(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 $G$?

(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 $G$?

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


Let $\; \langle \hspace{-0.02 in}G,\hspace{-0.04 in}+,\hspace{-0.04 in}\mathcal{T}\hspace{.03 in}\rangle \;$ be a $\hspace{.02 in}\big(\hspace{-0.03 in}$$\hspace{.03 in}\text{T}_{\hspace{-0.02 in}0}$$\hspace{-0.03 in}\big)$ topological abelian group, and let $0$ be its identity element.
Assume that for all index sets $\hspace{.025 in}I$,$\:$ for all functions $\: \hspace{.04 in}f : I\to G \:$, $\;$ if

$\big[$for all open subsets $U$ of $G$, $\:$ if $\: 0\in U \:$ then there exists a finite subset
$J$ of $\hspace{.02 in}I$ such that for all elements $i$ of $\hspace{.02 in}I$, $\:$ if $\: i \not\in J \:$ then $\: \hspace{.04 in}f(i\hspace{.02 in}) \in U$ $\big]$

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


Does it follow that for all open subsets $U$ of $G\hspace{-0.02 in}$, if $\: 0\in U \:$ then
there exists an open subgroup  $H$ of $G$ such that $\: H\subseteq U \;$?

 

(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 $G$?