# Arens-Fort Space is T5

## Theorem

Let $T = \left({S, \tau}\right)$ be the Arens-Fort space.

Then $T$ is a $T_5$ space.

## Proof

Let $A, B \subseteq S$ such that $A$ and $B$ are separated.

Let $p = \left({0, 0}\right)$.

If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both open and the problem is solved.

Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be separated.

Without loss of generality, suppose $p \in A$.

Then $p \notin B$ so $B$ is open.

$B \cup \left\{{p}\right\}$ is closed by the definition of the Arens-Fort space.

$B \subseteq B^- \subseteq B \cup \left\{{p}\right\}$

where $B^-$ is the closure of $B$.

So $B$ is closed or $p$ is in $B^-$.

But then $p \in A$, by hypothesis.

So $A \cap B^- \ne \varnothing$ and so $A$ and $B$ are not separated.

So $B$ must be closed.

Therefore $\complement_S \left({B}\right)$ is an open set such that $A \subseteq \complement_S \left({B}\right)$, and the $T_5$ separation axiom is satisfied.

$\blacksquare$