标题： 集合划分的Hilton-Milner定理模拟 显示英文标题

标题： An Analogue of Hilton-Milner Theorem for Set Partitions

摘要： 设$\mathcal{B}(n)$表示所有$[n]$的集合划分的集合。 假设$\mathcal{A} \subseteq \mathcal{B}(n)$是一个非平凡的$t$-相交的集合划分族，即$\A$的任意两个成员至少有$t$个共同的块，但不存在固定的$t$个大小为一的块属于它们中的每一个。 证明了对于足够大的$n$，它依赖于$t$，\[ |\mathcal{A}| \le B_{n-t}-\tilde{B}_{n-t}-\tilde{B}_{n-t-1}+t \]，其中$B_{n}$是第$n$个贝尔数，$\tilde{B}_{n}$是$[n]$的集合划分的数量，这些划分没有大小为一的块。 此外，当且仅当$\mathcal{A}$与\[ \{P \in \mathcal{B}(n): \{1\}, \{2\},..., \{t\}, \{i\} \in P \textnormal{for some} i \not = 1,2,..., t,n \}\cup \{Q(i,n)\ :\ 1\leq i\leq t\} \]等价，其中$Q(i,n)=\{\{i,n\}\}\cup\{\{j\}\ :\ j\in [n]\setminus \{i,n\}\}$。这是集合划分的Hilton-Milner定理的类似结果。 显示英文摘要

摘要： Let $\mathcal{B}(n)$ denote the collection of all set partitions of $[n]$. Suppose $\mathcal{A} \subseteq \mathcal{B}(n)$ is a non-trivial $t$-intersecting family of set partitions i.e. any two members of $\A$ have at least $t$ blocks in common, but there is no fixed $t$ blocks of size one which belong to all of them. It is proved that for sufficiently large $n$ depending on $t$, \[ |\mathcal{A}| \le B_{n-t}-\tilde{B}_{n-t}-\tilde{B}_{n-t-1}+t \] where $B_{n}$ is the $n$-th Bell number and $\tilde{B}_{n}$ is the number of set partitions of $[n]$ without blocks of size one. Moreover, equality holds if and only if $\mathcal{A}$ is equivalent to \[ \{P \in \mathcal{B}(n): \{1\}, \{2\},..., \{t\}, \{i\} \in P \textnormal{for some} i \not = 1,2,..., t,n \}\cup \{Q(i,n)\ :\ 1\leq i\leq t\} \] where $Q(i,n)=\{\{i,n\}\}\cup\{\{j\}\ :\ j\in [n]\setminus \{i,n\}\}$. This is an analogue of the Hilton-Milner theorem for set partitions.