# Compact Space is Countably Compact

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## Theorem

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

Then $T$ is countably compact.

## Proof

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

Then by definition every open cover of $S$ has a finite subcover.

So every countable open cover of $S$ has a finite subcover.

Hence by definition $T$ is countably compact.

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties