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I'm not that familiar with theoretical math in general (I studied engineering), but I recently ended up down a theoretical rabbit hole that led me to the following question:

Is there some type of well known property (e.g. locally compact, locally connected, regular) of a complete, Hausdorff topological ring $R$ that guarantees the following property:

Let $\sum_{i=1}^{\infty} r_i$ unconditionally converge in $R$, where $r_i \in R$. Given any open set $S$ containing $0_R$ (the additive identity of the ring) there exists an open set $S'$ containing $0_R$ such that, given any finite subset $F$ of $S'$ and any sequence $f_i \in F$, $\sum_{i=1}^{\infty} f_i r_i$ is in $S$.

This seems to be true if $R$ equals the real numbers with the usual topology, which I think I've proven:

Let $r = \sum_{i=1}^{\infty} |r_i|$ (we know a series of real numbers converges absolutely if it converges unconditionally) and choose $S$ to be the open ball of radius $\epsilon > 0$ centered at the origin. If $r=0$, any choice of an open set $S'$ will do, so we'll move on to the harder case.

If $r\ne0$, then let $S'$ be the open ball of radius $\frac{\epsilon}{r}$. Thus for any elements $f_i \in S'$, $|f_i| < \frac{\epsilon}{r}$. This leads to the conclusion that $|\sum_{i=1}^{\infty} f_i r_i| \le \sum_{i=1}^{\infty} |f_i r_i| = \sum_{i=1}^{\infty} |f_i||r_i| < \sum_{i=1}^{\infty} \frac{\epsilon}{r}|r_i| =\frac{\epsilon}{r} \sum_{i=1}^{\infty} |r_i| = \frac{\epsilon}{r}r = \epsilon$, assuming $|\sum_{i=1}^{\infty} f_i r_i|$ converges in the first place. In our case, we know the sum converges because, by assumption, the $f_i$ come from a finite set and $\sum_{i=1}^{\infty}r_i$ converges unconditionally.

It seems like the same would apply to the ring of complex numbers as well. But when does this property apply to complete Hausdorff topological rings in general?

(Just a note: it turns out that in a complete Hausdorff abelian topological group, for any unconditionally convergent series, the series' subseries also converge, as stated in the section "Unconditionally convergent series" at https://en.wikipedia.org/wiki/Series_(mathematics). Thus as $F$ is finite, unless I am making a mistake, $\sum_{i=1}^{\infty} f_i r_i$ also converges. The question is just if it converges in $S$.)

Alex Ravsky
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dch
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    Can you recall a definition of an unconditionally convergent series in a topological ring? – Alex Ravsky Jun 04 '20 at 03:52
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    Unconditionally convergent series are series that converge to the same element, even if the order of the terms being added are permuted (https://en.wikipedia.org/wiki/Absolute_convergence#Rearrangements_and_unconditional_convergence). For the reals, unconditional convergence is equivalent to absolute convergence, but in general topological rings, no such absolute value exists. The unconditional part is important, because even for the reals we cannot prove this theorem unless a series converges absolutely, because then we can choose $f_i = \pm 1$ in a way that "blows up" the sum. – dch Jun 06 '20 at 00:27
  • I sent your question to a specilaist. – Alex Ravsky Jun 06 '20 at 09:49
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    Wow thanks so much! And thanks for noticing the error in my proof, you're right! I'll edit that – dch Jun 07 '20 at 16:39

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I wrote “A note on unconditionally convergent series in a complete topological ring” answering your question. Namely, there is shown a topological ring which is a Banach space (and so a connected complete metric space) can fail to have the required property. On the other hand, a topological ring $R$ has the required property provided $R$ is locally compact Hausdorff or $R$ has a base at the zero consisting of open ideals and the additive topological group of $R$ is sequentially complete.

Alex Ravsky
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    I'm confused about the step $\sum{S'r_i} = S' \sum{r_i}$? What if the $r_i$ are not the zero element but sum to the zero element? That doesn't mean that the left side will only contain the zero element correct? – dch Jul 24 '20 at 18:25
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    @dch I updated the answer. I’m studying topological algebra for twenty years, and an answer for your question is my first chance to find its application. So, if you wish, you can provide a proper reference to your research and I’ll add it to the note. – Alex Ravsky Sep 18 '20 at 11:47
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    Wow, thanks so much! I think it'll take me a little while to understand this, but it looks very comprehensive. – dch Sep 25 '20 at 14:16
  • @dch Thank you for your kind words and for the question. This paper can be included into my habilitation thesis and my scientific consultant, Taras Banakh is interested in the question, now especially in the case when $R$ is a commutative Banach algebra. So now we are trying to extend the paper results. Feel free to provide us comments, questions, remarks etc. related to the paper. – Alex Ravsky Sep 25 '20 at 14:40