When can an isometry be constructed between to metric spaces?

Question

I wanted to show that the space $\mathcal{S} := \{ x = (x_k)_{k \in \mathbb{N}} \subset \mathbb{C}\}$ with the metric \begin{equation*} d: \mathcal{S} \times \mathcal{S} \to \mathbb{R}, \ (x,y) \mapsto \sum_{k \in \mathbb{N}} \frac{1}{2^k} \frac{| x_k - y_k |}{1 + | x_k - y_k |} \end{equation*} is complete.

What I did was to show that if a sequence $(x^{(n)})_{n \in \mathbb{N}} \subset \mathcal{S}$ converges to $x \in \mathcal{S}$ in $(S, d)$ if and only if $x_k^{(n)} \to x_k$ for all $k \in \mathbb{N}$ in $(\mathbb{C}, d_2)$, where $d_2$ is the standard euclidean norm.

I then took a Cauchy sequence $(x^{(n)})_{n \in \mathbb{N}} \subset \mathcal{S}$ and showed that $(x^{(n)}_k)_{n \in \mathbb{N}}$ is a Cauchy sequence in the complete metric space $(\mathbb{C}, d_2)$ for all $k \in \mathbb{N}$. All those coordinate-wise "subsequence" therefore converge and therefore, also $(x^{(n)})$ converges, finishing the proof.

My question is if this can be done in an easier fashion by constructing an isometry between $(\mathcal{S},d)$ and $(\mathbb{C}, d_2)$ ( cardinality of $\mathbb{N}$ times?) or if in the context of sequence spaces this is not possible-

Context A simpler example. Let $X := (0,1]$ a set with the metric $d(x,y) := \left| \frac{1}{x} - \frac{1}{y} \right|$. One can show that $(X,d)$ is complete by defining the isometry $$ \Phi: (X,d) \to ([1, \infty), d_2), \ t \mapsto \frac{1}{t} $$ Since $([1, \infty), d_2)$ is complete, we know that $(X,d)$ is, too. I wanted to use the same principle but struggled to come up with a suitable $\Phi$.

Answer 1

The fact that a Cauchy sequence in $S$ converges "co-ordinate - wise" does not, I think, finish the proof. See especially part $(II-b)$ below.

$(I). $ Let $(\,(X_k,d_k)\,)_{k\in \Bbb N}$ be a sequence of non-empty complete metric spaces.

For each $k$ and for $u,v\in X_k$ let $e_k(u,v)=d_k(u,v)(1+d(u,v))^{-1}.$ Then $e_k$ is a complete metric on $X_k$ (and $e_k$ generates the same topology as $d_k $). Later it will be useful that $e_k(u,v)< 1.$

Let $S=\prod_{k\in \Bbb N}X_k.$ For $x=(x_k)_{k\in \Bbb N}$ and $y=(y_k)_{k\in \Bbb N}$ in $S$ let $d(x,y)=\sum_{k\in \Bbb N}2^{-k}e_k(x_k,y_k).$

$(II-a).$ Suppose $(x(n))_{n\in \Bbb N}$ is a $d$-Cauchy sequence in $S,$ where $x(n)=(x_{k,n})_{k\in \Bbb N}.$ Then for each $k,$ the sequence $(x_{k,n})_{n\in \Bbb N}$ is $e_k$-Cauchy in $X_k.$

Otherwise for some $k_0\in \Bbb N$ and some $r\in \Bbb R^+$ we would have $\lim_{n\to \infty}\sup_{n\le m<m'}e_{k_0}(x_{k_0,m},x_{k_0,m'})\ge r.$ But $d(x(m),x(m'))\ge 2^{-k_0}e_{k_0}(x_{k_0,m},x_{k_0,m'})$ so we would have $\lim_{n\to \infty}\sup_{n\le m<m'}d(x(m),x(m'))\ge 2^{-k_0}r>0,$ contrary to the Cauchy condition on $(x(n))_{n\in \Bbb N}.$

Now since each $e_k$ is complete, there exists (unique) $y_k\in X_k$ such that $\lim_{n\to \infty}e_k(x_{k,n},y_k)=0.$

Let $y=(y_k)_{k\in \Bbb N}.$

$(II-b).$ Given $\epsilon >0,$ take $k_1\in \Bbb N$ with $2^{-k_1}<\epsilon /2.$ Now take $n_1\in \Bbb N$ such that $\forall n\ge n_1\,(e_k(x_{k,n},y_k)<\epsilon /2)$ holds for every $k\le k_1.$

(For example, for each $k\le k_1$ take $M(k)\in \Bbb N$ such that $n\ge M(k)\implies e_k(x_{k,n},y_k)<\epsilon /2, $ and let $n_1=\max \{M(k): k\le k_1\} .)$

Then for $n\ge n_1$ we have $d(x(n),y)=A+B$ where $$A=\sum_{k=1}^{k_1}2^{-k}e_k(x_{k,n},y_k)<\sum_{k=1}^{k_1}2^{-k}(\epsilon /2)<\epsilon /2 $$

$$\text { and }\quad B=\sum_{k=1+k_1}^{\infty}2^{-k}e_k(x_{k,n},y_k)<\sum_{k=1+k_1}^{\infty}2^{-k}\cdot 1=2^{-k_1}<\epsilon /2$$

so $n\ge n_1\implies d(x(n),y)<\epsilon.$

So $d$ is a complete metric.

BTW the topology on $S$ generated by $d$ is equal to the (Tychonoff) product topology.

In your Q let each $X_n=\Bbb C$ and $d_n(u,v)=|u-v|.$