Compactness of any Countable Product of Compact Metric Spaces in Product Topology without Using Tychonoff's Theorem

Abstract

For infinite products of compact spaces, Tychonoff's theorem asserts that their product is compact, in the product topology. Tychonoff's theorem is shown to be equivalent to the axiom of choice. In this paper, we show that any countable product of compact metric spaces is compact, without using Tychonoff's theorem. The proof needs only basic and standard facts of compact metric spaces and the Bolzano-Weierstrass property. Moreover, the component spaces need not be assumed to be copies of the same compact metric space, and each component space can be an arbitrary nonempty compact metric space independently. Total boundedness together with completeness of a metric space implies its compactness. Completeness of a product of complete spaces is easily inferred from the completeness of each component. Total boundedness therefore suffices to prove the compactness of a product space consisting of countably (infinitely) many nonempty compact component spaces. The countable infiniteness is needed in the proof to exhibit a standard metric that gives rise to the product topology. Any such metric topology for the product arises as exhibited, and they are all equivalent to the product topology. The requirement of summability of the sequences restricts the scope of the result to countably infinite products. In summary, the product space obtained by taking the product of any sequence of nonempty compact metric spaces in the product topology is shown to be compact, using only the basic and standard facts of compact metric spaces. In conclusion, compactness of the product of a countably infinitely many nonempty compact metric spaces can be proved within Cantor's set theory, without using the axiom of choice and Tychonoff's theorem.