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.