Let us say that a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ is $T\_1$ if for $x\neq y$ there is $e\in E$ such that $e\cap\{x,y\} = \{x\}$. 


Note that for any $T\_1$-[space](https://en.wikipedia.org/wiki/T1_space) $(X,\tau)$ the topology $\tau$ contains the cofinite topology (the collection of sets with finite complement, along with $\emptyset$).


With $T\_1$-hypergraphs, the situation does not seem as pleasant as this.


**Question.** If $H=(V,E)$ is a $T\_1$-hypergraph, is there $E\_1\subseteq E$ such that $(V, E\_1)$ is $T\_1$, but whenever $E\_2$ is a proper subset of $E\_1$, then $(V,E\_2)$ is no longer $T\_1$?



$T_1$-spaces vs $T_1$-hypergraphs
Counterexample:


For $a\in\mathbb Q$ let $L\_a=\{x\in\mathbb Q:x\lt a\}$, $R\_a=\{x\in\mathbb Q:x\gt a\}$.


Let $H=(\mathbb Q,E)$ where $E=\{L\_a:a\in\mathbb Q\}\cup\{R\_a:a\in\mathbb Q\}$.


Clearly $H$ is a $T\_1$-hypergraph. Consider any $E\_1\subseteq E$ such that $(\mathbb Q,E\_1)$ is $T\_1$. Choose $a\in\mathbb Q$ so that $L\_a\in E\_1$. Let $E\_2=E\_1\setminus\{L\_a\}$, a proper subset of $E\_1$. I claim that $E\_2$ is $T\_1$.


Assume for a contradiction that $E\_2$ is not $T\_1$. Then there exist $x,y\in\mathbb Q$ such that $x\ne y$ and there is no $e\in E\_2$ such that $e\cap\{x,y\}=\{x\}$. Since $E\_1$ is $T\_1$, we must have $L\_a\cap\{x,y\}=\{x\}$, i.e., $x\lt a\le y$. Choose $z\in\mathbb Q$ so that $x\lt z\lt a$. Since $E\_1$ is $T\_1$, there is some $L\_b\in E\_1$ such that $L\_b\cap\{x,z\}=\{x\}$, *i.e.*, $x\lt b\le z\lt a\le y$. But then $L\_b\in E\_2$ and $L\_b\cap\{x,y\}=\{x\}$, a contradiction.


  
**P.S.** Here is an example where the edges of the hypergraph are finite sets: $H=(V,E)$ where $V=\omega=\{0,1,2,3,\dots\}$ and
$$E=\{\{0,1\},\{0,2\},\{0,3\},\{0,4\},\dots\}\cup\{\{1,2\},\{1,2,3\},\{1,2,3,4\},\dots\}.$$


