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Let $ (X_1, d_1) $ and $ (X_2, d_2) $ be metric spaces and $ (X_1^*,d_1^*), (X_2^*,d_2^*) $, respectively, their completions. If $ X_1 $ and $ X_2 $ are homeomorphic, then so are $ X_1^* $ and $ X_2^*$.

Is that statement true in general?

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The classical example is that $\Bbb R$ and $(0,1)$ with the usual topology are homeomorphic, but one is complete and the other is not. – Pedro Tamaroff May 10 '14 at 22:51
Completions depend on more than the topological structure, but less than the full metric structure. What they depend on is precisely the uniform structure ( – Qiaochu Yuan May 11 '14 at 0:42
(This is one of several indications that the textbook definition of the category of metric spaces is wrong. Implicitly the morphisms are usually taken to be continuous maps, but to capture metric structure they should in fact be the distance-decreasing maps. In particular, completion is a functor on the latter category but not the former.) – Qiaochu Yuan May 11 '14 at 0:42
up vote 6 down vote accepted

No. Take $X_1:=(0,1)$ and $X_2:=\Bbb R$, both with the standard metric.

In $X_1$ there will be Cauchy sequences converging to the endpoints, but not in $X_2$.

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I've reached my daily vote limit (+1). Another way to see that these two are not homeomorphic is to see that one is compact while the other is not. – math.n00b May 10 '14 at 22:51
@math.n00b what is compact? – Ittay Weiss May 10 '14 at 23:02
@IttayWeiss: The completion of $X_1$ is $[0,1]$ which is compact, while the completion of $X_2$ is $\mathbb{R}$ which is not compact because it's unbounded. My comment was a bit ambiguous, sorry for that. – math.n00b May 10 '14 at 23:04
@math.n00b :) now it all makes sense. – Ittay Weiss May 11 '14 at 0:47

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