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We say $X$ is a Vitali set if there exists a countably dense subgroup, $\Gamma$, of the additive group $\mathbb{R}$, such that $X$ is a selector of the partition of $\mathbb{R}$ canonically associated with the equivalence relation $x \in \mathbb{R}$ & $y \in \mathbb{R}$ & $x - y \in \Gamma$.

Let $V$ be a Vitali set and let $r \in \mathbb{R}$. Is $V \cup (V \oplus r)$ a Vitali set where $V \oplus r$ = {$x + r : x \in V$ }?

A slightly easier question, perhaps, is the special case $V$ is the Vitali set with respect to the countably dense subgroup $\mathbb{Q}$ and with the restriction $r \in \mathbb{Q}$.

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    $\begingroup$ Am I missing something? No proper superset of a selector is itself a selector. $\endgroup$
    – Emil Jeřábek
    Commented Jun 13, 2011 at 12:38
  • $\begingroup$ Sorry perhaps I phrased it badly... I mean does there exist another countably dense subgroup of the additive group $\mathbb{R}$, $\Gamma_{1} \neq \Gamma$, such that $V \cup (V \oplus r)$ is a selector for the partition associated with this group... $\endgroup$
    – George Lazou
    Commented Jun 13, 2011 at 14:13
  • $\begingroup$ It does not hold in general when one further demands $\Gamma_1\subseteq\Gamma$: then it is easy to see that either $\Gamma=\Gamma_1$ (which is only possible in the trivial case when $V+r\subseteq V$), or $[\Gamma:\Gamma_1]=2$, and there are groups (such as $\mathbb Q$) that have no subgroup of index $2$. I'm not sure how to prove the same when $\Gamma_1$ is allowed to be arbitrary, but it seems very likely that the answer is the same. $\endgroup$
    – Emil Jeřábek
    Commented Jun 13, 2011 at 18:14
  • $\begingroup$ Thanks Emil... do you have a heuristic argument why the answer should be the same for arbitrary $\Gamma_{1}$ or is it just a gut feeling... I'm not saying I disagree, just still not 100% sure... I am wondering whether adding some extra restrictions on the selector may make it possible... i.e. whether one could choose a selector such that there exists $\Gamma_{1}$ that works... $\endgroup$
    – George Lazou
    Commented Jun 14, 2011 at 12:17
  • $\begingroup$ The heuristic argument is simply that when we enlarge $V$, there is no obvious way of constructing $\Gamma_1$ other than shrinking the $\Gamma$ we already have. In principle it's possible that some completely different $\Gamma_1$ incomparable with $\Gamma$ could do the job, but then we would basically have to construct it from scratch without using the information that $V$ is a Vitali set. $\endgroup$
    – Emil Jeřábek
    Commented Jun 14, 2011 at 13:59

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If $\Gamma$ has a subgroup $\Gamma_1$ of index 2, so $\Gamma = \Gamma_1 \cup (\Gamma_1 + r)$ for any $r \in \Gamma \backslash \Gamma_1$, and $V$ is a Vitali set with respect to $\Gamma$, then $V \cup (V + r)$ is a Vitali set with respect to $\Gamma_1$. There are dense subgroups of $\mathbb R$ that have subgroups of index 2, e.g. $\alpha {\mathbb Z} + (2 \beta) {\mathbb Z}$ is a subgroup of index 2 of $\alpha {\mathbb Z} + \beta {\mathbb Z}$ where $\alpha$ and $\beta$ are linearly independent over $\mathbb Q$.

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  • $\begingroup$ Thanks Robert, that's really helpful... out of interest why did you delete your initial example of $\mathbb{Q} + \alpha\mathbb{Z}$ where $\alpha$ is an irrational? $\endgroup$
    – George Lazou
    Commented Jun 13, 2011 at 19:49

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