A note on the union-closed sets conjecture

A collection A of finite sets is closed under union if A, B ∈ A implies that A ∪ B ∈ A. The Union-Closed Sets Conjecture states that if A is a union-closed collection of sets, containing at least one non-empty set, then there is an element which belongs to at least half of the sets in A. We show that if q is the minimum cardinality of ∪A taken over all counterexamples A, then any counterexample A has cardinality at least 4q − 1. A collection A of finite sets is closed under union if A, B ∈ A implies that A ∪ B ∈ A. The Union-Closed Sets Conjecture (also called Frankl’s Conjecture) states that if A is a union-closed collection of sets, containing at least one nonempty set, then there is an element which belongs to at least half of the sets in A. The conjecture dates from 1979 and is generally attributed to Peter Frankl [2]. Some authors place the further condition on A that it does not contain the empty set. The results of this note apply to either version of the conjecture. A number of necessary conditions for a counterexample have been established, including that if A is a counterexample then |A| ≥ 37 [4] and that | ∪ A| ≥ 12 [1] where ∪A = ∪A∈AA. See [1] and [3] and their bibliographies for more background. In this note we show that if q is the minimum cardinality of ∪A taken over all counterexamples then for any counterexample A we have |A| ≥ 4q − 1. It is known that q ≥ 12 [1] so this implies that a counterexample has cardinality at least 47. 266 IAN ROBERTS AND JAMIE SIMPSON We say that A is a basis set in A if it is not the union of other sets in A. Suppose that A is a counterexample with |A| = 2n + 2 for some integer n. Then no element occurs in more than n of its sets. If we remove a basis set from A, then we get a smaller collection which is also a counterexample. A counterexample of minimum size must therefore contain an odd number of sets. Let q be the minimum cardinality of ∪A taken over all counterexamples A. We will obtain a lower bound for |A| in terms of q. Henceforth A will be a minimal counterexample to the conjecture with |A| = 2n + 1. Set S = ∪A and note that |S| ≥ q. We define Ax = {A ∈ A : x ∈ A} Ax = {A ∈ A : x ∈ A} Cx = ∪Ax C = {Cx : x ∈ S} D = A\{S}\C Dx = {A ∈ D : x ∈ A}. Clearly, Cx is an element of A for each element x of S. Since A is a minimal counterexample there will be elements which occur in exactly n of the sets in A (otherwise we could remove a basis set to give a smaller counterexample). We set H = {x : |Ax| = n}. Let B be the collection of basis sets in A. If a and b are elements of A, then we say that a dominates b if whenever b occurs in a set A in A then a also occurs in that set. This is equivalent to saying b ∈ Ca. We will assume that A contains no mutually dominating pair of elements, since if such a pair existed then we could replace it with a single element without altering the size of A. Note that by this assumption the sets Ca are distinct. Lemma 1 If a ∈ H, then H ⊆ Ca and if x ∈ H, then H\{x} ⊆ Cx. Proof: Let y ∈ H. We show that y ∈ Ca where a ∈ H. Since y occurs in n sets and a occurs in fewer than n sets, there exists A in A such that y ∈ A and a ∈ A which implies y ∈ Ca. It follows that H ⊆ Ca. For the second part let y ∈ H, y = x. We are assuming that x and y do not dominate each other. So there exists A ∈ A such that y ∈ A, x ∈ A so A ⊆ Cx and therefore y ∈ Cx. Lemma 2 No set Ca is a basis set. Proof: Suppose that Ca is a basis set. If a ∈ H then by Lemma 1 A\{Ca} is a unionclosed collection in which no element occurs more than n − 1 times, contradicting the minimality of A. If, on the other hand, a ∈ H, then choose a basis set B which contains a. Then by Lemma 1 A\{B, Ca} is a union-closed collection of 2n−1 sets in which no element occurs more than n− 1 times, again contradicting the minimality of A. THE UNION-CLOSED SETS CONJECTURE 267 Theorem 3 If x ∈ H and a ∈ H, then |Dx| = n − q and |Da| < n − q. Proof: Let x ∈ H. By Lemma 1, x occurs in all sets Cb with b = x as well as in S, so x occurs in |Ax| − q = n − q sets in D. Now assume, without loss of generality, that |Da| is maximal for a ∈ S\H. . If there exists b = a such that a ∈ Cb, then b dominates a and there exists a basis set which contains b but not a. By Lemma 2 this belongs to D so b occurs more often in D than a which is a contradiction. If, on the other hand, a ∈ Cb for all b = a then a occurs in q sets of A\D. If a occurs in n− q or more sets in D then it occurs in at least n sets in A which means a ∈ H. Theorem 4 If A is a minimal counterexample, then |A| ≥ 4q − 1. Proof: Consider Ax and Ax for some x ∈ H. Then |Ax| = n, |Ax| = n + 1 and |Dx| = n − q by Theorem 3. Since Ax is union-closed, there exists an element a in at least half its sets. By Theorem 3 a occurs in at most n− q sets in D. By Lemma 1, each set in A\D except Cx contains x so Ax ⊆ D ∪ {Cx}. Thus a is in at most n − q + 1 sets in Ax. Hence |Ax| ≤ 2(n − q + 1) which implies n ≥ 2q − 1. Thus |A| = 2n + 1 ≥ 4q − 1. Corollary 5 If A is a minimal counterexample, then |A| ≥ 47. Proof: Bošnjak and Marković [1] have recently shown that a minimal counterexample to the conjecture has q ≥ 12. Acknowledgement We thank the referee for many suggestions.