If two metrics $d_i$ on the same set $X$ have the same Cauchy sequences (ie. if a sequence is Cauchy for the first metric, it is also Cauchy for the other one and vice versa), can we conclude that the mapping:
$f: \left(X,d_1\right) \rightarrow \left(X,d_2\right) : x \rightarrow x$
is uniform continuous?
My attempt at a solution:
If the Cauchy sequences are the same, the convergent sequences are also the same, and therefore $d_1$ and $d_2$ are topological equivalent. That means that $f$ is continuous. However, I fail at proving the uniform continuity, nor can I find a counterexample.
Any help would be appreciated!
Best Answer
$X=\mathbb{R}$, $d_1(x,y)=|x-y|$, $d_2(x,y)=|x^3-y^3|$ is a counterexample.