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

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  • $\begingroup$ How can we prove that the Cauchy sequences are the same if it's not stated by the problem ? $\endgroup$ – argiriskar Dec 19 '18 at 14:58
  • $\begingroup$ Does this arise from the fact that $d_2(x,y)=d_1(f(x),f(y)) ?$ $\endgroup$ – argiriskar Dec 19 '18 at 14:59
  • $\begingroup$ @argiriskar Well, that the Cauchy sequences for $d_1$ are precisely the convergent sequences, is will known. And so a sequence $(x_n)$ is Cauchy for $d_2$ if and only if $x_n^3$ converges (that's basically your observation, above). But since both $x\mapsto x^3$ and the inverse $x\mapsto x^{1/3}$ are continuous functions, $x_n^3$ converges if and only if $x_n$ converges. $\endgroup$ – Harald Hanche-Olsen Dec 19 '18 at 20:18

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