# Is a successor to a successor to the trivial group topology on an Abelian group, totally bounded?

Is there an example of an Abelian group $G$ and group topologies $\cal S$ and $\cal T$ on it such that $\cal S$ is an immediate successor to the trivial topology on $G$ (i.e there is no other group topologies between them w.r.t $\subseteq)$) and $\cal T$ is an immediate successor to $\cal S$; and $\cal T$ is not totally bounded? (Not part of this question but: Is it at least countably bounded ($\omega$-narrow)?)

A group topology $\tau$ on $G$ is totally bounded if for every nonempty open set $U$ in $\tau$, there is a finite $F\subseteq G$ with $UF=G$, or equivalently, if a completion of its right uniformity is compact. A totally bounded group topology can be non-Hausdorff.

• By a trivial group topology you mean the discrete topology or the antidiscrete topology? – Alex Ravsky Sep 4 '15 at 1:54
• @AlexRavsky The indiscrete one with only two open sets. – Minimus Heximus Sep 4 '15 at 1:56