0
$\begingroup$

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?

  • 2
    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.2012-03-18
  • 6
    I don't understand your last sentence at all.2012-03-18
  • 3
    A Vitali set is a set, not a question, so I guess the answer has to be no.2012-03-18

1 Answers 1