Is it possible to have a homeomorphism between a complete metric space and an incomplete one? If so, what examples can be given?
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As a side note, if there is a homeomorphism to a complete metric space, then there exists a metric to the other set as well that makes it a complete metric space. Being homeomorphic to a complete metric space is called "Completely metrizable". – T. Eskin Feb 07 '13 at 06:59
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Example of Homeomorphism Between Complete and Incomplete Metric Spaces, preservation of completeness under homeomorphism, Is Completeness intrinsic to a space? – Martin Sleziak May 14 '13 at 07:21
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A simple example is the homeomorphism $$(0, 1) \longrightarrow \Bbb R,$$ $$x \mapsto \tan( \pi(x - \tfrac{1}{2})).$$ Here we give both $(0,1)$ and $\Bbb R$ the usual Euclidean metric. $\Bbb R$ is of course complete, but $(0,1)$ is not complete since the Cauchy sequence $\{1/n\} \subset (0,1)$ doesn't converge in $(0,1)$.
Henry T. Horton
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Consider $X=\mathbb R$ and $Y=(-1,1)$ with the standard metric. Then $f(x)=\frac{x}{|x|+1}$ is a homeomorphism $X\to Y$.
Hagen von Eitzen
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Yes, possible.
The easiest is $\Bbb R\to (0,1)$ with the usual metrics. $\Bbb R$ is complete, though in some limit sense it still misses its endpoints, which are effectively missing in $(0,1)$.
Berci
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