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I know that completeness itself is not a topological property because a complete and a not complete metric space can be homeomorphic, e.g. $\Bbb R$ and $(0,1)$.

However, both $\Bbb R$ and $(0,1)$ are locally complete (each point has a neighborhood that is complete under the induced metric). As all examples I know of are of this form, the naturally occuring next question is

Question: Is being locally complete a topological property?

Or the other way around: are there metric spaces which are homeomorphic, but one is locally complete and the other one is not?

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    $\begingroup$ @Thompson That terminology is not used in the question. $\endgroup$ – John Gowers May 30 '17 at 18:58
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    $\begingroup$ How is $(0,1)$ locally complete? $\endgroup$ – ngenisis May 30 '17 at 19:10
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    $\begingroup$ @ngenisis Any $x\in(0,1)$ has a small closed interval $[x-\epsilon,x+\epsilon]\subset(0,1)$. This interval is complete. $\endgroup$ – M. Winter May 30 '17 at 19:11
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    $\begingroup$ Ah, I'm accustomed to neighborhoods being open by definition. $\endgroup$ – ngenisis May 30 '17 at 19:14
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    $\begingroup$ @ngenisis Oh. I learned that a neighborhood just contains an open set that contains the point. $\endgroup$ – M. Winter May 30 '17 at 19:28
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The irrational numbers are not locally complete, but they are homeomorphic to the Baire space $\mathbb N^{\mathbb N}$, which can be given a metric turning it into a complete metric space.

(For example, endow $\mathbb N$ with the discrete metric and set $d(s, t) = \sum_{i=0}^\infty\frac{1}{2^i}d(s_i,t_i)$).

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    $\begingroup$ +1. And for the OP, thinking about this counterexample can lead to considering local compactness, which is of course a topological property (and one which Baire space lacks). $\endgroup$ – Noah Schweber May 30 '17 at 20:21
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    $\begingroup$ I would be glad if you can show the "obvious" homeomorphism between the irrationals and $\Bbb N^{\Bbb N}$ (with what topology?). I am a bit confused whether the example below should show me the complete or incomplete metric. $\endgroup$ – M. Winter May 31 '17 at 16:56
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    $\begingroup$ It's not obvious (I never used that word), just well known. Essentially, you map an irrational number to the coefficients in its continued fraction representation. See the Wikipedia article I linked to for details. $\endgroup$ – John Gowers Jun 1 '17 at 14:44
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    $\begingroup$ The topology on $\mathbb N^{\mathbb N}$ is the product topology. Equivalently, it is the topology induced by the metric I provided, which makes the space into a complete metric space. $\endgroup$ – John Gowers Jun 1 '17 at 14:48
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Another way of proving that the irrationals can be made complete with respect to a metric $d$ which is equivalent to the usual one consists in providing such a metric. This can be donne as follows: let $(q_n)_{n\in\mathbb N}$ be an enumeration of the rationals. Then, if $x,y\in\mathbb{R}\setminus\mathbb Q$, define$$d(x,y)=|x-y|+\sum_{k=1}^\infty2^{-k}\inf\left(1,\left|\max_{i\leqslant k}\frac1{|x-q_i|}-\max_{i\leqslant k}\frac1{|y-q_i|}\right|\right).$$

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