Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

When I was first learning about the axiom of choice, a very helpful and fun exercise was to try to find a rule that maps every nonempty subset of $\mathbb R$ to one of its members. No matter how hard you try, it was impossible to find one. (If you try the same for the nonempty subsets of $\mathbb N$, it is pretty easy, just take the smallest member of each set.)

My first question is, has it been proved in some formal way that such a choice function cannot be completely described with a finite number of symbols? (I think this is equivalent to the undecidability of the existence of a choice function on $\mathcal P (\mathbb R) \setminus \{ \emptyset \}$ in ZF.)

Another question is, in ZF with the negation of AC, is it provable that there exists no choice function on $\mathcal P (\mathbb R) \setminus \{ \emptyset \}$ ? (My guess is negative.)

share|improve this question
1  
There might be a thread about this somewhere on MO. –  Qiaochu Yuan Aug 22 '10 at 11:40
    
@Qiaochu: I skimmed through the questions with the AC tag, but I couldn't find it. –  AgCl Aug 22 '10 at 11:54
1  
I guess you mean "in ZF" instead of "in ZF with the negation of AC". (An in ZF you cannot prove such a function does not exist, assuming the consistency of ZF+AC, right?) –  Mariano Suárez-Alvarez Aug 22 '10 at 13:09
    
@Mariano: Your statement in parenthesis perfectly makes sense to me, i.e. I can see that it can't be proven in ZF. But if one works in ZF+neg(AC), can anything be proved about it? –  AgCl Aug 22 '10 at 13:34
    
What is neg(AC)? –  Mariano Suárez-Alvarez Aug 22 '10 at 13:51

2 Answers 2

up vote 3 down vote accepted

If there is a definable wellordering of $\mathbb{R}$, then every real number is ordinal definable. Conversely, if every real number is ordinal definable, then there is a definable wellordering of $\mathbb{R}$. Thus the assumption that there is a definable wellordering of $\mathbb{R}$ is equivalent to $\mathbb{R} \subseteq \mathrm{OD}$.

This assumption is completely independent of AC in the sense that all four combinations AC + $\mathbb{R} \subseteq \mathrm{OD}$, AC + $\mathbb{R} \not\subseteq \mathrm{OD}$, ¬AC + $\mathbb{R} \subseteq \mathrm{OD}$, ¬AC + $\mathbb{R} \not\subseteq \mathrm{OD}$ are relatively consistent with ZF! The details of this are somewhat involved, see Thomas J. Jech's The Axiom of Choice for an introduction to the various methods to prove independence from ZF.

share|improve this answer
    
I was planning on buying this book, now I know I will! Thanks! –  Asaf Karagila Aug 23 '10 at 23:15
    
Thanks! That's very interesting. If I correctly understand, that means, the answer to my first question is positive, and to my second question is negative. The statement that R does not have a definable well-ordering is not provable in all of ZF, ZF + AC and ZF + ¬AC, and this implies that it is true. (Because if it was false, it would obviously be provable.) So this constitutes a proof (in a possibly larger system than all mentioned) that nobody can come up with such an ordering/choice function. –  AgCl Aug 23 '10 at 23:46

In Kunen's book on set theory, he leads one through an exercise where one creates a model of ZF (assuming ZF is consistent) in which the power set of the natural numbers is not well orderable. Since one can prove that the power set of the naturals is in bijective correspondence with the reals, in this model, the real numbers cannot be well orderable.

(I don't have Kunen at home with me now, and it's been a year or two since I even glanced at the relevant section. If I remember, I'll update the reference next time I have a copy of the book nearby).

However, there can (I think, but I'm not positive) be models where choice fails and the reals are well orderable. Note that choice refers to ALL sets, so one simply needs to rig things to fail at REALLY huge cardinal sizes but have choice work at lower levels.

Finally, there are models where choice functions are definable: Godel's universe $L$. For example, in $L$, EVERY set is definable. That is, given a set $X$, there is a first order formula (with parameters) $\phi$ such that $x\in X$ iff $\phi(x)$. Now, a function from $X$ to $Y$ (including a choice function), is nothing but a special subset of $X\times Y$, and hence will be definable in $L$.

I learned about this last paragraph from assorted posting by Joel David Hamkins over at MO. For example, see http://mathoverflow.net/questions/23393/set-theory-and-vl

edit I realize that this doesn't seem to address your question directly. However, it's true that for any set $X$, the powerset of $X$ has a choice function iff $X$ can be well ordered. See Levy's book "Basic Set Theory" on page 160 for details.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.