Is it true that $X $ is complete iff $Y$ is complete.

Question

Let $X ,Y $ be metric spaces and let $f : X\to Y $ be a bijective function. If $f,f ^ {-1 }$ are continuous, is it true in general that $X $ is complete if and only if $Y $ is complete?

I know that if we replace "continuous" by "uniformly continuous" the result is true.

I feel that the answer to the above problem is FALSE.

Efforts:

Consider $ X=\mathbb{ R } \text{ and } Y=(-1,1).$

Consider the homemorphism given by $$f:\mathbb{R}\to (-1,1),\quad x\longmapsto \dfrac{x}{1+\mid x\,\,\mid}\,\,$$

Now $R$ is complete but $Y=(-1,1)$ is not.

Am I right? I am aware that I have to fill in the details; I just want to see whether I am going in right direction or not.

Answer 1

You are right. To complete your argument the only thing you need is $f^{-1} (x)=\frac x {1-|x|}$ which is also continuous. For another argument note that $\arctan \,x$ is a hoemomorphism between $\mathbb R$ and $(-\pi /2,\pi /2)$. The former is complete, the latter is not.

Answer 2

Being complete is not a topological property but is a property of metric spaces.

If $X$ and $Y$ are both metrizable topological spaces, $d$ is a metric on $X$ such that metric space $(X,d)$ is complete and $f:X\to Y$ is a homeomorphism then $f$ induces a metric $d'$ on on $Y$ by $d'(a,b)=d(f^{-1}(a),f^{-1}(b))$ and also metric space $(Y,d')$ will be complete.

It is quite well possible that for topological space $X$ we can find different metrics (both inducing the topology) such that one of them makes $X$ a complete metric space and the other make $X$ a not complete metric space.