On the maximum size of $B_3$-free families

Abstract

A family $\mathcal{G}$ of sets is a weak copy of the poset $(P,\leqslant)$ if there exists a bijection $ι:P\rightarrow \mathcal{G}$ with $ι(p)\subset ι(q)$ whenever $p\leqslant q$. $\mathcal{G}$ is a strong copy if $ι(p)\subset ι(q)$ if and only if $p\leqslant q$ holds. A family is weak (strong) $P$-free if it does not contain any weak (strong) copies of $P$. For a poset $P$, let $e(P)$ ($e^*(P)$) denote the most number of middle layers of $2^{[n]}$ that does not contain a weak (strong) copy of $P$. Ellis, Ivan, and Leader were the first to show the existence of posets $P$ for which there exists a positive real $\varepsilon_P$ such that $La(n,P)\ge (e(P)+\varepsilon_P)\binom{n}{\lfloor n/2}$ and $La^*(n,P)\ge (e^*(P)+\varepsilon_P)\binom{n}{\lfloor n/2}$ holds, where $La(n,P)$ ($La^*(n,P)$) denotes the maximum size of a weak (strong) $P$-free family $\mathcal{F}\subseteq 2^{[n]}$. More precisely, they showed that $P=B_d$ are such posets for all $d\ge 4$, where $B_d$ is the Boolean lattice ordered by inclusion. Very recently, Tompkins showed that the diamond $B_2$ is also such a poset. In this short note, we apply his method to settle the case of the last Boolean poset $B_3$. We show that there exists a positive $\varepsilon$ such that $$La^*(n,B_3)\ge La(n,B_3)\ge La(n,D_6)\ge (3+\varepsilon)\binom{n}{\lfloor n/2\rfloor},$$ where $D_6$ is the poset on eight elements $a<b_1,\dots,b_6<c$.