[Math] Two metric space with same Cauchy sequences

Question

Question:

If d,e are equivalent metrics on X, then $\left ( X,d \right )$ and $\left ( X,e \right )$ have the same Cauchy sequences.

I've first assumed to the contrary that the metric spaces have different Cauchy sequences.

Any hint is appreciated.

Answer 1

This is false.

On $\mathbb R$ let $d(x,y)=|x-y|$ and $e(x,y)=|\tan^{-1}x - \tan^{-1}y|.$ Then $d$ and $e$ are equivalent, that is, they generate the same topology. But $(n)_{n\in \mathbb N}$ is an $e$-Cauchy sequence and not a $d$-Cauchy sequence.

What is true is that two metrics $d,e$ are equivalent iff they have the same convergent sequences. A convergent sequence is a Cauchy sequence that has a limit point in the space. Note that $(n)_{n\in \mathbb N}$ in the example above, is an $e$-Cauchy sequence but it has no limit point in $\mathbb R.$

Metrics $d,e$ are called uniformly equivalent if and only if there exist positive $k_1,k_2$ with $k_1d(x,y)\leq e(x,y)\leq k_2 d(x,y)$ for all $x,y.$ Uniformly equivalent metrics are equivalent metrics, and do have the same Cauchy sequences.