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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!

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up vote 2 down vote accepted

Let $h(x)$ be any homeomorphism of $[0,\infty)$ which is not uniformly continuous, such as $h(x)=x^2$. Then define $d_1(x,y)=|x-y|$ and $d_2(x,y)=|h(x)-h(y)|$.

More generally, if $(X,d_1)$ is a metric space, and $h:X\rightarrow X$ is a homeomorphism but not uniformly continuous, then you can define $d_2(x,y)=d_1(h(x),h(y))$.

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Thanks for the quick reply! – Bill Carson Apr 27 '12 at 20:17

$X=\mathbb{R}$, $d_1(x,y)=|x-y|$, $d_2(x,y)=|x^3-y^3|$ is a counterexample.

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