Suppose we have a non complete metric space. Can it be homeomorphic to its metric completion?
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2What are your thoughts? – Hmm. Jun 04 '16 at 11:56
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As a note on why completeness isn't topological by itself - http://math.stackexchange.com/q/1565350/128967 – snulty Jun 04 '16 at 12:01
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3How about the sum of infinitely many copies of $\mathbb R\setminus{0}$ (i.e., $(\mathbb R\setminus {0})\times \mathbb N$)? That is homeomorphic to its completion, which is the sum of infinitely many copies of $\mathbb R$. – John Gowers Jun 04 '16 at 12:12
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@Donkey_2009 post it as an answer. – Henno Brandsma Jun 04 '16 at 17:17
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There are also connected Riemannian examples. – Moishe Kohan Jun 04 '16 at 22:13
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That sounds interesting. Can you please elaborate? – Asaf Shachar Jun 05 '16 at 00:35
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Just take $R^2$ minus a discrete countably infinite subset. Then modify the flat metric by rescaling via a certain function so that the new metric is complete at infinity and at each puncture except for one. – Moishe Kohan Jun 05 '16 at 03:54
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Consider the metric space $\left(\mathbb R\setminus\{0\}\right)\times\mathbb Z\subset\mathbb R^2$
This metric space is not complete and its completion is $\mathbb R\times\mathbb Z$. The two spaces are homeomorphic. I hope that's easy to see, but here's an explicit pair of continuous inverses:
$f\;\colon\;\left(\mathbb R\setminus \{0\}\right)\times\mathbb Z\to\mathbb R\times \mathbb Z$ given by:
$$ f(x, n)=\begin{cases} (\log(x), 2n) & x>0 \\ (\log(-x), 2n+1) & x<0 \end{cases} $$
$g\;\colon\;\mathbb R\times\mathbb Z\to\left(\mathbb R\setminus\{0\}\right)\times\mathbb Z$ given by \begin{align} g(y, &2k) = (e^y, k)\\ g(y, 2k&+1) = (-e^y, k) \end{align}
You can check for yourself that these functions are continuous and that their compositions are the identity maps.
Henno Brandsma
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John Gowers
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are $f$ and $g$ continuous for the metric ? (what is the metric by the way ?) – reuns Jun 05 '16 at 14:05
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