Given any collection $\mathcal{C} = \{E_1, E_2, ..., E_m\}$ of finite and nonempty discrete sets, is there a set $I$ such that $$ \forall E_k \in \mathcal{C}, \; E_k \cap I \neq \emptyset,$$ and $$ \forall i \in I, \; \exists E_k \in \mathcal{C}, E_{k} \cap I = \{i\} $$ ?

For example, for $E_1 = \{1,2\}$, $E_2 = \{2,3\}$ and $E_3 = \{1,3\}$ with $m = 3$, a possible $I$ is $\{1,2\}$.

I've been trying to answer this question ever since stumbling upon it in my research, under a different form. I'm hardly a real mathematician so I submitted this problem to some nearby theoretical computer scientists hoping they'd tell me it was trivial, but no such luck.

All I've found until now are a few of rather trivial observations. To begin with, the second property to be verified by $I$ is equivalent to there being an injection $g$ from $I$ to $\{1, ..., m\}$ such that $E_{g(i)} \cap I = \{i\}$ for all $i \in I$. Also, the problem can be formulated as a graph problem using a bipartite graph with $m$ vertices on one side representing the sets in the collection and vertices on the other side for each element in the union set $\bigcup_{i=1}^m E_k$, but graph theory isn't my forte either. Finally, notice that we only really need to consider the collections where $E_1, ..., E_m$ also verify the following properties:

- any one set $E_j$ is not included in any other $E_k$;
- there exists no element $e$ in exactly one set $E_k$, in other words $e$ is either in none or in at least two sets of $\mathcal{C}$.