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

Abstract

For a poset $P$, let $e(P)$ ($e^{\ast }(P)$) 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 ${\epsilon }_P$ such that $La(n,P)\ge (e(P)+{\epsilon }_P)\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$ and $L{a}^{\ast }(n,P)\ge (e^{\ast }(P)+{\epsilon }_P)\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2}\right)$ hold, where $La(n,P)$ ($L{a}^{\ast }(n,P)$) 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 $\epsilon$ such that $$L{a}^{\ast }(n,B_3)\ge La(n,B_3)\ge La(n,D_6)\ge (3+\epsilon )\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right),$$ where $D_s$ is the poset on $s+2$ elements $a<b_1,\dots ,b_s<c$. Consider the intervals $I_m=[2^{m-1}-1,2^m-2]$, $I_m^{\ast }=[(\genfrac{}{}{0ex}{}{m-1}{\lfloor \frac{m-1}{2}\rfloor })+1,(\genfrac{}{}{0ex}{}{m}{\lfloor \frac{m}{2}\rfloor })]$. It is known that for values $s$ in the major initial parts of $I_m$ and $I_m^{\ast }$, one has $La(n,D_s)=(m+o(1))\left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n}{2}\rfloor }\right)$ and $L{a}^{\ast }(n,D_s)=(m+o(1))\left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n}{2}\rfloor }\right)$. The above equalities do not hold for the largest elements of the intervals, thus there exist $s_m\in I_m,s_m^{\ast }\in I_m^{\ast }$ such that for $s\in I_m$ we have $La(n,D_s)=(m+o(1))\left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n}{2}\rfloor }\right)$ if and only if $s<s_m$ and for $s\in I_m^{\ast }$ we have $L{a}^{\ast }(n,D_s)=(m+o(1))\left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n}{2}\rfloor }\right)$ if and only if $s<s_m^{\ast }$. Modifying previous constructions, we obtain upper bounds on $s_m$ and $s_m^{\ast }$.