Show that if \(X\) is compact and \(A\subseteq X\) is closed then \(A\) is compact.
Solution
Goal: Show that if \(X\) is compact and \(A \subseteq X\) is closed, then \(A\) is compact.
1. Let \(\theset{U_\alpha}\) be an open cover of \(A\).
Arbitrary open cover of \(A\) (by open subsets of \(X\)).
2. \(X \setminus A\) is open in \(X\).
\(A\) is closed.
3. \(\theset{U_\alpha} \cup \theset{X \setminus A}\) is an open cover of \(X\).
On \(A\), the \(U_\alpha\) cover; on \(X \setminus A\), the set \(X \setminus A\) covers.
4. There is a finite subcover of \(X\): some \(U_{\alpha_1}, \ldots, U_{\alpha_n}\) together with (possibly) \(X \setminus A\).
\(X\) is compact and 3 is an open cover.
5. \(\theset{U_{\alpha_1}, \ldots, U_{\alpha_n}}\) is a finite cover of \(A\).
The finite subcover from 4 covers \(A\) after discarding \(X \setminus A\) (which contains no points of \(A\)); the remaining \(U_{\alpha_j}\) still cover \(A\) since any \(a \in A\) lies in one of the subcover members, which must be a \(U_{\alpha_j}\) because \(a \notin X \setminus A\).
6. Q.E.D.
<1>1–<1>5 show every open cover of \(A\) has a finite subcover.