# Proof: if $K\subset M$ is compact and $A\subset K$ is closed, then $A$ is compact [duplicate]

Possible Duplicate:
Compactness is closed-hereditary - significance of closed property?

Proposition let $(M,d)$ be a metric space where $K\subset M$ is compact and $A\subset K$ is closed. Then $A$ is compact.

Proof let $\{ G_\alpha \}_{\alpha\in I}$ be a open cover of $A$, then $A \subset \cup_{\alpha\in I} G_\alpha$. Since $A^c$ is open and $K\setminus A \subset A^c$, $$K \subset \left( \bigcup_{\alpha\in I} G_\alpha \cup A^c\right) \quad (1)$$ and, since $K$ is compact, $\exists N\in\mathbb{N}$, $\exists \alpha_1,...,\alpha_N \in I$: $$A\subset K \subset \left( \bigcup_{i\in I}^N G_{\alpha_i} \cup A^c\right) \quad (2)$$ Then, $A$ is compact.

Question Why $A$ is necessarily closed? Step (1) is also true if $A$ is open since $\cup_{\alpha\in I} G_\alpha$ covers $A$, and $A\cup A^c = M$.

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## marked as duplicate by Austin Mohr, Henry T. Horton, Davide Giraudo, Zhen Lin, AndrewDec 7 '12 at 21:47

"Since $A^c$ is open..." – Chris Eagle Dec 7 '12 at 19:16
I am fairly sure I answered this sort of question not long ago. – Asaf Karagila Dec 7 '12 at 19:19
Yes, I answered this question. – Asaf Karagila Dec 7 '12 at 19:32

You need $A^c$ to be open for $\{G_{\alpha}\}_{\alpha \in I} \cup \{A^c\}$ to be an open cover of $K$.
You need the collection $\mathscr{U}=\{M\setminus A\}\cup\{G_\alpha:\alpha\in I\}$ to be an open cover of $K$; if $A$ isn’t closed, $M\setminus A$ isn’t open. If $\mathscr{U}$ isn’t an open cover of $K$, there’s no guarantee that it has a finite subcover, i.e., that you can manage your step $(2)$.
For a concrete example, let $M=\Bbb R$, $K=[0,3]$, and $A=(1,2)$: $K$ is compact, and $A$ isn’t.