If \(X\) is \(T_2\) and \(A\subseteq X\) is compact, then \(A\) is closed.
Solution
1. Every \(x \in X \setminus A\) has an open neighborhood disjoint from \(A\).
For each \(a \in A\), since \(X\) is Hausdorff and \(x \neq a\), choose disjoint open sets \(U_a \ni x\) and \(V_a \ni a\). The sets \(V_a\) cover \(A\), so by compactness \(A \subseteq V_{a_1} \cup \cdots \cup V_{a_k}\) for some \(a_1, \ldots, a_k \in A\). Put \(U = U_{a_1} \cap \cdots \cap U_{a_k}\), an open set containing \(x\). Since \(U \subseteq U_{a_i}\) and \(U_{a_i} \cap V_{a_i} = \varnothing\) for each \(i\), \(U\) is disjoint from \(V_{a_1} \cup \cdots \cup V_{a_k} \supseteq A\).
2. Q.E.D.
By step 1, \(X \setminus A\) is a union of open sets, hence open, so \(A\) is closed.