# Follow up question on union of disjoint Vitali sets…

Since I haven't received a satisfactory answer to my initial question I'm going to ask a somewhat weaker one...

This time we say $X$ is a Vitali set in the closed interval $[0, 1]$ with respect to $\mathbb{Q}$ if $X$ is a selector of the partition of $[0, 1]$ canonically associated with the equivalence relation $x \in \mathbb{R}$ & $y \in \mathbb{R}$ & $x - y \in \mathbb{Q}$.

Recall a set $A$ has the property of Baire iff it can be represented in the form $F \triangle Q = (A \cup B) - (A \cap B)$, where $F$ is closed and $Q$ is of the first category (i.e. can be represented as a countable union of nowhere dense sets).

Let $r \in [0, 1]$, does there exist a Vitali set in the closed interval $[0, 1]$ with respect to $\mathbb{Q}$, $V$, such that $V \cup (V \oplus r)$ has the property of Baire where $V \oplus r$ = {$x + r : x \in V$ & $x + r \leq 1$} $\cup$
{$x + r - 1 : x \in V$ & $x + r > 1$}

The motivation for this is that no Vitali set has the property of Baire so an answer in the affirmative would answer my previous question (modified for the closed interval $[0, 1]$ and $\Gamma = \mathbb{Q}$) in the negative. However my guess is that there is no such Vitali set, but in which case I am interested in the proof which might begin by splitting the $V$ into a set of the first category and a null set and might say something about the spacing of elements of $V$.

One further question that would be of great help is if anyone knows any other properties of Vitali sets other than being non-measurable in the Lebesgue sense and not having the property of Baire.

-

No, there does not exist such a $V$.
Let $W = V \cup (V \oplus r)$, and suppose $W = F \Delta Q$. Note that for $s \in {\mathbb Q}$, $W \cap (W \oplus s)$ is nonempty if and only if $s$ or $s-r$ or $s+r$ is an integer. But if $F$ contained an interval of positive length, $W \cap (W \oplus s)$ would be nonempty for all sufficiently small $|s|$. Thus $F$ is nowhere dense, and $W$ is of first category.
But this is impossible, because $[0,1] = \bigcup_{r \in {\mathbb Q}} (W \oplus r)$.
1) there is a dense set of $r$ such that $V \cap (V \oplus r) = \emptyset$.
2) there is a countable set $S$ such that $\bigcup_{r \in S} (V \oplus r) = [0,1)$.