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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.

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2 Answers 2

up vote 9 down vote accepted

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.

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With regards to the countable case: It shows that you have continuum many models of ZFC. But I'm interested in the number of non-isomorphic models of some fixed complete extension T of ZFC. With regards to the uncountable case. That should do the trick (I believe that since $\text{ZFC}\subset{T}$ and $T$ defines a linear order on its $V_{\alpha}$s. Am I correct?) –  Danul G Jan 14 at 15:55
    
Ah, that would need more thought... –  Andres Caicedo Jan 14 at 16:02
    
Thank you. I didn't catch an error in my question till I looked at your solution. I misquoted the theorem from Shelah. It should be for all uncountable $\kappa$, there are $2^{\kappa}$ non-isomorphic models of size $\kappa$. I have that the conclusion holds for $\kappa\geq{\aleph_{0}}$, which is clearly false. I will fix it. You argument does establish the result for uncountable $\kappa$. We can just look at the "standard" finite ordinals of the model of $T$ and use those as witnesses to the order property. I think that my previous idea for establishing the order property also works. –  Danul G Jan 14 at 16:21
    
Danul, I have now addressed the case of arbitrary $T\supseteq\mathsf{ZFC}$. I would much like to hear of simpler (or, at least, older) approaches. –  Andres Caicedo Jan 14 at 17:38
    
Thank you. The answer for the countable case is a lot harder than I thought it would be. –  Danul G Jan 14 at 19:34

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.

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I'm having a little bit of trouble seeing the compactness argument... –  Danul G Jan 14 at 19:34
    
Add a new constant $a$ and add axioms saying $n\in a$ and $n\notin a$ for each actual natural number $n$, depending on whether $n\in A$ or not, respectively. This is finitely consistent with $T$, and so it has a model $M$. The interpretation of $a$ in $M$ agrees with $A$ on the standard part. –  JDH Jan 14 at 19:35
    
Ah, this is definitely more reasonable! Thanks. –  Andres Caicedo Jan 14 at 19:51
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@DanulG The way I like to think of this is via primes: Given $A\subseteq\mathbb N$, by compactness there is a model of your theory with a "natural number" $d$ such that the $n$-th prime divides $d$ (for $n$ standard) iff $n\in A$. The advantage of this little extra work is that it makes $A$ the standard part of a set definable in the arithmetic part of your model, so this also applies to extensions of $\mathsf{PA}$. –  Andres Caicedo Jan 14 at 20:21
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A version of this (E2) appeared as a qualifying exam problem at UW Madison: math.wisc.edu/~miller/old/qual/2011-8.pdf –  hot_queen Jan 15 at 2:19

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