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 *transversal basis* is a set $B\subseteq V$ such that $|B\cap e|\leq 1$ for all $e\in E$. Setting $I\_B:=\{e\in E:B\cap e\neq \emptyset\}$, we say that that a transversal basis $B$ is **good** if for all transversal bases $B\_1$ with $I\_{B}\subseteq I\_{B\_1}$ we have $I\_B=I\_{B\_1}$.


**Question.** Does every hypergraph $H=(V,E)$ have an good transversal basis?



Finding a good transversal basis
Let $V=\mathbb Z,E=\{\{m\in\mathbb Z:m\geq n\}:n\in\mathbb Z\}$. It's clear every transversal basis has at most one element, but no basis $\{k\}$ is good, since $\{k+1\}$ intersects more sets from $E$.


