A question related to "Topology induced by the completion of a topological group"

Question

I am sorry if I mistake the answer posted on the question Topology induced by the completion of a topological group.

Stated as in that thread, let $G$ be a topological abelian group with a countable fundamental system of neighborhoods, $\widehat{G}$ the completion of $G$. Then I also want to know the induced group topology on $G$. Then I find that thread.

My question comes, perhaps it is not a mathematics question (sorry).

Did the answer

For each neighborhood $N$ of zero in $G$, define a neighborhood $\widetilde{N}$ in $\widehat{G}$ consisting of those equivalence classes for which all sequences in the class are eventually in $\widehat{G}$. This is a base (of neighborhoods of zero) for the new topology.

define $\widetilde{N}$ as the set $\{\alpha\in \widehat{G}\;|\; \text{ if } (x_n)\in\alpha,\; \text{ then almost all } x_n \in N \}$?

If so, I don't clear why those sets can generate a group topology on $\widehat{G}$. As in the comment, why $\widetilde{N}$ is nonempty? How do we know every representative of the class of constant sequence $\alpha=[(a,a,a,\ldots)]$ ($a\in N$) will eventually lie in $N$.

After some efforts (much more searching or something else), I find that if we define $N^{\prime}=\{\xi\;|\; \exists (x_n) \;\text{ with } [(x_n)]=\xi\; \text{and almost all } x_n\in N\}$, then we can check (not hard) those sets can generate a group topology on $\widehat{G}$ and with this topology $G\to \widehat{G}$ is continuous.

A reference is B.L.van de Waerden, chapter 20, topological algebra, Algebra II. Which first defines a group topology on the group of Cauchy sequences, then the group topology on $\widehat{G}$ is the quotient topology of the Cauchy sequences modulo the null Cauchy sequences. A fundamental system of $0$-neighborhoods of $\widehat{G}$ is exactly generated by those $N^{\prime}$ mentioned above.

Thanks!

Answer 1

It seems the following.

Let $\mathcal N$ be a base of all neighborhoods (not necessarily open) at the zero of the group $G$, $\widetilde{\mathcal N}=\{\widetilde{N}:N\in\mathcal N\}$ and $\mathcal N’=\{N’:N\in\mathcal N\}$. We can easily prove that the family ${\mathcal N’}$ is a base of neighborhoods (not necessarily open) at the zero of the group $\widehat{G}$ for a group topology $\tau$. So to prove that the family $\widetilde{\mathcal N}$ is a base of neighborhoods (not necessarily open) at the zero of the group $\widehat{G}$ for the group topology $\tau$, if suffices to show that for each neighborhood $N\in\mathcal N$ there exist neighborhoods $M,K\in\mathcal N$ such that $M’\subset \widetilde{N}$ and $\widetilde{K}\subset N’$. Since $\widetilde{N}\subset N’$ we may put $K=N$. Let $M\in\mathcal N$ be a neighborhood of the zero such that $M+M\subset N$. We claim that $M’\subset \widetilde{N}$. Indeed, let $\alpha\in M’$ be an arbitrary equivalence class. Then there exist a sequence $(x_n)\in\alpha$ and a number $n_0$ such that $x_n\in M$ for each $n>n_0$. Let $(y_n)\in\alpha$ be an arbitrary sequence. Since $(x_n)\sim (y_n)$, that is the sequence $(y_n-x_n)$ converges to the zero, there exists a number $n_1\ge n_0$ such that $y_n-x_n\in M$ for each $n>n_1$. Then $y_n=(y_n-x_n)+x_n\in M+M\subset N$ for each $n>n_1$. Since this property has an arbitrary sequence $(y_n)\in\alpha$, we see that $\alpha\in \widetilde{N}$.