Union-free Hypergraphs and Probability Theory

Notice that the upper bound is an immediate consequence of (I<;J) :s;; 2n. To prove the lower bound we use an algebraic construction which is a modification of a construction of Babai and S6s [1]. How a family of sets can fail to have the union-free property? There are essentially two possibilities: (a) there are four distinct sets A, B, C, D with Au B =CuD. (b) there are three distinct sets A, B, C with Au B = Au C. We call families for which (a) never holds weakly union-free, and those for which (b) never holds cancellative (the second name indicates that Au B =Au C implies B = C). We denote by F(n)(G(n)) the maximum number of subsets of an n-set in a weakly union-free (cancellative) family, respectively. Our main result is the following: