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Nov 11, 2014 at 17:51 comment added Minimus Heximus @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.
Nov 11, 2014 at 6:48 comment added Andreas Thom Do you know any pair $\mathcal A, \mathcal B$ of non-discrete topologies and $A \in \mathcal A,B \in \mathcal B$ with $A \cap B = \{0\}$?
Oct 2, 2014 at 17:48 history asked Minimus Heximus CC BY-SA 3.0