exactly N **. (If it is not, then we can prove the existence of the required xe S\T exactly the same way as in the previous case.) If |N**|=1, i.e., N **= {y} for some ye S then since all the sets in M\A intersects N **, they must contain y and so N **= {y }e H V4 , a contradiction. So |N**|>1 which implies, by Theorem 4.2, that there is (at least one) subset of N **=M ', say K, which is inM VI and so intersects every set from M. If M '=N ** is completely in T then so is every subset of it, that is, KcT. The sets ofA are contained by S\T and there is at least one set, say A, of A which is inM as well. We have that KcT, AcS\T, and K rY IV 0, a contradiction. Therefore, M '= N ** cannot be completely contained by T, that is, M 'n(S\T)?i0 again. This gives us the existence of x with the required property. Thus, we have now an xe S\T such that the degree of x in H\A is k\H \A |- £. On the other hand, xe S\T implies that the degree of x in A is i\A | and hence dH (x) = {\H |- { follows. □ This completes the proof that Chvatal's conjecture holds for H, but we still have to prove that the induction statement is true for H as well. If M does not contain all maximal sets of H, then let N f be any base of H not in M. Then M cH \{N j} trivially holds. If M contains all maximal sets ofH, then let x be any element of S satisfying (x)= i\H |- {. By Lemma 4.6, there exists a maximal set N, of H such that x g N (. It is trivial to show that M u { N tu {x }} is an intersecting family. We will prove that H u { N tu {x }} is a hereditary hypergraph. Suppose that it is not the case. Then H does not contain some subsets H bH2 ,...,Hn of Nsu{x} (n>l). But H contains all subsets of N f and thus xeH,- (l< i< n ). It is easy to see thatW = 73 (tfu{N)u{x}})u{H 1,H 2 ,...,Hn} is a hereditary hypergraph and the degree of x in H ' is (x)+n+l = |(|H |+2n+l) > |(|H |+n+l) = ||JT '| which contradicts Berge's theorem. Thus H u {N ,u {x }} is a hereditary hypergraph and we have completely proven the theorem. □ 4.4 The generalizations for hereditary multi-hvpergraphs We can also give the generalizations of Theorems 4.1, 4.2, 4.3 for hereditary multi-hypergraphs, which again will be only kinds of corollaries of the similar theorems for hereditary hypergraphs. Theorem 4.12 If H is a hereditary multi-hypergraph on a finite set S and w(JT )= [||H |], then u>(H )=d(H ), where [.] denotes the integral part of a number. Again, Theorem 4.12 is a trivial corollary of the following two results. (We recall that in a hereditary multi-hypergraph H with multiplicity function f an He H was called a semi-base of H if f(H)>f(H') for every proper superset H ’ of H.) Theorem 4.13 Let H be a hereditary multi-hypergraph on a finite set S, satisfying ||H |) if and only if it arises in the following way. There is a maximum intersecting family M ' in P(M) such that an edge of H is in M if and only if some subset of it is in M' (and then all copies of this edge are in M , of course). It implies that u>(H) = d(H) holds. Theorem 4.14 If H cP(S) is a hereditary multi-hypergraph on a finite set S, |H \ is odd, and M cH is an intersecting subfamily of cardinality |( |f/ |- 1 ), then there exists either a semi-base B( of H (which has to be a maximal element of H as well) such that no copy of B) is contained by M (and so we can delete a copy of B, from H resulting a smaller hereditary multi-hypergraph having the same intersecting subfamily M) or a B 2 cS such that adding one copy of Bz to H and M gives another hereditary multi-hypergraph and another intersecting subfamily of it, respectively. It implies that d>(H )=d(H ) holds again. 74 Before the proofs of these theorems we give a lemma, whose idea has been already used earlier to prove some results about hereditaiy multi-hypergraphs. Lemma 4.15 If H is a hereditary multi-hypergraph on the finite set S, f is its multiplicity function, then the hypergraph H\ defined by //)= {H c S :f(H )> l} is a hereditary hypergraph and the multi-hypergraph H 2 defined by the multiplicity function f(H)=max {0,f(H )-l} for every HcS is a hereditaiy multi-hypergraph. Proof If AcBe//,, then f(A) ^ f(B) > 1 and so A z H x as well, proving that H\ is a hereditary hypergraph. If AcBcS, then f(A) > f(B) and so f(A )-l> f(B )-l. If f(B)-l>0, then f(A)-l>0 as well, and so f(A)=f(A)-l^f(B)-l=f’(B). If f(B)-l<0, then f (A)2:0=f (B). These prove that H z is really a hereditary multi-hypergraph. □ Proof of Theorem 4.13 We will prove this result by induction on |H |. Consider the hereditary hypergraph - hereditary multihypergraph decomposition of H into II , and I I z, given by Lemma 4.15. L etM be any intersecting subfamily of H of cardinality i\H\ (we have at least one of it by the assumption of the theorem), and let M t be the underlying hypergraph of it, and thus an intersecting subfamily of H L e tM 2 be the intersecting multi-hypergraph given by deleting a copy of every edge of M from M. Then M z is an intersecting subfamily of H z. Here \M\\+\MZ\= |M |= £|H | = |(|Jt/ 1|+|flr2|) and by Theorems 2.1 and 2.6 |A/t|< || H/\ and |M z\< i\H z\. Thus, we have that \M{\= ? \H ,| and |M 2|= ||H 2|. Now we can use Theorem 4.2 for H , and we may assume that the statement of this theorem is true for H 2, since |H 2|<|W |. Let M t be the intersection of the bases of H , and M 2 be the intersection of the semi-bases of H z. Then M ,e M , by Theorem 4.2 and M 2eA f 2 by induction, thus both M| and M z are elements of M . It implies that M ,n M 2 is not empty. On the other hand, it is easy to see that an edge of H is a semi-base if and only if it is a base of H \ or a semi-base of H z (it may be both at the same time). Thus, M = M !& M 2 is not empty, and it is the intersection of the semi-bases of H. Now, by Theorem 2.10, Lemma 4.16 For every N cM M contains either N or M\N. Proof If M does not contain any of N and M\N, then neither do M , and M 2. But both N and M \N are subsets of M, and M 2 as well. If M t does not contain N, then, by Theorem 4.2, it contains M t\N. If M z does not contain M \N , then, by induction, it contains M2 \(M\N). Therefore, both MAN and M 2 \(M \N ) are edges of M. But (M l\N )n (M 2 \(M\N))=(M)nM 2 )\M =0, a contradiction. Thus,M really has to contain one of N and M\N. □ Now, by Corollary 1.3, M really contains a maximal intersecting subfamily M 1 of P(M), and thus, being maximal, it contains all copies of every edge of H which has a subset in M'. To finish the proof of this theorem it is enough to show that this later intersecting subfamily, say N, has cardinality ?\H | since then M = N and every intersecting subfamily of H having the structure described in the theorem has maximum cardinality as well. Let be the underlying hypergraph of N and N z be the multi-hypergraph obtained by deleting one copy of every edge of N from N. Then |/V,|= i\H\\ by Theorem 4.2 and |W2|= i\H z\ by induction, and thus \N |= |/V,|+|Af2|= k\Hz\= i\H |, the required result. □ Proof of Theorem 4.14 Again, we will prove the theorem by induction on \H |. Consider the decomposition of H into H x and H z and the decomnposition of M into M t and M z, as it were given in Lemma 4.15 and in the proof of Theorem 4.13, where is a hereditary hypergraph, H z is a hereditary multi-hypergraph, Af, and M z are intersecting subsystems ofH\ and H z, respactively. Easy counting shows that either \M\\= ||H ,| and |M 2|= g(|H 2 |-1) or |M ,|= |(|H ,|-1 ) and \M 2|= ||H 2|. In the first case, by induction, either there exists a semi-base B, of H z (which is also a semi-base ofH) such thatM z does not contain any copy of this edge or we can add a new edge to M z and H z such that they preserve their intersecting and hereditary properties, respectively. In the second case we can use Theorem 4.3, and either we can delete an edge of H\ which is not contained in M | or we can add an edge to H i and M\ preserving their hereditary and intersecting properties, respectively. In any of these cases, we can perform the same operation on H and M, which proves the structure part of the theorem. If we can delete a copy of an edge of H which is not in M , then in the smaller hereditary multi-hypergraph we can use Theorem 4.13, which gives a star of cardinality \M |. Thus, in H we also have a star of this cardinality, proving u(H )=d(H). If we can add a new edge to II and M preserving the hereditary and intersecting properties, respectively, then in the larger hereditary multi-hypergraph we can find a star of cardinality \M |+1 , and so in H there is a star of cardinality \M |, which again proves that w (// )=d(H ). □ CHAPTER 5 A MORE GENERAL CONJECTURE Since 1972, when Chvatal published his conjecture, mathematicians have tried to formulate and prove more general, and in some cases more specific conjectures. One possibility to prove the conjecture is to try to apply some kind of induction, like it is used in the proof of some special cases. The following proposition, which can be partly found in Schonheim's paper [20], was suggested by these trials to apply induction. Proposition 5.1 If there is a hereditary hypergraph H which gives a counterexample for Chvatal's conjecture, then the minimal counterexamples satisfy the following properties. (Here, by minimal counterexample we mean a hereditary hypergraph H for which Chvatal's conjecture does not hold such that Chvatal's conjecture is valid for every H'cH hereditary hypergraph.) (i) There are no two bases of H whose union is S, the underlying set of H. (ii) There are no two bases of H whose intersection is empty. (iii) There is no base of H which contains all the points ofH of maximum degree. (iv) Every intersecting subfamily of H (of maximum cardinality) contains all the bases of H. Proof We will prove that if Chvatal's conjecture is true for every hereditary hypergraph H 'cH and H does not satisfy any of (i) - (iv), then Chvatal’s conjecture is true for H as well. If there are two bases H, and H 2 of H such that H ,u H2 =S, then H ,u 0 e /T = {H : H e H }. Now, we can use the same induction step that was used in Section 2.2 to give another proof for Serge's theorem. Let H :R eH }, then, by Lemma 2.2, H\H\ =HZ is also a hereditary hypergraph. If M is an intersecting subfamily of H, then M z = H zr M is also an intersecting subfamily, and so by our assumption,\MZ\ < d(Hz). 77 78 On the other hand, by Lemma 2.3, H\ consists of pairs of the edges ofH whose intersection is empty, which implies that M S = H (n A / contains at most one edge from every pair and thus \M (| < ?|H ,|. It is easy to see that d (H ) = d{H 2) + k\H\\, and then |M | = |M,| + \MZ\ <, d(f/2) + £|//,| = d(// ). If there are two bases H ( and H 2 of H whose intersection is empty, then any intersecting subfamily M of H contains at most one of these bases, say H,. Then H ' = H \{H ,} is another hereditary hypergraph, M is an intersecting subfamily of H ' as well, and thus, by our assumption, \M |< d(H \{H (}) < d(H ). If H is a base of H such that xe H for every x with d(x) = d (H ), then delete H again from H, resulting another hereditary hypergraph H For any intersecting subfamily M of H, we have that |M \{H }| < d( H '), and thus, \M | ^ \M \{H}| + 1 ^ d(H ') + 1 = d(H). The case (iv) was proven in Section 4.1, where we justified that Conjecture 4 is really equivalent to Chvatal's conjecture. □ We note here that (ii) in the above proposition is clearly a special case of (iv), since if there are two bases with empty intersection, then there is an intersecting subfamily (in fact, all of them) which does not contain all the bases. Another trial to prove the conjecture led to the following, stonger conjecture. Conjecture 5 If H is a hereditary hypergraph and M is an intersecting subfamily of H , then every edge of M contains a point whose degree in H is at least |M |. Actually, while Chvatal's conjecture claims only that there is a point xe S (where S is the underlying set of H ) such that d(x) > |M |, this conjecture states that there is such a point in every edge of M. We can prove the following statement, similar to Proposition 5.1, about this conjecture. Proposition 5.2 If there is a hereditary hypergraph H, an intersecting subfamily M of H , and an edge M of M which give a counterexample for the above conjecture, i.e. for every xeM , d(x)<|M |, then the minimal counterexample satisfies the following properties. (Where by minimal counterexample we mean a triple H, M, and M such thatH is minimal among the counterexamples.) (i) M contains all the bases ofH. (ii) There is no proper subset of M in M. 79 Proof Like in the previous case, we will prove that if Conjecture 5 is true for every such a triple (M 1, M H ') where H 'c H, and the triple (M, M , H ) does not satisfy (i) and (ii), then the conjecture is true for (M ,M ,H ) as well. If there is a base H, of H which is not inM, then deleting this edge from H results in a hereditary hypergraph with fewer edges than H, where we have the same intersecting subfamily M and its edge M , and the degrees of the points are smaller or the same as in H. By our assumption, M has a point of appropriate degree, and then the same point will satisfy the conjecture in the larger hereditary hypergraph H. If N is a real subset of M and Ne M , then let M* be a base containing M and consider the intersecting subfamily M t=A /\{M '} in the smaller hereditary hypergraph H t=H\{M'}. By induction, N has a point of degree at least |flf||=|M \{M '}|=|M | - 1 in H u and since M'dN, this point has degree one more, i.e. degree \M | in H =H tu { M ’} satisfying the required property. □ This proposition implies that it would be enough to prove Conjecture 5 for the maximal sets (i.e., bases) of the intersecting subfamilies. In fact, it seems to be true that even the average of the degrees of the points of a maximal set in an intersecting subfamily M is at least |M |. The following, much weaker version of the above conjecture assures that we at least have a point in every edge of an intersecting family M such that the degree of this point is relatively large, depending on the cardinality of M, and, in particular, it goes to infinity as |M | goes to infinity. This result turns out to be useful in the next chapter, for the infinite case of Chvatal's conjecture, though we will give there another proof of a simplest version of it, using infinite techniques. Theorem 5.3 If H is a hereditary hypergraph and M is an intersecting subfamily of H with \M |=m, then every edge of M contains a point of degree at least m/log 2 m. Proof Let M be any edge of M, and denote k the cardinality of M. There are m-1 other edges of M intersecting M, so there has to be a point x in M which is contained by (m -l)/k of these edges. It implies that the degree of x is (m -l)/k+l in M, and thus <%(x)> [(m -l)/k]+l. On the other hand, there are 2 k _ 1 subsets of M which contain x, so dH (x) > 2k_1. Now, it is enough to prove that max{ [(m -lj/k j+ l^ '1} > m/log 2 m. If [(m -l)/k]+l> 80 m/log2 m, we are done. Otherwise, since k > 1, m/k < [(m -l)/k] + 1 < m/log 2 m, and so log2m < k, which implies that m < 2k. Now 2k _ 1 > m/2 > m/log2m if m>4, we are done in this case. If the cardinality of M, m is less than four, the situation is very simple, and the statement of the theorem can be checked easily by cases. □ Note that during the proof of Theorem 5.3 we used the hereditary property of H only to obtain that every subset of M is contained by H. Thus, the same proof works for hereditary multi-hypergraphs, giving the following result. Theorem 5.4 If H is a hereditary multi-hypergraph and M is an intersecting subfamily of H with |M |=m, then every edge of M contains a point of degree at least m/log2 m. □ CHAPTER 6 THE INFINITE CASE 6.1 Introduction In this chapter, we will investigate the infinite version of Chvatal's conjecture. In order to state it, we need some definitions and notations. Throughout this chapter, all lower case Greek letters (except w) will denote (usually infinite) cardinals. R 0 will denote the first infinite cardinal, i.e. the cardinality of the positive integers, while c will be used to denote the cardinality of the real numbers. We will identify the positive integer n and the set {0,1, ..., n-1}, as usual, and 0 will denote the cardinality of the empty set as well. We mention here that a cardinal K is the set of all of the cardinals less than K. Similarly for the finite case, |S| will denote the cardinality of the set S. The definitions of infinite hypergraphs, hereditary hypergraphs and pairwise intersecting hypergraphs (or of these objects in general) are exactly the same as the definitions of these objects in the finite case, with the exception that now the underlying set S may be of infinite cardinality as well. We will say that a hypergraph is infinite if its underlying set is infinite, with underlying set here meaning the union of the edges of the hypergraph. A multi-hypergraph, in general, on the (not necessarily finite) set S is defined by the multiplicity function f:P(S) —> k, where k is any cardinal, and thus, the set of the cardinals less thank. An HcS is said to be an element of the multi-hypergraph H if and only if f(H)>0. The multiplicity of an HcS in H is f(H). A multi-hypergraph H is hereditary if for every H eH and H'oH, f(H')>f(H) holds. Since a hypergraph H is a set (of sets) itself, |H | will denote its cardinality. The cardinality of a multi-hypergraph H on the set S given by the multiplicity function f is defined by |H I=2 hcS The two important parameters of finite (multi-)hypergraphs, u>(H ) and d(H), 81 82 should be defined here differently, since in the infinite case we may not have the maximum of the degrees of the points in S, for example. If H is any (multi-)hypergraph, then w (f/) denotes the supremum of the cardinalities of the pairwise intersecting subsystems of H, d//(x) (or sometimes d(x), as it was used in the finite case) denotes the degree of an element x of S, i.e. the cardinality of the set of those edges of H which contain x, and d(H ) denotes the degree of the (multi-)hypergraph, defined to be the supremum of the cardinals % (x) for the elements x of S. We mention here that the supremum of a set of cardinals, being equal to the union of these cardinals, always exists. With the help of these notations we can formulate the general version of Chvatal's conjecture, which includes both the finite and infinite cases. Conjecture 6 If H is a (finite or infinite) hereditary (multi-) hypergraph, then w ( //) = d (H). Obviously, this conjecture would imply the original conjecture of Chvatal. On the other hand, as we will see later, Conjecture 6 can be split into two parts and handled independently, considering the finite and the infinite cases, respectively. There are some similarities and some differences between these versions of the conjecture. First of all, since a star is a pairwise intersecting (multi-)hypergraph in the infinite case as well, w (W ) ^ d (H ) holds for every (multi-)hypergraph (finite or infinite, hereditary or not hereditary) H, and thus it would be enough to show the other inequality for hereditary (multi-)hypergraphs to prove Conjecture 6 . Also, one can find infinite hypergraphs with the property u>(H) > d(H ). One example of this can be found by taking the union of infinitely many copies (taken on disjoint underlying sets) of a finite hypergraph H with the property u(H ) > d (H ). In this case, K 0>u>(H)>d(H). We can also find examples where u>(H) > K 0 > d(H ). These are not so trivial, so we will show one here. Let An be the subset of the positive integers defined for every n > 1 in the following way: A n = Bn u Cn, where Bn = {2n k + 2 n _ 1 -1 : k is a positive integer} and Cn = {2 n _ 1 - 1 , 2n_1, 2" ' 1 +1, ... , 2n -2} if n > 1, C, = 0. Consider the hypergraph H defined on the positive integers whose edges are the sets A n, 83 n=l, 2,... . It is easy to see that the sets Bn are pairwise disjoint as well as the sets Cn. Thus any positive integer is in no more than two of the sets An. On the other hand, 2e B, and thus, 2e A,, and 2e C2 and thus, 2e A 2, which implies that d(2) = d(H ) = 2. Since the set Cn contains {(2n-2)-(2n' 1 - l)} + l = 2” ' 1 consecutive positive integers, it contains at least one of the form 2m k + 2 r a _ 1 - 1 for every 1 < m < n, and thus Cn intersects every Bm for 1 < m < n. Therefore, An intersects every Am for 1 < m < n, which implies that H is a pairwise intersecting hypergraph itself, and so ) = \H \ = K 0 > 2 = d(H ). We also mention two of the differences between the finite and infinite versions of Chvatal's conjecture. In the finite case, using Theorem 2.1 or Corollary 2.9, we saw that for every hereditary (multi-)hypergraph d(H ) < w(W ) <\H |. In the infinite case, the following example shows that u>(H) =\H | may hold. Let H be the hereditary hypergraph on the underlying set of the positive integers consisting of all of the subsets of the positive integers. Then, as is well-known, |H |= c and dH (n) = c for every positive integer n, which implies that in this case d(H ) = u>(H ) = \H |. O f course, every intersecting subfamily M of a hereditary hypergraph H is strictly contained by H; this example only shows, that\M\ = \H\ may occur. In the finite case, the statement that d (H) = u>(H) for a (multi-)hypergraph H is equivalent to the fact there is a star among the maximal intersecting subfamilies of H. In the infinite case, this is not true, i.e., the first statement does not imply the second one. First of all, it may even happen that there are no maximal intersecting subfamilies of H i.e., w (H ) is a limit cardinal and we have intersecting subfamilies of H only for cardinals less than o>(H), but the supremum of these cardinals is u>(H). Or it can be the case that u (H ) is a limit cardinal again, but there is an intersecting subfamily of H with cardinality u>(H ). On the other hand, there is a star in H with cardinality k for every cardinal k < w(H ), but there is not one having cardinality w(JZ ). We will encounter these difficulties in this chapter. The other implication, of course, is still valid, that is, if there are intersecting subfamilies of H of maximum cardinality and there is a star among 84 them, then d(H ) = u (H ) holds. We will show another major difference between the finite and infinite versions of Chvatal's conjecture, namely, that while the first case is still open, we can prove the second one. Moreover, the infinite version of Conjecture 5 will be proven as well. 6.2 If there is an intersecting subfamily of H of infinite cardinality In this section, we will handle the case when there is an intersecting subfamily of H of infinite cardinality. This assumption is stronger than the assumptionw(H ) > N 0, as we saw in the previous section. Theorem 6.1 If H is an infinite hereditary hypergraph and H has an intersecting subfamily of infinite cardinality, then Chvatal's conjecture holds for H, i.e. subfamily of H of infinite cardinality, then for every M e Af the supremum of the degrees of the points of M , counted in H, is at least as large as M| |. Proof Let H be a hereditary hypergraph, M an intersecting subfamily of H with \M\> K o, and M be any edge of M. Let \M | and |M| be denoted by K and X, respectively. Thus we have that K > N 0. C a s e 1 ( X < K ) In this case M has X elements and there are k edges of H intersecting M , since every set from M intersects M. This implies that the sum of the degrees of the elements of M in M (and thus, also inH ) is at least \M |= k. We have X < k cardinals (which are all at least one, since every element of M, being in M , has degree at least one) whose sum is at least k. Let p. denote the supremum of these cardinals, then the sum of them is exactly max{ |i,A, }. Now, k > X and max{ (I, X.} > k imply that |i > K, which is the required result. We note that if K is a regular cardinal (for example, if 85 it is not a lim it cardinal), then it can not be the sum of X < K cardinals which are all less than K, and thus, in this case there must be an element of M with cardinality at least k. Case 2 ( X > k ) Since H is a hereditary hypergraph, it contains every subset of M. Let x be any element of M. As in the finite case, we can pair the subsets of M such that exactly one of the sets of every pair contains x, by taking the pairs {N ,M \N } for every NcM . This implies again that half of the subsets of M contain x, and since X > K > X 0, we have 2f subsets of M containing x. Since all of these subsets are in H, dH (x) > 2^ > X >: K, that is,every element of M has degree larger than k, which obviously gives the required result. □ Proof of Theorem 6 .1 We will distinguish two cases again. Case 1 (There is an intersecting subfamily M ofH with cardinality oj(H )). Then, by Theorem 6.2, the supremum of the degrees of every element of this particular intersecting subfamily M is at least \M \ = d>(H), and so d(H ) > X (7 /) obviously. But, as we mentioned earlier, d (H) < w(H) is true in general, which implies that d(H)= u(H). Case 2 (There is no intersecting subfamily M ofH with cardinality w(H )). In this case, only the supremum of the cardinalities of the intersecting subfamilies of H is w(H ). Therefore, w (H ), which will be denoted by k, is a limit cardinal. Moreover, from the assumptions of the theorem, k > X0. We have to prove that the supremum of the degrees of the points of H is at least K, as well, which will imply the theorem together with the last sentence of the previous case. Let X be any cardinal less than k. It is enough to prove that there is a point x of H such that d(x) > X. Since K is a limit cardinal andX is less than k, we can find an infinite cardinal n with the property X < \i < K. The supremum of the cardinalities o f the intersecting subfamilies of H is K, thus, we have an intersecting subfamily M with \M \> (I > X 0. Applying the previous theorem to this M, we get 86 that the supremum of the degrees of the elements of any M e ilf is at least \M | > |i. But X < |i, and so we will have an element of degree X, which we wanted to prove. □ Again, during the proof of Theorems 6 .1 and 6.2 it is enough to assume that H is an infinte hereditary multi-hypergraph with the required properties, and thus these proves serve as proves of the following theorems. Theorem 6.3 If H is an infinite hereditary multi-hypergraph and H has an intersecting subfamily of infinite cardinality, then u>(//) = d (H ) holds forH. □ Theorem 6.4 If H is an infinite hereditary multi-hypergraph and M is an intersecting subfamily of H of infinite cardinality, then for every M e M the supremum of the degrees of the points of M, counted in H, is at least as large as \M |. □ 6.3 The general infinite case In this section, we aim to prove the stronger version of Theorem 6.1, replacing the assumption that we have an intersecting subfamily of infinite cardinality with the weaker one that weaker version of it. We will prove this result here in another way, using Kbnig's lemma, which is well-known to be very useful in this type of situations. Proposition 6.5 For every infinite sequence of finite hereditary hypergraphs H hH z, H 3, ..., H n, ... ,such that u (H n) -» «*>, d(tfn) goes to infinity as well. Proof If the statement of the theorem is not true, then there is an infinite sequence of finite hereditary hypergraphs H n, n=l,2,... and a positive integer k such that u>(// n) goes to infinity as n —> °°, but d(H n) < k for every n. It implies that if we take all of the finite hereditary hypergraphs H defined on the positive integers with the property that d(H) < k, and pick up in all of them the intersecting subfamilies M, forming the pairs (HM ), the size of M can be as large as we want. 87 For every pair (HJM), we will reduce H by substituting it with the hereditary closure of M, denoted by H H ' is trivially a subhypergraph of H, and thus d(H ') < k, and H' contains the intersecting subfamily M as well. Now, the pairs (H 'M ) are uniquely determined by M. Let s be the smallest positive integer such that k < 2s. If there were any edge M of M of cardinality larger than s, then the points of this edge would be in at least 2 s subsets of it, implying that the degree of every element of M would have degree at least 2s > k in H ', a contradiction. Thus, every set from M has cardinality at most s, and the underlying S set of H ', being equal to the union of the edges of M , has cardinality at most s>|M |. With respect to every ordering M h M 2, ... of the edges of M, renumber the points of S on every possible way, such that the points of M, are the first |M,| positive integers, the points which are in M 2 but not in M, are the next |M 2 \M|| positive integers, the points which are in M 3, but neither in M 2 nor in M , are the next |M 3 \(M 1u M 2)| positive integers, etc. With this procedure, we obtain finitely many intersecting hypergraphs M t for every positive integer t such that |Mt| = t. The underlying set of the hereditary closure H\ of M l is a subset of {1 , 2 ,... ,ts}, and d(H\) < k. We define an infinite graph on the following way. The vertices of the graph are partitioned into infinitely (more exactly into K 0 many) levels, L 0, L,, L 2, .... The level L 0 consists of the empty set. Each level Lt , for t > 0, consists of the previously obtained intersecting families M v There will be an edge of the graph joining the intersecting family M t of the t 111 level with the intersecting family M M of the level L t.j if and only if M t. j is a subfamily of M t. We have finitely many vertices of the graph on each level, and, since for every M t in the graph we can obtain a subhypergraph of M t still in the graph by deleting the last edge of M t in the given order, then there is an edge of the graph from every M t of every level Lt (t > 0) to the level Lt_j. We apply KOnig's lemma now, obtaining an infinite path in the graph containing exactly one vertex from every level. This infinite path corresponds to an infinite sequence 88 M 0, M ,, ... of intersecting hypergraphs such that 0 = M 0 c M ( c M z c .... All of these hypergraphs are defined on the set of the positive integers, |M t| = t, and the hereditary closure H 't of M t satisfies d (// 't) < k. Let M be the union of these intersecting hypergraphs and H be the hereditary closure of M . Then H is obviously the union of the hereditary closures H 'tof the hypergraphs M t, and we have H\cH\c H \ c ... as well.The cardinality of M is X 0, M is trivially intersecting, and M cH, which imply that u>(// ) = X 0. If H has any point of degree at least k, then pick up k edges of H containing this element, and for each of these edges take an H 't which contains the edge. If m is the maximum of the indices of these hereditary hypergraphs, then this point of H has degree at least k in H 'm already. But H ' is the hereditary closure of M m, and thus d(H ’m) < k, a contradiction. We obtained that every point of H has degree less than k, and so d (H ) < k. On the other hand, as we saw, H has an intersecting subfamily of infinite cardinality, and these together contradict Theorem 6.1. Our assumption, that there is an infinite sequence of finite hereditary hypergraphs violating the statement of the proposition led to a contradiction, and so the proposition is proven. p Now, we are ready to prove our final results. Theorem 6 . 6 If H is an infinite hereditary hypergraph and w(H) > X 0, then w(Zf ) = d(H ). Proof If H has an intersecting subfamily of infinite cardinality, then the statement follows from Theorem 6.1. Thus, we may suppose that H does not have an intersecting subfamily of infinite cardinality. Then w ( //) = X 0 and this together with fact that eveiy subfamily of an intersecting family is intersecting again, imply that we have intersecting subfamilies M of H with |M |= m for every positive integer m. I f / / had an edge of infinite cardinality, say H e // with |H|= K > X B, then every point in H would be contained by 2K > k > X 0 subsets of H, and thus these points ofH 89 would have degree at least 2K > N 0, a contradiction. Thus every edge of H is finite. We want to prove that there is a point of H of degree at least n for every positive integer n, since it implies that d (H ) = K 0. Since u>(H )= N 0, we have arbitrarily large finite intersecting subfamilies in H. LetM|,M2, ..., M h ... be a sequence of intersecting subfamilies of H such that \M f| > i for every i=l,2,.... Let H j be the hereditary closure of M j for every i=l,2,... . Then the H ,'s are finite herditary hypergraphs and u>(// () goes to infinity. Thus, d(H () goes to infinity as well, by Proposition 6.5, i.e. there is an H j whose degree is at least n. But all of these H ,*s are obviously subhypergraphs of H, and so the point having degree at least n in H {has degree at least n in H as well. 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