# Compactness implies countably compactness

Let X be compact. Then X is countably compact.
My thinking is like this: Let X be a topological space. Since compact, then every open cover hava a finite subcover. Hence it is true for countable open cover.
Is this proof true?

• a finite set is always countable. – Zest Mar 28 at 20:20
• @Zest : True, but why do you mention it? Was it mentioned in a comment that was deleted? "Countably compact" means "every countable open cover has a finite subcover", not "every open cover has a countable subcover" -- is that what you were thinking? That's the Lindelof property. – MPW Mar 28 at 20:32

## 1 Answer

You are exactly correct.

In particular, countable compactness follows from compactness because every countable open cover is an open cover.