关于克莱特曼一个结果的推广

Ryan R. Martin, Balázs Patkós
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引用次数: 0

摘要

克莱特曼的一个经典结果确定了$nequiv 0,-1$(mod $s$)情况下不包含成对不相交的不同集合$F_1,F_2,\dots,F_s$的集合族$mathcal{F}\subseteq 2^{[n]}$中子集的最大数目$f(n,s)$。卡托纳和纳吉确定了$n$元素集合中不包含$A_1,A_2,\dots,A_t,B_1,B_2,\dots,B_t$且$\bigcup_{i=1}^t A_i$ 和$\bigcup_{i=1}^t B_i$ 的子集族的最大大小。在本文中,我们考虑的问题是在一个没有集合$F^1_1的族$mathcal{F}/subseteq 2^{[n]}$ 中求取最大数目$vex(n,K_{s\times t})$、\dots,F^1_t,\dots,F^s_1,\dots,F^s_t$ 这样$G_j=\bigcup_{i=1}^tF^j_i$ $j=1,2,\dots,s$是成对不相交的。如果 $nequiv -1$ (mod $s$)适用于所有$t$,我们将确定$2^n-vex(n,K_{s/times t})$的渐近线;如果 $nequiv 0$ (mod $s$),$t/ge 3$,我们将确定$2^n-vex(n,K_{s/times t})$的渐近线,并证明在后一种情况下,$t=2$子情况的渐近线不同于$t=1$和$t/ge 3$子情况。
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On a generalization of a result of Kleitman
A classical result of Kleitman determines the maximum number $f(n,s)$ of subsets in a family $\mathcal{F}\subseteq 2^{[n]}$ of sets that do not contain distinct sets $F_1,F_2,\dots,F_s$ that are pairwise disjoint in the case $n\equiv 0,-1$ (mod $s$). Katona and Nagy determined the maximum size of a family of subsets of an $n$-element set that does not contain $A_1,A_2,\dots,A_t,B_1,B_2,\dots,B_t$ with $\bigcup_{i=1}^t A_i$ and $\bigcup_{i=1}^t B_i$ being disjoint. In this paper, we consider the problem of finding the maximum number $vex(n,K_{s\times t})$ in a family $\mathcal{F}\subseteq 2^{[n]}$ without sets $F^1_1,\dots,F^1_t,\dots,F^s_1,\dots,F^s_t$ such that $G_j=\bigcup_{i=1}^tF^j_i$ $j=1,2,\dots,s$ are pairwise disjoint. We determine the asymptotics of $2^n-vex(n,K_{s\times t})$ if $n\equiv -1$ (mod $s$) for all $t$, and if $n\equiv 0$ (mod $s$), $t\ge 3$ and show that in this latter case the asymptotics of the $t=2$ subcase is different from both the $t=1$ and $t\ge 3$ subcases.
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