Identical Cauchy sequences and continuity.

Question

Find a set $X$ and two metrics $d$ and $m$ on $X$ such that the Cauchy sequences of $(X,d)$ and $(X,m)$ are identical and the identity map from $(X,d)$ to $(X,m)$ is continuous but not uniformly continuous.

My attempt: if we take $X=\mathbb{R}$ and the identity map $f:\mathbb{R} \rightarrow \mathbb{R}$ and the metrics are equivalent, then we get our result.(?)

A question: Uniform continuity means that the cauchy sequences of the first metric space maps to the second. How can it not be uniformly continuous?

Answer 1

Let $X=\mathbb R$, $d(x,y)=|x-y|$ and $m(x,y)=|x^{2}-y^{2}|+|x-y|$. Use the fact that d-Cauchy sequences are bounded to show that Cauchy sequences under $d$ and $m$ are the same. If the identity map from $(X,d)$ to $(X.m)$ is uniformly continuous then, given $\epsilon >0$ there exists $\delta >0$ such that $|x^{2}-y^{2}|+|x-y| <\epsilon$ whenever $|x-y| < \delta$ $\,\,\,$ (1). But this would imply that $x \to x^{2}$ is uniformly continuous function (when both domain and range are given the metric $d$) which you know is false. [ You can take $x=n,y=n+\frac 1 n$ with large $n$ to get a contradiction to (1)].

Answer 2

Take $X=\mathbb R$, let $d$ be the usual distance and define$$m(x,y)=\bigl\lvert x^3-y^3\bigr\rvert.$$Then $(X,d)$ and $(X,m)$ have the same Cauchy sequences, but $\operatorname{Id}\colon(X,d)\longrightarrow(X,m)$ is not uniformly continuous (but it is continuous).