Does a uniformly continuous function map bounded sets to bounded sets? I'm talking about a uniformly continuous function on an arbitrary metric space here. In $\mathbb{R}$ I use the following argument : 
Suppose $A$ is a bounded set such that $f(A)$ is unbounded. We can choose a sequence $(f(x_n))$ in $f(A)$ such that $|f(x_n)|>n$. Now $X=(x_n)$ is a bounded sequence and hence admits a convergent subsequence $Y=(x_{n_k})$. But $Y$ is a Cauchy sequence whose image under $f$ (a uniformly continuous function) is not Cauchy. 
But for this argument to work in an arbitrary metric space, I need the set $A$ to be totally bounded; so that every sequence in $A$ will have a Cauchy subsequence.  
So how do we prove it? Is this even true?
 A: Let $X$ be $\mathbb N$ with the metric $d(n,m)=\frac {|n-m|} {1+|n-m|}$  and$Y$ be $\mathbb N$ with the metric $D(n,m)=|n-m|$ Let $f$ be the identity function . Can you verify that this is a counter-example?
[Take $\delta=\frac {\epsilon}{1+\epsilon}$ in the definition of uniform continuity and take the bounded set as $\mathbb N$]. 
A: YES, image of a totally bounded set under a uniformly continuous map is totally bounded.
Indeed, let $\,A\,$ be a totally bounded set in a metric space $\,X.\ $ Let $\,f:X\to Y\,$ be a uniformly continuous function. Consider the induced map $\,g:X'\to Y'\,$ of completion of $\,X\,$ into the completion $\,Y.\,$ Then the closure $\,A'\,$ of $\,A\,$ in
$\,X'\,$ is compact hence $\,g(A')\subseteq Y'\,$ is compact. Thus $\,f(A)\,=\,g(A)\cap Y\,$ is totally bounded.   Great!
A: Take as domain $X$ of the function $f$ the integers $\mathbb Z$, with the discrete metric
$$
d_X(x,y)=\begin{cases}
0 & \text { if } x=y, \\
1 & \text { otherwise.}\\
\end{cases}
$$
As co-domain $Y$ take the integers $\mathbb Z$ with the standard metric
$$d_Y(x,y) = \lvert x-y\rvert.$$
Then
$$f: X\rightarrow Y; \;f(x):=x$$ is uniformly continuous (just take $\delta=\frac12$ for any $x \in X$ and $\epsilon > 0$ given). $A=X$ is bounded, but $f(A) = \mathbb Z$ is not (under the metric in $Y$).
