I´ve been dealing with topologically equivalent metrics for a while, using the usual definition, that $d$ and $d'$ are topologically equivalent iff they have the same open sets. However, there is another equivalent definition that says that $d$ and $d'$ are topologically equivalent if both $id_{1}:(X,d)\to(X,d')$ and $id_{2}:(X,d)\to(X,d')$ are continuous. Where $X$ is the metric space were both $d$ and $d'$ are defined. However, when trying to use this definition, I don't know If what Im doing is correct. I'll give two examples:
Decide wether the following are true or false (with $d$, $d'$ topologically equivalent):
a)If $(x_n)$ converges with $d$ then $(x_n)$ converges with $d'$
b)If $(x_n)$ is Cauchy with $d$ then $(x_n)$ is Cauchy with $d'$
For a) I've used the continuity of $id_1$ in the following manner: Given $\epsilon>0$ there exist $\delta>0$ such that $d(x,y)<\delta$ implies $d'(x,y)<\epsilon$. Now I can easily show that $(x_n)$ converges with $d'$.
EDIT
b) We want to see that if $(x_n) $ is Cauchy with $ d$ then it is with $ d'$. That is, I want to see that given $\epsilon_{0} > 0$ there exist $ n_0$ such that $ d'(x_n, x_m) < \epsilon $ if $ n, m \geq n_0$.
Since $ id_1$ is continuous, given $\epsilon_0$ there exist a $\delta_0> 0$ such that $ d (x_n, x_m) <\delta_0$ implies $ d'(x_n, x_m)<\epsilon_0$.
Since $(x_n) $ is Cauchy in $(X, d) $ we have that given $\delta_0 > 0$ there exist $ n_1\in \mathbb {N} $ such that $ d (x_n, x_m)<\delta_0 $ if $ n, m \geq n_0$.
Then taking $ m, n \geq n_1$ we have that $ d (x_n, x_m)<\delta_0$ implying $ d'(x_n, x_m)<\epsilon_0$.
However I know this proof is wrong, since this would imply completeness is topologically preserved and thats not true!