Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $r \in \mathbb{Q}$.

share|improve this question
2  
Am I missing something? No proper superset of a selector is itself a selector. –  Emil Jeřábek Jun 13 '11 at 12:38
    
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... –  George Lazou Jun 13 '11 at 14:13
    
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. –  Emil Jeřábek Jun 13 '11 at 18:14
    
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... –  George Lazou Jun 14 '11 at 12:17
    
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. –  Emil Jeřábek Jun 14 '11 at 13:59

1 Answer 1

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 $\mathbb Q$.

share|improve this answer
    
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? –  George Lazou Jun 13 '11 at 19:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.