# Compactness is closed-hereditary - significance of closed property?

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 $F is compact. Question What is the significance of$F$being closed, I don't see any mention of that fact being made in the proof? If we had$F$is an open subset where would the proff - You used$X-F$is open so that you have a new open cover. – Hui Yu Nov 24 '12 at 14:44 Cheers, overlooked that. – sonicboom Nov 24 '12 at 15:25 A space where every subspace is compact is a called a noetherian topological space. – Zhen Lin Nov 24 '12 at 16:50 ## 1 Answer In the general case there are spaces where every subspace is compact. However in metric spaces this is not true. For example$[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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