Finite Topological Space is Compact/Proof 2

Theorem

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

Then $T$ is compact.

Proof

From Power Set of Finite Set is Finite, the power set of $S$ is finite.

Since the topology $\tau$ is by definition a set of subsets of $S$, it follows that $\tau$ is finite as well.

Let $\VV$ be an open cover of $S$.

By definition $\VV \subseteq \tau$ and so is also a finite set.

From Set is Subset of Itself, $\VV \subseteq \VV$.

Thus by definition $\VV$ is a finite subcover of $\VV$.

The result follows by definition of compact.

$\blacksquare$