Closed Subspace of Compact Space is Compact

Theorem

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

Let $C = \struct {H, \tau_H}$ be a subspace of $T$.

Let $C$ be closed in $T$.

Then $\struct {H, \tau}$ is compact.

Let $T$ be a compact space.

Let $C$ be a closed subspace of $T$.

Let $\UU$ be an open cover of $C$.

We have that $C$ is closed in $T$.

It follows by definition of closed that $T \setminus C$ is open in $T$.

So if we add $T \setminus C$ to $\UU$, we see that $\UU \cup \set {T \setminus C}$ is also an open cover of $T$.

As $T$ is compact, there exists a finite subcover of $\UU \cup \set {T \setminus C}$, say $\VV = \set {U_1, U_2, \ldots, U_r}$.

This covers $C$ by the fact that it covers $T$.

Suppose $T \setminus C$ is an element of $\VV$.

Then $T \setminus C$ may be removed from $\VV$, and the rest of $\VV$ still covers $C$.

Thus we have a finite subcover of $\UU$ which covers $C$.

Hence $C$ is compact.

$\blacksquare$