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AC is enough to guarantee the existence of both Bernstein Sets and Vitali Sets...

However is the existence of Vitali Sets strictly weaker than that of Bernstein Sets?

What about the other way round?

flag	asked Jul 29 2011 at 13:51 George Lazou 295●1●6

	Are these your sets? Vitali set: one element chosen from each coset of $\mathbb R$ modulo $\mathbb Q$. Bernstein set: a set $E \subseteq \mathbb R$ such that both $E$ and its complement meet each uncountable closed set. – Gerald Edgar Jul 29 2011 at 14:48
	A bit more general: Vitali Set: one element chosen from each coset of $\mathbb{R}$ modulo a countable dense subgroup of $\mathbb{R}$, Bernstein Set: a set $E \subset \mathbb{R}$ such that both $E$ and its complement are totally imperfect in $\mathbb{R}$. – George Lazou Jul 29 2011 at 17:39

Liang Yu · Answer 1 · 2012-05-21 00:41:47Z UTC

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For your second definition of Vitali set, I have a weak partial answer. Namely the existence of a Bernstein set does not imply the existence of a $T$-Vitali set. The answer can be found in logic blog maintained by Andre Nies:
http://dl.dropbox.com/u/370127/Blog/Blog2012.pdf

Note that a Turing degree does not need to be an addition group.

I don't know whether the existence of a Vitali set implies the existence of a Bernstein set. But it is not difficult to see, under $ZF+DC$, that there is a Vitali set (if it exists) which contains a perfect subset.

For you first definition of Vitali set, I have no idea.

edited May 21 2012 at 0:41

answered May 20 2012 at 2:14

Liang Yu
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Very nice! The specific version proved in the link is: In ZF, the existence of Bernstein sets does not imply the existence of $T$-Vitali sets, where $A$ is $T$-Vitali iff $|A\cap{\mathbf x}|=1$ for each Turing degree ${\mathbf x}$. This follows from Wei Wang, Liuzhen Wu, and Liang Yu, "Cofinal maximal chains in the Turing degrees", Proc. Amer. Math. Soc., to appear, where it is proved that the following is consistent: ZF+DC + there is an $\omega_1$-cofinal chain of Turing degrees + there is no well-ordering of the reals. – Andres Caicedo May 20 2012 at 2:59

Andres, thanks for making my answer clearer. I was surprised George's question seems unknown. I don't even know whether there is a $ZF+DC$ model in which there is no well ordering of reals but there is a Vitali set (in the George's first version sense). – Liang Yu May 20 2012 at 5:51

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