English

Two Results on Union-Closed Families

Combinatorics 2017-08-07 v1

Abstract

We show that there is some absolute constant c>0c>0, such that for any union-closed family F2[n]\mathcal{F} \subseteq 2^{[n]}, if \mbox{F(12c)2n|\mathcal{F}| \geq (\frac{1}{2}-c)2^n}, then there is some element i[n]i \in [n] that appears in at least half of the sets of F\mathcal{F}. We also show that for any union-closed family F2[n]\mathcal{F} \subseteq 2^{[n]}, the number of sets which are not in F\mathcal{F} that cover a set in F\mathcal{F} is at most 2n12^{n-1}, and provide examples where the inequality is tight.

Keywords

Cite

@article{arxiv.1708.01434,
  title  = {Two Results on Union-Closed Families},
  author = {Ilan Karpas},
  journal= {arXiv preprint arXiv:1708.01434},
  year   = {2017}
}

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13 pages