Completion (construction Atiyah MacDonald chapter 10)

Question

Following Atiyah MacDonald (Chapter 10: Completions), let $G$ be a topological abelian group. We assume that $0 \in G$ has a fundamental system of neighborhoods consisting of subgroups $G = G_0 \supseteq G_1 \supseteq … \supseteq G_n \supseteq … $ Suppose $(x_k)$ is a Cauchy sequence in $G$. Then it is claimed, that the image of $x_k$ in $G/G_n$ is ultimately constant. Why is this the case?

Answer 1

Saying that $(x_n)$ is Cauchy means that, for every $k$, there exists $N$ such that, for $m,n>N$, $x_m-x_n\in G_k$.

Thus, for $m,n>N$, $x_m+G_k=x_n+G_k$. This is the same as saying that the image of $(x_n)$ in $G/G_k$ is eventually constant.

Answer 2

Definition: A Cauchy sequence in $G$ is a sequence $(x_k)$ of elements of $G$ such that, for any neighborhood $U$ of $0$, there exists an integer $s(U)$ with the property that $x_m-x_l \in U$ for all $m,l \geq s(U)$.

My idea: Let $\phi: G \rightarrow G/G_n$ denote the projection sending $x_k$ to $x_k +G_n$.
Choose $U = G_n$ $\Rightarrow x_m-x_l \in G_n$ for all $m,l \geq s(G_n)$

$\Rightarrow x_m+G_n = x_l +G_n$ for all $m,l \geq s(G_n)$

$\Rightarrow \phi(x_m) = \phi(x_l)$ for all $m,l \geq s(G_n)$

$\Rightarrow$ For all $r \geq s(G_n)$ we have $\phi(x_r) = \lambda_n$

Answer 3

Notice that $G_{n}$ is a neighbourhood of $0$. And by the definition of Cauchy sequence in $G$, $\forall U$ neighbourhood of $0$, $\exists N$ great enough such that $x_{n} - x_{m} \in U$, $\forall n,m \geqslant N$. So $x_{n} - x_{m} \in G_{n} $ $\forall n,m \geqslant N$, for some $N$.

Now, recall that for the map $\phi : G \to G/G_{n}$, $g \mapsto g + G_{n}$, ker$\phi = G_{n}$. Hence $(x_{s} - x_{q}) + G_{n} = 0 + G_{n} \implies x_{s} + G_{n} = x_{q} + G_{n}$.