【摘要】 <正> 设S是n元集合,（s）是S的子集系,并满足条件: （1） 对任意A₁,…,A_k∈（S）,存在A_i与A_j使得（i≠j）,称为无k个子集两两不相交的子集系,或简称为G_3k子集系;特别地,又满足: （2） 若,则存在A₁,…,A_k-1∈（S）使得A_i∩A_j=φ,0≤i≠j≤k-1,称为极大的无k个子集两两不相交的子集系,或简称为极大的G_3k子集系。更多还原

【Abstract】 Let S be a finite set with n elements, and k(s) be a maximal family of subsets of S no k of which are pairwise disjoint, i.e for any A1, … Ak∈ k(s), there exist A, and Aj(i≠j) such that At ∩ Aj≠ φ ; and for A0(?)S, A0 ∈ k(S), there exist A1’, … Ak-1’ in k(s) such that A1’, … Ak-1’ and A0 are pairwise disjoint. How large can k(s) be? An unper bound on size of k(S) has been obtained by kleitman (see [ 1 ]), but lower bound remains open. P.Erdos and D. Kleit-man asked if the smallest size of a maximal family of subsets of S no k of which are pairwise disjoint is equal to 2n-2m-k (see [2’]) This paper answers the question nogatively, and the smallest size of k(s) is determined. The main results can be described as follows.Theorom l min, where the minimum is taken over all maximal family of subsets of S no k of which are pairwise disjoint.Theorem 2 .更多还原