# Discrete Space is Compact iff Finite

## Theorem

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

Then $T$ is compact if and only if $S$ is a finite set, thereby making $\tau$ the finite discrete topology on $S$.

## Proof

### Necessary Condition

Let $T$ be a compact discrete space.

Aiming for a contradiction, suppose $T$ is infinite.

Let $\mathcal C$ be the cover for $S$ defined as:

- $\mathcal C = \left\{{\left\{{x}\right\}: x \in S}\right\}$

As $S$ is an infinite set then so is $\mathcal C$.

Let $\mathcal C'$ be a proper subset of $\mathcal C$.

Then:

- $\exists y \in S: \left\{{y}\right\} \notin \mathcal C'$

and so $\mathcal C'$ is not a cover for $S$.

So by definition $\mathcal C'$ is not a subcover of $\mathcal C$.

So $\mathcal C$ can have no finite subcover.

Hence by definition $T$ can not be compact.

By Proof by Contradiction it follows that of $T$ is compact then $S$ must be finite.

$\Box$

### Sufficient Condition

Let $T$ be a finite discrete space.

Then from Finite Topological Space is Compact it follows that $T$ is a compact space.

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): Exercise $5.10.2$