Does Vitali set imply the axiom of choice

Question

I know that the construction of Vitali set needs the axiom of choice, but this only states that $AC \implies V$. Is it also true that $V \implies AC$?

If $\neg AC \implies \neg V$, then what contradiction results from $\neg AC \wedge V$?

Answer 1

Not even remotely.

First of all if $\Bbb R$ can be well-ordered then we can easily dispense with the need for the axiom of choice in order to prove the existence of a Vitali set.

And of course, we can destroy pretty much all form of choice while keeping the continuum well-ordered (except the obvious things which can be stated as "such and such happens below the continuum").

But this is not just the thing. It is consistent that there are Vitali sets when even the axiom of countable choice fails, and the continuum cannot be well-ordered.

In Cohen's first model, there exists a set $D\subseteq\Bbb R$ which is infinite, but has no countably infinite subset. So the axiom of choice fails, and $\Bbb R$ cannot be well-ordered. But in that model, if $F$ is a family of well-orderable sets, then $F$ admits a choice function. One can observe that a Vitali set is a choice function from equivalence classes which are all countable. Therefore in Cohen's first model we can prove the existence of Vitali sets, yet the axiom of countable choice fails badly below the continuum.