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