# Closed Subspace of Compact Space is Compact

## Theorem

That is, the property of being compact is weakly hereditary.

## Proof

Let $T$ be a compact space.

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

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

Since $C$ is closed, 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 is 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 $T$.

If $T \setminus C$ is an element of $\VV$, then it can be removed from $\VV$ and the rest of $\VV$ still covers $C$.

Thus we have a finite subcover of $\UU$ which covers $C$, and hence $C$ is compact.

$\blacksquare$