# Closure of a compact space always compact?

Is the closure of a compact subspace of a topological space always compact?

I want to say no, but i can't think of/find any counterexamples. I think its probably true with added conditions, like Hausdorff property? Because the closure is always closed and closed sets in a compact hausdorff space are compact.

Any counterexamples for a non-Hausdorff space?

• Are you asking "Is the closure of a compact subspace of a topological space always compact"? – Lee Mosher Aug 19 '17 at 21:25
• This is true in the Hausdorff setting, since there compact sets are closed, and closed sets are their own closure: math.stackexchange.com/questions/83355/… – Kaj Hansen Aug 19 '17 at 21:48
• Also see this. – user 170039 Aug 20 '17 at 4:15

Consider the space $X = \mathbb{Z}$ equipped with the topology where the non-empty open sets are precisely those containing $0$.
As a finite set $\{ 0 \}$ is compact. Further we have that $\overline{\{0\}} = X$ and $X$ is not compact since $\bigcup_{x \in X} \{0,x\}$ is an open cover with no finite subcover.