(This question has been on [math.SE](http://math.stackexchange.com/q/734052/57159) for over a week and has not gotten any answers.)
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Let $\; \langle \hspace{-0.02 in}G,\hspace{-0.04 in}+,\hspace{-0.04 in}\mathcal{T}\hspace{.03 in}\rangle \;$ be a $\hspace{.02 in}\big(\hspace{-0.03 in}$[$\hspace{.03 in}\text{T}_{\hspace{-0.02 in}0}$](http://en.wikipedia.org/wiki/Kolmogorov_space)$\hspace{-0.03 in}\big)$ topological abelian group, and let $0$ be its identity element.
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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
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$J$ 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)$](http://en.wikipedia.org/wiki/Series_%28mathematics%29#Abelian_topological_groups)$ \:$ exists.
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Does it follow that for all open subsets $U$ of $G\hspace{-0.02 in}$, if $\: 0\in U \:$ then
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there exists an open subgroup $H$ of $G$ such that $\: H\subseteq U \;$?
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