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In my recent paper with Makoto Kikuchi,

we proved the following theorem.

Theorem. All countable models $\langle M,\in^M\rangle\models\text{ZFC}$, whether well-founded or not, have the same inclusion relation $\langle M,\subseteq^M\rangle$, up to isomorphism.

And the same is true for much weaker theories, such as KP and even finite set theory, provided that one excludes the $\omega$-standard models without any infinite sets and also the $\omega$-standard models with amorphous sets.

The proof proceeds by proving that for most of the models of set theory $\langle M,\in^M\rangle$, the corresponding inclusion relation $\langle M,\subseteq\rangle$ is an $\omega$-saturated model of what we have called set-theoretic mereology, which is the theory of an unbounded atomic relatively complemented distributed lattice. Those are the basic facts of $\subseteq$ in set theory and that is a complete, finitely axiomatizable, decidable theory. Since the theory is complete, and the resulting models are $\omega$-saturated, it follows by the back-and-forth method that all countable saturated models are isomorphic.

But the possibility of amorphous sets throws a wrench in the works. For most models of set theory, we prove that the inclusion relation is $\omega$-saturated; but when the model is $\omega$-standard and has an amorphous set, then it is not $\omega$-saturated, and so this is an irrirating obstacle to the general phenomenon. What we want to know is whether the ZF models provide just one or many different isomorphism types for the inclusion relation.

Question. Do all countable $\omega$-standard models of ZF with an amorphous set have isomorphic inclusion relations?

Kikuchi and I state in the paper that we believe that the answer to this question will come from an understanding of the Tarski/Ersov invariants combined with a knowledge of the models of ZF with amorphous sets.

Joel David Hamkins
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  • Joel, what happens when you add a proper class of "pairwise-generic" amorphous sets compared to adding just a single one? – Asaf Karagila Nov 20 '17 at 10:52
  • That is a good idea, and I don't know what happens to the subset relation when you do this. It is a good candidate model for another isomorphism type. – Joel David Hamkins Nov 20 '17 at 12:13
  • Well, what happens when you have the one amorphous set? – Asaf Karagila Nov 21 '17 at 09:40
  • If you add an amorphous set to an $\omega$-standard model of ZFC, then the subset relation is definitely not isomorphic to the original model, since the property that all subsets are either finite or co-finite is revealed by the subset relation. – Joel David Hamkins Nov 21 '17 at 11:12
  • Right, right. But start with a standard model (just to omit this $\omega$ from the notation everywhere). And add a single amorphous set to it. That gives you an "almost minimal" infinite cardinal, which I suspect is what you mean by this fact. That you get an $(\omega,\omega)$-gap in the cardinals with the lower chain being $\omega$ itself. But, I mean, what else do you get there which "bothers" the containment structure? – Asaf Karagila Nov 21 '17 at 11:20
  • (If at some point you want to move this to somewhere with more than 600 characters, do feel free. :-)) – Asaf Karagila Nov 21 '17 at 11:24
  • Well, if you add one amorphous set $A$, then of course there are infinitely many disjoint copies of it ${n}\times A$. And also $A\times A$ is not amorphous, but it becomes amorphous in the quotient of the Boolean algebra by the ideal generated by the atoms. This is related to the Tarski invariants for a Boolean algebra. – Joel David Hamkins Nov 21 '17 at 11:35
  • Okay, I'm not entirely sure that I understand that last part completely. What are the atoms here? And how would $A\times A$ become amorphous in the ideal generated by the atoms? – Asaf Karagila Nov 21 '17 at 11:38
  • The atoms are the singletons. If $A$ is amorphous, then in the quotient by the ideal of finite sets, $A$ becomes an atom, since it is not finite, but every subset is either finite or finitely different from $A$. What happens with $A\times A$? Every subset is either finite or cofinite on every section, and uniformly on cofinitely many sections. I think this shows that $A\times A$ lasts for two steps in the Tarskian quotient process, which is what I meant by saying $A\times A$ is amorphous after one step. – Joel David Hamkins Nov 21 '17 at 11:51
  • So you basically consider what happens with inclusion modulo finite sets in the universe, and there there will be an atom again which is $A\times A$. That's the gist I understand from this, is that right? – Asaf Karagila Nov 21 '17 at 11:56
  • Yes, you iterate this, and the number of steps it takes, plus the number of atoms left at the last step, is an isomorphism invariant (the Tarski invariants). – Joel David Hamkins Nov 21 '17 at 12:16
  • My idea is to understand these invariants in connections with a model of ZF + an amorphous set. If we can show that the invariants are the same, then they are all isomorphic. OK, it's not a Boolean algebra, but only a relatively complemented distributed lattice, and in this case these are the Ersov invariants, but the idea is the same. – Joel David Hamkins Nov 21 '17 at 12:17
  • So, if there is an amorphous set, then essentially $A^A$ gives you a sort of atom at the limit step? (Or maybe $A^\omega$? Or $A^{<\omega}$?) But now, if you have countably many "sufficiently different amorphous sets", you can mix and match them to get different kind of structures, no? – Asaf Karagila Nov 21 '17 at 12:21
  • Yes, that is the right idea. We need to understand what happens in the iterated quotient process. What survives and for how long? – Joel David Hamkins Nov 21 '17 at 12:24
  • What do you do at limits? Or is there a proof that this process can only go at most $\omega$ steps or something? – Asaf Karagila Nov 21 '17 at 12:25
  • I think it is explained in Chang-Keisler. In the general result, it is an invariant for the theory, not the isomorphism type, but for some kinds of Boolean algebras, it is an isomorphism invariant. I am confused about whether it goes transfinite, but since you have an increasing sequence of ideals, it would seem natural to quotient by the union of them. – Joel David Hamkins Nov 21 '17 at 12:33

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No, except in trivial cases.$\newcommand{\ck}{\omega_1^{\mathrm{ck}}} \newcommand{\nmereo}{n_{\mathrm{mereo}}} \DeclareMathOperator{MT}{MT} \DeclareMathOperator{rk}{rk} \DeclareMathOperator{rks}{rks}$ If $\nmereo$ is the cardinality of the set of isomorphism classes of reducts of countable models of ZF to their inclusion relation, then $\nmereo\neq 2.$

A suitable notion of rank is studied by Mendick and Truss [1]. A simpler permutation model and further connections are discussed in Truss [2, Section 4], which cites [3] for this algebraic definition. For each Boolean algebra $B$ define the following transfinite sequence of ideals.

  • $I_0(B)$ is $\{0\}$
  • $I_{\alpha+1}(B)$ is the ideal of elements $x\in B$ such that the equivalence class $x+I_\alpha(B)$ is a finite sum of atoms in $B/I_\alpha(B)$
  • $I_\beta(B)=\bigcup_{\alpha<\beta}I_\alpha(B)$ at limit ordinals $\beta>0$

The rank $\rk(B)$ is $-1$ if $B$ is the zero ring, or else the least ordinal $\alpha$ such that $I_{\alpha+1}(B)=B,$ if it exists, and otherwise $\infty.$ The Mendick-Truss rank $MT(x)$ of a set $x$ is the rank of the powerset $\mathcal P(x)$ with its usual Boolean algebra structure.

(Beware that [3] uses the similar but different quantity $\delta(B)=\rk(B)+1,$ i.e. the Cantor-Bendixson rank of the Stone space. “Rank” is an overloaded term, but this use for Boolean algebras appears in at least one paper [4]. These ideals are what you get by applying Stone duality to Cantor-Bendixson derivatives. They're more appropriate for studying powersets than the Tarski-Ershov ideal of sums of atomic and atomless elements: every element of a powerset is atomic.)

We can look at the rank of powersets within a model or externally, and it is useful if the two notions agree. Let $|p|$ denote the order type of a code $p$ in Kleene’s $\mathcal O,$ to be concrete. Then for any $\omega$-standard model $M$ of ZFC, $|p|^M$ is isomorphic to $|p|.$ So statements about ranks below $\ck$ will be computed correctly: $M\Vdash \rk(B)=|p|$ if and only if $\rk(B)=|p|.$ The following argument only needs to compare ranks of at most $\omega.$

For a model $M$ of ZFA, define $\rks(M)$ to be the least ordinal $\alpha$ such that the rank of $(\mathcal P(x)^M,\subseteq^M)$ is less than $\alpha$ for all $x\in M.$ So at least for models of ZF, $\rks(M)$ is an invariant of the inclusion reduct. We can do a similar computation for class models $M$ adequate for the Boolean algebra structure on powersets, though we might get $\infty$ if there is no such ordinal. This justifies the convenient approach of working within a model, instead of reasoning externally to a model.

Assuming that there is an $\omega$-standard countable model of ZFC, the two different inclusion reducts will come from:

  1. An $\omega$-standard countable model of ZF with $\rks(M)>\omega$
  2. An $\omega$-standard countable model of ZF with $\rks(M)=\omega$

We’ll make use of permutation models. So I should explain how to get a model of pure set theory instead of just ZFA. I think the best approach is to appeal to embedding theorems. See Note 103 in “Consequences of the Axiom of Choice” for the statements; the proofs only use the usual techniques of forcing and inner models, so they should work in the setting of countable $\omega$-standard models. For (1) it’s easy because we just need a property to hold of one powerset algebra, which is exactly the kind of thing the Jech-Sochor embedding theorem allows. For (2) we can use the fact that if a set $x$ admits a surjection to $\omega$ then its MT rank is $\infty$ [1, Lemma 1.3]. A statement about powersets of sets that do not admit a surjection to $\omega$ is “surjectively boundable” so Pincus’s embedding theorem applies.

For (1), apply Jech-Sochor with a permutation model containing a set of MT rank $>\omega.$ [1, Theorem 3.1] [2, Section 4].

For (2), apply Pincus’s embedding theorem with the first Fraenkel model $\mathcal N1.$ This has a set of urelements $A,$ with full permutation group $G.$ The action of $G$ extends by abuse of notation to the whole of $\mathcal N1.$

Any $x$ breaks up as a disjoint union of non-empty sets $$x\cong \bigcup_{i\in\alpha} x_i$$ with a family of functions $f_i$ with $\operatorname{dom}(f_i)\subseteq A^{k_i}$ for some $k_i\in\omega$ and $\operatorname{rng}(f_i)=x_i.$ I don’t need any stronger properties of this model.

This decomposition is standard, but I don’t know of a suitable reference for this particular case. We can work in the full universe $V(A)$ where choice holds. Each set $x\in\mathcal N1$ is supported by a finite sequence $\xi\in A^{<\omega},$ meaning that $x$ is fixed by the stabilizer group $G_\xi\subset G$ of automorphisms that fix $\xi.$ Enumerate the $G_\xi$-orbits of $x$ by $x_i,$ $i<\alpha.$ Pick $y_i\in x_i$ for each $i<\alpha.$ Pick $\eta_i\in A^{<\omega}$ such that $y_i$ is supported by $\eta_i.$ Define $f_i=\{(\pi \eta_i,\pi y_i) : \pi\in G_\xi\}.$ Then the family $(f_i)_{i<\alpha}$ is fixed by $G_\xi$ and each $f_i$ is a surjection to $x_i.$

If $\alpha$ is infinite then there is easily seen to be a surjection $x\to\omega.$ In this case $\rk(x)=\infty$ [1, Lemma 1.3]. If $\alpha$ is finite then $\MT(x)$ is finite because $\MT(A)=1$ and the class of sets of finite MT rank is closed under finite unions, products, subsets, and quotients: the first two operations are covered by [1, Corollary 1.5] and [1, Theorem 1.9], and as [1, page 2] says "[...] It is also easy to see by transfinite induction that the image of any [MT] rank $\alpha$ set under a function has rank $\leq \alpha,$ and hence that a subset of a rank $\alpha$ set has rank $\leq \alpha$."

This shows $\rks(\mathcal N1)\leq \omega.$ And $\MT(A^k)=k$ by [1, Theorem 1.9], so $\rks(\mathcal N1)$ is exactly $\omega.$

I've written up some thoughts on trying to show $\nmereo\geq\aleph_0$ (assuming $\nmereo>1$), but didn't prove anything. You might want to try asking Mendick and Truss for their advice.

[1] Mendick, G. S.; Truss, J. K., A notion of rank in set theory without choice, Arch. Math. Logic 42, No. 2, 165-178 (2003). ZBL1025.03047.

[2] J. K. Truss, The axiom of choice and model-theoretic structures, submitted. http://www1.maths.leeds.ac.uk/~pmtjkt/preprints.html

[3] Day, G. W., Superatomic Boolean algebras, Pac. J. Math. 23, 479-489 (1967). ZBL0161.01402.

[4] Bonnet, Robert; Rubin, Matatyahu, A thin-tall Boolean algebra which is isomorphic to each of its uncountable subalgebras, Topology Appl. 158, No. 13, 1503-1525 (2011). ZBL1229.54045.

Harry West
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