Show that a finite product or union compact spaces is again compact.
Solution
(1) Finite Union of Compact Spaces: Let \(X = \bigcup_{i=1}^n K_i\), where each \(K_i\) is a compact topological space (or compact subspace of a larger space). Let \(\mathcal{U} = \{U_\alpha\}_{\alpha \in A}\) be an open cover of \(X\). For each \(i \in \{1, \ldots, n\}\), \(\mathcal{U}\) is an open cover of \(K_i\). Since \(K_i\) is compact, there exists a finite subcover \(\mathcal{U}_i = \{U_{i, 1}, \ldots, U_{i, m_i}\} \subseteq \mathcal{U}\) such that: \[ K_i \subseteq \bigcup_{j=1}^{m_i} U_{i, j}. \] Then the union of these finite collections: \[ \mathcal{U}' = \bigcup_{i=1}^n \mathcal{U}_i = \bigcup_{i=1}^n \{U_{i, 1}, \ldots, U_{i, m_i}\} \] is a finite subcollection of \(\mathcal{U}\). Moreover: \[ X = \bigcup_{i=1}^n K_i \subseteq \bigcup_{i=1}^n \bigcup_{j=1}^{m_i} U_{i, j}. \] Thus \(\mathcal{U}'\) is a finite subcover of \(X\), so \(X\) is compact.
(2) Finite Product of Compact Spaces: It suffices to prove that the product of two compact spaces \(X\) and \(Y\) is compact (by induction, any finite product \(\prod_{i=1}^n X_i\) is compact).
Let \(\mathcal{U} = \{U_\alpha\}_{\alpha \in A}\) be an open cover of \(X \times Y\). 1. Tube Lemma: For each fixed \(x \in X\), the slice \(\{x\} \times Y \cong Y\) is compact. The collection \(\mathcal{U}\) covers \(\{x\} \times Y\). By compactness of \(Y\), there is a finite subcollection \(\{U_{\alpha_1}, \ldots, U_{\alpha_k}\} \subseteq \mathcal{U}\) such that: \[ \{x\} \times Y \subseteq U_x := \bigcup_{j=1}^k U_{\alpha_j}. \] By the Tube Lemma, there exists an open neighborhood \(V_x\) of \(x\) in \(X\) such that: \[ V_x \times Y \subseteq U_x = \bigcup_{j=1}^k U_{\alpha_j}. \]
Covering \(X\): The collection \(\{V_x\}_{x \in X}\) forms an open cover of \(X\). Since \(X\) is compact, there exists a finite subcover \(\{V_{x_1}, V_{x_2}, \ldots, V_{x_m}\}\) of \(X\), so \(X = \bigcup_{i=1}^m V_{x_i}\).
Finite Subcover of \(X \times Y\): For each \(i \in \{1, \ldots, m\}\), \(V_{x_i} \times Y\) is covered by a finite subcollection \(\mathcal{U}_{x_i} \subseteq \mathcal{U}\). Then: \[ X \times Y = \left( \bigcup_{i=1}^m V_{x_i} \right) \times Y = \bigcup_{i=1}^m (V_{x_i} \times Y) \subseteq \bigcup_{i=1}^m \bigcup_{U \in \mathcal{U}_{x_i}} U. \] The union \(\bigcup_{i=1}^m \mathcal{U}_{x_i}\) is a finite subcollection of \(\mathcal{U}\) covering \(X \times Y\). Hence \(X \times Y\) is compact.