A [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ consists of an non-empty set $V$ and a collection $E\subseteq {\cal P}(V)\setminus \{\emptyset\}$ of non-empty subsets of $V$. A *transversal* of $H$ is a set $T\subseteq V$ such that $|T\cap e| = 1$ for all $e\in E$. 


It is easy to see that transversals need not exist: Take $V = \{0,1,2\}$ and let $E$ be the collection of $2$-element subsets of $V$.


A *pseudotransversal* is a set $T\subseteq V$ such that $T\cap e\neq \emptyset$ for all $e\in E$. We call $$G\_T:=\{e\in E:|T\cap e| =1\}$$ the set of *good* members of $E$ with respect to $T$. We say that that a pseudotransversal $T$ is *optimal* if for all pseudotransversals $T\_1$ with $G\_T\subseteq G\_{T\_1}$ we have $G\_T=G\_{T\_1}$.


**Question.** Does every hypergraph $H=(V,E)$ allow for an optimal pseudotransversal?



Optimal pseudotransversals
No, here is an example of a hypergraph with no optimal transversal.


Let $V=\omega$ (the set of nonnegative integers), and let 
$$E=\{\{0\}\} \cup\{\{0,n\};n\ge 1\} \cup \{\{i; i\ge n\}; n\in \omega\}.$$


In other words, $E$ consists of the singleton $\{0\}$, all pairs containing $0$, and all intervals $[n,\infty]$.


If $T$ is a pseudotransversal, then $0\in T$, and so $T$ intersects all the pairs $\{0,n\}$. Also $T$ must be infinite to intersect all the intervals $[n,\infty]$. But now for every nonzero $n\in T$, the set $T\setminus \{n\}$ is still a pseudotransversal, and $G\_{T\setminus \{n\}} \setminus G\_T$ contains the pair $\{0,n\}$.


