Fort Space is T5

Theorem

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.

Then $T$ is a $T_5$ space.

Let $A, B \in \tau_p$ such that $A$ and $B$ are separated.

If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both open.

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

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

Now suppose $B$ were not closed.

Then $B$ is infinite by definition of Fort space.

But every open set containing $p$ is cofinite in $S$.

So $p$ would be in $B^-$, where $B^-$ is the closure of $B$.

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

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

So $B$ must be closed.

Therefore $\relcomp S B$ is an open set such that $A \subseteq \relcomp S B$, and the $T_5$ separation axiom is satisfied.

$\blacksquare$

1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $2$