Compact Sets in Fortissimo Space

Theorem

A subset of a Fortissimo space is compact if and only if it is finite.

Proof

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

Necessary Condition

By Finite Topological Space is Compact, all finite sets are compact in $T$.

$\Box$

Sufficient Condition

We prove the contrapositive.

Let $A$ be an infinite subset of $S$.

Let $C$ be a countably infinite subset of $A$ that does not contain $p$.

For each $x \in A$, $C \setminus \set x$ is a countably infinite set.

Hence $\relcomp S {C \setminus \set x} = \relcomp S C \cup \set x$ is open in $T$.

Thus $\CC = \set {\relcomp S C \cup \set x: x \in A}$ is an open cover of $A$.

However, each subset of the $\CC$ contains exactly $1$ element in $C$, so a finite subset of $\CC$ can only contain a finite number of elements in $C$.

Therefore $\CC$ has no finite subcover.

Hence $A$ is not compact.

This shows that if a subset of a Fortissimo space is compact, it must be finite.

$\blacksquare$