I'm working on Richard P. Stanley's Algebraic Combinatorics and have been stuck on this problem for a while:
Let $S_1, S_2,..., S_{k}$ be finite sets with $\#S_1 = \#S_2 = ... = \#S_{k}$. Let $P$ be the poset of all sets $T$ contained in some $S_{i}$, ordered by inclusion. In symbols, \begin{equation} P = 2^{S_1} \cup 2^{S_2} \cup ... \cup 2^{S_{k}} \end{equation} where $2^{S}$ denotes the set of all subsets of $S$. Is $P$ always rank-unimodal?
I think I've made progress, but I'm not sure where to go right now. I think that $P$ is always rank unimodal and I suspect that the "middle" level is always the largest, and I'm trying to show this inductively; the condition holds for all cases of size 1, then 2, and so on, and if it works for $k$ sets of some size, it works for $k + 1$ of that size. The poset of subsets of a single set is rank-unimodal, and the poset formed by multiple singleton sets is too, so this provides base cases.
Given a collection of sets that we wish to show satisfy the property, if no set is contained within the union of the others, every set must have an element not contained in any other set. For set $S_{i}$, call one such element $a_{i}$. Ignoring each $a_{i}$, the condition must be satisfied by the sets by our inductive hypotheses. We now also have $k {\#S \choose n}$ additional sets of each size $n$, and adding these retains rank-unimodality (and possibly the middle level being the largest, but I don't know if it's necessarily true).
If some set is contained in the union of the others, WLOG let $S_{k}$ be such a set. From here, I have had trouble counting "new" sets of each size and relating the sizes of levels.
Any hints or help would be appreciated!