Are there totally bounded group topologies $\mathcal S$ and $\mathcal T$ on $\Bbb Q$ such that for some open sets $A\in\mathcal S$ and $B\in \mathcal T$ we have $A\cap B=\{0\}$?



Totally bounded group topologies on $\Bbb Q$ with trivial intersection of two neighborhoods
Two non-discrete group topologies which together generate the discrete topology are sometimes called *transversal*. It is proved in **"On transversal group topologies"** by Dikranjan, Tkachenko and Yaschenko (see Theorem 3.13) that no totally bounded group topology on an infinite abelian group admits a transversal group topology. In particular, the answer to your question is no, even if you only require one of the topologies to be totally bounded. On the other hand it is proved in the same paper that any infinite abelian group admits infinitely many pairwise transversal group topologies (none of which is totally bounded, of course). 


