I am trying to construct a bijection function from vitali set to the real number. As you may know vitali set has been defined as following.
Define an equivalence relation on $\\R$ by $$x\sim y\iff x-y\in{\\Q}.$$ by choosing exactly one element from each class we will get a vitali set. So, define a function from $V$ to $\\R$ as following by
$f(v)=[x]$ where $\\R=\bigcup_{x\in{\\R}}[x]$. It is clear injective and we can define another injective function from $\bigcup_{x\in{\\R}}[x]$ to $V.$Thus, byCantor-Schroder-Bernstein, we will have a bijective function from $V$ to $\\R.$
Is this correct? Any suggestion will be appreciated.
$V\subseteq \mathbb R,$ so clearly we have an injective map one way, so the nontrivial part is to find an injective map $\mathbb R\to V.$ However, constructing explicit maps is rarely the way to go... it's usually much more expedient to rely on a few fundamental theorems about cardinality.
First, notice that we have an explicit bijection between $V\times \mathbb Q$ and $\mathbb R$ given by $(v,q)\to v+q.$ In cardinal arithmetic terms, this means $$ |V|\cdot |\mathbb Q| = |\mathbb R|.$$ So if we have at our disposal the basic rules of cardinal arithmetic, i.e. that $\lambda \cdot \kappa = \max(\lambda,\kappa),$ then we immediately get $|V|=|\mathbb R|.$ (There are some niceties involving axiom of choice here - in fact this cardinality argument would be dead in the water without it - but we already need a good amount of choice to construct the Vitali set, so it's safe to say we can assume it freely.)
The way this cardinal arithmetic statement is usually proved (adapted and slightly simplified for this special case) is as follows: First use AC to put $V$ and $\mathbb Q$ into one-to-one correspondence with the least ordinals they can be put into one-to-one correspondence with (i.e. their cardinals). Call $|V|=\kappa$ and of course $|\mathbb Q| = \omega.$
Then the problem of showing there is an injective function $V\times \mathbb Q\to V$ is reduced to showing there is an injective function $\kappa\times\omega\to \kappa.$ To see this, consider the order type of the dictionary well-ordering on $\kappa\times\omega.$ Assume for induction that the statement is true for all infinite cardinals less than $\kappa$ (there is the standard "diagonal path" construction for the base case $\omega\times \omega\to \omega$). Then for any $\langle\alpha,n\rangle \in \kappa\times \omega,$ with $\alpha$ infinite, we have an injection $\alpha\times \omega\to \alpha$ so that $|\alpha\times \omega |\le |\alpha| < \kappa.$ This guarantees that since the initial segment less than $\langle \alpha,n\rangle$ is a subset of $\alpha \times \omega,$ it has order type less than $\kappa.$ Thus the order type of $\kappa\times \omega$ is $\le \kappa,$ and the well ordering gives an injective function $\kappa\times\omega\to \kappa.$