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.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $1$