3

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.

Elías Guisado Villalgordo
  • 3,209
user371231
  • 2,481
  • 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. – silvascientist 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

2 Answers2

2

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

Paul Sinclair
  • 43,643
  • I need to check if it works, though looks like it will work. – user371231 Nov 21 '18 at 07:02
  • 1
    $\def\N{\mathbb{N}}$In general, ${z_{nn}}{n\in\N}$ is not Cauchy in $G$. Consider for instance $G=\mathbb{N}$ with the discrete topology and ${z{nm}}{m\in\N}\subset G$, where $z{nm}=\delta_{nm}(-1)^n$, and denote $z_n=[(z_{nm})_m]\in\hat{G}$. Then $z_n=0$ in $\hat{G}$, for all $n$, but $z_{nn}=(-1)^n$ is not Cauchy in $G$. – Elías Guisado Villalgordo Sep 07 '23 at 14:37
  • Obviously you meant $G = \Bbb Z$. But yes, I failed to consider the freedom allowed by equivalence classes. – Paul Sinclair Sep 07 '23 at 16:35
1

(As I commented on Paul Sinclair's answer, their hint doesn't work to prove completeness of the completion.)

I will assume first-countability.

We imitate the proof of the completeness of the completion of a metric space. First-countability will give us a countable local basis at zero, that will play the role of the open balls in the metric case.

Lemma (Completeness Criterion). Let $G$ be a first-countable abelian topological group. Suppose $A\subset G$ is dense and that every Cauchy sequence of $G$ contained in $A$ converges in $G$. Then every Cauchy sequence in $G$ converges.

Proof. Let $(x_n)\subset G$ be Cauchy. Pick a local basis $U_n\subset G$ at zero. We can suppose $U_n$ is decreasing (intersect the first $n$ open sets and replace the $n$th open by this). We can also suppose $-U_n=U_n$ (replace $U_n$ by $U_n\cap(-U_n)$). For each $n$, pick $y_n\in A\cap (x_n+U_n)$. We claim that $(y_n)$ is Cauchy. Pick $N$. There is $M$ such that $U_M+U_M+U_M\subset U_N$ (for the sum $G\times G\times G\to G$ is continuous). Let $k$ be such that $x_n-x_m\in U_M$ for all $n,m\geq k$. Then, for $n,m\geq \max(k,M)$, \begin{align*} y_n-y_m&\in x_n+U_n-x_m-U_m\\ &=x_n-x_m+U_n+U_m\\ &\subset x_n-x_m+U_M+U_M\\ &\subset U_M+U_M+U_M\\ &\subset U_N. \end{align*} By hypothesis, there is $y\in G$ with $y_n\to y$. Now, we claim that $x_n\to y$. Pick $N$. There is $M$ such that $U_M+U_M\subset U_N$. There is $k$ such that $y_n-y\in U_M$ for all $n\geq k$. Hence, for $n\geq\max(k,M)$, \begin{align*} x_n-y&\in y_n-y+U_n\\ &\subset U_M+U_M\\ &\subset U_N.&\square \end{align*}

Lemma. Let $G$ be a first-countable abelian topological group. Then every Cauchy sequence in $\hat{G}$ converges.

Proof. We will apply the completeness criterion with $A=\phi(G)$, where $\phi:G\to\hat{G}$ is the map that sends $x\in G$ to the class of the sequence constantly $x$. Density of $\phi(G)$ is proven in Remark 4.2. Suppose then $\phi(x_n)\in\phi(G)\subset \hat{G}$ is Cauchy in $\hat{G}$. We claim that $(x_n)$ is Cauchy in $G$: Let $U\subset G$ be an open neighborhood of zero. There is $k$ such that $\phi(x_n)-\phi(x_m)\in\hat{U}$ for all $n,m\geq k$. In particular, $x_n-x_m\in U$ for all $n,m\geq k$. Next, we claim that $\phi(x_n)\to[x_n]$. This is proven in the linked Remark above. $\square$

Elías Guisado Villalgordo
  • 3,209