Supersaturation, counting, and randomness in forbidden subposet problems

TL;DR

This paper investigates the maximum size of families in the Boolean lattice avoiding certain poset patterns, proposing strengthened conjectures and proving them for specific classes, with implications for counting and random settings.

Contribution

It formulates three strengthened conjectures in forbidden subposet problems and proves them for specific classes, advancing understanding of extremal and probabilistic aspects.

Findings

Largest P-free families are asymptotically determined by e(P)

Number of P-free families grows exponentially with size

Largest P-free family size in random subsets matches deterministic bounds

Abstract

In the area of forbidden subposet problems we look for the largest possible size $La(n,P)$ of a family $\mathcal{F}\subseteq 2^{[n]}$ that does not contain a forbidden inclusion pattern described by $P$. The main conjecture of the area states that for any finite poset $P$ there exists an integer $e(P)$ such that $La(n,P)=(e(P)+o(1))\binom{n}{\lfloor n/2\rfloor}$. In this paper, we formulate three strengthenings of this conjecture and prove them for some specific classes of posets. (The parameters $x(P)$ and $d(P)$ are defined in the paper.) $\bullet$ For any finite connected poset $P$ and $\varepsilon>0$, there exists $\delta>0$ and an integer $x(P)$ such that for any $n$ large enough, and $\mathcal{F}\subseteq 2^{[n]}$ of size $(e(P)+\varepsilon)\binom{n}{\lfloor n/2\rfloor}$, $\mathcal{F}$ contains at least $\delta n^{x(P)}\binom{n}{\lfloor n/2\rfloor}$ copies of $P$. $\bullet$ The number of $P$-free families in $2^{[n]}$ is $2^{(e(P)+o(1))\binom{n}{\lfloor n/2\rfloor}}$. $\bullet$ For any finite poset $P$, there exists a positive rational $d(P)$ such that if $p=\omega(n^{-d(P)})$, then the size of the largest $P$-free family in $\mathcal{P}(n,p)$ is $(e(P)+o(1))p\binom{n}{\lfloor n/2\rfloor}$ with high probability.