I'm wondering if there are any non-trivial sufficient conditions (or even just a google-search-friendly name for) for the following scenario:
I am given a set family $\mathcal{F}$ on ground set $E$ (such that each $F \in \mathcal{F}$ has $F \subseteq E$), such that $\bigcap_{F\in \mathcal{F}} F = \emptyset$. I want to find some $U \subsetneq E$ such that for some integer $m$, $|U \cap F| = m > 0$ for every $F \in \mathcal{F}$.
Obviously this is not always possible; I am looking for conditions on $\mathcal{F}$ to guarantee existence of some such $U$.
Some additional assumptions I can make about $\mathcal{F}$ include:
- $\mathcal{F}$ is a Sperner family
- $\emptyset \notin \cal{F}$.
- Let $ \cal{I}$ be the downward closure of $\cal{F}$ (all subsets of each $F \in \cal{F}$, including $F$ itself). Suppose I have $F_1, F_2 \in \cal{F}$ and $I_1, I_2 \in \cal{I}$. If the multiset union $F_1 \uplus F_2 = I_1 \uplus I_2$, then $I_1, I_2 \in \cal{F}$.