# Fortissimo Space is not Compact

## Theorem

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

Then $T$ is not a compact space.

## Proof

From the definition, $T$ is a compact space if every open cover of $T$ has a finite subcover.

Let $N \subset S$ be a countably infinite set such that $p \notin N$.

Then $\relcomp S N$ is open in $T$ by the definition of the Fortissimo space.

Let $q \in S$ such that $q \ne p$.

As $p \in \relcomp S {\set q}$ it follows that $\set q$ is open in $T$.

Then:

$\displaystyle S = \relcomp S N \cup \paren {\bigcup_{q \mathop \in N} \set q}$

Let $\CC$ be defined as:

$\CC := \set {\set q: q \in N} \cup \set {\relcomp S N}$

Then from the above, $\CC$ is an open cover of $S$.

But all the sets of $\CC$ are pairwise disjoint.

Therefore $\CC$ can have no finite subcover.

Hence the space cannot be compact.

$\blacksquare$