# Totally bounded group topologies on $\Bbb Q$ with trivial intersection of two neighborhoods

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 $A\cap B=\{0\}$?

• You mean with $\mathcal S \neq \mathcal T$? – Andreas Thom Oct 4 '14 at 12:54
• @AndreasThom: The topologies must be different since they are totally bounded. – Ramiro de la Vega Oct 4 '14 at 15:52
• I see, isn't this quit obviously impossible -- basically since the product of two compact spaces is compact. How can $\mathbb Q$ be discrete in the product? Maybe I am missing something. – Andreas Thom Oct 4 '14 at 16:21
• @AndreasThom: I don´t understand your comment, what does the product topology have to do here? – Ramiro de la Vega Oct 4 '14 at 22:01
• I thought $\mathcal S$ and $\mathcal T$ correspond to compactifications of $\mathbb Q$. Then, $A \times B$ would be a neighborhood of $0$ in the product topology and $A \cap B$ would be the intersection of $A \times B$ with the diagonal embedding of $\mathbb Q$ into the product. If $A \cap B = \{0\}$, then the induced topology follows to be discrete - contradicting the compactness of the product. – Andreas Thom Oct 5 '14 at 9:27