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?
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$\begingroup$ 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. $\endgroup$– Gerald EdgarCommented Jul 29, 2011 at 14:48
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$\begingroup$ 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 $\mathbb{R}$. $\endgroup$– George LazouCommented Jul 29, 2011 at 17:39
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1 Answer
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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.
Added: A The Logic Blog is on the arXiv. According to a comment below, the following is a better reference than the dropbox link:
- Logic Blog, 2012, Andre Nies (editor), http://arxiv.org/abs/1302.3686
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.
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2$\begingroup$ 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. $\endgroup$ Commented May 20, 2012 at 2:59
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$\begingroup$ 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). $\endgroup$– 喻 良Commented May 20, 2012 at 5:51
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1$\begingroup$ The Logic Blog is on arXiv. The following is a better reference than the dropbox link Logic Blog, 2012, Andre Nies (editor), arxiv.org/abs/1302.3686 $\endgroup$– user94844Commented Jul 8, 2016 at 8:58
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$\begingroup$ Ralf and Mariam gave a complete answer to this question. See wwwmath.uni-muenster.de/u/rds $\endgroup$– 喻 良Commented Jul 8, 2016 at 13:00