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.
Completeness for product topolgy
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general-topology
metric-spaces
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0I 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? – 2012-11-19
1 Answers
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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 $X$.