# Optimal pseudotransversals

A 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?

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 $$\{0,n\}$$.