Theorem 5.14.4 (Tychonov). A product of quasi-compact spaces is quasi-compact.
Proof. Let I be a set and for i \in I let X_ i be a quasi-compact topological space. Set X = \prod X_ i. Let \mathcal{B} be the set of subsets of X of the form U_ i \times \prod _{i' \in I, i' \not= i} X_{i'} where U_ i \subset X_ i is open. By construction this family is a subbasis for the topology on X. By Lemma 5.12.15 it suffices to show that any covering X = \bigcup _{j \in J} B_ j by elements B_ j of \mathcal{B} has a finite refinement. We can decompose J = \coprod J_ i so that if j \in J_ i, then B_ j = U_ j \times \prod _{i' \not= i} X_{i'} with U_ j \subset X_ i open. If X_ i = \bigcup _{j \in J_ i} U_ j, then there is a finite refinement and we conclude that X = \bigcup _{j \in J} B_ j has a finite refinement. If this is not the case, then for every i we can choose an point x_ i \in X_ i which is not in \bigcup _{j \in J_ i} U_ j. But then the point x = (x_ i)_{i \in I} is an element of X not contained in \bigcup _{j \in J} B_ j, a contradiction. \square
Comments (0)