A question about the Vitali set.

Question

Let $X=[0,1)$ and we define on $X$ the relation $\sim$ in the following way:$$x\sim y\quad\iff\quad x-y\in\mathbb{Q}.$$ The relation $\sim$ is an equivalence relation. For all $\alpha\in[0,1)$ the equivalent class is $$E_{\alpha}:=\{y\in[0,1)\;|\;y-\alpha\in\mathbb{Q}\}.$$ The class $E_\alpha$ is countable for all $\alpha\in[0,1)$, indeed if $\alpha\in\mathbb{Q}\cap[0,1)$, $E_\alpha=\mathbb{Q}\cap [0,1)$, while if $\alpha\in\big(\mathbb{R}\setminus\mathbb{Q}\big)\cap [0,1)$, $y\in E_\alpha$ if and only if $y=(\alpha+q)$, where $q\in\mathbb{Q}\cap [0,1).$

Question 1. Is my reasoning correct to show that each class $E_\alpha$ is countable?

Therefore $$[0,1)=\bigcup_{\alpha\in[0,1)} E_\alpha.\tag1$$ From $(1)$ it is clear that exists an uncountable infinity of classes $E_\alpha.$

For the axiom of choice exists $V\subseteq [0,1)$ such that $$V\cap E_\alpha=\{x_\alpha\}\quad\text{for all}\;\alpha\in [0,1).$$ The set $V$ is called the Vitali set. Let $\{q_n\}_{n\in\mathbb{N}}$ an enumeration of $\mathbb{Q}\cap [0,1)$ and we define for all $n\in\mathbb{N}$ $$V_n:=\{x_\alpha+q_n\;\text{mod}\;1\;|\;x_\alpha\in V\}.$$I must prove that $$[0,1)=\bigcup_{n\in\mathbb{N}} V_n.$$ Let $x\in\bigcup_{n} V_n$ then $x\in V_n$ for same $n\in\mathbb{N}$, therefore $x=x_{\alpha}+ q_n\;\text{mod}\;1$, where $x_\alpha\in V$, then $x\in [0,1).$ Let's proceed with the other inclusion: let $x\in [0,1)$ then $x\in E_\alpha$ for same $\alpha\in [0,1)$, in particular, since the classes $E_\alpha$ are disjointed $\alpha$ is unique, then $$x-x_\alpha\in\mathbb{Q},$$ in particular, since $x\in [0,1)$ and $x_\alpha\in [0,1)$ we have that $$x-x_\alpha\in\mathbb{Q}\cap (-1,1)\tag2$$

Question. From $(2)$ how can I proceed to show inclusion?

Thanks!

Answer 1

1. It is not entirely correct. For instance, you claim that$$y\in E_\alpha\iff y=\alpha+q\text{, where }q\in\mathbb{Q}\cap[0,1).$$What if $y<\alpha$? Besides, there is no nedd to treat the cases $\alpha\in\mathbb Q$ and $\alpha\notin\mathbb Q$ separately.
2. Why induction? If $x-x_\alpha=q\in\mathbb{Q}$, then $q-\lfloor q\rfloor=q_n$ for some $n\in\mathbb N$ (since $q-\lfloor q\rfloor\in\mathbb{Q}\cap[0,1)$) and therefore $x\in V_n$.

Answer 2

One way to see countability of all classes is to note that the class of $x\in \mathbb{R}$ is just $[x]=\{x+q: q \in \mathbb{Q}\}\cap [0,1)$: every $x+q \in [x]$ is equivalent to $x$, as $x+q-x = q \in \mathbb{Q}$ and if $x \sim y$ then $q:=y-x \in \mathbb{Q}$ and $y = x+q \in [x]$.

So every class is in bijective correspondence with $\mathbb{Q}$ (via $q \to x+q$ for the class of $x$).

$[0,1)$ is a partition of classes so (assuming choice) there are $\mathfrak{c}=|\mathbb{R}|$ many different classes; if there are $\kappa \ge \aleph_0$ classes each of size $\aleph_0$, $|[0,1)|$ has size $\kappa \cdot \aleph_0=\kappa$ as well. So $\kappa=\mathfrak{c}$.

There is not the Vitali set but only a Vitali set (it depends on a lot of choices and every different choice for a representative of a class gives a different Vitali set).

To see that $[0,1) = \bigcup_n (V+\{q_n\} \pmod{1})$, let $x \in [0,1)$. Then $x$ is in $[x]$ and we have some unique $v \in [x] \cap V$ as a representative. So $x-v= q_k \pmod{1}$ for some $k$ so $x \in V+\{q_k\} \pmod{1}$.