Is locally completeness a topological property?

Question

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?

This result https://mathoverflow.net/questions/21954/locally-complete-space-is-topologically-equivalent-to-a-complete-space is a partial answer (the first hit when I googled "locally complete space") — Thompson, May 30 '17 at 19:03
@Thompson I know of this question. As far as I can tell it does not answer my question, it just says if there is a locally complete metric, then there is a complete one. Maybe I am missing how it can help here. Please explain. I understand this post, so one can refer to it in an answer. — M. Winter, May 30 '17 at 19:05
I see. I'm making a big mess of this so I think I'll bow out of trying to add anything useful! — Thompson, May 30 '17 at 19:09
@ngenisis Any $x\in(0,1)$ has a small closed interval $[x-\epsilon,x+\epsilon]\subset(0,1)$. This interval is complete. — M. Winter, May 30 '17 at 19:11
Ah, I'm accustomed to neighborhoods being open by definition. — , May 30 '17 at 19:14
@ngenisis Oh. I learned that a neighborhood just contains an open set that contains the point. — M. Winter, May 30 '17 at 19:28
@Thompson: maybe the idea in the question https://math.stackexchange.com/questions/1071275/quasi-cauchy-sequences-in-general-topology could be used to define a "complete topological space"? — Lehs, May 30 '17 at 19:35

John Gowers · Accepted Answer · 2017-05-30T20:17:02.570

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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 $d(s, t) = \sum_{i=0}^\infty\frac{1}{2^i}d(s_i,t_i)$).

edited May 30 '17 at 20:17

answered May 30 '17 at 20:10

John Gowers

24,959

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+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). – Noah Schweber May 30 '17 at 20:21
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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. – M. Winter May 31 '17 at 16:56
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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. – John Gowers Jun 01 '17 at 14:44
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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. – John Gowers Jun 01 '17 at 14:48

score 4 · Answer 2 · answered May 30 '17 at 21:14

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$$d(x,y)=|x-y|+\sum_{k=1}^\infty2^{-k}\inf\left(1,\left|\max_{i\leqslant k}\frac1{|x-q_i|}-\max_{i\leqslant k}\frac1{|y-q_i|}\right|\right).$$

Is locally completeness a topological property?

2 Answers2