Show that if \(X\) is Hausdorff and \(A\subseteq X\) is compact then \(A\) is closed.
Solution
Goal: Show that if \(X\) is Hausdorff and \(A \subseteq X\) is compact, then \(A\) is closed.
1. Fix \(x \in X \setminus A\); it suffices to find an open neighborhood of \(x\) disjoint from \(A\).
Then \(X \setminus A\) is a union of open sets, hence open, so \(A\) is closed.
2. For each \(a \in A\), choose disjoint open sets \(U_a \ni a\) and \(V_a \ni x\).
\(X\) is Hausdorff, and \(a \neq x\) since \(x \notin A\).
3. \(\theset{U_a}_{a \in A}\) is an open cover of \(A\), so it has a finite subcover \(U_{a_1}, \ldots, U_{a_n}\).
\(A\) is compact.
4. Define \(V := V_{a_1} \cap \cdots \cap V_{a_n}\); then \(V\) is open, contains \(x\), and is disjoint from \(A\).
4.1. \(V\) is open and \(x \in V\).
4.2. \(V \cap A = \emptyset\).
If \(y \in V \cap A\), then \(y \in U_{a_j}\) for some \(j\) (<1>3), but \(y \in V \subseteq V_{a_j}\) contradicts \(U_{a_j} \cap V_{a_j} = \emptyset\) (<1>2).
5. Q.E.D.
1 and 4 show every \(x \notin A\) has a neighborhood avoiding \(A\); hence \(A\) is closed.