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Let $G$ be an abelian topological group and let $\hat{G}$ be its completion, i.e. the group containing the equivalence classes of all Cauchy sequences of $G$. What exactly is the topology of $\hat{G}$?

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    For each neighborhood $N$ of zero in $G$, define a neighborhood $\hat{N}$ in $\hat{G}$ consisting of those equivalence classes for which all sequences in the class are eventually in $N$. This is a base (of neighborhoods of zero) for the new topology.2012-09-08
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    Another remark. Unless $G$ is metrizable, you cannot expect the "completion" by sequences to be complete in the sense of uniform space. So normally we would do completion by nets or by filters or similar.2012-09-08
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    @GEdgar: How do we know that this $\hat{N}$ will be nonempty?2012-09-08
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    It contains many constant sequences. For general $G$ it could happen that every cauchy sequence is eventually constant, so that the sequential completion is nothing new.2012-09-08
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    @GEdgar: I see, thank you.2012-09-08
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    @GEdgar: If you would like to post this as answer i will gladly vote and accept it.2012-09-09
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    @MSina: Do we really need a [completeness] tag?2013-03-03
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    @Manos: Out of curiosity, what equivalence relation are we looking at on $G$? And is $G$ metrizable, or how do you define Cauchy sequences?2013-03-10
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    See also [this thread](http://math.stackexchange.com/questions/311897/completion-as-a-functor-between-topological-rings). It's about topological rings, but the answer of Martin Brandenburg also fits to topological groups. I really recommend the book "General Topology" from Nicolas Bourbaki for further details.2013-03-10

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formerly a remark

For each neighborhood $N$ of zero in G, define a neighborhood $\hat{N}$ in $\hat{G}$ consisting of those equivalence classes for which all sequences in the class are eventually in $N$. This is a base (of neighborhoods of zero) for the new topology.

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    This construction of $\hat N$ is not well-defined. Consider $G = \mathbb{R}$ with its usual topology, $N = (-1,1)$. Then the Cauchy sequences $(1 - 1/2^n)_n$ and $(1 + (-1/2)^n)_n$ are equivalent, but the first one is an element of $\hat N$ while the second one is not. (Thanks to Tim Baumann for bringing this point to my attention and supplying this counterexample.)2016-01-15
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    @IngoBlechschmidt... Formulation says "all sequences in the class". So in this example, the class of $(1-1/2^n)$ is not in $\widehat{N}$.2016-01-15
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    Ah, okay! Thanks for the quick reply and sorry that I didn't read carefully enough.2016-01-15
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    @GEdgar: Under this topology can we prove that $\hat{G}$ is complete ? i.e., every Cauchy sequence is convergent ?2018-11-20