Countable Discrete Space is Sigma-Compact

Theorem
Let $T = \left({S, \vartheta}\right)$ be a topological space where $\vartheta$ is the discrete topology on $S$.

Let $S$ be a countable set, thereby making $\vartheta$ the countable discrete topology on $S$..

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

If $S$ is uncountable, then $T$ is not $\sigma$-compact.

Proof
We have that Singleton Sets in Discrete Space are Compact.

We also have that $S$ is the union of all its singleton sets:
 * $\displaystyle S = \bigcup_{x \in S} \left\{{x}\right\}$

As $S$ is countable, it is the union of countably many compact sets.

Hence the result, by definition of $\sigma$-compact.