The product of a sequence of completely metrizable spaces is completely metrizable.
Let be a sequence of completely metrizable spaces. For each , let denote a compatible metric.
Let have the product topology, which is given by the metric , where . I will show that this metric is complete.
Let be a Cauchy sequence in . Let . There exists , such that . Hence, for each , . Thence, is Cauchy with respect to . By p13(D), is complete since is, so (with respect to both ). This defines . Now, is it true that ?
Consider , the measure space of the natural numbers with counting measure. Let , and . Then , by the continuity of (see p13(E)). Also, for all and .
Let . Since is Cauchy, there exists , such that .
By Lebesgue’s Dominated Convergence Theorem, if ,
Hence , as desired. So the metric completely metrizes .