Is this map uniformly continuous? continuous? Let $s$ denote the metric space of all sequences of complex numbers with the metric 
$$d(x,y) \colon= \sum_{j=1}^\infty \frac{1}{2^j} \frac{\vert \xi_j(x) - \xi_j(y)\vert}{1+\vert \xi_j(x) - \xi_j(y) \vert} $$ 
for all $x, y \in s$. Here $\xi_j(x)$ denotes the $j$-th term of $x$. 
Let $k$ be a fixed natural number. Then is the map $T_k \colon s \to \mathbb{C}$ defined by 
$$T_k x \colon= \xi_k(x) \qquad \mbox{ for all } x \in s$$ 
uniformly continuous? continuous? 
 A: $T_k$ is indeed uniformly continuous.
(Note: I'll also write $x_j$ instead of $\xi_j(x)$.)
First, observe that
$$
\frac{\lvert x_k - y_k \rvert}{1 + \lvert x_k - y_k \rvert}
= 2^k \cdot \frac{1}{2^k} \frac{\lvert x_k - y_k \rvert}{1 + \lvert x_k - y_k \rvert}
\leq 2^k \sum_{j=1}^\infty \frac{1}{2^j} \frac{\lvert x_j - y_j \rvert}{1 + \lvert x_j - y_j \rvert}
= 2^k d(x,y).
$$
Now the function $f \colon [0,\infty) \to [0,\infty)$ defined by $f(t) = \frac{t}{1 + t}$
is strictly increasing: $f'(t) = \frac{1}{(1+t)^2} > 0$. Thus for any $\epsilon > 0$, the condition $d(x,y) < 2^{-k} f(\epsilon)$ implies $\lvert x_k - y_k \rvert < \epsilon$, completing the proof.

A more abstract topological approach if we just need to prove continuity:
Given a metric space $(X,\delta)$, the function $\delta' = \frac{\delta}{1+\delta}$ is a bounded metric which generates the same topology as $\delta$. It is further seen that the corresponding product topology on $\prod_{j=1}^\infty X$ is generated by the metric $\Delta(x,y) = \sum_{j=1}^\infty 2^{-j} \delta'(x_j,y_j)$. So the statement that the $T_k$ are continuous is merely one manifestation of the fact that the canonical projections of product spaces are always continuous.
A: I'll just write $x_i$ for the $i$th term of $x$.
Take an open set $S\subset\mathbb{C}$, and consider $S'=T_k^{-1}(S)$.
Take $x\in S'$, and say $x_k=v\in S$, and furthermore say $B(v,r_v)\subset S$ (since $S$ is open). Then consider $B(x,\frac{1}{2^k}\frac{r_v}{1+r_v})$. Take an $x'\in B(x,\frac{1}{2^k}\frac{r_v}{1+r_v})$. Then
$$d(x,x')=\sum_{i=0}^{\infty}\frac{1}{2^i}\frac{|x_i-x'_i|}{1+|x_i-x'_i|}<\frac{1}{2^k}\frac{r_v}{1+r_v}$$
So that in particular 
$$\frac{1}{2^k}\frac{|x_k-x'_k|}{1+|x_k-x'_k|}=\frac{1}{2^k}\frac{|v-x'_k|}{1+|v-x'_k|}<\frac{1}{2^k}\frac{r_v}{1+r_v}$$
So
$$\frac{|v-x'_k|}{1+|v-x'_k|}<\frac{r_v}{1+r_v}$$
From which we find, after some manipulations
$$|v-x'_k|<r_v$$
Which means that $x'_k\in B(v,r_v)\subset S$, so that $x'\in S'$, which means $ B(x,\frac{1}{2^k}\frac{r_v}{1+r_v})\subset S'$, which means that $S'$ is open, and so the inverse image of any open set is open, so $T_k$ is continuous.
