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Someone that can give me a proof of that a countable product of complete metric spaces is complete ?

Ninni
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  • It depends on the metric on product you choose. See also http://math.stackexchange.com/questions/194951/is-the-product-of-polish-spaces-polish. – Moishe Kohan Nov 29 '13 at 15:51
  • d(f,g) = sup_xd(f(x),g(x)) :) – Ninni Nov 29 '13 at 16:01
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    Then see this discussion (note that your "metric" is not a metric): http://math.stackexchange.com/questions/249129/countable-product-of-polish-spaces – Moishe Kohan Nov 29 '13 at 16:06

1 Answers1

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Countable product of completely metrizable spaces is again completely metrizable.

  1. Given a complete metric space $(Y,d)$, the metric $d'(x,y)= \min(1, d(x,y))$ is also complete since a sequence $(x_i)$ in $(Y,d)$ converges if and only if it converges in $(Y,d')$ and a sequence is Cauchy in $(Y,d)$ if and only if it is Cauchy in $(Y,d')$. Furthermore, the topologies on $Y$ given by $d, d'$ are the same (Theorem 4.3.8 in [E]).

  2. Suppose that $(X_i,\rho_i), i\in {\mathbb N}$ is a sequence of metric spaces. For sequences $x=(x_i)$, $y=(y_i)$ define $$ \rho(x,y)= \sum_{i=1}^\infty 2^{-i} \rho_i(x_i, y_i). $$ By Theorem 4.2.2 in [E], $\rho$ metrizes the product topology on $$ X=\prod_{i=1}^\infty X_i. $$

  3. Suppose that $(X_i,d_i)$ is a sequence of complete metric spaces. Then for $\rho_i= d'_i$, each $(X_i,\rho_i)$ is also complete and the metric $\rho$ as above on the product space $X$ is complete as well (Theorem 4.2.2 in [E]).

Thus, the product of completely metrizable spaces is also completely metrizable.

[E] R.Engelking, "General Topology", Sigma Series in Pure Mathematics, 1989.

Moishe Kohan
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