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It is known that the complex plane $\mathbb C$ contain dense connected (additive) subgroups with dense complement but each dense path-connected subgroup of $\mathbb C$ necessarily coincides with $\mathbb C$.

Problem. Does $(\mathbb C,+)$ contain a proper dense connected subgroup, which is a Borel subset of $\mathbb C$?

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  • $\begingroup$ Lovely question Taras! $\endgroup$ – Dominic van der Zypen Jul 17 '18 at 14:40
  • $\begingroup$ Can you show there is no connected dense proper $G_\delta$-subgroup of the plane? $\endgroup$ – D.S. Lipham Jul 20 '18 at 21:59
  • $\begingroup$ @David Each $G_\delta$-subgroup $G$ is Polish and hence is closed in the plane. If it is dense, then $G$ coincides with the plane. In this argument we do not use the connectedness of $G$. $\endgroup$ – Taras Banakh Jul 21 '18 at 6:38

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