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Let $(X_n, d_n)$ be a countable family of metric spaces. Show that the distance function on the product topology defined by $$d\big( (x),(y) \big) = \sum_{n\geqslant 1} 2^{-n} \frac{d_n(x_n,y_n)} {1+d_n(x_n,y_n)}$$ is complete, if all spaces $(X_n, d_n)$ are complete.

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That’s pretty straightforward; what have you tried? – Brian M. Scott Nov 19 '12 at 7:49
I first looked at every "coordinate" to get a coordinate-wise cauchy sequences. From this I defined $$x^*$$. Then I split the sum to get a finite one and an arbitrary small one. But how do I now that $$x^*$$ converge to something in the product? – Johan Nov 19 '12 at 8:34
up vote 1 down vote accepted

Here’s a guide; all you have to do is fill in the details.

Start with a Cauchy sequence in $X=\prod_nX_n$, say $\sigma=\langle x^n:n\in\Bbb N\rangle$, where $x^n=\langle x_k^n:k\in\Bbb N\rangle\in X$. Use the fact that $\sigma$ is $d$-Cauchy to show that $\langle x_k^n:n\in\Bbb N\rangle$ is $d_k$-Cauchy in $X_k$ for each $k\in\Bbb N$. Each $X_k$ is complete, so $\langle x_k^n:n\in\Bbb N\rangle\to y_k$ for some $y_k\in X_k$. Let $y=\langle y_k:k\in\Bbb N\rangle\in X$, and show that the sequence $\sigma$ converges to $y$ in $X$.

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