Working in $\mathbf{ZF}$ let $R$ be a proper class of ordered pairs that is well-founded. This means that for every non-empty set $a$ there is a set $b\in a$ such that $cRb\implies c\notin a$. Here $cRb$ is a notation for $\langle c,b\rangle\in R$ and $b$ is a so-called $R$-minimal element of $a$. If $R$ is local (i.e. collections $\{x\mid xRb\}$ are all sets) then it can be shown that also non-empty proper classes have $R$-minimal elements.

I encountered a proof that made the condition of being local redundant. It made use of an operation on classes that adds to each class a set that is contained in it (Bottom-operation) but the definition of this operation relied on the regularity axiom.

My question:

Is there a proof that every non-empty class has an $R$-minimal element that does not make use of the regularity axiom?

Another formulation:

If $R$ is a class of ordered pairs that is well-founded, then is it legal to apply $R$-induction if all axioms of $\mathbf{ZF}$ are accepted with exception of the axiom of regularity?

  • $\begingroup$ So the question is really in $\sf ZF-Reg$, without assuming the locality. $\endgroup$ – Asaf Karagila Jun 20 '15 at 18:14
  • $\begingroup$ @AsafKaragila Yes. If there is locality then it can be proved without REG. That proof makes use of $\overline R$ (the transitive closure wich is also local in that case). $\endgroup$ – drhab Jun 20 '15 at 18:23
  • $\begingroup$ Now reposted at mathoverflow.net/q/214775 $\endgroup$ – Emil Jeřábek Aug 20 '15 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.