Here is the full thread of comments for http://mathoverflow.net/questions/150459:

• 8 The definition, and therefore the question, makes no sense. ZF can prove only formulas, and sets of reals are not formulas. You could define that a formula $\phi(x)$ is a "definition of a choice-free well-orderable set" if ZF proves "there is a well ordering on ${x \in \mathbb{R} : \phi(x)}$", but this will not help, as the definition of $A$ and $B$ then still make no sense. In the metatheory, there are only countably many formulas, so the set of all definitions of c.f.w.o.s. is trivially countable, whereas from within the theory, you cannot express the property of being definable by ... Emil Jeřábek 1d ago
• 1 ... a first-order formula, hence the informal collection of sets defined by a formula that is c.f.w.o. is not a set, and as such you cannot speak about its cardinality. Even the collection of all definable sets in a model is not preserved by elementary equivalence, and it models where it happens to be an internal set after all, it may well be anything from countable to the full powerset of the reals. Emil Jeřábek 1d ago
• 1 And of course, one set in a model can be definable by two different formulas, one of which may be c.f.w.o., and the other one not. In fact, every definable set has a definition that is not c.f.w.o. Emil Jeřábek 1d ago
• What do you mean "find"? I'm getting confused by your edit, because sets are semantical objects for set theory. This means that in a given universe some sets of reals will be well-orderable. You don't use the axiom of choice to "find" these well-orders, they exist. Is you are talking about definable subsets that's a whole other thing. I think that the right question, and indeed this is what I interpreted from the question originally, is asking for the set $A={X \subseteq \mathbb{R} | X \text{\ can be well-ordered}}$ and asking what can we prove about the cardinality of $A$ in ZF, [cont.] Asaf Karagila 1d ago
• @EmilJeřábek: How can one define the "clear" notion of a choice free well-orderable set? Is Asaf's answer meaningless too? If not, is it answer of a question different from my question? If yes, what is that question? Saint Georg 1d ago
• 1 My answer, was, it seems (and I agree with Emil) to a slightly different question. About what is provably true about ${X \subseteq \mathbb{R} | X \text{\ can be w.o.}}$, and about its complement. Note, to your edit, that sets are the semantical objects in set theory. If $A$ is a set of reals in a model $M$ either it can or cannot be well-ordered, and the axiom of choice says nothing about it. If we want to ask whether or not every definable (with real parameters?) set of real numbers can be well-ordered, that's another question (whose answer is similar to mine), which admits consistency results. Asaf Karagila 1d ago
• 12 Saint Georg: The appropriate way to react to Emil's criticism is to take some time to digest it, not trying to argue that your question makes sense. Over the past two days or so, you have asked 5 questions and at least 3 had deep flaws demonstrating lack of research. Perhaps you should ask questions on Math.StackExchange until you reach the point that your questions meet the standards expected by the MathOverflow community. François G. Dorais♦ 1d ago 9h ago
• @FrançoisG.Dorais: Was my comment "trying to argue that my question makes sense"? I simply asked about the probable problem because in the first view it seems that Asaf had no problem to understand my question and I had no problem to understand his answer. I asked them (Asaf & Emil) to illustrate the problem more and I received useful explanations. Saint Georg 1d ago