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Consider sets of Vitali's type in models of $\mathsf{ZF}+\mathsf{GCH}$ where $V \neq L$. Are there sets of Vitali's type in both $L$ and $V \backslash L$? If so, is there any way one can distinguish the constructible sets of Vitali's type from the nonconstructible sets of Vitali's type?

By a set of Vitali's type it is meant a subset of $\mathbb{R}$ containing exactly one element of every equivalence class of the relation $x - y \in G$, where $G$ is some fixed countable subgroup of $( \mathbb{R} , + )$.

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  • $\begingroup$ Could you clarify what is meant by a set of Vitali type? For example, I think if $G$ is any countable subgroup of $(\mathbb{R}, +)$, then choosing representatives under the equivalence relation $x - y \in G$ will give a "Vitali set relative to $G$". The rationals are one instance. $\endgroup$
    – Avshalom
    Commented Dec 30, 2014 at 18:52
  • $\begingroup$ @Avshalom: Exactly so. What you wrote is the definition of sets of Vitali's type, relative to $G$. As you point out, $\mathbb Q$ is the usual subgroup of $($$\mathbb R$,$+$$)$ used to define "Vitali sets" . My question to you is, does any choice for $G$ make the set of representatives under the equivalence relation $x$$-$$y$$\in$$G$ nonmeasurable with respect to the Lebesgue measure? The phrase "sets of Vitali's type" is used as a synonym for "Vitali sets". $\endgroup$
    – Thomas Benjamin
    Commented Dec 30, 2014 at 22:04
  • $\begingroup$ (With respect to $\mathbb Q$, that is....) $\endgroup$
    – Thomas Benjamin
    Commented Dec 30, 2014 at 22:11
  • $\begingroup$ Let $G$ be a countable subgroup. If $x_E$ is the equivalence class of $x \in \mathbb{R}$, then $\mathbb{R} = \bigcup_{g \in G}(x_E + g)$; now apply $\sigma$-additivity and translation invariance of Lebesgue measure. Will that work? The argument will apply more generally. $\endgroup$
    – Avshalom
    Commented Dec 30, 2014 at 22:26
  • $\begingroup$ If so, then the countable subgroup $\langle G, \tau \rangle$, where $\tau$ is a generic real, should give rise to new Vitali sets that are not in the ground model too. $\endgroup$
    – Avshalom
    Commented Dec 31, 2014 at 0:26

1 Answer 1

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There are several things to say that may answer your question.

  • It is consistent with ZFC$\pm$GCH that the set of all real numbers in $L$ is countable in $V$, in which case every set of reals in $L$ has measure zero.

  • It can happen that $V$ and $L$ have precisely the same sets of reals, yet $V\neq L$, because they disagree only very high up. In this case, of course there can be no set of reals in $V-L$.

  • Meanwhile, if by a set of "Vitali type" you mean a set that selects one member of each equivalence class under the relation of translation by rationals, then if $L$ has such a set, then it must have all the reals (since every real differs by a rational from a member of it), and so $\mathbb{R}^V=\mathbb{R}^L$.

  • Also, one can force to add a new Vitali set without adding reals, and so one can have ZFC+GCH+$V\neq L$ with Vitali sets in both $L$ and in $V-L$.

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  • $\begingroup$ @Prof. Hamkins: Thank you for that very nice answer--it is very helpful. The case that interests me most is the last one, that is, the case where one forces to add a new Vitali set without adding reals so that one can have $ZFC+GCH$+$V$$\neq$$L$ with Vitali sets in both $L$ and $V-L$. In this case, is there a way to distinguish the Vitali sets in $L$ from those in $V-L$? I ask this question to see if it makes sense to speak of an 'evolving continuum'--that is, a continuum that grows larger to meet the needs of evolving mathematical practice. $\endgroup$
    – Thomas Benjamin
    Commented Dec 30, 2014 at 15:34
  • $\begingroup$ Actually, I should really say that evolving mathematical practice might force one to construct a new Vitali set not in $L$--sorry. As for the 'evolving continuum' one could always form new Vitali sets by adding reals so as not to violate CH, right? $\endgroup$
    – Thomas Benjamin
    Commented Dec 30, 2014 at 15:56

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