The product of a sequence of completely metrizable spaces is completely metrizable.

*Proof*.

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 .

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I prefer to cut off the n-th metric at 1 instead of considering d’. This is to say, consider d”(x,y)=min( d(x,y), 1). Comparisons between different metrics for the same space can be interesting. Say two metrics are equivalent if one is between two constant multiples of the other. Can you produce many inequivalent metrics for the same given topology?

Sorry it has taken me a while to respond. I have been on break for a week, and went down some wrong paths before that. But I was making my formulas overcomplicated; it appears that, given a metric space with an unbounded metric , is an infinite family of inequivalent metrics that give the same topology on .