One-to-one mapping from Vitali set to $\mathbb{R}$ The Vitali set $V$ is an elementary example of a set of real numbers that is not Lebesgue measurable. Since $V= \mathbb{R}/\mathbb{Q}$, where $\mathbb{R}$ denotes the set of real numbers and set $\mathbb{Q}$ of rational numbers is countable, $V$ must be uncountable. Can anyone provide an example of 1-to-1 mappings from $V$ to $\mathbb{R}$?
 A: The question seems to be: Assume that there is a Vitali set $V$. Is there an explicit bijection between $V$ and $\mathbb R$?
The answer is yes, by an application of the Cantor-Schröder-Bernstein theorem: there is an explicit injection from $\mathbb R$ into $\mathbb R/\mathbb Q$ (provably in ZF, this requires some thought, or see the answers to this question), and there is an injection of $\mathbb R/\mathbb Q$ into $\mathbb R$, since the former is in bijection with $V$, which is a subset  of $\mathbb R$. 
From these two injections, the Cantor-Schröder-Bernstein theorem explicitly gives you a bijection.
A: It is quite unlikely anyone can describe such a bijection.  There are two problems.  One is that there is not just one Vitali set, even as a subset of $[0,1)$.  You pick one representative from each equivalence class, the set of numbers that are related by adding or subtracting a rational, but there is no description of which one you pick.  You can overcome this by making a bijection between the classes and the reals.  The second problem is that anything created by the axiom of choice cannot be easily described.  We know that defining a Vitali set requires AC because under ZF it is consistent that all sets of reals are measurable.  We know such a bijection exists because of the cardinality argument you allude to, but that does not mean we can exhibit one.
A: There is a slight error in your post. The Vitali set is not $\mathbb R / \mathbb Q$, but some choice of a system of representatives of $\mathbb R / \mathbb Q$. Hence, there is a bijection $V \longrightarrow \mathbb R / \mathbb Q$ taking $x$ to its equivalence class $x + \mathbb Q$. We therefore need only define a bijection between $\mathbb R$ and $\mathbb R / \mathbb Q$. It's a consequence of (in fact equivalent to) the axiom of choice that every vector space has a basis. In fact, any linearly independent subset can be extended to a basis. $\mathbb R$ is a $\mathbb Q$ vector space, so by choice pick some basis $B$ of $\mathbb R$ as a $\mathbb Q$ vector space and say that $1 \in B$. Then every element in $\mathbb R$ can be written uniquely as a sum $\sum_{b \in B} q_b b$ where $q_b \in \mathbb Q$ and all but finitely many $q_b = 0$. Hence, take two reals $\sum q_b b$, $\sum p_b b$. Their difference is rational (i.e. they map to the same element of $\mathbb R / \mathbb Q$) iff $\sum (q_b - p_b) b$ is rational. This occurs iff $q_b = p_b$ for all $b \neq 1$. Hence, the set $B - \{1\}$ maps bijectively to a basis for the $\mathbb Q$ vector space $\mathbb R / \mathbb Q$. Of course, $|B|$ must be infinite, so $|B - \{1\}| = |B|$ and $\mathbb R \cong \mathbb R / \mathbb Q$ as $\mathbb Q$ vector spaces. In particular, they have the same cardinality.
