# Closed subsets of compact spaces are compact.

I'm having trouble understanding a point in the following standard argument to show that closed subsets of compact spaces are compact.

The proof usually goes as follows.

Let $$X$$ be compact and let $$C$$ be closed. Let $$\{U_\alpha \}$$ be some open cover of $$C$$. Now, $$\{U_\alpha \}$$ with $$X \setminus C$$ is an open cover of $$X$$. As such, we get a finite open cover of $$X$$ from this collection, from which we get finite subcover of $$C$$.

The point I am having trouble understanding is that if $$\{U_\alpha \}$$ is an open cover of closed set $$C$$, it does not seem necessary that all $$U_\alpha$$ is open in $$X$$. The example I am thinking of is $$[0,1]$$ in $$\mathbb{R}$$. I believe there is an open cover(with respect to subspace topology of $$[0,1]$$) of $$[0,1]$$ with $$[0,0.6)$$ and $$(0.4,1]$$. However, both of these are not open sets in $$\mathbb{R}$$. So I am troubled by the point that $$\{U_\alpha \}$$ with $$X \setminus C$$ forms an open cover of $$X$$.

• The cover $\mathcal{U}$ is open in $X$, not in the subspace topology. Commented Dec 31, 2020 at 19:59
• @BrianM.Scott the thing that confuses me is then, shouldn't the cover be open in $C$, since we want to prove that $C$ is a compact subset, and it should be compact with respect to the subspace topology of $C$?
– Phil
Commented Dec 31, 2020 at 20:02
• It doesn’t matter, for the reason given in leoli1’s answer. (The author whom you’re quoting probably should have made that explicit.) Commented Dec 31, 2020 at 20:05

You are right, if $$U$$ is open in $$C$$ then it is not necessarily open in $$X$$. However there will be an open subset $$U'$$ of $$X$$ such that $$U=C\cap U'$$. Then you can simply replace the $$U_\alpha$$ with the $$U_\alpha'$$.

• Thanks so much!
– Phil
Commented Dec 31, 2020 at 20:12