Can cardinality be defined with essentially no practical restriction on non-well-ordered combinatorics or ill-foundedness of sets?

Question

Question: Can we have a model of $ZF-\text {Regularity}$ where there exist an ordinal $\kappa$ such that $H_{\kappa}$ exists and $H_{\kappa}$ is not equinumerous to any well founded set?

The motivation for this question comes in connection with defining Cardinality under some situations beyond Regularity and Choice. Especially in connection with the following anti-foundation axiom:

$\text{Anti-Foundation axiom: }$ Every set is subnumerouse to some iterative power of some $H_\kappa$ set, where $\kappa$ is an ordinal.

where $H_\kappa = \{x| x \text{ is hereditarily subumerous to } \kappa\}$

Where $x \text{ is hereditarily subnumerous to } \kappa $ is defined as: $$ \forall y \in TC(\{x\} ) \exists f (f: y \rightarrowtail \kappa)$$

Where $TC$ stands for the "transitive closure function" defined in the usual manner.

Iterative powers $P^i(x)$ are defined recursively as:

$P^0(x) = x$

$P^j(x) = \bigcup (\{P (P^i(x))| i < j\})$

Using this as an anti-foundation axiom would enable us to define a notion for Cardinality that covers more sets than does Scott's definition of Cardinality.

$\text{Define:} $ Card(x) is the set of all subsets of the first supernumerous to $x$ iterative power of the nearest $H_{\kappa}$ set to $x$, that are equinumerous to $x$.

The distance of $x$ from $H_{\kappa}$ is the minimal ordinal $i$ such that $P^i(H_{\kappa})$ supernumerous to $x$.

Of all $H_{\kappa}$ sets that lie at the least distance from $x$, the one with the least $\kappa$ value is the "Nearest $H_{\kappa}$ set to $x$".

The idea is that there is no combinatorial restriction on what constitutes the cardinality of an $H_{\kappa}$ set, so this definition can work without having a practical kind of restriction over the non-well ordered, non-well founded realm. While Scott's definition can only define cardinality for sets as long as those are equinumerous to some well founded set. Which is in some sense restrictive outside of choice.

Answer 1

It depends on what "well-founded set" means.

The most natural interpretation, in my opinion, is: $x$ is well-founded iff the transitive closure of $x$ is well-founded with respect to $\in$. Note that this means exactly that $x$ is a pure set! If this is what we mean, the answer to your question is yes: we can have a model of ZF-Reg in which there are only countably many pure sets but uncountably many Quine atoms (= sets satisfying $a=\{a\}$).

Why does such a model exist? Roughly, we can build the $L$-hierarchy on any "starting set" of Quine atoms $A$ - let $L^A_0=A$, $L^A_\lambda=\bigcup_{\alpha<\lambda} L^A_\alpha$ for $\lambda$ limit, and $L^A_{\alpha+1}=L^A_\alpha\cup\mathcal{P}_{def}(L^A_\alpha)$. Letting $W_A$ be the $L$-hierarchy on $A$ continued until we get a model of ZF-Reg, we can show that the pure part of $W_A$ is countable regardless of whether $A$ is; now simply take an uncountable set of Quine atoms.

On the other hand, if by "well-founded set" you simply mean a set which is well-founded under $\in$, then the answer is definitely no: letting $x$ be a set of cardinality $>\kappa$ which is disjoint from every element of $H_\kappa$, we have that $H_\kappa$ is in bijection with $J:=\{h\cup x: h\in H_\kappa\}$, but $J$ is clearly well-founded in this weaker sense.