If two metric spaces are topologically equivalent (homeomorphic) imply that they are complete?

Question

My definition of topologically equivalent is: every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit and vice versa.

I know that

Let $(X,d_A)$ and $(X, d_B)$ be two strongly equivalent metric spaces, if $(X, d_A)$ is complete then $(X, d_B)$ is complete.

I was thinking that if these spaces were just topologically equivalent yet would the statement be true?

My answer is no.

But I can't think of a counter example.

Answer 1

Consider $(\Bbb R,d)$ , where $d$ is the usual metric and $(\Bbb R,\rho)$ , where $\rho$ is the metric defined by $$\rho(x,y)=\vert \arctan x-\arctan y \vert$$ Then $(\Bbb R,d)$ is topologically equivalent to $(\Bbb R,\rho)$(How ?), but one is complete while the other is not (How?)

Try to fill the gaps!

Answer 2

$(-1,1)$ is topologically equivalent to $\mathbb R$ via the homeomorphism $x \to \frac x {1-|x|}$. $(-1,1)$ is not complete but $\mathbb R$ is complete. [Usual metric on both spaces].