Does descending chain imply well ordering without axiom of choice?

Question

I can prove no descending chain implies well ordering. Take any subset $S$, assume it has no least element, then we can construct a function $f: S \to S$ such that $f(x) < x$, then use recursion theorem to prove there is a descending chain.

However the construction of $f$ requires axiom of choice. Is this needed? Or is there proof without AC?

@bof You're absolutely right - and if you turn this comment into an answer, I'll upvote it and delete my own (which I've made CW since I'm basically just copying you). — Noah Schweber, Dec 26 '20 at 23:59
You're right. See Herrlich, Axiom of Choice, $\S 4.3$ and the references therein. — Maurizio Barbato, Sep 25 '23 at 13:13

score 3 · Answer 1 · answered Dec 26 '20 at 23:59

3

To remove this from the unanswered queue:

Bof's comment is exactly right. In Cohen's original model of $\mathsf{ZF+\neg AC}$, we have a set of reals $X$ (namely the set of Cohen reals added to the model) which is not well-ordered but has no infinite descending sequence.

answered Dec 26 '20 at 23:59

Noah Schweber

245,398

Would you mind pointing me to a reference for this? Thanks. – subobject_classifier Apr 17 '21 at 08:06

Does descending chain imply well ordering without axiom of choice?

1 Answers1