Let us say that a 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 $(X,\tau)$ 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 $T_1$?

  • $\begingroup$ Thanks for putting this effort in! $\endgroup$ Commented Mar 1, 2019 at 13:19

1 Answer 1



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 $$E=\{\{0,1\},\{0,2\},\{0,3\},\{0,4\},\dots\}\cup\{\{1,2\},\{1,2,3\},\{1,2,3,4\},\dots\}.$$

  • $\begingroup$ In the second line, wouldn't some notation like $R_a$ be better for the latter set? Because right now it's unclear if $L_c$ would mean the latter or the former. $\endgroup$
    – Wojowu
    Commented Feb 26, 2019 at 11:43

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