Number of types in simple theories

Question

1. Let $T$ be the theory of random graphs. I want to count the number of 1-types in this theory, i.e $|S_1(T)|.$

My claim is this: $|S_1(T)|=1$, because every two points have the same types by extension axiom. (I cannot prove my claim precisely!)

We know the theory of random graph is simple. So there is a following natural question.

2. If $T$ is an arbitrary simple theory, what can we say about the number of $n$-types. In particular what can we say about the size of $S_1(T)$?

I have no idea about question 2.

Answer 1

Looking at $S_1(T)$ usually tells you very little about the theory $T$. This is because $S_1(T)$ only captures the behavior of formulas with one variable and no parameters, so it says nothing about how elements of a model relate to other elements. It's much more interesting to look at types in multiple variables ($S_n(T)$ for all $n$) or types with parameters ($S_1(A)$ where $A\subseteq M\models T$).

Here are some properties that are captured by the spaces $S_n(T)$. Assume the language is countable.

• The Ryll-Nardzewski theorem says that $T$ is $\aleph_0$-categorical if and only if $S_n(T)$ is finite for all $n$.
• A theory is called small if $S_n(T)$ is countable for all $n$. A theory is small if and only if it has a countable saturated model, and every small theory also has a countable prime model.

Smallness and $\aleph_0$-categoricity are essentially orthogonal to Shelahian dividing lines like stability and simplicity, e.g. there are $\aleph_0$-categorical theories on both sides of all the major dividing lines. (The one exception to this orthogonality that I can think of is that $\omega$-stability implies smallness.)

Instead, the Shelahian dividing lines tend to have interactions with the spaces $S_1(A)$ of types with parameters (and often the local type spaces $S_\varphi(A)$ where we only consider $\varphi$-types). For example, the following are equivalent (again, assume a countable language): (1) $T$ is stable, (2) there is some cardinal $\kappa$ such that $|S_1(A)|\leq \kappa$ for all $A$ with $|A|\leq \kappa$, (3) for all formulas $\varphi(x;y)$, $|S_\varphi(A)|\leq |A|$.

This characterization of stability generalizes somewhat nicely to NIP theories: If $T$ is NIP, then for all formulas $\varphi(x;y)$, $|S_\varphi(A)|\leq \mathrm{ded}(|A|)$, where $\mathrm{ded}(\kappa)$ is the supremum of the cardinalities $|L|$ over all linear orders $L$ which contain a dense set of cardinality $\kappa$. Now it is consistent that $\mathrm{ded}(\kappa) = 2^\kappa$ for all $\kappa$, in which case this theorem tells us nothing. But it is also consistent that $\mathrm{ded}(\kappa) < 2^\kappa$ for some cardinals $\kappa$, and in such a model of set theory, the condition $|S_\varphi(A)|\leq \mathrm{ded}(|A|)$ characterizes NIP.

Simplicity also has a counting types characterization, but it's more complicated to state, and it involves counting partial types (closed sets in the space $S_1(A)$). Given cardinals $\kappa$ and $\lambda$, let $\mathrm{NT}(\kappa,\lambda)$ be the supremum of the cardinalities $|P|$ over all families $P$ of pairwise-contraditory partial types in one variable of cardinality $\leq\kappa$ over a set of parameters of cardinality $\leq \lambda$. Then $T$ is simple if and only if $\mathrm{NT}(\kappa,\lambda) \leq \lambda^{|T|}+2^\kappa$ for all cardinals $\lambda$ and $\kappa$.

See the paper The number of types in simple theories by Casanovas for much more on this theme.

Answer 2

You are right that the random graph has precisely one $1$-type. This follows directly from the fact that the random graph has quantifier elimination.

We can generally not say something about the cardinality of $S_1(T)$, even for stable $T$. For example, take the theory of $(\mathbb{Z}, +, 0, 1)$, which is stable. In this theory congruence modulo $n$, which I will write as $\equiv_n$, is definable for any $n \in \mathbb{N}$. Let $p_1, p_2, \ldots$ enumerate all the primes. Then for any $I \subseteq \mathbb{N}$ we define: $$ \Sigma_I(x) = \{ x \equiv_{p_i} 0 : i \in I \} \cup \{x \not \equiv_{p_i} 0 : i \not \in I\}. $$ For any such $I$ we have that $\Sigma_I(x)$ is finitely satisfiable: any finite part of $\Sigma_I(x)$ is realised by the product of the primes that appear in that finite part. Clearly, for $I \neq I'$ we have that $\Sigma_I(x)$ and $\Sigma_{I'}(x)$ are incompatible types. So we find at least $2^{\aleph_0}$ many $1$-types. That is as bad as it can get generally, because for any theory $T$ there are at most $2^{|T|}$ many $1$-types.