Bernstein sets, Well-Ordering theorem vs Axiom of Choice

Question

In the construction of Bernstein sets (see here), is it necessary to use the well-ordering theorem? Why can't you just use the Axiom of Choice to pick two points?

Answer 1

Yes, in the proof you link to, the well-ordering of $\mathbb R$ can be replaced by a choice function on the set of all infinite (or even continuum-sized) subsets of $\mathbb R$. It's essentially just a stylistic choice by the author which one to use. Certainly it doesn't have any deep significance, since the well-ordering theorem and the axiom of choice imply each other.

On the other hand it seems to be essential to the proof that the $F_\alpha$s are well-ordered according to the initial ordinal of size continuum (thus in particular on the fact that the continuum can be well-ordered), which is what guarantees that the subsets we choose from are always nonempty.

Answer 2

The well-ordering theorem just gives us a "reasonable" method of choosing the elements in the construction. If you have a choice function on the subsets of $\Bbb R$ then it suffices.

However one can show the following theorem holds in $\sf ZF$.

$X$ can be well-ordered if and only if there exists a choice function on $\mathcal P(X)\setminus\{\varnothing\}$.

It should be remarked that the use of the axiom of choice/well-ordering is essential, and there are models where the axiom of choice fails and the real numbers cannot be well-ordered, and there are no Bernstein sets.