关于不含\(B_3\)的族的最大规模
On the maximum size of $B_3$-free and $D_s$-free families
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中文总结 AI 辅助
本文通过应用 Tompkins 的方法，解决了 $B_3$ 的最大弱自由族问题，证明了其大小下界与布尔格 $B_3$ 和钻石 poset $D_6$ 的关系。
中文摘要 AI 辅助
如果存在一个双射\(\iota:P\rightarrow \mathcal{G}\)，使得当\(p\leqslant q\)时，\(\iota(p)\subset \iota(q)\)，则集合族\(\mathcal{G}\)是偏序集\((P,\leqslant)\)的弱副本。如果\(\iota(p)\subset \iota(q)\)当且仅当\(p\leqslant q\)成立，则\(\mathcal{G}\)是强副本。如果一个族不包含任何\(P\)的弱（强）副本，则称其为弱（强）\(P -\)自由的。对于偏序集\(P\)，设\(e(P)\)（\(e^*(P)\)）表示\(2^{[n]}\)中不包含\(P\)的弱（强）副本的中间层的最大数量。Ellis、Ivan和Leader首先证明存在偏序集\(P\)，使得存在正实数\(\varepsilon_P\)，满足\(La(n,P)\ge (e(P)+\varepsilon_P)\binom{n}{\lfloor n/2\rfloor}\)和\(La^*(n,P)\ge (e^*(P)+\varepsilon_P)\binom{n}{\lfloor n/2\rfloor}\)。最近Tompkins证明了菱形\(B_2\)也是这样的偏序集。本文应用其方法解决最后一个布尔偏序集\(B_3\)的情况。证明存在正的\(\varepsilon\)，使得\(La^*(n,B_3)\ge La(n,B_3)\ge La(n,D_6)\ge (3+\varepsilon)\binom{n}{\lfloor n/2\rfloor}\)，其中\(D_6\)是具有八个元素\(a < b_1,\dots,b_6 < c\)的偏序集。
英文摘要
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 $s^*_m$.
