Topological proof that a Vitali set is not Borel This question is purely out of curiosity, and well outside my field — apologies if there is a trivial answer. Recall that a Vitali set   is a subset $V$ of $[0,1]$ such that the restriction to $V$ of the quotient map $\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Q}$ is bijective. It follows easily from the definition that $V$ is not Lebesgue measurable. Now suppose you don't know the Lebesgue measure (which, after all, is not that easy to construct). Is there a topological proof that $V$ is not a Borel subset of $\mathbb{R}$?
 A: Sometimes a convenient substitute for Lebesgue measurability is the property of Baire. Just like Lebesgue measurability, the class of sets with this property is a $\sigma$-algebra containing the open subsets - indeed, open sets clearly have property of Baire, this class is closed under countable unions since meager sets are closed under countable unions, and it's closed under complements essentially since the boundary of an open set is nowhere dense.
It of course follows that every Borel set has the property of Baire. It remains to show the Vitali set doesn't have a property of Baire. Indeed, assume it was, then either $V$ is meager, or it is comeager in some open interval. The former cannot hold, since $[0,1]$ is contained in countable union of translates of $V$, so would be meager itself. In the other hand, if it were comeager in some open interval $I$, then for any rational $q$ the intersection $I\cap(V+q)$, since it's contained in $I\setminus V$, is meager, from which we would conclude $V$ is meager.
This might be a slightly overly detailed answer, but I did try to make it reasonably self-contained - it indeed is quite easier than if we had to build up the theory of Lebesgue measure from scratch!
