Cauchy sequences of two metric spaces with only different distance functions

Question

Consider two metric spaces $(X,d)$ and $(X,d_1)$. If given that for a fixed natural number $N$ we have $$\large{d_1(x,y) = \frac{d(x,y)}{N}, \forall x,y \in X}$$, is there any relation between the Cauchy sequences of the two metric spaces? If yes, how do we get the relation?

Answer 1

The two metric spaces have exactly the same Cauchy sequences. To see that, you can show in general that a uniformly continuous function between metric spaces preserves Cauchy sequences. The two metric structures you are looking at are uniformly equivalent, meaning there is a bijective bi-uniform function between them (in this case it's just the identity function which does the job). The result follows.

Of course, you can also show directly that any Cauchy sequence in one space is also Cauchy in the other. This is quite easy in this case by simply following the definitions.

Answer 2

Every Cauchy sequence in $(X,d)$ is a Cauchy sequence in $(X,d_1),$ and vice-versa. To show, let be $(x_n)_{n\in \mathbb{N}}\subset (X,d).$ So, for each $\varepsilon > 0$, there is $n_0\in \mathbb{N}$, such that $m,n \geq n_0$, we have

$$d(x_m,x_n)\leq \frac{\varepsilon}{N},$$

with $N\in\mathbb{N}$ fixed. Since

$$d_1(x,y)=\frac{d(x,y)}{N}$$

for every $x,y \in X$ and $N\in\mathbb{N}$ fixed, we have

$$d_1(x_m,x_n)=Nd(x_m,x_n)\leq N\frac{\varepsilon}{N}=\varepsilon.$$

The other sense is analogue.