Let $\Bbb Z$ be embedded in the circle group $\Bbb T$ by an irrational rotation and regard $\Bbb Z$ as a subgroup/subspace of the topological group $\Bbb T$.

Are there group topologies $\mathcal A$ and $\mathcal B$ on $\Bbb Z$ such that each nonempty $A\in \mathcal A$ and each nonempty $B\in \mathcal B$ is dense in $\Bbb T$ while there are $A_0\in \mathcal A$ and $B_0\in \mathcal B$ with $A_0\cap B_0=\{0\}$?

  • $\begingroup$ Do you know any pair $\mathcal A, \mathcal B$ of non-discrete topologies and $A \in \mathcal A,B \in \mathcal B$ with $A \cap B = \{0\}$? $\endgroup$ – Andreas Thom Nov 11 '14 at 6:48
  • $\begingroup$ @AndreasThom No. I hoped two different irrational rotations may be an example. But usual topology on $\Bbb T$ is totally bounded so its subspaces $\Bbb Z$ are totally bounded and cannot be traversal. $\endgroup$ – Minimus Heximus Nov 11 '14 at 17:51

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