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 $A\cap B=\{0\}$?
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$\begingroup$ You mean with $\mathcal S \neq \mathcal T$? $\endgroup$– Andreas ThomCommented Oct 4, 2014 at 12:54
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$\begingroup$ @AndreasThom: The topologies must be different since they are totally bounded. $\endgroup$– Ramiro de la VegaCommented Oct 4, 2014 at 15:52
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1$\begingroup$ 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. $\endgroup$– Andreas ThomCommented Oct 4, 2014 at 16:21
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$\begingroup$ @AndreasThom: I don´t understand your comment, what does the product topology have to do here? $\endgroup$– Ramiro de la VegaCommented Oct 4, 2014 at 22:01
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1$\begingroup$ 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. $\endgroup$– Andreas ThomCommented Oct 5, 2014 at 9:27
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1 Answer
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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).
answered Oct 4, 2014 at 15:50
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$\begingroup$ Moreover, Protasov and Zelenyuk showed that there is a family of continuum many such group topologies (see Theorem 5.10 from their paper "Complementable topologies on abelian groups", Siberian Math. Jour., 42:3 (May – June 2001), 550-560, (in Russian)). $\endgroup$ Commented Jun 28, 2017 at 6:07