A sharp bound for the product of weights of cross-intersecting families

Two families A and B of sets are said to be cross-intersecting if each set in A intersects each set in B. For any two integers n and k with 1 6 k 6 n, let ([n] 6k ) denote the family of subsets of {1, . . . , n} of size at most k, and let Sn,k denote the family of sets in ([n] 6k ) that contain 1. The author recently showed that if A ⊆ ( [m] 6r ) , B ⊆ ( [n] 6s ) , and A and B are cross-intersecting, then |A||B| 6 |Sm,r||Sn,s|. We prove a version of this result for the more general setting of weighted sets. We show that if g : ( [m] 6r ) → R+ and h : ( [n] 6s ) → R+ are functions that obey certain conditions, A ⊆ ( [m] 6r ) , B ⊆ ( [n] 6s ) , and A and B are cross-intersecting, then ∑


Introduction
Unless otherwise stated, we shall use small letters such as x to denote elements of a set or non-negative integers or functions, capital letters such as X to denote sets, and calligraphic letters such as F to denote families (i.e.sets whose elements are sets themselves).It is to be assumed that arbitrary sets and families are finite.We call a set A an r-element set, or simply an r-set, if its size |A| is r.For a set X, the power set of X (that is, the family the electronic journal of combinatorics 23(4) (2014), #P4.45 of all subsets of X) is denoted by 2 X , the family of all r-element subsets of X is denoted by X r , and the family of all subsets of X that have at most r elements is denoted by X r .The set of all positive integers is denoted by N. For any m, n ∈ N with m < n, the set {i ∈ N : m i n} is denoted by [m, n].We abbreviate [1, n] to [n].
We say that a set A intersects a set B if A and B contain at least one common element.A family A of sets is said to be intersecting if for every A, B ∈ A, A and B intersect.Families A 1 , . . ., A k are said to be cross-intersecting if for every i and j in [k] with i = j, each set in A i intersects each set in A j .
For x ∈ X and F ⊆ 2 X , we denote the family {F ∈ F : x ∈ F } by F(x).If F(x) = ∅, then we call F(x) the star of F with centre x.A star of a family is the simplest example of an intersecting subfamily.
One of the most popular endeavours in extremal set theory is that of determining the size of a largest intersecting subfamily of a given family F.This took off with [27], which features the classical result, known as the Erdős-Ko-Rado (EKR) Theorem, that says that if r n/2, then the size of a largest intersecting subfamily of [n]   r is the size n−1 r−1 of every star of [n]  r .There are various proofs of the EKR Theorem, two of which are particularly short and beautiful: Katona's [38], introducing the elegant cycle method, and Daykin's [23], using the fundamental Kruskal-Katona Theorem [39,41].Various generalizations and analogues have been obtained; of particular note are the results in [40,29,50,1].The EKR Theorem inspired a wealth of results that establish how large a system of sets can be under certain intersection conditions; see [24,30,28,17,33].
For intersecting subfamilies of a given family F, the natural question to ask is how large they can be.For cross-intersecting families, two natural parameters arise: the sum and the product of sizes of the cross-intersecting families.It is therefore natural to consider the problem of maximizing the sum or the product of sizes of k cross-intersecting subfamilies (not necessarily distinct or non-empty) of a given family F. In [19], this problem is analysed in a general way, and it is shown that for k sufficiently large it reduces to the problem of maximizing the size of an intersecting subfamily of F. Solutions have been obtained for various families, as outlined in [19,9].Wang and Zhang [49] solved the maximum sum problem for an important class of families that includes [n]   r and many others, elegantly combining the method in [10,11,12,20,14] and the no-homomorphism lemma [3,21].The solution for [n]   r had been obtained by Hilton [35] and is the first result that addressed the cross-intersection problem described above.Pyber [47] solved the maximum product problem for [n]   r (see also [43,6]).The maximum product problem for [n]  r has been solved in [9], which actually provides the solution to the more general problem where the cross-intersecting families do not necessarily come from the same family.
s , and A and B are cross-intersecting, then the electronic journal of combinatorics 23(4) (2014), #P4.45 and equality holds if A = {A ∈ [m]  r : 1 ∈ A} and B = {B ∈ [n]  s : 1 ∈ B}.We consider the more general setting where the sets are assigned weights (positive real numbers).The weight of a family is the sum of weights of its members, and the objective is to maximize the product of weights of the cross-intersecting families.Before stating our main result, we need some additional definitions and notation.We also point out that, as explained in the next section, a product result such as Theorem 1, where the product is maximum when the cross-intersecting families are stars with a particular center, automatically yields an EKR-type result and generalizes to one for k 2 crossintersecting families.
For any i, j ∈ [n], let δ i,j : 2 [n] → 2 [n] be defined by be the compression operation defined by The compression operation was introduced in the seminal paper [27].The paper [30] provides a survey on the properties and uses of compression (also called shifting) operations in extremal set theory.All our new results make use of compression operations.
A family H is said to be hereditary if for each H ∈ H, all the subsets of H are in H. Thus, a family is hereditary if and only if it is a union of power sets.The family r (which is 2 [n] if r = n) is an example of a hereditary family that is compressed.We mention that one of the central problems in extremal set theory is a conjecture of Chvátal [22] that claims that at least one of the largest interesting subfamilies of any hereditary family H is a star of H; a similar conjecture for levels of H is made and partially solved in [15], and generalizes [37,Conjecture 7].
Let R + denote the set of positive real numbers.For a non-empty family F, a function w : F → R + (a weight function), and a subfamily A of F, the sum A∈A w(A) (of weights of sets in A) is called the w-weight of A. With a slight abuse of notation, the w-weight of A is denoted by w(A).Note that if A is empty, then w(A) is the empty sum, which is 0 by convention.
The following is our main result, which we will prove in Section 3.
and ∅ = H ⊆ 2 [n] such that G and H are hereditary and compressed.Let g : G → R + and h : H → R + be functions such that (a) g(G)  In [9], it is proved that the result also holds if g(G) = 1 = h(H) for every G ∈ G and every H ∈ H; Theorem 1 is the special case where G = [m]  r and H = [n]  s .The proof of Theorem 2 is based on induction, compression, and a subfamily alteration.The method can be summarized as follows.We use induction on m + n.We take A and B to be such that the product of their weights is maximum.The challenging part is the case m = n.The first problem that arises is that we can have a set A ∈ A and a set B ∈ B that intersect only in n; in this case, we cannot simply remove n and apply the induction hypothesis.Thus, we consider two alterations: removing A from A and adding B\{n} to B, and removing B from B and adding A\{n} to A. This yields two new pairs of cross-intersecting families.The second problem is that the product of the weights of a new pair obtained in this way may become smaller.The critical part of the proof is the observation that if we assume that this happens for both pairs, then we can construct a new pair of cross-intersecting families for which the product of weights is larger than that for A and B (unless we have the trivial case m = n = 2), hence contradicting the initial assumption.

Applications of Theorem 2
We will show that Theorem 2 yields cross-intersection results for integer sequences and for multisets.
We represent a sequence a 1 , . . ., a n by an n-tuple (a 1 , . . ., a n ), and we say that it is of length n.We call a sequence of positive integers a positive sequence.We call (a 1 , . . ., a n ) an r-partial sequence if exactly r of its entries are positive integers and the rest are all zero.Thus an n-partial sequence of length n is positive.A sequence (c 1 , . . ., c n ) is said to be increasing if c 1 . . .c n .We call an increasing positive sequence an IP sequence.
We call {(x 1 , y 1 ), . . ., (x r , y r )} a labeled set (following [16]) if x 1 , . . ., x r are distinct.For any IP sequence c = (c 1 , . . ., c n ) and any r ∈ [n], let L (r) c be the family of all labeled sets {(x 1 , y x 1 ), . . ., (x r , y xr )} such that {x 1 , . . ., x r } ∈ [n]   r and y c , we obtain that L Moreover, if c 1 3 and d 1 3, then equality holds if and only if Thus, Theorem 3 is equivalent to the following: for c, d, r and s as in Theorem 3, if It is immediate from Theorem 3 that c and d do not need to be increasing as long as there exists p ∈ intersect, and hence we can take c .The case where 3 c 1 + d 1 5 seems to require special treatment and remains a problem to be investigated.However, for the special case where c = d and r = n, we easily obtain from Theorem 3 the sharp bound for all values of c 1 , given by the following.
This result is also proved in Section 4. It was first established in the preliminary version [13] of this paper.An alternative proof has been obtained by Pach and Tardos [46].Sum versions of Theorems 3 and 4 are given in [20] and [16], respectively (see also [49]).The EKR problem for L c and L (r) c has been widely studied, and several results have been obtained.The EKR-type version of Theorem 4 (that is, the solution to the problem of maximizing the size of an intersecting subfamily of L c ) is given in [4,42,16] (for c 1 = c n , this is given in a stronger form in [31,2,32]).The EKR problem for L (r) c has been solved [24,36,5]; see [24,25,8,26,7] for c 1 = c n , [36] for c 1 2, and [5] for c 1 = 1.The special case c 1 = c n of Theorem 4 was treated by Moon [45] (for c 1 3), Tokushige [48] (for c 1 4) and Zhang [51] (for c 1 4) via an induction argument, an eigenvalue method and Katona's cycle method, respectively.Allowing c to be increasing appears to be a significant relaxation for the product problem.Our approach is based on the idea of the electronic journal of combinatorics 23(4) (2014), #P4.45 generalizing the setting enough for induction to work.The setting of Theorem 2 not only allows us to deal with the more general problem for L (r) c , but also to obtain Theorem 3, where the cross-intersecting families can come from different families.
Our second application of Theorem 2 is a cross-intersection result for multisets.
A multiset is a collection A of objects such that each object possibly appears more than once in A. Thus the difference between a multiset and a set is that a multiset may have repetitions of its elements.We can uniquely represent a multiset A of positive integers by an IP sequence (a 1 , . . ., a r ), where a 1 , . . ., a r form A. Thus we will take multisets to be IP sequences.For A = (a 1 , . . ., a r ), the support of A is the set {a 1 , . . ., a r } and will be denoted by S A ; thus, S A is the set of distinct elements of A. For any n, r ∈ N, let M n,r denote the set of all multisets (a 1 , . . ., a r ) such that a 1 , . . ., a r ∈ [n]; thus M n,r = {(a 1 , . . ., a r ) : a 1 . . .a r , a 1 , . . ., a r ∈ [n]}.An elementary counting result is that With a slight abuse of terminology, we say that a multiset A intersects a multiset B if A and B have at least one common element, that is, if S A intersects S B .A set A of multisets is said to be intersecting if every two multisets in A intersect, and k sets A 1 , . . ., A k of multisets are said to be cross-intersecting if for every i, j ∈ [k] with i = j, each multiset in A i intersects each multiset in A j .
In Section 5, we prove the following result.
s , and A and B are cross-intersecting, then Moreover, if u = 0 and v = 0, then the bound is attained if and only if for some a ∈ EKR-type results for multisets have been obtained in [44,34].To the best of the author's knowledge, Theorem 5 is the first cross-intersection result for multisets.It is an analogue of the product version in [47,43] of the EKR Theorem.
As indicated in Section 1, the results above imply EKR-type theorems.In general, if I ⊆ F, k 2, and the sum or the product of sizes of k cross-intersecting subfamilies then I is a largest intersecting subfamily of F. Indeed, the cross-intersection condition implies that every two sets A and B in I intersect (as A ∈ A 1 and B ∈ A 2 ), and by taking an intersecting subfamily A of F, and setting Similarly, if the sum or the product of weights is maximum when As also indicated in Section 1, the results above generalize for k 2 families.For example, applying the line of argument in the proof of [18,Theorem 1.2] to Theorem 2 yields the following generalization of Theorem 2.
the electronic journal of combinatorics 23(4) (2014), #P4.45 Theorem 6.Let k 2, n 1 , . . ., n k ∈ N, and u 1 , . . ., u k ∈ {0} ∪ R + such that u i + u j 2 for every i, j ∈ such that H i is hereditary and compressed, and let h i : H i → R + be a function such that (a) h i (H) (1 + u i )h i (H ) for every H, H ∈ H i with H H , and (b) h i (δ p,q (H)) h i (H) for every H ∈ H i and every p, q ∈ [n i ] with p < q.If A 1 , . . ., A k are cross-intersecting families such that

Moreover, equality holds if and only if
We simply observe that a i a j and that if A 1 , . . ., A k are cross-intersecting, then any A i and A j with i = j are cross-intersecting.Thus, if, for example, A 1 , . . ., A k are as in Theorem 6, . We now start working towards the proofs of Theorems 2, 3, 4 and 5.

Proof of the main result
This section is dedicated to the proof of Theorem 2.
For the extremal cases, we shall use the following lemma.
Lemma 7. Let H be a compressed subfamily of 2 [n] , and let w : We need to use the following well-known properties of compressions.It is straightforward that for i, j ∈ [n] and A ⊆ 2 [n] , Moreover, we have the following.
the electronic journal of combinatorics 23(4) (2014), #P4.45 Lemma 8. Let A and B be cross-intersecting subfamilies of 2 [n] .(i) For any i, j ∈ [n], ∆ i,j (A) and ∆ i,j (B) are cross-intersecting subfamilies of 2 [n] .(ii) If r, s ∈ [n], A ⊆ [n]  r , B ⊆ [n]  s , and A and B are compressed, then for any A ∈ A and any B ∈ B.
A proof of Lemma 8 is essentially given in [18, Section 2] (see also [30]).The only difference is that in [18], part (ii) is proved for A ⊆ [n]   r and B ⊆ [n]  s ; however, the argument carries forward for A ⊆ [n]  r and B ⊆ [n]  s .Proof of Theorem 2. We use induction on m + n.The basis is m + n = 2 with m = n = 1, in which case the result is trivial.Now consider m + n > 2. We may assume that m n.If m = 1, then the result is trivial too, so we consider m 2. If at least one of G and H is {∅}, then we trivially have g(A)h(B) = 0 = g(G(1))h(H(1)).Thus, we will assume that G = {∅} and H = {∅}, meaning that each of G and H contain at least one nonempty set.Since G and H are hereditary and compressed, we clearly have {1} ∈ G and {1} ∈ H.So g(G(1)) > 0 and h(H(1)) > 0. Let A ⊆ G and B ⊆ H such that g(A)h(B) is maximum under the condition that A and B are cross-intersecting.Since G(1) and H(1) are cross-intersecting, it follows that g(A)h(B) g(G(1))h(H(1)) > 0. ( We will first show that we may assume that A and B are compressed.By Lemma 8(iii), we can apply left-compressions to A and B simultaneously until we obtain two compressed cross-intersecting families A We now show that we may also work with A * and B * for the purpose of establishing the second part of the theorem (that is, the characterization of the extremal structures for u = 0 = v).Suppose that A * = G(c) and
Define H 0 = {H ∈ H : n / ∈ H} and H 1 = {H\{n} : n ∈ H ∈ H}.Define G 0 , G 1 , A 0 , A 1 , B 0 and B 1 similarly.Since A, B, G and H are compressed, we clearly have that A 0 , A 1 , B 0 , B 1 , G 0 , G 1 , H 0 and H 1 are compressed.Since G and H are hereditary, we clearly have that G 0 , G 1 , H 0 and H 1 are hereditary, G 1 ⊆ G 0 and , and hence we obtain the result immediately from the induction hypothesis.The same occurs if H 1 = ∅.So we assume that G 1 and H 1 are non-empty.Since G 1 ⊆ G 0 and H 1 ⊆ H 0 , G 0 and H 0 are non-empty too.Obviously, we have . By (b) and (d), we have the following consequences.For any A, B ∈ H 0 with For any C ∈ H 0 and any i, j For any A, B ∈ H For any C ∈ H 1 and any i, j Thus, we have shown that properties (b) and (d) are inherited by h 0 and h 1 .
Along the same lines, Suppose m < n.Clearly, A and B 0 are cross-intersecting.Since m < n, no set in A contains n, and hence A and B 1 are cross-intersecting.Thus, by the induction hypothesis, Together with ( 6) and ( 7), this gives us the electronic journal of combinatorics 23(4) (2014), #P4.45 This establishes the first part of the theorem for m < n, and we now verify the second part for this case.By (1), equality holds throughout in (9).Thus, in (8), we actually have equality.Suppose u = 0 and v = 0.Then, by the induction hypothesis, for each j ∈ {0, 1} we have A = G(a j ) and B j = H j (a j ) for some a j ∈ [m] such that g(G(a j )) = g(G(1)) and h j (H j (a j )) = h j (H j (1)).So g(G(a 0 )) > 0, and hence G(a 0 ) = ∅.Thus, since G is hereditary, {a 0 } ∈ A. Since A and B are cross-intersecting, B ⊆ H(a 0 ).Since G(a 0 ) and H(a 0 ) are cross-intersecting, it follows by the choice of A and B that A = G(a 0 ) and B = H(a 0 ).Thus, since g(A)h(B) = g(G(1))h(H(1)), and since Lemma 7 gives us g(G(a 0 )) g(G(1)) and h(H(a 0 )) h(H( 1)), we have g(G(a 0 )) = g(G(1)) and h(H(a 0 )) = h(H( 1)).
Now suppose m = n.Similarly to h 0 and h 1 , let g 0 : . Then properties (a) and (c) are inherited by g 0 and g 1 in the same way (b) and (d) are inherited by h 0 and h 1 as shown above; that is, similarly to (2)-( 5), we have the following.For any A, B ∈ G 0 with For any C ∈ G 0 and any i, j For any For any C ∈ G 1 and any i, j Similarly to ( 6) and ( 7), we have Clearly, A 0 and B 0 are cross-intersecting, and, since n = m, so are A 0 and B 1 , and also A 1 and B 0 .
Suppose that A 1 and B 1 are not cross-intersecting.Then there exists which is a contradiction as A and B are cross-intersecting.
We have therefore shown that By a similar argument, Since Since (17), A and B are cross-intersecting.By (18), A and B are cross-intersecting.Since G and H are hereditary, and since A 1 ∈ A ⊆ G and B 1 ∈ B ⊆ H, we have A 1 ∈ G and B 1 ∈ H, and hence A , A ⊆ G and B , B ⊆ H.
Let x = g(A) and x 1 = g(A 1 ).Let y = h(B) and y 1 = h(B 1 ).We have By the choice of A and B,

So we have
the electronic journal of combinatorics 23(4) (2014), #P4.45 Suppose g(A 1 ) and h(B 1 ) h(B 1 ).We therefore have x x 1 + g(A 1 ) 2x 1 and y y 1 + h(B 1 ) 2y 1 .By (20), we have (since we are given that u + v 2), and hence equality holds throughout.Thus x = 2x 1 and y = 2y 1 .Consequently, we have  .By (19), the sets {1}, . . ., {n} are all in A, and obviously no proper subset of [n] intersects each of these sets.Thus, by the cross-intersection condition, B 1 is the only set that is in B. So Since {1}, . . ., {n} ∈ A and A 1 = {n}, we have and hence g(A 1 ) g(A)/n.Since g(A ) = g(A) − g(A 1 ), g(A ) n−1 n g(A).Thus g(A )h(B ) Lemma 12. Let c be an IP sequence (c 1 , . . ., c n ) and let r ∈ [n].Let w : [n]  r → N such that for each A ∈ [n]  r , Then: (i) w(A) (c 1 − 1)w(A ) for any A, A ∈ [n]  r with A A .(ii) w(δ i,j (A)) w(A) for any A ∈ [n]  r and any i, j ∈ [n] with i < j.
We now prove Theorem 3, and then we prove Theorem 4.
the electronic journal of combinatorics 23(4) (2014), #P4.45 Let H = [n]  s .Let w : H → N such that for each H ∈ H, So C ⊆ G, D ⊆ H, and by Corollary 10, C and D are cross-intersecting.We have Now and similarly w(J ) = |Y|.Together with ( 23) and (24), this gives us |A||B| |X ||Y|.Suppose |A||B| = |X ||Y|.Then all the inequalities in ( 22)-( 24) are equalities.The equalities in (22)  Finally, suppose c 1 = 2. Let mod * be the usual modulo operation with the exception that for any a, b ∈ N, (ba) mod * a is a rather than 0. Let θ : L c → L c such that θ(E) = {(i, (j + 1) mod * c i ) : (i, j) ∈ E} for each E ∈ L c .Clearly, θ is a bijection, and θ(E) ∩ E = ∅ for each E ∈ L c .Thus, since A and B are cross-intersecting, θ(C) / ∈ B for each C ∈ A, and hence 5 Proof of Theorem 5 In this section, we use Theorem 2 to prove Theorem 5. Recall that for any n, r ∈ N, M n,r denotes the set {(a 1 , . . ., a r ) : Similarly, |B| h(F).

Theorem 4 .
If c is an IP sequence, A, B ⊆ L c , and A and B are cross-intersecting, then H)) w(H) for every H ∈ H and every i, j ∈ [n] with i < j.Then w(H(a)) w(H(1)) for each a ∈ [n].Proof.Let a ∈ [n].Let D = ∆ 1,a (H(a)).Since H is compressed, D ⊆ H. Thus it is immediate from the definitions of D and w that w(D) w(H(a)).The result follows if we show that D ⊆ H(1).Let D ∈ D. If D / ∈ H(a), then D = δ 1,a (H) = H for some H ∈ H(a), and hence 1 ∈ D. Suppose D ∈ H(a).If we assume that δ 1,a (D) / ∈ H(a), then we obtain D / ∈ ∆ 1,a (H(a)), contradicting D ∈ D. Hence δ 1,a (D) ∈ H(a).Thus, since a ∈ D and a ∈ δ 1,a (D), 1 ∈ D.
* and B * such that |A * | = |A| and |B * | = |B|.Since G and H are compressed, A * ⊆ G and B * ⊆ H. From (b) we obtain g(A) g(A * ) and h(B) h(B * ).By the choice of A and B, we actually have g(A) = g(A * ) and h(B) = h(B * ).
Then g(G(c)) > 0 and h(H(c)) > 0. So G(c) = ∅ and H(c) = ∅.Thus, since G and H are hereditary, {c} ∈ A * and {c} ∈ B * .So {a} ∈ A for some a ∈ [m], and {b} ∈ B for some b ∈ [n].Since A and B are cross-intersecting, we have a = b, A ⊆ G(a) and B ⊆ H(a).Since G(a) and H(a) are cross-intersecting, it follows by the choice of A and B that A = G(a), B = H(a), and g and g(A 2 ) g(A 1 ).Similarly, B 2 ∈ H and h(B 2 ) h(B 1 ).Since A 1 = ∅, we have |A 1 | 2, and hence |A 2 | 2. Since G is hereditary and I A 2 ∈ G, I ∈ G and g(I) (1+u)g(A 2 ).Similarly, I ∈ H and h(I) (1+v)h(B 2 ).Let C = {I, A 2 } and D = {I, B 2 }.So C ⊆ G and D ⊆ H. Also, C and D are cross-intersecting.We have

c 1 3 and d 1 3 .
By Lemma 12 and Theorem 2, equality in(24) gives us that for some p ∈[m] ∩ [n], C = G(p) and D = H(p).It follows that A * = {L ∈ L (r) c : (p, 1) ∈ L} and B * = {L ∈ L (s) d : (p, 1) ∈ L}.By Lemma 11, A = {L ∈ L (r) c : (p, q) ∈ L} and B = {L ∈ L (s) d : (p, q) ∈ L} for some q ∈ [c p ] ∩ [d p ].So A is a star of L (r)c with centre (p, q), and B is a star of L (s) d with centre (p, q).Now clearly X is a star of L (r) c of maximum size, and Y is a star of L (s) d of maximum size.Thus, since |A||B| = |X ||Y|, |A| = |X | and |B| = |Y|.So A is a star of L (r) c of maximum size, and hence we must have c p = c 1 .Similarly, d p = d 1 .the electronic journal of combinatorics 23(4) (2014), #P4.45 Proof of Theorem 4. Since |A| |L c | and |B| |L c |, the result is trivial if c 1 = 1.If c 1 3, then the result is given by Theorem 3.
for every H, H ∈ H with ∅ = H H ,