## Product of a sequence of completely metrizable spaces is completely metrizable (p13 (C))

The product of a sequence of completely metrizable spaces is completely metrizable.
Proof.
Let $(X_n)$ be a sequence of completely metrizable spaces.  For each $X_n$, let $d_n$ denote a compatible metric.

Let $X = \prod_{n=0}^{\infty} X_n$ have the product topology, which is given by the metric $d(x,y) = \sum_{n=0}^{\infty}2^{-n-1}d_n'(x_n,y_n)$, where $d_n' = \frac{d_n}{1 + d_n}$.  I will show that this metric $d$ is complete.

Let $(x_n)$ be a Cauchy sequence in $X$.  Let $\epsilon > 0$.  There exists $N$, such that $m,n \ge N \Rightarrow d(x_m, x_n) = \sum_{k=0}^{\infty} 2^{-k-1} d_k'(x_m(k),x_n(k)) < \epsilon$. Hence, for each $k$, $d_k'(x_m(k), x_n(k)) < 2^{k + 1}\epsilon$.  Thence, $(x_n(k))_{n=0}^{\infty}$ is Cauchy with respect to $d_k'$.  By p13(D), $d_k'$ is complete since $d_k$ is, so $x_n(k) \to x(k)$ (with respect to both $d_k, d_k'$).  This defines $x \in X$.  Now, is it true that $(x_n) \to x \in (X,d)$?

Consider $(\mathbf{N}, \mu)$, the measure space of the natural numbers with counting measure.  Let $m \in \mathbf{N}$, and $f_n(k) = 2^{-k-1}d_k'(x_m(k),x_n(k))$.  Then $\lim_{n \to \infty} f_n(k) = 2^{-k-1}d_k'(x_m(k),x(k)) := f(k)$, by the continuity of $d_k'$ (see p13(E)).  Also, $|f_n(k)| \le 2^{-k-1} := g(k)$ for all $k,n$ and $\int_{\mathbf{N}} g \text{ d} \mu = 1$.

Let $\epsilon > 0$. Since $(x_n)$ is Cauchy, there exists $N$, such that $m, n \ge N \Rightarrow \int_{\mathbf{N}} f_n \text{ d} \mu < \epsilon$.
By Lebesgue’s Dominated Convergence Theorem, if $m \ge N$,
$\epsilon \ge \lim_{n \to \infty} \int_{\mathbf{N}} f_n d \mu = \int_{\mathbf{N}} f d \mu = \sum_{k = 0}^{\infty} 2^{-k-1}d_k'(x_m(k),x(k)) = d(x_m,x).$

Hence $(x_n) \to x$, as desired.  So the metric $d$ completely metrizes $X$.

Sorry it has taken me a while to respond. I have been on break for a week, and went down some wrong paths before that. But I was making my formulas overcomplicated; it appears that, given a metric space $X$ with an unbounded metric $d$, $\{ d_n : d_n(x , y ) = (d(x,y))^{1/n} \text{ for all } x,y \in X, n \in \mathbb{N} \}$ is an infinite family of inequivalent metrics that give the same topology on $X$.