I am a bit confused about the cardinality of the Vitali sets.
Just a quick background on what I gather about their construction so far:
We divide the real interval $[0,1]$ into an uncountable number of disjoint classes in such a way that two numbers $x$ and $y$ are in the same class if their difference x-y is a rational number. Therefore the equivalence relation on the real numbers would be: $x\sim y\iff x-y\in\mathbb Q$.
The cardinality of our classes is equivalent to the cardinality of $\mathbb Q$, so the classes have countably infinite elements.
How many of these classes are there? As each only has countably infinite elements, but each element of $[0,1]$ lies in one class, there are uncountably infinite classes.
Now, using the axiom of choice, we take one element out of each class and put it in the set V. This is a Vitali set.
Hopefully this is all good at this point.
So each Vitali set should have uncountably many elements (because we have uncountably many classes), but there should be countably many Vitali sets (because each class has countably many elements) correct?
But then why does Wikipedia state that "There are uncountably many Vitali sets…"?
Best Answer
You have $\mathfrak c=2^{\aleph_0}$ classes, each of them is of the size $\aleph_0$.
You get the set $V$ by choosing one element from each class. How many possibilities are for such choice? Precisely $$\aleph_0^{\mathfrak c}=2^{\mathfrak c}$$ possibilities. (Size of each class raised to the number of classes - you are basically counting the maps from the set of classes, for each class you have countably many possibilities.)
This cardinal number is definitely uncountable: $2^{\mathfrak c}>\mathfrak c>\aleph_0$.
The above cardinal equality holds since $$2^{\mathfrak c} \le \aleph_0^{\mathfrak c} \le \mathfrak c^{\mathfrak c} = (2^{\aleph_0})^{\mathfrak c} = 2^{\aleph_0\cdot\mathfrak c} = 2^{\mathfrak c},$$ so by Cantor-Bernstein Theorem all cardinal numbers in this chain of inequalities are the same.