Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\hat{G}$?

share|cite|improve this question
3  
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
3  
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
2  
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
1  
@MSina: Do we really need a [completeness] tag? – Asaf Karagila Mar 3 '13 at 18:16
up vote 4 down vote accepted

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.

share|cite|improve this answer
    
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
1  
@IngoBlechschmidt... Formulation says "all sequences in the class". So in this example, the class of $(1-1/2^n)$ is not in $\widehat{N}$. – GEdgar Jan 15 at 16:27
    
Ah, okay! Thanks for the quick reply and sorry that I didn't read carefully enough. – Ingo Blechschmidt Jan 15 at 16:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.