Cauchy Convergence in Sequence space induces convergence on elements of sequence I am working on the following problem:

Consider the space of sequences (bounded and unbounded) with the metric given by:
  $$d(x,y) = \sum_{i=1}^{\infty}\frac{1}{2^i}\frac{|x^{(i)}-y^{(i)}|}{1+|x^{(i)}-y^{(i)}|}$$
Where $(x^{(i)})$ are elements in the sequence $x$. Show that $z_n \rightarrow z$ implies that $z_n^{(i)} \rightarrow z^{(i)}$

The answer key says:

Fix any $j$, then for all $\epsilon > 0$ there exists an $N$ such that:
  $$\frac{1}{2^i}\frac{|x_n^{(j)}-x^{(j)}|}{1+|x_n^{(j)}-x^{(j)}|} \leq d(x_n,x) < \frac{1}{2^j}\frac{\epsilon}{1+\epsilon}$$ 
  Thus for all $n>N$ we have $|x_n^{(j)}-x^{(j)}|<\epsilon$ 

I don't understand where the second inequality is coming from: $d(x_n,x) <\frac{1}{2^j}\frac{\epsilon}{1+\epsilon}$. Thoughts?
 A: According to the definition, for all $\delta>0$ there exists an $N$ such that for all $n>N$ we have that $d(x_n,x)<\delta$. They just choose $\delta=\frac{1}{2^j}\frac{\epsilon}{1+\epsilon}$, which is also small for small $\epsilon$, as $\frac{\epsilon}{1+\epsilon}<\epsilon$. Now because $f(x)=\frac{x}{1+x}$ is a strictly increasing function for $x\geq 0$ we have that $|x_n^{(j)}-x^{(j)}|<\epsilon$.
A: For a more conceptual explanation, notice that $d$ induces the product topology on $\mathbb{R}^\mathbb{N}$:
Let $x = (x_n)_{n=1}^\infty \in \mathbb{R}^\mathbb{N}$ and let $B_d(x, r)$ be an open ball with respect to the metric $d$. We'll show that there exists a basis element of the product topology containing $x$ and contained in $B_d(x, r)$.
Indeed, let $n_0 \in \mathbb{N}$ be such that $\sum_{n=n_0+1}^\infty \frac1{2^n} < \frac{r}2$ and consider:
$$U = B\left(x_1, \frac{\frac{r}4}{1-\frac{r}4}\right) \times B\left(x_2, \frac{\frac{r}4}{1-\frac{r}4}\right) \times \cdots \times B\left(x_{n_0}, \frac{\frac{r}4}{1-\frac{r}4}\right) \times \mathbb{R} \times \mathbb{R}\times \cdots$$
Obviously $x \in U$, and $U$ is a basis element of the product topology. Notice that we also have $U \subseteq B_d(x, r)$. Indeed, for $y = (y_n)_{n=1}^\infty \in U$ we have:
$$d(x, y) = \sum_{n=1}^\infty \frac{1}{2^n}\frac{\left|x_n - y_n\right|}{1 + \left|x_n - y_n\right|} < \sum_{n=1}^{n_0}\frac{1}{2^n}\underbrace{\frac{\overbrace{\left|x_n - y_n\right|}^{<\frac{\frac{r}4}{1-\frac{r}4}}}{1 + \left|x_n - y_n\right|}}_{< \frac{r}4} +  \frac{r}2 < \frac{r}{2} + \frac{r}2 = r$$
Conversely, let $x \in U_1 \times \cdots \times U_n \times \mathbb{R} \times \mathbb{R} \times \cdots$ be a basis element in the product topology containing $x = (x_n)_{n=1}^\infty \in \mathbb{R}^\mathbb{N}$, where $U_1, \ldots, U_n$ are open sets in $\mathbb{R}$.
For $i \in \{1, \ldots, n\}$, since $x_i \in U_i$, there exists $r_i < 1$ such that $B\left(x_i, r_i\right) \subseteq U_i$.
Consider $r = \min\left\{\frac{r_1}{r_1 + 1}, \frac{1}{2}\frac{r_2}{r_2 + 1}, \ldots, \frac{1}{2^n}\frac{r_n}{r_n + 1}\right\}$, and lets show that $B_d(x, r) \subseteq U_1 \times \cdots \times U_n \times \mathbb{R} \times \mathbb{R} \times \cdots$. Let $y = (y_n)_{n=1}^\infty \in B_d(x, r)$. For any $i \in \{1, \ldots, n\}$ we have:
$$\frac{1}{2^i}\frac{\left|x_i - y_i\right|}{1 + \left|x_i - y_i\right|} \le 
  \sum_{n=1}^\infty \frac{1}{2^i}\frac{\left|x_i - y_i\right|}{1 + \left|x_i - y_i\right|} = d(x, y) < 
 r\le \frac{1}{2^i}\frac{r_i}{r_i + 1}$$
so $\left|x_i - y_i\right| < r_i$ which implies $y_i \in U_i$.
Now you can show an even stronger claim:
A sequence $(x_n)_{n=1}^\infty$ converges in $\mathbb{R}^\mathbb{N}$  with respect to $d$ if and only if all coordinate sequences $\left(x^{(i)}\right)_{n=1}^\infty$ converge in $\mathbb{R}$.
Have a look here: Convergence in product topology
