Number of Non-isomorphic models of Set Theory

Question

Assume that the meta theory allows for model theoretic techniques and handling infinite sets etc (The meta theory itself is, informally, "strong as ZFC"). Also assume that I'm studying ZFC inside this meta-theory and that I'm working with the assumption that ZFC is consistent.

Let $ZFC\subset{T}$, where $T$ is a complete theory. I would like to know the number of non-isomorphic models of $T$ for some given $\kappa\geq{\aleph_{0}}$. I feel as if this should be the maximum number possible, i.e. $2^{\kappa}$.

My idea for showing this was to use Shelah's theorem that says that unstable theories have $2^{\kappa}$ many models for any $\kappa>{\aleph_{0}}$ (which establishes the result for all uncountable $\kappa$). But in order to do this I first need to show that $T$ is unstable. However I'm not sure about how to do this (or even if I'm on the right track).

Any help would be much appreciated.

Answer 1

You get $2^\kappa$ many models for any uncountable $\kappa$. This follows immediately from the fact that $\mathsf{ZFC}$ (trivially) defines an infinite linear order, and this is all that's needed to ensure unstability.

(The same argument shows that already much weaker theories, such as $\mathsf{PA}$, are unstable.)

This gives the result for $\kappa$ uncountable. For $\kappa$ countable you also have the maximum number of nonisomorphic models of $\mathsf{ZFC}$. To see this, use the incompleteness theorem to show that you can recursively label the nodes of the complete binary tree as $T_s$, $s\in 2^{<\omega}$, so that $T_\emptyset=\mathsf{ZFC}$ (or $\mathsf{PA}$, if you prefer), $T_s$ is consistent for each $s$, $T_s\subsetneq T_t$ for $s\subsetneq t$, and each $T_{s{}^\frown\langle i\rangle}$, for $i=0,1$, is obtained from $T_s$ by adding a single new axiom (that depends on $s$, of course). Now, for each $x\in 2^\omega$, let $T_x$ be any consistent complete theory extending $\bigcup_n T_{x\upharpoonright n}$. These are $2^{\aleph_0}$ pairwise incompatible theories, all extending $\mathsf{ZFC}$ (or $\mathsf{PA}$), and all having countable models. (Examples such as the theory of $(\mathbb Q,<)$ show that the argument must be different for countable models.)

The paragraph above shows that there are $2^{\aleph_0}$ incompatible extensions of $\mathsf{ZFC}$, and therefore $2^{\aleph_0}$ non-isomorphic countable models of $\mathsf{ZFC}$. The result is also true for a fixed complete extension $T$, but the argument seems harder. A proof follows from the following, but most likely there are easier approaches:

Consider first the paper

Ali Enayat. Leibnizian models of set theory, J. Symbolic Logic, 69 (3), (2004), 775–789. MR2078921 (2005e:03076).

In it, Enayat defines a model to be Leibnizian iff it has no pair of indiscernibles. He shows that there is a first order statement $\mathsf{LM}$ such that any (consistent) complete extension $T$ of $\mathsf{ZF}$ admits a Leibnizian model iff $T\vdash\mathsf{LM}$. He also shows that $\mathsf{LM}$ follows from $\mathrm{V}=\mathsf{OD}$, and that any (consistent) complete extension of $\mathsf{ZF}+\mathsf{LM}$ admits continuum many countable nonisomorphic Leibnizian models.

Now consider the paper

Ali Enayat. Models of set theory with definable ordinals, Arch. Math. Logic, 44 (3), (2005), 363–385. MR2140616 (2005m:03098).

In it, Enayat defines a model to be Paris iff all its ordinals are first order definable within the model. He shows that any (consistent) complete extension of $\mathsf{ZF}+\mathrm{V}\ne\mathsf{OD}$ admits continuum many countable nonisomorphic Paris models.

These two facts together imply the result you are after (and more, of course). An earlier result of Keisler and Morley, that started the whole area of model theory of set theory, shows that any countable model of $\mathsf{ZFC}$ admits an elementary end-extension. (This fails for uncountable models.) It may well be that an easy extension of this is all that is needed to prove the existence of continuum many non-isomorphic countable models of any fixed (consistent) $T\supset\mathsf{ZFC}$, but I do not see right now how to get there. The Keisler-Morley theorem alone does not seem to suffice, in view of Joel Hamkins's beautiful answer to this question.

Answer 2

As Andres requested, here is a somewhat easier argument for the case of countable models, showing that every consistent extension $T\supset\text{ZFC}$ has continuum many countable models.

Fix the theory $T$. A simple compactness argument shows that for any $A\subset\mathbb{N}$, we can find a countable model $M$ of $T$ in which $A$ arises as the standard part of a set of natural numbers in $M$. Each such $M$ realizes only countably many such sets $A$, and so there must be continuum many distinct models of $T$ .

This argument can be described as follows: any given set of natural numbers can arise in the standard system of a countable model — the set of reals arising as the standard parts of sets in the model — and the standard system of any model uses up at most countably many reals. So there must be continuum many pairwise non-isomorphic models of the theory.