- #1

radou

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## Homework Statement

One needs to show that a countable product of topologically complete spaces is topologically complete in the product topology.

## The Attempt at a Solution

A space X is topologically complete if there exists a metric for the topology of X relative to which X is complete.

So, let X = ∏Xi, where (Xi, di) are complete metric spaces (di denotes the bounded metric deduced from the metric on Xi).

My first guess is the metric that induces the product topology on X, given with:

D(x, y) = sup{di(xi, yi)/i}, where d is the standard bounded metric. Of course, a little variation is that we consider di for each i, i.e. the metric in Xi.

Now, let xn be a Cauchy sequence in X. Given ε > 0, there exists N such that for all m, n >= N, we have D(xn, xm) < ε. In particular, for every m, n, and for every i, we have:

di(xni, xmi)/i <= D(xn, xm) < ε. Let i be fixed. We find that di(xni, xmi) <= i D(xn, ym) < i ε, for all n, m >= N. This implies that the sequence xni is a Cauchy sequence in Xi, hence it converges to xi. I assert that xn converges to the point x of X, where x = (xi) (a bit stupid notation, but I think it's obvious what I mean).

This follows from the fact that for ε > 0, we can find n, m such that D(xn, xm) < ε/2, whenever m, n >= N. By what is written above, for every i, and m, n >= N, we have di(xni, xmi) < i ε/2. If we fix n and i, and let m --> ∞, we have di(xni, xi) <= ε/2. This holds for any i, and for n >= N. Hence D(xn, x) <= ε/2 < ε, whenever n >= N. So, xn --> x.

I think this should work, since I studied some similar proofs in the book and applied them to this problem.