Supersaturation, counting, and randomness in forbidden subposet problems

preprint2020arXivOpen access

Abstract

In the area of forbidden subposet problems we look for the largest possible size La(n,P)La(n,P) of a family F⊆2[n]\mathcal{F}\subseteq 2^{[n]} that does not contain a forbidden inclusion pattern described by PP. The main conjecture of the area states that for any finite poset PP there exists an integer e(P)e(P) such that La(n,P)=(e(P)+o(1))(n⌊n/2⌋)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)x(P) and d(P)d(P) are defined in the paper.) ∙\bullet For any finite connected poset PP and ε>0\varepsilon>0, there exists δ>0δ>0 and an integer x(P)x(P) such that for any nn large enough, and F⊆2[n]\mathcal{F}\subseteq 2^{[n]} of size (e(P)+ε)(n⌊n/2⌋)(e(P)+\varepsilon)\binom{n}{\lfloor n/2\rfloor}, F\mathcal{F} contains at least δnx(P)(n⌊n/2⌋)δn^{x(P)}\binom{n}{\lfloor n/2\rfloor} copies of PP. ∙\bullet The number of PP-free families in 2[n]2^{[n]} is 2(e(P)+o(1))(n⌊n/2⌋)2^{(e(P)+o(1))\binom{n}{\lfloor n/2\rfloor}}. ∙\bullet For any finite poset PP, there exists a positive rational d(P)d(P) such that if p=ω(n−d(P))p=ω(n^{-d(P)}), then the size of the largest PP-free family in P(n,p)\mathcal{P}(n,p) is (e(P)+o(1))p(n⌊n/2⌋)(e(P)+o(1))p\binom{n}{\lfloor n/2\rfloor} with high probability.

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