Empty Set is Compact Space

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Then the empty set $\varnothing$ is a compact subspace of $T$.

Proof
Recall the definition of compact subspace:
 * $\left({\varnothing, \tau_\varnothing}\right)$ is compact in $T$ every open cover $\mathcal C \subseteq \tau_\varnothing$ for $\varnothing$ has a finite subcover.

The only open cover for $\varnothing$ that is contained in $\varnothing$ is $\left\{{\varnothing}\right\}$ itself.

This has only one finite subcover, and that is $\left\{{\varnothing}\right\}$.

This is a finite subcover.

Hence the result, by definition of compact subspace.