Example of topological group with an uncountable fundamental system of neighborhoods of zero I'm interested in topological (abelian) groups whose topology is translation invariant and defined by a filtration of subgroups.
To be more specific : let $G$ be an abelian group, and $(G_i)_{i\in I}$ be a filtration of subgroups of $G$, where $I$ is an ordered set. 
That means that if $i \leq j$ then $G_j \subset G_i$.
One can define a topology (invariant under translation) on $G$ by taking the family $(G_i)_{i \in I}$ as fundamental system of $0$-neighborhood. 
This topology endow $G$ with a structure of topological group. 
Question : Is there an example of an abelian group $G$ such that its topology is given by a uncountable filtration of subgroups (and such that no countable filtration fits) ?
 A: I propose to you a usual and general example. Let $I$ be a directed ordered set and $\{H_i:i\in I\}$ be a family of abelian groups, $G=\prod_{i\in I} H_i$, and for each $i$ let $G_i$ [$G_i’$] be a group $\prod_{k\in I\, k\not > i}\{0\}\times \prod_{k\in I\, k> i} H_k$ [$\prod_{k\in I\, k\not\ge  i}\{0\} \times \prod_{k\in I\, k\ge  i} H_k$]. The family $\{G_i\}$ is a filtration of subgroups of $G$ and in can be chosen as a base at the zero of a group (and, hence, translation invariant) topology $\tau$ [$\tau’$] on the group $G$. It has the following properties. The topology $\tau$ is Hausdorff, the topology $\tau’$ is not Hausdorff provided the set $I$ has a largest element $l$ and the group $H_l$ is non-zero. If the set $I$ has uncountable cofinalty (that is there is no countable subset $J$ of $I$ such that for each element $i\in I$ there exists $j\in J$ such that $i\le j$) and the family $\{H_i:i\in I\}$ consists of non-zero groups then both topologies $\tau$ and $\tau’$ are not first-countable.
