Let $X\neq \emptyset$ be a set and let ${\cal S} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ be a collection of non-empty subsets of $X$. We say $C\subseteq X$ is a *choice set* for ${\cal S}$ if $|C\cap s| = 1$ for all $s\in S$. Moreover, we say that 


1. $D\subseteq X$ is *shy* if $|D\cap s| \leq 1$ for all $s\in S$; and
2. $E\subseteq X$ is *gregarious* if $|E \cap s| \geq 1$ for all $s\in S$.


Trivially, $\emptyset$ is shy, and $X$ is gregarious. Zorn's Lemma shows that every shy set is contained in a choice set. (It is straightforward to check that a union of a chain of shy sets is shy; and that a maximal shy set is a choice set.)


But on the other hand, not every gregarious set contains a choice set: Let $X = [0,1]$, $E = X$ and let ${\cal S} = \{]0, \frac{1}{n}[: n\in \mathbb{N}, n\geq 1\}$.


I find this a curious asymmetry that "from below" there is always a choice set, but not necessarily "from above"!


However: Let $D\subseteq X$ be such that $|D\cap s|$ is finite for all $s\in {\cal S}$.


**Question.** Does $D$ contain a choice set in that case?



Choice sets from above and below
Let $\cal S=\{\{1,2\},\{2,3\},\{3,1\}\}$.


Then $\cal S$ has no choice set, whatsoever.


So there is no asymmetry -- not every shy set is contained in a choice set, and not every gregarious set contains a choice set.


