Countable product of complete metric spaces Someone that can give me a proof of that a countable product of complete metric spaces is complete ? 
 A: Countable product of completely metrizable spaces is again completely metrizable. 


*

*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]).  

*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
$$
X=\prod_{i=1}^\infty X_i.
$$

*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. 
