Show that a compact subset of a Hausdorff space is closed.
Solution
1. Let \(A\) be a compact subset of a Hausdorff space \(X\).To prove that \(A\) is closed, it suffices to prove that \(X\setminus A\) is open.
A subset of a topological space is closed exactly when its complement is open.
2. Fix \(x\in X\setminus A\).For every \(a\in A\), choose disjoint open sets\[ U_a\ni a, \qquad V_a\ni x. \]
Since \(x\notin A\), one has \(x\ne a\) for every \(a\in A\). Because \(X\) is Hausdorff, the distinct points \(a\) and \(x\) admit disjoint open neighborhoods \(U_a\) and \(V_a\).
3. There are finitely many points \(a_1,\dots,a_m\in A\) such that\[ A\subseteq U_{a_1}\cup\cdots\cup U_{a_m}. \]
The family \[ \{U_a:a\in A\} \] is an open cover of the compact set \(A\). Compactness therefore gives a finite subcover.
4. The set\[ V=V_{a_1}\cap\cdots\cap V_{a_m} \]is an open neighborhood of \(x\) disjoint from \(A\).
It is open as a finite intersection of open sets, and it contains \(x\) because every \(V_{a_i}\) contains \(x\).
Suppose \(y\in V\cap A\). By 3, \(y\in U_{a_i}\) for some \(i\). But \(y\in V\subseteq V_{a_i}\) as well, contradicting \[ U_{a_i}\cap V_{a_i}=\varnothing. \] Thus \(V\cap A=\varnothing\).
5. Therefore \(A\) is closed in \(X\).
By 4, every point \(x\in X\setminus A\) has an open neighborhood contained in \(X\setminus A\). Hence \(X\setminus A\) is open. By 1, \(A\) is closed.