# 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.

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 $C$.

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.

$\blacksquare$

## Also see

## Sources

- 1953: Walter Rudin:
*Principles of Mathematical Analysis*: $2.35$ - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $5$: Compact spaces: $5.6$: Compactness and Constructions: Proposition $5.6.1$ - 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction