# Topology induced by the completion of a topological group

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. – GEdgar Sep 8 '12 at 16:06
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. – GEdgar Sep 8 '12 at 16:07
@GEdgar: How do we know that this $\hat{N}$ will be nonempty? – Manos Sep 8 '12 at 16:11
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. – GEdgar Sep 8 '12 at 16:12
@MSina: Do we really need a [completeness] tag? – Asaf Karagila Mar 3 '13 at 18:16

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.
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.) – Ingo Blechschmidt Jan 15 at 16:12
@IngoBlechschmidt... Formulation says "all sequences in the class". So in this example, the class of $(1-1/2^n)$ is not in $\widehat{N}$. – GEdgar Jan 15 at 16:27