Theorem. If \(X\) is a Hausdorff space that is locally compact and paracompact, then each component of \(X\) has a countable basis.
Proof. If \(X_0\) is a component of \(X\), then \(X_0\) is locally compact and paracompact. Let \(\mathcal{C}\) be a locally finite covering of \(X_0\) by sets open in \(X_0\) that have compact closures. Let \(U_1\) be a nonempty element of \(\mathcal{C}\), and in general let \(U_n\) be the union of all elements of \(\mathcal{C}\) that intersect \(\overline{U}_{n-1}\). Show that \(\overline{U}_n\) is compact, and the sets \(U_n\) cover \(X_0\).
Solution
Closures are taken in \(X_0\).
1. \(\overline{U}_n\) is compact for every \(n \ge 1\).
1.1. \(\overline{U}_1\) is compact.
\(U_1 \in \mathcal{C}\), and every member of \(\mathcal{C}\) has compact closure.
1.2. If \(\overline{U}_{n-1}\) is compact, then only finitely many members \(V_1, \ldots, V_m\) of \(\mathcal{C}\) meet \(\overline{U}_{n-1}\).
By local finiteness, each \(x \in \overline{U}_{n-1}\) has an open neighborhood \(W_x\) that meets only finitely many members of \(\mathcal{C}\). Finitely many of these, \(W_{x_1}, \ldots, W_{x_k}\), cover the compact set \(\overline{U}_{n-1}\). A member of \(\mathcal{C}\) that meets \(\overline{U}_{n-1}\) meets some \(W_{x_i}\), and there are only finitely many such members.
1.3. Q.E.D.
1.1. If \(\overline{U}_{n-1}\) is compact, step 2 gives \(U_n = V_1 \cup \cdots \cup V_m\), so\[\overline{U}_n = \overline{V}_1 \cup \cdots \cup \overline{V}_m\]is a finite union of compact sets, hence compact.
2. \(A = \bigcup_{n=1}^\infty U_n\) is nonempty and open in \(X_0\).
\(A\) contains the nonempty set \(U_1\), and each \(U_n\) is a union of sets open in \(X_0\).
3. \(A\) is closed in \(X_0\).
Let \(x \in \overline{A}\). Since \(\mathcal{C}\) covers \(X_0\), some \(V_0 \in \mathcal{C}\) contains \(x\). The open set \(V_0\) meets \(A\), so it meets \(U_{n-1}\) for some \(n \ge 2\), hence meets \(\overline{U}_{n-1}\). By the definition of \(U_n\), \(V_0 \subseteq U_n \subseteq A\), so \(x \in A\).
4. Q.E.D.
Step 1 shows that each \(\overline{U}_n\) is compact. By steps 2 and 3, \(A\) is a nonempty open and closed subset of the connected space \(X_0\), so \(A = X_0\): the sets \(U_n\) cover \(X_0\).