Packing the Boolean lattice with copies of a poset

2020 
Let $P$ be a partially ordered set. We prove that if $n$ is sufficiently large, then there exists a packing $\mathcal{P}$ of copies of $P$ in the Boolean lattice $(2^{[n]},\subset)$ that covers almost every element of $2^{[n]}$: $\mathcal{P}$ might not cover the minimum and maximum of $2^{[n]}$, and at most $|P|-1$ additional points due to divisibility. In particular, if $|P|$ divides $2^{n}-2$, then the truncated Boolean lattice $2^{[n]}-\{\emptyset,[n]\}$ can be partitioned into copies of $P$. This confirms a conjecture of Lonc from 1991.
    • Correction
    • Source
    • Cite
    • Save
    • Machine Reading By IdeaReader
    15
    References
    0
    Citations
    NaN
    KQI
    []