Countable Space is Sigma-Compact

Theorem

Let $T = \struct {S, \tau}$ be a topological space, where $S$ is a countable set.

Then $T$ is $\sigma$-compact.

Proof

From Finite Space Satisfies All Compactness Properties, for every $p \in S$, $\set p$ is compact.

Then $\ds S = \bigcup_{p \mathop \in S} \set p$ is a countable union of compact sets.

Thus by definition $T$ is $\sigma$-compact.

$\blacksquare$