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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 Lebesgue Measurable" when I write "and therefore it's not Lebesgue 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.

KCd
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Robert Frost
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  • Note $\mathbb Z[\frac12]$ is countable, so it is Lebesgue measurable and therefore not a Vitali set. On the other hand, $\Bbb R/\big(\Bbb Z[\frac12]\big)$ is not a subset of $\mathbb R$, so it is not a Vitali set. – GEdgar Sep 14 '23 at 09:20
  • I know of Lebesgue. But you talk about this other guy, Lebesque ... :) – GEdgar Sep 14 '23 at 09:22
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    I'm unaware of a convention that calls Lebesgue non-measurable sets "Vitali sets". As far as I know, only the ones that intersect each class of $\Bbb R/\Bbb Q$ in exactly one point are called Vitali sets. – Sassatelli Giulio Sep 14 '23 at 09:25
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    @GEdgar I'm guessing the OP means a set of representatives of $\mathbb{R}/\left(\mathbb{Z}\left[\frac{1}{2}\right]\right)$ chosen in $[0,1]$ for example. – Joel Cohen Sep 14 '23 at 09:30
  • @GEdgar fixed, I believe. – Robert Frost Sep 14 '23 at 11:40
  • Related: https://math.stackexchange.com/questions/4512874/ – Robert Frost Sep 14 '23 at 12:22

1 Answers1

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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.

Robert Frost
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