# Modified Fort Space is Compact

## Theorem

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.

Then $T$ is a compact space.

## Proof

Let $\CC$ be an open cover of $T$.

Then $\exists U \in \CC: a \in U$.

Then by definition of modified Fort space, $U$ is cofinite.

That is, $S \setminus U$ is finite.

Then $S \setminus U$ is covered by a finite subcover of $\CC$.

Hence, by definition, $T$ is compact.

$\blacksquare$