to show a function is bounded when domain is bounded

Question

How to show that $f$ : $S$ -> $R$ is uniformly continuous and $S \subset R $ is bounded then $f$ must be bounded.

I have tried using the theorem which is states the three statements to be equivalent.But could not suceed.

Answer 1

If $f$ is uniformly continuous, it can be extended to a continuous function on $\overline S$. This closure is compact, so the extension of $f$ is bounded on it. Therefore $f$ is bounded.

In answer to the comment below:

Let $x\in {\overline S}$. Then there is a sequence $x_n$ in $S$ so $x_n\to x$. Since $f$ is uniformly continuous, you can show that $\{f(x_n)\}$ is Cauchy. By the completeness of the real numbers it has a limit we dub ${\overline f}(x)$. An appropriate uniqueness argument shows that there is only one possible ${\overline f} (x)$. You can then argue to show that ${\overline f}$ is continous on $\overline S$. Note that uniform continuity is critical to this enterprise. I hope this set of hints helps.