TITLE: How to show that Vitali set can't be nowhere dense in $[0,1)$
QUESTION [6 upvotes]: I saw a comment mentioning that it can be shown "a Vitali set cannot be nowhere dense, nor even meager" by Baire category theorem. But I don't know how. In particular, assuming $\bf{AC}$, the cardinality of a Vitali set is $\mathfrak c$ (continuum). How to represent it as the union of a countable collection of closed sets with empty interior?

REPLY [9 votes]: Since a Vitali set $V$ includes one member of $x + \mathbb{Q}$ for all $x \in \mathbb{R}$ it follows that $\bigcup_{q \in \mathbb{Q}} ( q + V ) = \mathbb{R}$, and this is a countable union.  By the Baire Category Theorem $\mathbb R$ is not a countable union of nowhere dense sets, and so at least one of these is not nowhere dense.  But since they are just translates of each other, none is nowhere dense; in particular $V$ is not nowhere dense.