Nested Chain Order

Abstract:

      Let $X_{1},X_{2},\ldots,X_{k}$ be $k$ disjoint subsets of $S$ with the same cardinality $m$. Define $H(m,k)=\{X\subseteq S: X \not\subseteq X_i$ for $1\leq i\leq k\}$ and $P(m,k)=\{X\subseteq S: X \cap X_i \neq \emptyset$ for at least two $X_i$'s\}. Suppose $S=\bigcup_{i=1}^k X_i$, and let $Q(m,k,2)$ be the collection of all subsets $K$ of $S$ satisfying $|K\cap X_{i}|\geq 2$ for some $1 \leq i \leq k$. For any two disjoint subsets $Y_1$ and $Y_2$ of $S$, we define ${\cal F}_{1,j}=\{X\subseteq S: \mbox{either $|X\cap Y_1|\geq 1$ or $|X\cap Y_2|\geq j$}\}$. It is obvious that the four posets are graded posets ordered by inclusion. In this paper we will prove that the four posets are nested chain orders.