Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


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.