Based on this answer. Working in ZF-Reg., is it possible to have a non-well founded set that is strictly supernumerous to any well founded set?

If that is possible, then can this be extended indefinitely? That is lets consider the well founded realm as the hierarchy whose base (the first stage) is the empty set, and for the pusposes of the development I'm intending here lets phrase the base set of that hierarchy (which is the empty set) as the set of all sets that are hereditarily strictly subnumerous to the empty set, now can we have a hierarchy whose base is the set of all herediarily strictly subnumerous to 2, sets, such that this base set is strictly supernumous to every set in the first hierarchy (whose base is empty), and that in turn every set in that hierarchy itself is strictly subnumerous to the set of all hereditarily strictly subnumerous to 4, sets, and so on... That is, for every hierarchy whose base is the set of all hereditarily strictly subnumerous to $x$ sets, then all sets in that hierarchy are strictly subnumerous to the set of all hereditarily strictly subnumerous to $y$ sets, for any set $y$ that is strictly supernumous to a set $k$ that is strictly subnumreous to $x$

Of course the idea is that each base set (except the emptyset ) is non-well founded, it can contain sets having cyclic membership, and that what cause them to have their cardinality unleashed in such a manner.

Can we have such a model of ZF-Reg.??

Of course we need to add to ZF-Reg. the axiom that for every set $x$ there is a set of all sets hereditarily strictly subnumerous to $x$. (which is a theorem of ZF (Holmes)).


1 Answer 1


No. Hartogs' theorem does not depend on the axiom of regularity. Once you have Replacement, and every well-ordered set is isomorphic to a von Neumann ordinal, which is a well-founded set, you cannot have that.

If you remove Replacement, sure. Start with a set of atoms, $A$, of size $V_{\omega+\omega+1}$, then cut off the universe at $V_{\omega+\omega}(A)$.

  • $\begingroup$ Why we don't take the size of $A$ to be $V_{\omega+\omega}$? $\endgroup$ Commented Apr 11 at 17:01
  • $\begingroup$ It's not well founded... I don't understand your question. $\endgroup$
    – Asaf Karagila
    Commented Apr 12 at 15:54
  • $\begingroup$ I meant of size $V_{\omega+\omega}$, sorry for the typo. $\endgroup$ Commented Apr 12 at 16:30
  • $\begingroup$ If we started with $A$ strictly larger, how would that come to be? $\endgroup$
    – Asaf Karagila
    Commented Apr 12 at 16:31
  • 1
    $\begingroup$ Oh, I see what you're asking for now. Sure, that can also work. But I was aiming for correct, not optimal. $\endgroup$
    – Asaf Karagila
    Commented Apr 12 at 16:35

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