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We know that, given a first countable abelian topological group $G$, the sum of two Cauchy sequences gives yet another Cauchy sequence (see, e.g., this answer).

For those wondering, we say that a sequence $(x_n)$ in $G$ is Cauchy if for any neighborhood $U$ of $0$, there exists an integer $s = s(U)$ such that $x_n - x_m \in U$ whenever $n,m \geq s$.

However, I don't see where the countability hypothesis is used. I know that it is required, as there exist counterexamples (take $G = \mathbb{R}$ with the metric $d(x,y) = |\arctan(x)-\arctan(y)|$ and the Cauchy sequences $(x_n),(y_n)$ with $x_n = n, \, y_n = (-1)^n - n$) (This is actually first-countable, as a metric space).

Update: I am essentially wondering why Atiyah-Macdonald restrict themselves to groups where $0$ has a countable neighborhood basis when talking about these concepts. Let's include a picture:

Picture

Can anyone enlighten me?

Thank you in advance.

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Alex Provost
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2 Answers2

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No, first countable is not needed for the Cauchyness of sum of Cauchy sequences, the given proof uses only that for each open neighborhood $U$ there is a $V$ such that $V+V\subseteq U$, which is basically the continuity of addition.

Instead, your counterexample is rather tricky! They are Cauchy sequences with respect to the given metric, but they are not Cauchy as defined from the topological group structure (I guess the usual addition is used)..

Anyway, $\Bbb R$ is first countable.

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Berci
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    By definition of the uniform structure on an abelian group, $(x,y) \mapsto x-y$ and $y \mapsto -y$ are uniformly continuous, hence so is $(x,y) \mapsto x+y$. Uniformly continuous functions preserve Cauchy-sequences/nets/filters. – Martin May 06 '13 at 22:02
  • @Martin Okay, this is really confusing to me. Is there any reason why Atiyah-Macdonald restrict themselves to topological groups where $0$ admits a countable neighborhood basis when defining these concepts? I thought this was the reason but apparently I'm wrong. – Alex Provost May 06 '13 at 22:21
  • I also see why my "counterexample" is silly, since any metric space is first countable... – Alex Provost May 06 '13 at 22:22
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    @AlexP. The reason is that they later want to describe the completion using sequences, I think. The completion of a general topological group is a bit of a mess. – Martin May 06 '13 at 22:28
  • @Martin I included the relevant picture in my original post. I really don't see where the first countability hypothesis simplifies anything, though. We can simply define the completion as the set of equivalence classes of Cauchy sequences. – Alex Provost May 06 '13 at 22:30
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    @AlexP. Yes, I think you can give a meaningful definition for every topological abelian group, as you suggested. The resulting group should be sequentially complete (I haven't checked). However, completeness of a uniform space is a stronger requirement than just sequential completeness. Probably their choice is a mixture of two things: avoid technicalities with uniform spaces and cover the interesting situations for the applications they have in mind. – Martin May 06 '13 at 23:00
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The reason why Atiyah and MacDonald restrict to first-countable topological abelian groups is because otherwise to construct the completion one would need to use the theory of filters (and other things I haven't ever studied, see Bourbaki, General Topology, Chap. III, §3), in which case it is “a bit of a mess,” as Martin said on a comment in Berci's answer.

One leverages the first-countability property to allow $\hat{G}$ to be set-theoretically described as the set of Cauchy sequences in $G$ modulo the equivalence relation stated on the screenshot from the OP. If you want to see the first-countability hypothesis in action, one actually needs to invoke it to show that

  1. $\hat{G}$ has a well-defined topology (see proof of Lemma 3 here),
  2. the sum $\hat{G}\times\hat{G}\to\hat{G}$ is continuous (see here),
  3. if $A$ a first-countable topological ring, then the product $\hat{A}\times\hat{A}\to\hat{A}$ is well-defined and it is continuous (see Lemmas 3 and 5 here), and
  4. the scalar product $\hat{A}\times\hat{M}\to\hat{M}$ is well-defined and it is continuous, for $M$ a first-countable topological $A$-module (see ibid., Lemmas 8, 9 and 10).

All these four facts are left unproven by Atiyah, MacDonald.

Elías Guisado Villalgordo
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