Well-Ordering Theorem without Axiom of Regularity

Question

Personally, I am not fond of the Axiom of Regularity. Some alternative models in set theory use the negation of the Axiom of Regularity as an axiom (non well-founded theories). I am curious if the Well-Ordering Theorem is provable in this theories — I greatly suspect that the answer is no but would appreciate any kind of insight in this matter.

Answer 1

There is no actual appeal to the axiom of regularity in the usual proof of the well-ordering theorem, assuming the axiom of choice.

Instead we invoke the power set and replacement axioms in quite a strong way. The transfinite recursion is not applied on $\in$, but rather on the ordinals. And even without the axiom of regularity (at least in the presence of replacement), we have the same ordinal structure.

Of course, if you reject the axiom of choice as well, then this is no longer true. If you reject replacement, then things become slightly trickier, but we can use separation instead. If you reject that as well, then perhaps you should list the axioms you allow.

It should be noted that Zermelo in his initial set theory didn't have neither replacement nor regularity. But it sufficed to prove the well-ordering theorem just the same.

(Something that might be proved useful is A.R.D Mathias' paper "The Strength of the Mac Lane Set Theory" in which there is a table of set theories, their assumptions and some consequences, there he points out that the equivalence of the axiom of choice and the well-ordering theorem is provable in a weak system he denotes as $\sf M_0$.)