Prove that the product of two compact spaces is compact.
Show that if \(A\) and \(B\) are compact, then so is \(A\times B\).
Solution
1. If \(X=\varnothing\) or \(Y=\varnothing\), then \(X\times Y\) is compact.
In either case \(X\times Y=\varnothing\). The empty space is compact because every open cover of it has the empty finite subcover.
Henceforth assume \(X\) and \(Y\) are nonempty.
2. Cover of a compact slice \(\{x\} \times Y\):
2.1. Let \(\mathcal{A}\) be an arbitrary open cover of \(X \times Y\).
2.2. Fix a point \(x \in X\). The slice \(\{x\} \times Y\) is homeomorphic to \(Y\), and is therefore compact.
2.3. For each \(y \in Y\), the point \((x, y) \in X \times Y\) belongs to some open set \(U_{x,y} \in \mathcal{A}\).
2.4. By the definition of the product topology, there exist open sets \(V_{x,y} \subseteq X\) and \(W_{x,y} \subseteq Y\) such that\[(x, y) \in V_{x,y} \times W_{x,y} \subseteq U_{x,y}.\]
2.5. The family \(\{W_{x,y} : y \in Y\}\) is an open cover of \(Y\).
2.6. Since \(Y\) is compact, there exist finitely many points \(y_1, y_2, \dots, y_{k(x)} \in Y\) such that\[Y = \bigcup_{i=1}^{k(x)} W_{x, y_i}.\]
3. Construction of a tube neighborhood \(N_x \times Y\) (The Tube Lemma):
3.1. Define\[N_x = \bigcap_{i=1}^{k(x)} V_{x, y_i}.\]
3.2. Since \(N_x\) is a finite intersection of open sets containing \(x\), \(N_x\) is an open neighborhood of \(x\) in \(X\).
3.3. For each \(i \in \{1, \dots, k(x)\}\), \(N_x \subseteq V_{x, y_i}\), so \(N_x \times W_{x, y_i} \subseteq V_{x, y_i} \times W_{x, y_i} \subseteq U_{x, y_i}\).
3.4. Thus the tube \(N_x \times Y\) is covered by finitely many elements of \(\mathcal{A}\):\[N_x \times Y = N_x \times \left( \bigcup_{i=1}^{k(x)} W_{x, y_i} \right) = \bigcup_{i=1}^{k(x)} (N_x \times W_{x, y_i}) \subseteq \bigcup_{i=1}^{k(x)} U_{x, y_i}.\]
4. Extraction of a finite subcover for \(X \times Y\):
4.1. The collection \(\{N_x : x \in X\}\) forms an open cover of \(X\).
4.2. Because \(X\) is compact, there exist finitely many points \(x_1, x_2, \dots, x_m \in X\) such that\[X = \bigcup_{j=1}^m N_{x_j}.\]
4.3. Taking products with \(Y\):\[X \times Y = \left( \bigcup_{j=1}^m N_{x_j} \right) \times Y = \bigcup_{j=1}^m (N_{x_j} \times Y) \subseteq \bigcup_{j=1}^m \bigcup_{i=1}^{k(x_j)} U_{x_j, y_i}.\]
4.4. The collection\[\mathcal{A}' = \left\{ U_{x_j, y_i} : 1 \le j \le m, \; 1 \le i \le k(x_j) \right\} \subseteq \mathcal{A}\]is a finite subcollection of \(\mathcal{A}\) that covers \(X \times Y\).
5. Conclusion:
Every open cover \(\mathcal{A}\) of \(X \times Y\) has a finite subcover \(\mathcal{A}'\). Therefore \(X \times Y\) is compact.