# Is $\hat{G}$ is complete with respect to the induced topology of $G$?

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.

• Normally you'd have to consider Cauchy nets for full completeness. You seem to only want sequential completeness? – Henno Brandsma Nov 20 '18 at 17:34
• Is the completion not the completion of the uniform structure on $G$? – Robert Thingum Nov 20 '18 at 19:39
• 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. – Sir Jective Nov 20 '18 at 20:52
• Right, I didn't mean to imply that sequences were sufficient. Just wanted to clarify what the completion was. – 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 $$s = \{z_{nn}\}_{n\in \Bbb N}$$.