Israel Journal of Mathematics

, Volume 219, Issue 1, pp 431–448 | Cite as

On the number of union-free families



A family of sets is union-free if there are no three distinct sets in the family such that the union of two of the sets is equal to the third set. Kleitman proved that every union-free family has size at most (1+o(1))( n/2 n ). Later, Burosch–Demetrovics–Katona–Kleitman–Sapozhenko asked for the number α(n) of such families, and they proved that \({2^{\left( {\begin{array}{*{20}{c}} n \\ {n/2} \end{array}} \right)}} \leqslant \alpha \left( n \right) \leqslant {2^{2\sqrt 2 \left( {\begin{array}{*{20}{c}} n \\ {n/2} \end{array}} \right)\left( {1 + o\left( 1 \right)} \right)}}\) They conjectured that the constant \(2\sqrt 2 \) can be removed in the exponent of the right-hand side. We prove their conjecture by formulating a new container-type theorem for rooted hypergraphs.


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© Hebrew University of Jerusalem 2017

Authors and Affiliations

  1. 1.Department of MathematicsUniversity of Illinois at Urbana-ChampaignUrbana, IllinoisUSA

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