We say a metric space X is completely metrizable if there a metric that induces its usual topology and X with that metric is complete. I´ve seen, for example, that (0,1) is completely metrizable and the argument usually is "it is homeomorphic to a complete space".
But is it true that if X, Y are metric spaces, Y with complete with the metric $d_{Y}$ and $f$ homeomorhism between X and Y, then there is a metric in X which turns the space into a complete one?
My thoughts were on the metric $d_{X}(x,y)=d_{Y}(f(x),f(y))$. But I don´t see why X has to be complete with this?
I would really apreciate any help on this, thank you in advance!
So, as @Jakobian was saying, if $x_{n}$ in $X$ is a cauchy sequence with respect to the metric $d_{X}$, $d_{X}(x,y)=d_{Y}(f(x),f(y))$, then $f(x_{n})$ would be cauchy in the complete metric space $Y$! So, it converges to some point y in $Y$, that's of the form $f(x)$ for some $x\in X$. As $f^{-1}$ is continuous and these are metric spaces, $f(x_{n}) \rightarrow f(x)$ implies $(f^{-1}of)(x_{n}) \rightarrow (f^{-1}of)(x)$ that is: $x_{n} \rightarrow x$.
This proves every cauchy sequence converges, therefore, $X$ is complete!