# Canonical Vitali set

The axiom of choice implies that there is a subset $E$ of $\mathbb R$ that are Vitali sets : this means that $E$ is a transversal with respect to the additive subgroup $\mathbb Q$ of $\mathbb R$, i.e. the natural projection $p:{\mathbb R} \to \frac{{\mathbb R}}{\mathbb Q}$ becomes bijective when restricted to $E$.

It is also known that there are models of ZF where no Vitali sets exists.

My question is in-between : is there a formula $\phi$ of set theory such that it is provable in ZF that “If there is a Vitali set then $\phi$ defines a unique Vitali set”.

• I do not understand what you mean by "no such $E$ exists". Do you also require that $E$ is uncountable? $E=\{\sqrt p\mid p$ prime$\}$ is an example of an infinite set, no two of its elements a rational distance apart. Perhaps the requirement is that $p\upharpoonright E$ is again onto? – Andrés E. Caicedo Nov 23 '13 at 16:29
• @Andrés: I think that it’s clear that Ewan has in mind that $E$ should be a transversal for the family of cosets, and that bijective is intended to be bijective with $\Bbb R/\Bbb Q$. – Brian M. Scott Nov 23 '13 at 16:41
• @BrianM.Scott Indeed, that’s exactly what I meant. – Ewan Delanoy Nov 23 '13 at 17:02
• Ewan, you seem to require that $E$ is some canonical Vitali set. I don't think this is possible, even if you do have the axiom of choice. One of the things that make Vitali sets so... "non-constructive" is exactly their non-canonical nature. – Asaf Karagila Nov 23 '13 at 17:21
• @AsafKaragila In your opinion, is it impossible to construct a “canonical example” for anything that needs the axiom of choice ? – Ewan Delanoy Nov 23 '13 at 17:48

No, there cannot be a formula that always defines a Vitali set whenever one exists. In fact we can say something stronger: There is a model of $\mathsf{ZFC}$ with no definable Vitali set.

Assume that $\mathsf{ZF} + \mathsf{DC}$ holds but there is no Vitali set (this follows, for example, from "$\mathsf{ZF} + \mathsf{DC} + {}$every set of reals has the Baire property," which is consistent relative to $\mathsf{ZFC}$ by a theorem of Shelah.) Force with $\text{Col}(\omega_1,\mathbb{R})$ to add a well-ordering of the reals. Because the forcing is countably closed and $\mathsf{DC}$ holds in the ground model, no reals are added. The generic extension has a well-ordering of its reals, so it has a Vitali set. But it cannot have a definable Vitali set because the forcing is homogeneous, so every subset of the ground model that is definable in the forcing extension is an element of the ground model.

Remark: As Andreas Blass points out in the comments, there is a more direct construction assuming the existence of an inaccessible cardinal. Let $\kappa$ be inaccessible and let $g \subset \text{Col}(\omega,\mathord{<}\kappa)$ be a $V$-generic filter. Then in the generic extension $V[g]$ every definable set of reals is Lebesgue measurable by a theorem of Solovay (in fact, every set of reals that is ordinal-definable from a real parameter is Lebesgue measurable, has the Baire property, etc.) So in this generic extension there can be no definable Vitali set.

• Very nice trick! – Asaf Karagila Nov 24 '13 at 1:46
• @Asaf Thanks.$\,\,$ – Trevor Wilson Nov 24 '13 at 1:48
• You can also use Shelah's theorem about $\sf ZF+DC+BP$ rather than $\sf DC_{\omega_1}$. When it comes to adding reals, countable closure and $\sf DC$ are more than enough. – Asaf Karagila Nov 24 '13 at 2:03
• This construction via the Solovay model seems to be three steps, of which the last two are unnecessary. The steps are (1) Lévy-collapse all ordinals below an inaccessible $\kappa$ to be countable (so $\kappa$ becomes $\aleph_1$), (2) pass to an inner model like HOD$(\mathbb R)$, and (3) force to add a well-ordering of the reals. After step (1), you already have a model of ZFC where all HOD$(\mathbb R)$ sets of reals are Lebesgue measurable, so there's no Vitali set. – Andreas Blass Nov 24 '13 at 2:42
• Trevor, the same point as Andreas made about Solovay's model can be made about Shelah's model as well. @Ewan: Shelah has a very difficult paper called "Can you take Solovay's inaccessible away?" – Asaf Karagila Nov 24 '13 at 8:15

Something much weaker is false: It is consistent with $ZFC$ that every (ordinal) definable collection of sets of reals consists solely of Lebesgue measurable sets of reals or has the same size as the set of all subsets of continuum. This (and a category analogue of it) is due to Harvey Friedman.

• I am not sure if you'd care about it but a determinacy analogue of this also holds using a very similar forcing - I learned this from Woodin. – hot_queen Nov 25 '13 at 6:48