# A continuous bijection between two complete metric spaces that is not a homeomorphism.

Suppose $$X$$ and $$Y$$ are two metric spaces and $$f: X\to Y$$ be a continuous bijection.

Now my question is does the completeness of $$X$$ and $$Y$$ implies $$f$$ to be a Homeomorphism?

My idea. First of all I try to prove $$f$$ to be a closed map assuming $$X$$ and $$Y$$ be a complete metric space. But this idea didn't work.

I know if $$X$$ is given to be compact then whether or not $$Y$$ complete given initially $$f$$ becomes a homeomorphism. But that is not the case here. So I try to find a counter example.

I take $$Y=\Bbb{R}$$ and try to choose $$X$$ to be a non compact but closed subset of $$\Bbb{R}$$ (and $$\Bbb{R}^2$$) but the problem is in that situation the bijections I found was not continuous. Also I cannot found any example beyond the metric spaces $$\Bbb{R}$$ or $$\Bbb{R}^2$$ as my $$X$$.

Can any one help me to figure out how to construct an counter example here. Thanks ...

• For the record, if $X$ is compact then there is no "whether or not $Y$ is complete", because $Y$ will be complete just by the fact that there is a continuous bijection.
– user562983
Sep 11 '18 at 6:27
• Yes of course..... Sep 11 '18 at 6:28
• Between linear normed spaces any such bijective linear map will be a homeomorphism because of the open mapping theorem which holds for complete linear spaces. Sep 11 '18 at 8:35

The identity map from $\mathbb R$ with discrete metric into $\mathbb R$ with usual metric is a continuous bijection which is not a homeomorphism. Both spaces are complete.
First, observe that $[0,1)$ is homeomorphic to $[0,\infty)$ and clearly $[0,\infty)$ is complete because is a closed subset of a complete metric space ($\mathbb{R}$). Take $f:[0,\infty)\to[0,1)$ such homeomorphism. Consider the function $g:[0,1)\to\mathbb{S}^1$ defined by $g(x)=(\cos(2\pi x),\sin(2\pi x))$. It's not to hard to prove that $g$ is continuous and biyective. Thus, $g\circ f:[0,\infty)\to\mathbb{S}^1$ is continuous and biyective function and the domain is complete but $g\circ f$ is not an homeomorrphism because $\mathbb{S}^1$ is compact and $[0,\infty)$ isn't.