Let \(\ts{X_\alpha}\) be an indexed family of nonempty spaces.
Show that if \(\prod X_\alpha\) is locally compact, then each \(X_\alpha\) is locally compact and \(X_\alpha\) is compact for all but finitely many values of \(\alpha\).
Prove the converse, assuming the Tychonoff theorem.
Solution
Goal: Prove the necessary and sufficient conditions for the Cartesian product \(\prod_{\alpha \in J} X_\alpha\) of non-empty topological spaces to be locally compact.
1. Part (a): Necessity.If \(X = \prod_{\alpha \in J} X_\alpha\) is locally compact, then each \(X_\alpha\) is locally compact and all but finitely many \(X_\alpha\) are compact.Proof:
1.1. Let \(\mathbf{x} = (x_\alpha)_{\alpha \in J} \in X\). Since \(X\) is locally compact, \(\mathbf{x}\) has a compact neighborhood \(C \subseteq X\).
1.2. By definition of neighborhood, there exists an open set \(U \subseteq X\) such that \(\mathbf{x} \in U \subseteq C\).
1.3. In the product topology, \(U\) contains a basic open set \(V = \prod_{\alpha \in J} V_\alpha\) containing \(\mathbf{x}\), where each \(V_\alpha\) is open in \(X_\alpha\) and \(V_\alpha = X_\alpha\) for all \(\alpha \in J \setminus F\), where \(F \subset J\) is a finite subset.
1.4. For every index \(\beta \in J\), the canonical coordinate projection \(\pi_\beta: X \to X_\beta\) is continuous, open, and surjective.
1.5. The image \(C_\beta := \pi_\beta(C)\) is a compact subset of \(X_\beta\) because continuous maps preserve compactness.
1.6. Furthermore, \(x_\beta \in V_\beta = \pi_\beta(V) \subseteq \pi_\beta(C) = C_\beta\), and \(V_\beta\) is open in \(X_\beta\), so \(C_\beta\) is a compact neighborhood of \(x_\beta\) in \(X_\beta\).
1.7. Since \(x_\beta\) was arbitrary, each space \(X_\beta\) is locally compact.
1.8. For every index \(\alpha \in J \setminus F\), \(V_\alpha = X_\alpha\), which implies:\[X_\alpha = V_\alpha = \pi_\alpha(V) \subseteq \pi_\alpha(C) \subseteq X_\alpha \implies X_\alpha = \pi_\alpha(C) = C_\alpha.\]
1.9. Thus \(X_\alpha\) is compact for every \(\alpha \in J \setminus F\), so all but finitely many \(X_\alpha\) are compact.
2. Part (b): Sufficiency (Converse).If each \(X_\alpha\) is locally compact and \(X_\alpha\) is compact for all \(\alpha \notin \{\alpha_1, \dots, \alpha_n\}\), then \(\prod_{\alpha \in J} X_\alpha\) is locally compact.Proof:
2.1. Let \(\mathbf{x} = (x_\alpha)_{\alpha \in J} \in X\).
2.2. For each \(k \in \{1, \dots, n\}\), local compactness of \(X_{\alpha_k}\) provides a compact neighborhood \(C_{\alpha_k} \subseteq X_{\alpha_k}\) of \(x_{\alpha_k}\), which contains an open neighborhood \(U_{\alpha_k} \subseteq C_{\alpha_k}\) of \(x_{\alpha_k}\).
2.3. For all indices \(\alpha \notin \{\alpha_1, \dots, \alpha_n\}\), define \(C_\alpha = X_\alpha\) and \(U_\alpha = X_\alpha\). By hypothesis, each such \(C_\alpha\) is compact and open in itself.
2.4. Define \(C = \prod_{\alpha \in J} C_\alpha\) and \(U = \prod_{\alpha \in J} U_\alpha\).
2.5. By the Tychonoff Theorem, the product of compact spaces \(C\) is compact in \(X\).
2.6. Since \(U_\alpha = X_\alpha\) for all but finitely many \(\alpha\), \(U\) is a basic open set in the product topology on \(X\), and \(\mathbf{x} \in U \subseteq C\).
2.7. Thus \(C\) is a compact neighborhood of \(\mathbf{x}\) in \(X\).
2.8. Therefore \(X = \prod_{\alpha \in J} X_\alpha\) is locally compact. Q.E.D.