# Closed Subspace of Compact Space is Compact

## Contents

## Theorem

A closed subspace of a compact space 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 $\mathcal U$ 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 $\mathcal U$, we see that $\mathcal U \cup \left\{{T \setminus C}\right\}$ is also an open cover of $T$.

As $T$ is compact, there is a finite subcover of $\mathcal U \cup \left\{{T \setminus C}\right\}$, say $\mathcal V = \left\{{U_1, U_2, \ldots, U_r}\right\}$.

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

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

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

$\blacksquare$

## Also see

## Sources

- 1953: Walter Rudin:
*Principles of Mathematical Analysis*: $2.35$ - 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 1$ - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $5.6$: Compactness and Constructions: Proposition $5.6.1$