Empty Set is Compact Space

Theorem

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

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

Proof

Recall the definition of compact subspace:

$\struct {\O, \tau_\O}$ is compact in $T$ if and only if every open cover $\CC \subseteq \tau_\O$ for $\O$ has a finite subcover.

The only open cover for $\O$ that is contained in $\O$ is $\set \O$ itself.

This has only one finite subcover, and that is $\set \O$.

This is a finite subcover.

Hence the result, by definition of compact subspace.

$\blacksquare$