Let $X$ be a compact topological space, and $A$ a closed subspace. Show that $A$ is compact.

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In order to show that $A$ is compact. We need to show that for any open cover $U$ of $A$, it contains a finite open cover of $A$.

Let $U$ be a open cover of $A$.

For each $U_\alpha \in U$, there exists an open set $V_\alpha \in X$ so that $V_\alpha \cap A = U_\alpha$.

Here we denote the index set of $U$ by $I$, and $\alpha \in I$.

Now, set $V = \{V_\alpha\}_{\alpha \in I}$, then $V \cap \{X \setminus U\}$ is an open cover of $X$.

It contains a finite subcover $\{V_1, ..., V_n\} \cup \{X \setminus U\}$ of $X$,

hence $\{ V_1 \cap A, ..., V_n \cap A \}$ is a finite subcover of $A$.

Hence $A$ is compact.


That's an excellent proof, with exactly the details needed.

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  • $\begingroup$ @PandaMan: Only one thing to fix. Namely, you should have $\mathcal V\cup\{X\setminus A\}$ as your open cover of $X.$ You might also want to prove that $A$ is covered by $\{V_1\cap A,...,V_n\cap A\}.$ $\endgroup$ – Cameron Buie Mar 3 '14 at 20:01
  • $\begingroup$ Good points, @Cameron. I was lulled into a false complacency. :) $\endgroup$ – John Hughes Mar 3 '14 at 21:13

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