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Let us define an equivalence relation $\sim$ on $\mathbb{R}$ by saying that $x\sim y$ if $x-y\in \mathbb{Q}$? This equivalence relation partitions $\mathbb{R}$ into uncountably many equivalence classes. My question is, is it possible to construct a set which has exactly one element from each of these equivalence classes?

Can we define these elements explicitly? Or failing that, can we at least prove that there exists a definable subset of $\mathbb{R}$ which has this property? What about a Borel subset of $\mathbb{R}$?

Keshav Srinivasan
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1 Answers1

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Such a set is a Vitali set. It is non-measurable (and in particular not Borel). Solovay showed it is consistent with ZF (without Axiom of Choice) that all subsets of $\mathbb R$ are measurable. Therefore without some form of Axiom of Choice it is impossible to construct such a set and prove it has the desired property.

Robert Israel
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    That's not quite accurate. Solovay showed that ZF + DC + "All sets are measurable" is consistent relative to ZFC + an inaccessible, not relative to ZFC alone - and Shelah later showed that that was optimal. However, pre-Solovay it was known that ZF + "All sets are measurable" is equiconsistent with ZFC (although in the absence of DC, "measurable" is a bit funny ...); I believe this was proved by Truss. – Noah Schweber Mar 26 '20 at 19:14
  • Actually, after searching for a citation I can't back up my claim about pre-Solovay knowledge. – Noah Schweber Apr 14 '20 at 03:03