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?



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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.