On the maximum size of $B_3$-free and $D_s$-free families

Abstract

For a poset $P$, let $e(P)$ ($e^*(P)$) denote largest positive integer $k$ such that the union of the $k$ middle layers of $2^{[n]}$ does not contain a weak (strong) copy of $P$. Ellis, Ivan, and Leader showed the existence of posets $P$ for which there exists a positive real $\varepsilon_P$ such that $La(n,P)\ge (e(P)+\varepsilon_P)\binom{n}{\lfloor n/2\rfloor}$ and $La^*(n,P)\ge (e^*(P)+\varepsilon_P)\binom{n}{\lfloor n/2}$ hold, 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. Tompkins showed that the diamond $B_2$ is also such a poset. 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 $D_s$ is the poset on $s+2$ elements $a<b_1,\dots,b_s<c$. Consider the intervals $I_m=[2^{m-1}-1,2^m-2]$, $I^*_m=[\binom{m-1}{\lfloor \frac{m-1}{2}\rfloor}+1,\binom{m}{\lfloor \frac{m}{2}\rfloor}]$. It is known that for values $s$ in the major initial parts of $I_m$ and $I_m^*$, one has $La(n,D_s)=(m+o(1))\binom{n}{\lfloor \frac{n}{2}\rfloor}$ and $La^*(n,D_s)=(m+o(1))\binom{n}{\lfloor \frac{n}{2}\rfloor}$. The above equalities do not hold for the largest elements of the intervals, thus there exist $s_m\in I_m, s^*_m\in I^*_m$ such that for $s\in I_m$ we have $La(n,D_s)=(m+o(1))\binom{n}{\lfloor \frac{n}{2}\rfloor}$ if and only if $s<s_m$ and for $s\in I^*_m$ we have $La^*(n,D_s)=(m+o(1))\binom{n}{\lfloor \frac{n}{2}\rfloor}$ if and only if $s<s^*_m$. Modifying previous constructions, we obtain upper bounds on $s_m$ and $s^*_m$.