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 $\hat{G}$?
Topology induced by the completion of a topological group
14
$\begingroup$
abstract-algebra
general-topology
topological-groups
-
3For 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
-
3Another 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
-
0@GEdgar: How do we know that this $\hat{N}$ will be nonempty? – 2012-09-08
-
2It 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
-
0@GEdgar: I see, thank you. – 2012-09-08
-
0@GEdgar: If you would like to post this as answer i will gladly vote and accept it. – 2012-09-09
-
1@MSina: Do we really need a [completeness] tag? – 2013-03-03
-
0@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
-
0See 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
1 Answers
5
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.
-
0This 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
-
1@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
-
0Ah, okay! Thanks for the quick reply and sorry that I didn't read carefully enough. – 2016-01-15
-
0@GEdgar: Under this topology can we prove that $\hat{G}$ is complete ? i.e., every Cauchy sequence is convergent ? – 2018-11-20