Subspace of Finite Complement Topology is Compact

Theorem

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.

Then every topological subspace of $T$, including $T$ itself, is a compact space.

Proof

Let $T_H = \struct {H, \tau_H}$ be a subspace of $T$.

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

Let $U \in \CC$ be any set in $C$.

$U$ covers all but a finite number of points of $T_H$.

So for each of those points we pick an element of $\CC$ which covers each of those points.

Hence we have a finite subcover of $T_H$.

So by definition $T_H$ is a compact space.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $2$