Show that a compact set in a Hausdorff space is closed.
Solution
1. Let \(K\) be compact in a Hausdorff space \(X\), and fix \(x\notin K\).
We construct an open neighborhood of \(x\) disjoint from \(K\).
2. For every \(y\in K\), choose disjoint open sets \(U_y\ni x\) and \(V_y\ni y\).
This is the Hausdorff property.
3. Compactness gives \(y_1,\dots,y_r\) with \(K\subseteq V_{y_1}\cup\cdots\cup V_{y_r}\). Then \(U=U_{y_1}\cap\cdots\cap U_{y_r}\) is an open neighborhood of \(x\) disjoint from \(K\).
If a point of \(U\) lay in \(K\), it would lie in some \(V_{y_i}\), contradicting \(U\subseteq U_{y_i}\) and \(U_{y_i}\cap V_{y_i}=\varnothing\).
4. Thus \(X\setminus K\) is open and \(K\) is closed.
Every point of the complement has the neighborhood constructed in 3.