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

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?

• I think there is missing a part :definition is every convergent sequence in $(X, d_A)$ converges at $(X, d_B)$ to the same limit - and vice versa Commented Apr 7, 2019 at 11:56
$$(-1,1)$$ 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].
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?)