Scott's trick without the Axiom of Regularity

Question

Scott's trick is a method for constructing a set as a subset of a proper class. Specifically, let $A$ be a class (proper or otherwise), and let $$S(A)=\{x\in A:\forall y\in A,\operatorname{rank}(x)\le\operatorname{rank}(y)\}.$$

Then $S:\overline V\to V$ (where $\overline V$ represents the collection of all classes, a concept that cannot be formalized in ZF) satisfies three important properties: (1) $S(A)\subseteq A$, (2) $S(A)$ is a set, even if $A$ is not, and (3) $S(A)=\emptyset$ iff $A=\emptyset$.

I would like to generalize this construction to produce another function on $\overline V$ which satisfies the same properties, but does not require the Axiom of Regularity. The place where Regularity appears in the standard Scott's trick is in the proof of property (2), because assuming the negation of Regularity, there is a set $z$ with no rank, and a proper class of sets built from this set (say $z, \{z\}, \{\{z\}\}, \dots$ and continued by transfinite recursion), that also have no rank. Under the conventional definition of rank, $\operatorname{rank}(x)=\emptyset$ whenever $x$ has no rank, so if I take $A$ to be the class of sets containing $z$ described above, then all members of $A$ have the same rank, and thus $S(A)=A$ is also a proper class.

My question is if there is any way to construct a similar type of function that works even in such cases. Is it necessary to have a total order on all sets like $\operatorname{rank}$ in order for any such trickery to work? If so, I would guess that there is no way to do it without Regularity.

Answer 1

Some assumption is needed: It is consistent with $\mathsf{ZFA}$ that there is a proper class of atoms, and there is no class function $C:V\to V$ with the property that $C(X)=C(Y)$ iff $|X|=|Y|$. This is due independently to Gauntt and to Lévy. Jech's The axiom of choice gives references.

On the other hand, a positive answer is possible in some situations: Assume $\mathsf{ZF}^-+\mathsf{AFA}$, that is, $\mathsf{ZF}$ with foundation replaced by the antifoundation axiom due independently to Aczel and Forti-Honsell. One can prove that any model of this theory is completely determined by its well-founded part $\mathsf{WF}$, in the sense that if $(M,\in^M)$ and $(N,\in^N)$ are models of $\mathsf{ZF}^-+\mathsf{AFA}$ and $\mathsf{WF}^M\cong\mathsf{WF}^N$, then $M\cong N$.

Let me review what $\mathsf{AFA}$ says precisely. Recall that a graph is a pair ${\mathcal G}=(G,E)$ where $G$ is a set and $E$ is a binary relation on $G$.

A decoration of ${\mathcal G}$ is an assignment $\pi$ of a set to each element of $G$ in such a way that for every $x\in G$, $\pi(x)=\{\pi(y)\colon y \mathrel E x\}$.

An apg (accessible pointed graph) is a graph $(G,E)$ with a distinguished point $p\in G$ such that every element of $G$ is accessible form $p$, (i.e., there is a finite sequence $p,p_1,\dots,p_n,q$ such that $q\mathrel E p_n\mathrel E\dots\mathrel E p_1 \mathrel E p$).

A picture of a set $x$ is an apg ${\mathcal G}$ with distinguished node $p$ admitting a decoration $\pi$ such that $\pi(p)=x$.

Working in $\mathsf{ZF}^-$, one can prove that every set has a picture. Mostowski's collapsing theorem states that every well-founded graph has a unique decoration. This easily implies that every well-founded apg is a picture of a unique set.

$\mathsf{AFA}$ is the statement that every graph has a unique decoration.

It follows from $\mathsf{AFA}$ that every apg is a picture of a unique set, and that non-well-founded sets exist. Moreover, every apg is equivalent to an apg on a well-founded set, where two apgs are equivalent iff they are pictures of the same set.

For references on antifoundation, and proofs of its consistency and the above claims, see

Peter Aczel. Non-well-founded sets. CSLI Lecture Notes 14, Stanford University, Center for the Study of Language and Information, Stanford, CA, 1988. MR0940014 (89j:03039),

and

Marco Forti, and Furio Honsell. Axioms of choice and free constructions principles, I. Bull. Soc. Math. Belg. Vol 36 (b) fasc. 1 ser. b (1984), 69-79. MR0739920 (85f:03054).

Since every apg is equivalent to a well-founded apg, we can make do in $\mathsf{ZF}^-+\mathsf{AFA}$ and indirectly implement Scott's trick: Given any class $C$ consider the class $C_{\mathsf{WF}}$ of well-founded apgs that are pictures of sets in $C$. If $C$ is non-empty, select a subset of $C_{\mathsf{WF}}$ using Scott's trick. The collection of corresponding sets is a subset of the class $C$.