Question
Let any two real numbers be equivalent if they differ by a dyadic rational. Is any set of representatives of such a Vitali set?
My Musings
I initially thought it was because $\Bbb Z[\frac12]$ is dense in $\Bbb R$. Is that sufficient for the proof, or is more needed?
I think the defintion of a Vitali set is that it's a subset of $\Bbb R$ which is not Lebesgue measurable. Informally, I think because $\Bbb Z[\frac12]$ is dense in $\Bbb R$ that means we can find a set of representatives of $\Bbb R/\Bbb Z[\frac12]$ in any arbitrarily narrow segment of $\Bbb R$ and therefore it's not Lebesgue measureable.
But I'm conscious I'm waving away a lot of "not really understanding the definition of Lebesque Measurable" when I write "and therefore it's not Lebesque measureable".
EDIT
Ok I found a better definition of a Vitali set which I trust more:
A Vitali set is a subset $V$ of the interval $[ 0 , 1 ]$ of real numbers such that, for each real number $r$, there is exactly one number $v ∈ V$ such that $v − r$ is a rational number.
I'm currently trying to work out if this makes $\Bbb R/\Bbb Z[\frac12]$ a Vitali set but it looks doubtful. Tentatively, I think it gives that it can be a Vitali set if chosen appropriately, but it's not necessarily. Although I think the axiom of choice is still required to define it. But I'm trying and failing to come up with an example which is not a Vitali set.
...I guess I need a couple of numbers which differ by say $1/3$, right? And every set of representatives will have a pair differing by $1/3$, correct? So it can never be a Vitali set.
Every set of representatives of $\Bbb R/\Bbb Z[\frac12]$ will contain a pair of elements differing by a rational not in $\Bbb Z[\frac12]$ e.g. $\frac13$ and therefore it cannot be a Vitali set.
However by a theorem of Sierpinski every countable set with no isolated points is homeomorphic to all the others so (tentatively) the set is homeomorphic to a Vitali set.
Please comment / criticise as this is a very tentative answer.