Compact Space is Countably Compact

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