(This question has been on math.SE for over a week and has not gotten any answers.)

Let $G$ be a (T$_0$) topological abelian group, and let $0$ be its identity element.

Assume that for all index sets $I$ and all functions $f\colon I\to G$, if

$\big[$for each neighborhood $U$ of $0$, $f(i)\in U$ for all but finitely many $i$ $\big]$

then $\: $$\displaystyle\sum_{i\in I}\hspace{.03 in}f(i)$$ \:$ exists.

Does it follow that every neighborhood of $0$ contains an open subgroup $H$ of $G$?

  • $\begingroup$ Just commenting to say that I deleted my second answer. As Ricky Demer and Joseph Van Name pointed out, the counterexample didn't work. $\endgroup$
    – Nik Weaver
    May 18, 2014 at 19:29

1 Answer 1


The answer is yes if you additionally assume $G$ is first countable.

I claim that for any open neighborhood $U$ of $0$ there exists a symmetric open neighborhood $V$ of $0$ such that $nV \subseteq U$ for all $n \in {\bf N}$. This implies that $H = \bigcup_{n \in {\bf N}} nV$ is an open subgroup of $U$, as desired. To prove the claim, suppose it fails and let $U$ be an open neighborhood of $0$ such that every symmetric open neighborhood $V$ of $0$ satisfies $nV \not\subseteq U$ for some $n$. Let $(V_i)$ be a neighborhood base at $0$ and wlog (replacing $V_i$ with $V_i \cap -V_i$) assume each $V_i$ is symmetric. Then for each $i$ we can find a finite sequence $s_i$ of elements of $V_i$ whose sum is not in $U$. Concatenating the $s_i$ yields a sequence of elements which converges to $0$, but whose sum does not converge. To see the second assertion, let $W$ be a neighborhood of $0$ such that $W - W \subseteq U$ and let $z \in G$. For each $i$ let $x_i$ be the sum of the sequence $s_1\hat{\phantom{.}}s_2\hat{\phantom{.}}\cdots\hat{\phantom{.}}s_i$. If $x_i \in z + W$ then $x_{i+1} - x_i = \sum s_{i+1} \not\in U$, so that $x_{i+1} \not\in z + W$. Thus the sequence $(x_i)$ cannot converge to $z$, and $z$ was arbitrary. This contradicts the hypothesis on $G$, so the claim is proven.

  • $\begingroup$ I'm pretty sure addition is not continuous on "the real line with the cocountable topology". $\hspace{1.18 in}$ $\endgroup$
    – user5810
    May 8, 2014 at 4:29
  • $\begingroup$ @Ricky: you're right, I'll have to delete that part of my answer. $\endgroup$
    – Nik Weaver
    May 8, 2014 at 4:32

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