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I tried to solve it by cases: domain is a set of numbers; domain is an interval,;domain is a union of numbers and some intervals.

For the first case, I thought about arctanh is unbounded but its domain is bounded. To make it uniformly continuous, I can let Z be the domain.

For the second case, I think there does not exist a function like this.

For the third case, I am not sure if there exists a function satisfying all these conditions..

Did I think anything wrong for this question? or Could you give some idea or hint about that?

Thanks!

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If $f$ is uniformly continuous, and its domain $D$ is totally bounded, then $f$ must be a bounded function. So, to find the necessary counterexample, you need to be a little trickier...

Here is a counterexample: Consider $$ \begin{align} f:(\mathbb R,d) &\to(\mathbb R,|\cdot|)\\ x &\mapsto x \end{align} $$ Where $d(\cdot,\cdot)$ is the metric defined by $$ d(x,y)= \begin{cases} |x-y| & |x - y|<1\\ 1 & |x-y|\geq 1 \end{cases} $$ Then $f$ is uniformly continuous, its domain is bounded (but not totally bounded), and its image is unbounded.

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The image of a totally bounded subset under a uniformly continuous function is totally bounded. However, the image of a bounded subset of an arbitrary metric space under a uniformly continuous function need not be bounded: as a counterexample, consider the identity function from the integers endowed with the discrete metric to the integers endowed with the usual Euclidean metric.

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