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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
    Commented Jul 17, 2018 at 14:40
  • $\begingroup$ Can you show there is no connected dense proper $G_\delta$-subgroup of the plane? $\endgroup$
    – D.S. Lipham
    Commented Jul 20, 2018 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
    Commented Jul 21, 2018 at 6:38

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