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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 ...

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    $\begingroup$ 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. $\endgroup$ – Saucy O'Path Sep 11 '18 at 6:27
  • $\begingroup$ Yes of course..... $\endgroup$ – Indrajit Ghosh Sep 11 '18 at 6:28
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    $\begingroup$ 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. $\endgroup$ – Henno Brandsma Sep 11 '18 at 8:35
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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.

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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.

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    $\begingroup$ Very nice example... thank you so much $\endgroup$ – Indrajit Ghosh Sep 11 '18 at 6:26

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