Let $X\neq \emptyset$ be a set. We say ${\cal C} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ is a *cover* if $\bigcup {\cal C} = X$. A subset $D\subseteq X$ is a *choice set* for ${\cal C}$ if $|D\cap c| = 1$ for all $c\in C$. As [Bjørn Kjos-Hanssen](https://mathoverflow.net/users/4600/bj%c3%b8rn-kjos-hanssen) pointed out, [choice sets do not always exist](https://mathoverflow.net/a/285451/8628).


What is an example of a set $X$ and a cover ${\cal S}$ such that for no cover ${\cal S}\_0 \subseteq {\cal S}$ there is a choice set for ${\cal S}\_0$?



Subcovers without a choice set
Well order $X$ and at stage $\alpha$ let $d\_\alpha$ be the first element of $X\setminus \bigcup\_{\xi<\alpha}\bigcup\{S \in \mathcal S: d\_\xi \in S \}$ (if this set was empty we stop and let $\gamma=\alpha$). Then $D=\{d\_\alpha: \alpha <\gamma\}$ is a choice set for the subcovering $\mathcal S\_0=\{S \in \mathcal S: S\cap D \neq \emptyset \}$.


