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I am trying to get a better handle on the nature of Vitali sets, generated by choosing representatives in $[0,1)$ from the equivalence classes $\mathbb{Q} + r$ where $r \in \mathbb{R}$.

If $V$ is a Vitali set described above, I understand that $V$ contains only a single rational number and that it is uncountable. Also, its complement $[0,1) \sim V$ is uncountable since we have excluded from $V$ a collection of elements in $[0,1)$ associated to each of the uncountable number of elements in $V$.

However, I have been unable to find (or determine) if $V$ is dense in [0,1). Is this known?

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    There's more than one Vitali set (one for every choice of representatives of the equivalence classes); it would be better to refer to it as *a* Vitali set.2011-04-11
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    I didn't realize that the multiplicity of "Vitali sets" came from the choice of representatives; I had imagined this wording was due to the existence of other constructions that are Vitali-like. I'll modify the verbiage in the question to reflect this.2011-04-11

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Note that you get one Vitali set for each choice of representatives of the equivalence classes, so in fact there are many Vitali sets, not just one.

A Vitali set need not be dense in $[0,1)$. For example, instead of picking representatives that are in $[0,1)$, you can pick representatives that are in $[0,q)$ for any rational $q\gt 0$; in particular, you can make sure that your Vitali set is constrained to as small a part of $[0,1)$ as you care to specify (and you can translate the set by adding a constant rational to it, as well).

To see this, it suffices to show that for every real number $r$, there is a real number $s\in [0,q)$ such that $r-s\in\mathbb{Q}$. But this is easy: pick a rational $t\in [0,q)$. By the Archimedean property, there exists $N\geq 0$ such that $Nt\leq r\lt (N+1)t$. In particular, $0\leq r-Nt\lt t$, so letting $s=r-Nt$ gives the desired real number.

Since every real is equivalent to some real in $[0,q)$, you can always pick the class representatives to be in $[0,q)$ (instead of $[0,1)$). So you can ensure that you have a Vitali set is contained in $[0,q)$.

Similarly, if you select any interval $(a,b)$ contained in $[0,1)$ you can find a Vitali set that is contained in $(a,b)$, and in particular whose closure is not $[0,1]$ if $0\lt a\lt b\lt 1$.

In fact, you can find a Vitali set that has very small outer measure (any positive outer measure greater than $0$ and smaller than $1$ that you care to specify ahead of time): see for example JDH's answer to this question.

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    In fact, it is also possible to find a Vitali set of outer measure 1. See e.g. my sci.math posting from January 1997 on the subject "Vitali nonmeasurable" http://groups.google.com/group/sci.math/msg/bbf375f519887c232011-04-11
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    JDH's answer to the linked question handles the case of outer measure 1, too, and in that case it is dense. My answer to the linked question also links to Robert Israel's 1997 post.2011-04-12
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    On the other hand, it may be worth noting that thanks to the Baire category theorem, a Vitali set cannot be *nowhere* dense, nor even meager.2012-03-04