2
$\begingroup$

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 $T_1$?

$\endgroup$
  • $\begingroup$ Thanks for putting this effort in! $\endgroup$ – Dominic van der Zypen Mar 1 '19 at 13:19
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 Feb 26 '19 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.