Show that a finite product or union compact spaces is again compact.
Exercise
Solution
1. If \(K_1,\dots,K_r\) are compact subspaces of one space, then \(K_1\cup\cdots\cup K_r\) is compact.
An open cover restricts to an open cover of each \(K_i\). Choose a finite subcover for each; the union of these finitely many finite families covers the whole union.
2. If \(X\) and \(Y\) are compact, then \(X\times Y\) is compact.
Let \(\mathcal U\) cover \(X\times Y\). For each \((x,y)\) choose a basic rectangle \(A\times B\) containing it and lying in some member of \(\mathcal U\). Fixing \(x\), compactness of \(Y\) gives finitely many such rectangles covering \(\{x\}\times Y\); intersect their \(A\)-factors to obtain a neighborhood \(A_x\) of \(x\) whose entire strip \(A_x\times Y\) is covered by finitely many members of \(\mathcal U\). Compactness of \(X\) gives finitely many \(A_x\) covering \(X\), hence a finite subcover of \(X\times Y\).
3. Induction gives compactness of every finite product and finite union of compact spaces.
Apply 1 and 2 repeatedly.