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The posting here aims at defining a relative notion of well foundedness and thus of ranks, in relation to a base set, then tries to use Scott's trick to define a relative notion of cardinality to that base set as well.

Let's at first work in ZF-Reg. + the following anti-foundation axiom:

$\exists B \forall x \exists \alpha : x \in P^\alpha (B) $

Where:

$P^\emptyset (B)= B \\ P^{\alpha +1} (B)= P(P^\alpha(B)) \\P^\alpha = \bigcup_{\lambda<\alpha} P^\lambda(B) \text { for a limit } \alpha$

Define: $base(B) \iff \forall x \exists \alpha : x \in P^\alpha(B)$

Now we can define the notion of relative rank to a base set $B$, denoted by $\rho^B$ as: $$\rho^B(x)=\alpha \iff \forall \lambda (x \in P^\lambda (B) \Leftrightarrow \alpha \leq \lambda)$$. Now well-founded sets would here correspond to sets having ranks relative to the empty set. Formally: $$\forall x [wf(x) \iff \exists \alpha: \alpha= \rho^\emptyset(x)]$$. The idea here is that the definition of a set as what has a rank, need not be confined to the well founded sets! $B$ can be a set of Quine atoms for example, and so all of its elements would have zero ranks relative to $B$, while the empty set would have rank 1 relative to $B$.

Now one can define a notion of cardinality relative to a base set as:

$|x|^B = \{y: y \sim x \land \neg \exists z (z \sim y \land \rho^B(z) < \rho^B(y))\}$

For that notion of cardinality to be total we need the axiom:

$\exists B \forall x \exists y [\exists \alpha (\rho^B(y)=\alpha) \land x \sim y]$

Which is weaker than the above anti-foundation axiom.

Its obvious that the above line is just another use of Scott's trick, but through extending it to a world outside of regularity and choice, and outside of Coret's axiom of every set being equinumerous to some well founded set. However, here we'll have a modified Coret's axiom of every set being equinumerous to some $B$-well founded set, where the later is any set that has a rank relative to base set $B$.

Are there models of $\sf ZF-Reg.$ which violates the above axiom [the weaker one]?

It is indeed well-known that $\sf ZF-Reg.$ doesn't support a definable notion of cardinality that is total over the whole universe of discourse. But here the mentioned notion is a relative one, so is it the case that even this relative notion of cardinality cannot survive in $\sf ZF-Reg.$ as a total function?

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    $\begingroup$ I wouldn't call it an anti-foundation axiom, since for $B=\varnothing$ it is the axiom of foundation, whereas AFAs usually posit the existence of ill-founded sets. Also, if you have a proper class of Quine atoms, then obviously the axiom is false. $\endgroup$
    – Asaf Karagila
    Dec 18, 2020 at 18:44
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    $\begingroup$ Also, very clearly, if there is $B$ such that the universe is simply the iterated power sets of $B$, then you can again do Scott's trick using $B$ as a parameter. Since you already know that it is possible that there is no such definition, you should be able to confirm your own suspicions here. $\endgroup$
    – Asaf Karagila
    Dec 18, 2020 at 18:46
  • $\begingroup$ @AsafKaragila, my last question was in reference to the weaker axiom (the one I named as modified Coret's axiom) I mean to place this axiom instead the first stipulated axiom that I called as the anti-foundation axiom. So a proper class of Quine atoms won't affect it! So the universe is no longer iterated over $B$, no ever set in the universe is subnumerous to a the hierarchy raised over $B$. $\endgroup$
    – Zuhair
    Dec 18, 2020 at 19:00
  • $\begingroup$ The weaker axiom is a reasonable weakening of AC to allow Scott cardinals over an inner model to represent the structure of cardinals. Take a proper class of Quine atoms and the universe is generated by iterated power sets of sets of atoms, and consider the permutation model which partitions the class into countable sets and makes them all Dedekind-finite and pairwise incomparable. Then your weaker axiom fails. $\endgroup$
    – Asaf Karagila
    Dec 18, 2020 at 19:04
  • $\begingroup$ @AsafKaragila, when you said partition the class, did you mean the class of all Quine atoms? $\endgroup$
    – Zuhair
    Dec 18, 2020 at 19:10

1 Answer 1

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Start with a proper class of Quine atoms, $A$, and suppose that the universe is generated by iterated power sets over $A$ (the axiom of choice holds, and there are infinite sets of atoms, the nicest situation you can think of).

Now consider the permutation model obtained by writing $A=\{a_{\alpha,n}\mid\alpha\in\mathrm{Ord},n<\omega\}$, then considering permutations of $A$ moving set-many atoms and satisfying: $\pi(a_{\alpha,n})=a_{\alpha,m}$ for all $\alpha$. In other words, we partition $A$ into countable blocks and permute them independently.

Finally, consider the filter of groups given by finite stabilisers.

Let $A_\alpha=\{a_{\alpha,n}\mid n<\omega\}$, then in the permutation model $A_\alpha$ is Dedekind-finite for any $\alpha$, that is easy.

Now suppose $\alpha$ is an ordinal, and let $B_\alpha=A\setminus A_\alpha$. Then $A_\alpha$ is not equipotent to any set generated by any subset of $B_\alpha$. If it wasn't the case, there was some $X\subseteq B_\alpha$, an ordinal $\gamma$, and an injective $f\colon A_\alpha\to\mathcal P^\gamma(X)$. And it had a finite support $E$.

Since $E\cap A_\alpha$ is finite, pick some $a,b\in A_\alpha\setminus E$ and let $\pi=(a\ b)$. Trivially, $\pi\restriction\mathcal P^\gamma(X)$ is the identity. Now, $\pi A_\alpha=A_\alpha$ and $\pi f=f$, so $f(a)=\pi(f(a))=f(b)$, so $f$ is not injective.

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  • $\begingroup$ if $A$ is a proper class, how can one have iterated power sets over $A$? do you mean power classes over $A$ i.e. classes of all subsets of $A$. $\endgroup$
    – Zuhair
    Dec 19, 2020 at 7:22
  • $\begingroup$ No, I mean subsets. It's the union of all iterated power sets over all sets of atoms. $\endgroup$
    – Asaf Karagila
    Dec 19, 2020 at 9:34

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