Discrete Space is Compact iff Finite

Theorem

Let $T = \struct {S, \tau}$ 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 $\CC$ be the cover for $S$ defined as:

$\CC = \set {\set x: x \in S}$

As $S$ is an infinite set then so is $\CC$.

Let $\CC'$ be a proper subset of $\CC$.

Then:

$\exists y \in S: \set y \notin \CC'$

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

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

So $\CC$ can have no finite subcover.

Hence by definition $T$ can not be compact.

By Proof by Contradiction it follows that if $T$ 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): $5$: Compact spaces: Exercise $5.10: 2$