For a topological group $G$ and a given fundamental system of neighbourhoods of $G$ we can define the completion of G and we call it $\hat{G}$. The induced fundamental system of neighbourhoods of $\hat{G}$ is given by this Topology induced by the completion of a topological group.

Then Can we say that $\hat{G}$ is complete ? i.e., every Cauchy sequence in $\hat{G}$ is convergent ?

For this we assume that $\{z_n\}$ be any Cauchy sequence in $\hat{G},$ then given any open neighbourhood $\tilde{N}$ of $\hat{G}$ there exists an integer $k$ such that whenever $m,n \geq k,$ $z_m-z_n \in \tilde{N}.$ Then how can I show that $\{z_n\}$ is convergent ? That is we are looking for an element $s \in \hat{G}$ such that for any neighbourhood $\hat{P}$ of $\hat{G},$ $z_n \in s+\hat{P}$ for large $n$. In general will it hold ? I need some help. Thanks.

  • $\begingroup$ Normally you'd have to consider Cauchy nets for full completeness. You seem to only want sequential completeness? $\endgroup$ – Henno Brandsma Nov 20 '18 at 17:34
  • $\begingroup$ Is the completion not the completion of the uniform structure on $G$? $\endgroup$ – Robert Thingum Nov 20 '18 at 19:39
  • $\begingroup$ Yes, but if $G$ is not first-countable then the completion of $G$, as a uniform space, in general is not determined by Cauchy sequences alone. Sequential completeness only suffices if $G$ is first-countable. $\endgroup$ – Sir Jective Nov 20 '18 at 20:52
  • $\begingroup$ Right, I didn't mean to imply that sequences were sufficient. Just wanted to clarify what the completion was. $\endgroup$ – Robert Thingum Nov 20 '18 at 20:57

For simplicity, I'll ignore the very accurate point made in the comments that defining completeness in terms of sequences isn't sufficient in general, and assume that the topology on $G$ is first-countable, and therefore $\hat G$ is first-countable as well.

First, we need to correct your definition of Cauchy sequences and of convergence. For example, in $\Bbb R, \left\{\frac 1n\right\}$ is a Cauchy sequence (as is any convergent sequence), and $(1,2)$ is a neighborhood in $\Bbb R$, but there is not a $k \in \Bbb N$ such that for $n, m > k, \left(\frac 1m - \frac 1n\right) \in (1,2)$.

$\tilde N$ needs to be a neighborhood of $0$, not just any neighborhood of $\hat G$. Similarly, $\hat P$ is also a neighborhood of $0$ (and so $s + \hat P$ is a neighborhood of $s$).

Since $\{z_n\}_{n\in\Bbb N} \subset \hat G$, for each $n, z_n$ is some Cauchy sequence $\{z_{nm}\}_{m\in\Bbb N}$ in $G$. You can use the fact that $\{z_n\}_{n\in\Bbb N}$ is Cauchy in $\hat G$ to show that the diagonal sequence $\{z_{nn}\}_{n\in \Bbb N}$ is Cauchy in $G$.

Then $s = \{z_{nn}\}_{n\in \Bbb N}$.

  • $\begingroup$ I need to check if it works, though looks like it will work. $\endgroup$ – user371231 Nov 21 '18 at 7:02

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