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Let $G$ be a normed abelian group1 (which seems to be the most general place to study rearrangements of series). For a sequence $x_0, x_1, \ldots\in G$, define \begin{align*} \Sigma & := \Bigl\{ \sigma\in\operatorname{Symm}(\mathbb N) : \sum_i x_{\sigma(i)}\text{ converges} \Bigr\}\text{, and}\\ A & := \Bigl\{ \sum_i x_{\sigma(i)}\in G : \sigma\in\Sigma \Bigr\}\text. \end{align*}

Then, one usually defines $\sum_i x_i$ to be unconditionally convergent iff $\Sigma = \operatorname{Symm}(\mathbb N)$ and $|A| = 1$.

Question: Can it be that $\Sigma = \operatorname{Symm}(\mathbb N)$ but $|A| \ge 2$?

I know:

  1. Due to Riemann's theorem, this is impossible if $G = \mathbb R$ and thus, impossible too if $G$ is any finite dimensional normed vector space (possibly over $\mathbb C$).
  2. Such a series should not be convergent in norm (that is, $\sum_i\|x_i\| = \infty$).
  3. Thanks to Bruno B (in the comments), a sufficient condition to prevent this is to have that ensure the existence of a separating family of continuous group homomorphisms into a normed abelian group $H$ in which this is known to be impossible.

1A normed abelian group $G$ is an abelian group together with a "norm" $\|\cdot\|\colon G\to [0, +\infty)$ that is positive definite, satisfies triangle inequality (a.k.a. subadditive) and is invariant under negation.

Atom
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    Notice that you can replace the counterexample $G = \mathbb{K}$ ($\mathbb{K} =\mathbb{R}$ or $\mathbb{C}$) by $G = X$ for any real/complex normed vector space $X$, as the (continuous) dual $X^*$ of $X$ separates the points of $X$, i.e. given $x \neq y$ distinct elements of $X$ there exists $f : X \to \mathbb{K}$ such that $f(x) \neq f(y)$, hence you can apply such functionals $f$ to $\Sigma$ and $A$ and go back to the $\mathbb{K}$ case checking one pair of distinct elements of $A$ at a time. Seeing as the vector space structure is hardly used in what I just said, I thought this could help. 1/2 – Bruno B Sep 30 '24 at 14:06
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    Because you could try to see if you can have the same notion of duality to bring the study back to $\mathbb{R}$ or some other familiar abelian group(s) where you know the answer to your question. For example, maybe you could have the same notion for homomorphisms $G \to S^1$? I am not familiar enough with groups in their generality however. 2/2 – Bruno B Sep 30 '24 at 14:07
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    @BrunoB, that is a great insight! Many thanks. – Atom Oct 01 '24 at 06:09

1 Answers1

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Different general frameworks to study unconditionally convergent series are considered in our paper [BR]. In particular, in Section 1, if $G$ is an Abelian topological group and $\Sigma = \operatorname{Symm}(\mathbb N)$ then it is easy to show that the series $\sum_i x_i$ is unconditionally Cauchy. Then Proposition 1 implies that if $|A|\ge 1$ and $G$ is $T_0$ then $|A|=1$.

References

[BR] Taras Banakh, Alex Ravsky, On unconditionally convergent series in topological rings, Carpathian Math. Publ. 14 (2022) 266-288.

Alex Ravsky
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    Finally, the complete answer to the problem! Thanks to @BrunoB for putting this question in spotlight. – Atom Nov 08 '24 at 17:04
  • Btw, Alex, why do you require $G$ to be $T_0$ here? – Atom Nov 08 '24 at 17:05
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    @Atom Thanks for your kind words. We wrote this paper answering an other MSE question, and I am glad that it turned to be useful once more. The $T_0$ condition is needed to ensure that a sequence of elements of $G$ has a unique limit. If $G$ is not $T_0$ then it contains two distinct points $x,y$ such that each neighborhood of $x$ contains $y$ and vise versa. Then the series $x+0+0+\dots$ converges both to $x$ and to $y$, so $|A|\ge 2$. – Alex Ravsky Nov 09 '24 at 00:05
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    For instance, if $G$ is the two-element group $\mathbb Z_2$ endowed with the antidiscrete topology then any sequence of elements of $G$ converges to any element of $G$, so in this case $|A|=2$. – Alex Ravsky Nov 09 '24 at 00:05
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    Of course! I had falsely assumed that if a series unconditionally converges to $x$, then it can't converge to any other value (in my defense, this unconscious stance was probably induced because I was already in the setting of normed abelian groups, so that $T_0$-ness was already satisfied). – Atom Nov 09 '24 at 00:39