Fortissimo Space is T5

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Let $T = \struct {S, \tau_p}$ be a Fortissimo 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.

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

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

Now suppose $B$ were not closed.

Then $B$ is uncountable by definition of Fortissimo space.

But every open set not containing $p$ is co-countable 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.