A general question is:
Given a first order theory T, what can we say about the number of non-isomorphic models of T of size \lambda, as \lambda varies through the cardinals?
This is addressed by Shelah's classification theory. Historically, the first step was Morley's categoricity theorem (1965) which states:
- If T is a first-order theory in a countable language and T is categorical in some uncountable cardinal, then it is categorical in all uncountable cardinals.
The proof of this theorem, which was the content of the lectures, involves the ideas of stability, indiscernibles, saturated models, and splitting types. It's somewhat different from the one I recall from the basic model theory class I took many years ago.
To begin with, we can define a rank R on sets of formulas (i.e., types) with parameters in some large saturated model \mathcal{C} (the "monster model"). Let p be a set of formulas. Then, working in \mathcal{C},
- R(p)=-1 if p is not satisfiable.
- R(p)\ge 0 if p is satisfiable.
- R(p)\ge \alpha+1 if for any finite p_0\subseteq p, there is a formula \psi such that R(p_0\cup\{\psi\})\ge \alpha and R(p\cup\{\neg \psi\})\ge \alpha.
The rank R(p) is then defined inductively to be the minimum value so that all of the above conditions are satisfied. Note: this isn't what is known as Morley rank. Andrew Marks noticed that this rank is the Cantor-Bendixson rank on the Stone space of complete types.
Some properties of the rank:
- (Monotonicity) If p\vdash q then R(p)\le R(q).
- (Finite character) For every p, there is a finite p_0\subseteq p so that R(p)=R(p_0).
- Proof: If R(p)=\alpha, there is a finite p_0 witnessing R(p)\not\ge \alpha+1, and monotonicity gives the reverse inequality.
- (Invariance) If f is an automorphism of \mathcal{C}, then R(p)=R(f(p)).
- Proof is by induction on the rank.
Definition: T is \lambda-stable if for any A\subseteq C of size \lambda, |S^n(A)|\le \lambda. Here S^n(A) is the set of complete n-types over A, that is, maximal satisfiable sets of formulas with parameters over A and free variables x_1,\ldots, x_n.
Theorem: Let T be a theory in a countable language. Then T is \aleph_0-stable iff R(\{\bar{x}=\bar{x}\})<\infty, i.e., has some ordinal value. In fact, if these equivalent conditions hold, then R(\{\bar{x}=\bar{x}\})<\omega_1.
Proof: Suppose R(\{\bar{x}=\bar{x}\})\ge \omega_1. We will inductively construct a complete binary tree whose nodes are finite sets of formulas p_\eta for \eta\in 2^{<\omega} and whose branches will give 2^{\aleph_0} many different complete types, contradicting stability. For the construction, we will maintain that R(p_\eta)\ge\omega_1.
Let p_\emptyset=\{\bar{x}=\bar{x}\}. Given p_\eta, there are formulas \psi_{\eta,i}(\bar{x},\bar{a}_{\eta,i}) for each i<\omega_1 witnessing that R(p_\eta)\ge i+1. Using the fact that the language of T is countable together with \aleph_0-stability, there are \psi and t so that for an unbounded set of i<\omega_1, \psi=\psi_{\eta,i} and t=\mathrm{tp}(\bar{a}_{\eta,i}) (the set of all formulas satisfied by \bar{a}_{\eta,i}, over the finitely many parameters in p_\eta). Pick a_\eta realizing t. Then define
- p_{\eta\circ 0}=p_\eta\cup\{\psi(\bar{x},\bar{a}_\eta)\}
- p_{\eta\circ 1}=p_\eta\cup\{\neg\psi(\bar{x},\bar{a}_\eta)\}
Using invariance of rank under automorphisms and the fact that in \mathcal{C} for any two tuples realizing the same type there exists an automorphism sending one to the other, the ranks of these sets of formulas are at least i for an unbounded set of i<\omega_1, so they must be \ge\omega_1, allowing the construction to continue. The branches of the tree we construct can be extended to complete types, and the construction ensures that each of these complete types is distinct (for two different branches, look at the first place of disagreement).
For the converse, assume R(\bar{x}=\bar{x})<\infty and T is not stable. Then there are A\subseteq \mathcal{C}, |A|\le \lambda such that there are distinct \{p_i:i<\lambda^+\}\subseteq S^n(A). By assumption, for any i<\lambda^+, there is \alpha_i so that R(p_i)=\alpha_i. By finite character, there is q_i\subseteq p_i finite with R(q_i)=R(p_i)=\alpha_i. By counting, |\{q_i:i<\lambda^+\}|\le \omega. Let i\neq j be such that q_i=q_j, so p_i,p_j are different completions of q_i=q_j. This means there is a formula \psi with \neg\psi\in p_i and \psi\in p_j. This gives R(q_j)\ge R(p_j)+1, a contradiction to the choice of q_j. \Box.