Skip to main content
1 of 2
Harry West
  • 1.9k
  • 7
  • 13

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 models of ZFC 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
  • 1.9k
  • 7
  • 13