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Here's a proof that any closed subset of a compact metric space is compact -

Let $(X, d)$ be a compact metric space and let $F$ be a closed subset of $X$.

Let $U = \{U_i : i \in I \}$ be an open cover of $F$.

Now we can form $V = \{X\backslash F\} \cup U$ which is an open cover of $X$.

As $X$ is compact $\exists$ finitely many $V_is$ from $V$ such that $\cup_i^n V_i = X$

Those $V_is$ from $U$ will cover $F$ and hence $F is compact.


What is the significance of $F$ being closed, I don't see any mention of that fact being made in the proof? If we had $F$ is an open subset where would the proff

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You used $X-F$ is open so that you have a new open cover. –  Hui Yu Nov 24 '12 at 14:44
Cheers, overlooked that. –  sonicboom Nov 24 '12 at 15:25
A space where every subspace is compact is a called a noetherian topological space. –  Zhen Lin Nov 24 '12 at 16:50

1 Answer 1

In the general case there are spaces where every subspace is compact. However in metric spaces this is not true.

For example $[0,1]$ is compact, but consider the covering of $(0,1)$ defined as $$\left\{\left(\frac1{n+1},1\right)\mid n\in\Bbb N\right\}$$

There is no finite subcover, because every finite union is just a subinterval which doesn't contain its left endpoint (which is strictly positive).

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