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I want to verify that if $G$ is an abelian topological group then so is its completion $\widehat{G}$.

Note that $G$ is not necessarily a metric space. Hence we define a sequence $x_n$ to be Cauchy if for every neighborhood $U$ of $0$ there exists an $N$ such that for $n,m \geq N$, $x_n - x_m \in U$.

It's clear to me that the constant sequence $\bar{0}$ is Cauchy and hence in $\widehat{G}$. It's also clear that if $x_n$ is Cauchy then so is $-x_n$.

The part I'm not sure about is, if we define the group operation elementwise: $\overline{x} + \overline{y} = (x_n) + (y_n) = (x_n + y_n) = \overline{x+y}$, then I need to check two things: first that addition is well-defined, that is if $\overline{x} = \overline{x^\prime}$ and $\overline{y} = \overline{y^\prime}$ then $\overline{x} + \overline{y} = \overline{x^\prime} + \overline{y^\prime}$ and also that $\overline{x} + \overline{y}$ is Cauchy.

So I thought perhaps I could show that $\overline{x} + \overline{y}$ is Cauchy by doing something like this: $x_n + y_n = (x_n - x_m) + (x_m + y_i) + (y_n - y_i)$ but then I cannot say anything about $x_m + y_i$.

Would someone please show me how to verify these two properties? Thank you.

Edit

This is the relevant part in Atiyah-Macdonald:

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  • 1
    $x_n + y_n - (x_m + y_m) = (x_n - x_m) + (y_n - y_m)$ and by continuity of addition, given a 0-nhood $U$ there is a 0-nhood $V$ with $V + V \subseteq U$.2012-07-25
  • 2
    Not metrizable? You cannot use mere sequences. Must use something else: nets, filters, neighborhoods, etc.2012-07-25
  • 0
    Sequences should be enough if the topology comes from a subgroup filtration $G_1 \supset G_2\supset \cdots$. Otherwise I have no idea.2012-07-25
  • 0
    @GEdgar Yes, right. We are using a countable neighborhood basis.2012-07-26
  • 5
    In a (Hausdorff) topological group: if there is a countable neighborhood basis, then there is a metric for the topology.2012-07-26
  • 0
    There is a metric for the topology of the group, but not for the completion of the group. This will happen if and only if $\cap G_i = \{0\}$.2012-12-05

1 Answers 1