Converse Statement of Uniform Continuity

[CoachZ]

Recently, I proved that Given $$f:A \rightarrow \mathbb R$$ is uniformly continuous and $$(x_{n}) \subseteq A$$ is a Cauchy Sequence, then $$f(x_{n})$$ is a Cauchy sequence, which really isn't too difficult a proof, however I'm having issues with the converse statement... More specifically, Suppose $$A \subseteq \mathbb R$$ and $$f: A \rightarrow \mathbb R$$ with the property that it preserves Cauchy sequences in $$A$$, i.e. $$x_{n} \in A$$ and $${x_{n}}$$ is Cauchy, $${f(x_{n})}$$ is Cauchy, then prove that $$f$$ is uniformly continuous on $$A$$

My idea was to show by way of contradiction that the boundedness of $$A$$ cannot be removed, which would then imply that $$x_{n}$$ is not Cauchy if it were removed, which implies that $$f(x_{n})$$ is not Cauchy. From this point, I'm a bit stuck... I was under that impression that I could use Cauchy Criterion, which states that a sequence converges iff it is a Cauchy sequence, and since $$x_{n}$$ isn't a Cauchy sequence, then it doesn't converge, which implies that it cannot be continuous on all points in $$\mathbb R$$, which implies that it is not continuous on $$\mathbb R$$, therefore cannot be uniform continuous on $$\mathbb R$$.

Perhaps contrapositive would work better? Any suggestions?

 

[snipez90]

I think the best way to handle this is to look at the contrapositive, since that's a good way to generate sequences. From the perspective of the contrapositive, we can basically always pick two points x and y arbitrarily close while their images are a fixed distance apart. It shouldn't be too hard to show that this would imply that we can find a sequence that is Cauchy (eventually any two terms are arbitrarily close), but the image sequence is not Cauchy (images of terms are a fixed distance apart).

 

[LCKurtz]

What about f(x) = x²?

 

[snipez90]

Good point, you need A to be compact for this to work.