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If I have a homeomorphism between $X$ and $Y$. Suppose $X$ is complete and $Y$ is incomplete under the same metric, how can I make $Y$ complete using the fact that it is homeomorphic to $X$. Consider $\mathbb{R}$ and open interval $(0,1)$ how do I make the interval complete under usual metric. Of course if I add $0,1$ to the open interval it will be complete , but what my interest is how do you use the advantage of homeomorphism to make it complete. Thanks in advance.

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3 Answers 3

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In general, if $(X,d)$ is a complete metric space, $Y$ is a topological space and you have a homeomorphism $f:X\to Y$ you can define a metric $d_Y$ on $Y$ which induces the same topology such that $(Y,d_Y)$ is complete. In particular, if you let $d_Y(x,y)=d(f^{-1}(x),f^{-1}(y))$ you get such a metric.

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Using any homeomorphism $\mathbb R\cong (0,1)$ you can pull back the euclidean metric to get a complete metric on $(0,1)$ inducing the standard topology. Of course, this metric will not be the same as the non-complete metric coming from the canonical embedding of $(0,1)$ into $\mathbb R$.

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@ Rasmus , can you explain a bit more ? How do you actually do it ? It would be great if you can give a example with lots of explanation :) –  Theorem Apr 27 '12 at 14:56
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@Vedananda If you have a homeomorphism $\mathbb R\to (0,1)$, you can define the metric $d:(0,1)\to\mathbb R_{\geq 0}$ by letting $d(x,y)=|f^{-1}(x)-f^{-1}(y)|$. –  Alex Becker Apr 27 '12 at 15:09
    
@AlexBecker: Thanks for helping out. :) –  Rasmus Apr 27 '12 at 17:58

You can't. The issue is that completeness is a purely metric space idea and so should not (and is not, as your example shows) preserved under homeomorphism.

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@ Alex can you check this lecture video linkaround 6th minute . may be i understood wrongly . –  Theorem Apr 27 '12 at 14:46

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