# Minimal hypergraphs with respect to separation

Let $H = (V, E)$ be a hypergraph, that is $V$ is a set and $E \subseteq \mathcal{P}(V)$. We say that $H$ is $T_1$ if for $v\neq w$ there are $e_v, e_w \in E$ such that $v\in e_v, w\notin e_v, w\in e_w, v\notin e_w$.

Fix a set $V$. Let $E\in \mathcal{P}(\mathcal{P}(E))$ be $T_1$. Does the set \begin{eqnarray} \{E'\subseteq E: (V, E') \textrm{ is } T_1\} \end{eqnarray}

contain a minimal member with respect to $\subseteq$?

Motivation: It is well-known and easy to prove that a topological space $(X,\tau)$ is $T_1$ if and only if $\tau \supseteq \tau_{\textrm{cf}}(X)$ where $\tau_{\textrm{cf}}(X) = \{\emptyset\} \cup \{U\subseteq X: X\setminus U \textrm{ is finite}\}$. So the question of minimal $T_1$-topologies is uninteresting, but I'm not sure that the same quesion, when asked in the context of hypergraphs, is as simple as that.

Counterexamples: Let $V=\mathbb R$ and let $E$ be the base for the usual topology consisting of the open intervals, or the subbase consisting of the open rays of the form $(a,\infty)$ and $(-\infty,a)$.

• This is huge! Just $K_4$ and the "graph" on the same four vertices and edges all one-element sets would do. The "intersection" of these "graph" structures has empty edge set, hence obviously not $T_1$. Nov 20 '14 at 12:57
• I claim that on the four element set there are two hypergraph structures (the complete "honest" graph $K_4$ and the graph whose edges are all one-element sets) which are both $T_1$ but, on the other hand, their intersection is empty (which is obviously not $T_1$). Hence, there's no minimal $T_1$ structure. Nov 20 '14 at 20:22
• @AlexDegtyarev, I see. I don't think that by "minimal" the OP means a $T_1$ structure $E$ which is a subset of every $T_1$ structure; I think he means a $T_1$ structure $E$ such that no proper subset of $E$ is a $T_1$ structure. This is the usual meaning of "minimal element" in English; the other would be called "minimum" or "least element".
– bof
Nov 20 '14 at 23:05