Prove that if $(X_n,d_n)$ is complete, then $(Z,s)$ is complete. Let $\{(X_n,d_n):n \in \Bbb N\}$ be a sequence of complete metric spaces, and let $Z=\prod_{n=1}^\infty X_n$. For each $x=(x_n),y=(y_n) \in Z$, define a metric $s$ on $Z$ as follows:
$$s(x,y) = \sum_{n=1}^\infty \frac{1}{2^n} \cdot \frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}.$$
Show that $(Z,s)$ is complete.
Attempt:
Let $(x_k)$ be arbitrary Cauchy sequence in $(Z,s)$, where $x_k = (x_{nk}:n \in \Bbb N)$. Then, for any $\epsilon>0$, there exists $N \in \Bbb N$ such that for all $j,k \ge N$, we have
$$d_n(x_{nj},x_{nk}) \le \sum_{n=1}^\infty \frac{1}{2^n} \cdot \frac{d_n(x_{nj},x_{nk})}{1+d_n(x_{nj},x_{nk})} =: s(x_j,x_k) < \epsilon. \qquad (1)$$
(Since $\sum \frac{1}{2^n}$ is convergent, then $\lim \frac{1}{2^n}=0$ and so, $(\frac{1}{2^n})$ is bounded (by $1$). Thus, since $\sum \frac{1}{2^n} \frac{d_n}{1+d_n}< \epsilon$, then $\frac{d_n}{1+d_n}<\epsilon$.)
Hence, $(x_{nk})$ is a Cauchy sequence in $(X_n,d_n)$ for all $n \in \Bbb N$. Since each $(X_n,d_n)$ is complete, then $(x_{nk})$ is converges to $y_n$ for some $y_n \in X_n$. Define $y:=(y_n:n \in \Bbb N) \in Z$.
We want to show that $x_k \to y$ in $(Z,s)$.
Let $\epsilon>0$ be given. Since $x_{nk} \to y_n$ for each $n \in \Bbb N$, then there exists $M_n \in \Bbb N$ such that for all $n \ge M_n$, we have
$d_n(x_{nk},y_n) < \epsilon$. Let $M:= \max\{M_n: n \in \Bbb N\}$. Then $M \in \Bbb N$ and for each $n \ge M$, we have
\begin{align*}
s(x_k,y) := \sum_{n=1}^\infty \frac{1}{2^n} \cdot \frac{d_n(x_{nk},y_n)}{1+d_n(x_{nk},y_n)} &\le \sum_{n=1}^\infty \frac{1}{2^n} \cdot d_n(x_{nk},y_n) \\
&< \frac{\epsilon}{2} + \frac{\epsilon}{4} + \frac{\epsilon}{8} + \cdots \\
&= \epsilon \cdot \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \right) \\
&= \epsilon.
\end{align*}
Thus, any Cauchy sequence in $(Z,s)$ is convergent, so $(Z,s)$ is complete.
Is it correct? Also, I'm in doubt whether the (reason of) inequality on $(1)$ is correct. I got stuck here. Any help please? Thanks in advanced.
 A: The right idea, but a couple of comments:

*

*The second part of inequality (1) is true by definition of a Cauchy sequence in our new product space; however, the first inequality needs to be justified. This is the key step to show that each individual component is Cauchy.


*The proof of convergence uses the fact that $M = \max\{ M_n : n \in \mathbb{N} \}$. However, since we are dealing with an infinite sequence, this is not necessarily finite (i.e. $M = \sup \{ M_n : n \in \mathbb{N} \}$ could actually be $+\infty$).
To fix this, we just need to somehow split the problem to deal with the finite and infinite parts separately. Here are two hints for the two parts:

*

*Fix $n \in \mathbb{N}$ and $\epsilon_n > 0$. Let $\epsilon > 0$ be some number to be chosen later. Let's denote $d_n = d(x_{nj}, x_{nk})$ to simplify things. Clearly $2^{-n} \frac{d_n}{1 + d_n} \leq \sum_{n \geq 1} 2^{-n} \frac{d_n}{1 + d_n} < \epsilon$. Rearranging, we have $d_n < \frac{2^n \epsilon}{1 - 2^n \epsilon}$ (which makes sense as long as $2^n \epsilon < 1$). Now we can make the right-hand side less than $\epsilon_n$ by making $\epsilon$ small enough, and so $d_n < \epsilon_n$ for all $j, k > N_n$. This shows that the sequence is Cauchy in each of its $n$th components.


*Fix $\epsilon > 0$. Note that $\frac{d_n}{1 + d_n} \leq 1$, and $\sum_{n \geq 1} 2^{-n} = 1$. We can bound the metric part by 1 for $n$ large enough, say $n > N$, so that this half of the sum is less than $\epsilon / 2$. Now we only have to control the sum for $n \leq N$. This are now finitely many terms and we can control this sum to also be less than $\epsilon / 2$ by using the fact that each component is Cauchy (i.e. as you have shown, by choosing $\epsilon_n = \epsilon / 2$).
