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choose $r(i)$ such that it is irrational and from $[0,1]$.

$r_1 - r_2 = q\in\mathbb{Q}$ implies its in an equivalence class.

Seems the equivalence class for $r_1$ has countably infinite members.

choose $r_3$ such that $r_1 < r_3 < r_2$ and $r_3 - r_1$ does not equal some $q\in \mathbb{Q}$. The equivalence class for $r_3$ then seems to have countably infinite irrational numbers as members.

This continues until finally exhausting all $r$ such that $r_1 < r < r_2$. Supposedly, there is now an uncountable number of equivalence classes, each with countably infinite members.

Is a Vitali set then a question of what is the measure of $r_2 - r_1$ for real numbers that are arbitrarily close to each other?

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To answer the question in your title, yes since the equivalence class containing an element $r$ is $E_r = \{r+q: q \in \mathbb{Q}\}$. Hence, $E_r$ is a countable set. – user17762 Mar 18 '12 at 18:59
I don't understand your last sentence at all. – Nate Eldredge Mar 18 '12 at 19:11
A Vitali set is a set, not a question, so I guess the answer has to be no. – Robert Israel Mar 18 '12 at 19:29

To recap:

We define an equivalence relation on $[0,1]$ as follows: $r\sim s$ if and only if $r-s\in\mathbb{Q}$.

It is easy to verify that this is indeed an equivalence relation, and therefore partitions $[0,1]$ into equivalence classes.

Your first question is: how many elements does an equivalence class under this equivalence relation have?

The answer is indeed "denumerably many" (countably infinite, $\aleph_0$).

One way to see this is to note that for each $r\in[0,1]$, the equivalence class of $r$, $[r]$, satisfies: $${} [r] = \{s\in[0,1]\mid s\sim r\} \subsetneq \{r+q\mid q\in\mathbb{Q}\}.$$ To see the inclusion, note that if $s\sim r$, then $s-r\in\mathbb{Q}$; hence there exists $q\in\mathbb{Q}$ such that $s-r=q$, so $s=r+q$. Thus, every element in $[r]$ is in $\{r+q\mid q\in\mathbb{Q}\}$. Of course, the inclusion is proper, since for example $r+2\in \{r+q\mid q\in\mathbb{Q}\}$, but $r+2\notin[r]$ (since $r+2\notin [0,1]$). Therefore, $$\Bigl|[r]\Bigr|\leq \Bigl|\{r+q\mid q\in\mathbb{Q}\}\Bigr| = |\mathbb{Q}|=\aleph_0.$$ On the other hand, if $r\lt 1$, then let $\epsilon=1-r\gt 0$; by the Archimedean property, there exists $N\in\mathbb{N}$ such that $\frac{1}{N}\lt\epsilon$. Thus, for every $n\geq N$, $\frac{1}{n}\lt \epsilon$, so $r+\frac{1}{n}\in [0,1]$; thus, for every $n\geq N$, $r+\frac{1}{n}\in[r]$, so $$\aleph_0 = \Bigl|\{n\in\mathbb{N}\mid n\geq N\}\Bigr| = \Bigl|\{r+\frac{1}{n}\mid n\geq N\}\Bigr| \leq \Bigl|[r]\Bigr|\leq\aleph_0.$$ So the cardinality of $[r]$ is exaclty $\aleph_0$, as claimed.

Next: you can perform the process you describe: given $r_1\in [0,1]$, there must exist some $r_2\notin[r_1]$ (that is, $r_1$ is not related to $r_2$); without loss of generality say $r_1\lt r_2$. Since $[r_1]\cup[r_2]$ is countable, but $(r_1,r_2)$ is uncountable, there exists $r_3$, $r_1\lt r_3\lt r_2$ such that $r_3\notin[r_1]\cup[r_2]$ (that is, $r_3$ is not related to either $r_1$ or to $r_2$.

You can set up an induction and get a sequence $\{r_n\}$ with $n\in\mathbb{N}$, with, say, $r_1\lt \cdots \lt r_{n+1}\lt r_{n}\lt\cdots\lt r_2$.

But it is not clear what to do at this point. It's entirely possible that the limit of $r_n$ is $r_1$, so we may not be able to continue this transfinitely.

If you don't really care about the order, then (necessarily assuming at least a part of the Axiom of Choice, if not the whole thing) we can find an injection from an ordinal $\beta$ (necessarily uncountable) and the numbers in $(r_1,r_2)$, with the property that for every $x\in (r_1,r_2)$ there exists one and only one $\alpha\in\beta$ such that $[x] = [r_{\alpha}]$ (that is, an "ordinal list" of the equivalence classes in $(r_1,r_2)$. In fact, you can show that every real in $[0,1]$ is equivalent to some real in $[r_1,r_2]$, so this gives you an "ordinal list" of a set of representatives of the equivalence classes. So that $\{r_{\alpha}\}_{\alpha\in\beta}$ is a Vitali set in $[0,1]$.

A Vitali set is just a complete set of representatives of the equivalence relation. Using arguments such as the above, one can show that you can find a Vitali set that is contained in a subinterval of $[0,1]$ that is as small as you want (different sets may be required for different specifications of how small you want it).

But a Vitali set is not a question; a Vitali set is a set to which you cannot meaningfully assign any measure if you want the measure to be invariant under translation, $\sigma$-additive, and to assign to every interval $[a,b]$ the measure $b-a$.

Note that even in your construction, $r_1$ and $r_2$ are not "arbitrarily close": they are a fixed distance from each other, though you can begin by specifying any particular $\epsilon$, and you can find a Vitali set constructed along your lines in which $r_1$ and $r_2$ will be less than $\epsilon$ apart; but different $\epsilon$s will give you possibly different $r_1$ and $r_2$; two real numbers are never "arbitrarily close to one another", because if $x$ and $y$ are reals, then $|y-x|$ is a specific, fixed, number. That's exactly how close to one another they are.

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