For integers n ≥ s ≥ 2 let e(n, s) denote the maximum of |F|, where F is a family of subsets of an n-element set and F contains no s pairwise disjoint members. Half a century ago, solving a conjecture of Erdős, Kleitman determined e(sm − 1, s) and e(sm, s) for all m, s ≥ 1. During the years very little progress in the general case was made.In the present paper we state a general conjecture concerning the value of e(sm − l, m) for 1 < l < s and prove its validity for s > s 0 (l, m). For l = 2 we determine the value of e(sm − 2, m) for all s ≥ 5.Some related results shedding light on the problem from a more general context are proved as well.
IntroductionLet [n] := {1, 2, . . . , n} be the standard n-element set and 2[n] its power set. A subset F ⊂ 2[n] is called a family. For 0 ≤ k ≤ n we use the notationThe maximum number of pairwise disjoint members of a family F is denoted by ν(F ) and called the matching number of F . Note that ν(F ) ≤ n unless ∅ ∈ F .Two of the important classical results in extremal set theory are concerning the matching number.Definition 2. For positive integers n, k, s ≥ 2, n ≥ ks defineFor s = 2 both e(n, s) and e k (n, s) were determined by Erdős, Ko and Rado. e(n, 2) = 2 n−1 ,For m = n+1 s the family[n] ≥ m := {H ⊂ [n] : |H| ≥ m} does not contain s pairwise disjoint sets. Erdős conjectured that for n = sm − 1 one cannot do any better. Half a century ago Kleitman proved this conjecture and determined e(sm, s) as well. Kleitman [14]).e(sm, s)Note that e(sm, s) = 2e(sm − 1, s). In general, e(n + 1, s) ≥ 2e(n, s) is obvious, and, since the constructions of families that match the bounds are easy to provide, (3) follows from (4). For s = 2 both formulae give 2 n−1 , the easy-to-prove bound (1). In the case s = 3 there is just one case not covered by the Kleitman Theorem, namely n ≡ 1 ( mod 3). This was the subject of the PhD dissertation of Quinn [16]. In it a very long, tedious proof for the following equality is provided:Unfortunately, this result was never published and no further progress was made on the determination of e(n, s). Let us first make a general conjecture. Proof. Assume that P 1 , . . . , P s ∈ P(s, m, l) are pairwise disjoint. Then