Frankl–Akiyama conjecture on antichains in convex families

Let Bn\mathcal{B}_n be the Boolean lattice of subsets of [n][n]. For a family FBn\mathcal{F}\subseteq\mathcal{B}_n, define its convex closure by

F={SBnASB for some A,BF}.\overline{\mathcal{F}}=\{S\in\mathcal{B}_n\mid A\subseteq S\subseteq B\text{ for some }A,B\in\mathcal{F}\}.

A family is convex when F=F\mathcal{F}=\overline{\mathcal{F}}, and an antichain is a family of pairwise incomparable sets. Frankl–Akiyama conjecture. For every convex family FBn\mathcal{F}\subseteq\mathcal{B}_n, there exists an antichain AF\mathcal{A}\subseteq\mathcal{F} such that

AF(nn/2)2n.\frac{|\mathcal{A}|}{|\mathcal{F}|}\geq\frac{\binom{n}{\lfloor n/2\rfloor}}{2^n}.

The conjecture asserts that every convex subfamily of the Boolean lattice contains an antichain whose relative size is at least the maximum rank proportion in the whole Boolean lattice. Its status is not resolved by the supplied source context.

Sources & referencesView supporting material

Primary source

Andrew P. Dove and Jerrold R. Griggs, “Packing Posets in the Boolean Lattice”, arXiv:1309.6686 (2013).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.