Showing that the theory of an infinite descending $\subseteq$ chain has Morley rank 2

Question

Consider the language $\{P_i\mid i\in\omega\}$ and the theory asserting

1. $P_0\supseteq P_1\supseteq P_2\supseteq...$
2. $\neg P_0$ is infinite
3. Each $P_n\smallsetminus P_{n+1}$ is infinite

I want to show that, relative to this theory, the Morley rank of "$x=x$" is 2.

I think I am not having the right picture of what saturated models of this theory look like. The model that I use for guiding my intuition is the following: the domain is the rational numbers $\mathbb{Q}$, each $P_n$ is interpreted as $\{x\in\mathbb{Q}\mid x>n\}$. This picture allows one to see relatively easily that $\text{MR}(x=x)\geq 1$. This is because $\langle P_i(x)\wedge\neg P_{i+1}(x)\mid i\in \omega \rangle$ is a list of consistent formulas that are pointwise disjoint. However, I am having troubles finding the formulas to use to witness $\text{MR}(x=x)\geq 2$ and also why it is exactly 2.

Answer 1

I think I am not having the right picture of what saturated models of this theory look like. The model that I use for guiding my intuition is the following: the domain is the rational numbers $\mathbb{Q}$, each $P_n$ is interpreted as $\{x\in\mathbb{Q}\mid x>n\}$.

That's a great picture to have in mind, but it's not saturated, because it doesn't have any elements satisfying all the $P_n$. A countable model of this theory is determined up to isomorphism by the number of elements which satisfy all the $P_n$. This number can be $0$ (as in your model) or any positive natural number, or $\aleph_0$ (this is the saturated model).

This picture allows one to see relatively easily that $\text{MR}(x=x)\geq 1$. This is because $\langle P_i(x)\wedge\neg P_{i+1}(x)\mid i\in \omega \rangle$ is a list of consistent formulas that are pointwise disjoint.

These formulas actually witness $\mathrm{MR}(x=x)\geq 2$, because they each have Morley rank $\geq 1$. It's good to remember the following interpretations of small Morley ranks:

• $\mathrm{MR}(\varphi(x))\geq 0$ iff $\varphi(x)$ defines a non-empty set.
• $\mathrm{MR}(\varphi(x))\geq 1$ iff $\varphi(x)$ defines an infinite set.
• $\mathrm{MR}(\varphi(x))\geq 2$ iff $\varphi(x)$ can be partitioned into infinitely many infinite definable sets.

Explicitly, to see that $\mathrm{MR}(P_i(x)\wedge\neg P_{i+1}(x))\geq 1$, pick infinitely many distinct elements $(a_n)_{n\in \omega} \in M$ such that $M\models P_i(a_n)\land \lnot P_{i+1}(a_n)$ for all $n$. Then we have $\mathrm{MR}(x = a_n)\geq 0$ for all $n$, the formulas $\{x = a_n\mid n\in \omega\}$ are pairwise disjoint, and each formula $x = a_n$ implies $P_i(x)\wedge\neg P_{i+1}(x)$.

As for showing that $\mathrm{MR}(x=x)$ is exactly $2$, this is a bit trickier. Obtaining an upper bound on the Morley rank requires understanding all the definable sets (or, equivalently, all the complete types). As a strategy, you could prove (by quantifier elimination or an automorphism argument) that any infinite definable set is either

1. A finite union of sets of the form $\lnot P_0(x)$ or $P_i(x)\land \lnot P_{i+1}(x)$, plus or minus finitely many points, or
2. Some $P_n$, plus a set of the first form.

Then observe that (a) sets of the first form have Morley rank $1$, since they do not contain infinitely many disjoint infinite definable sets, and (b) no two sets of the second form are disjoint. So we cannot find an infinite family of disjoint sets of Morley rank $2$, and thus $x = x$ has Morley rank at most $2$.

A more efficient strategy is to use the equivalence between Morley rank and the Cantor-Bendixson rank on the space of types over the countable saturated model. This space has the realized types (Morley rank $0$), one non-realized type containing $\lnot P_0(x)$ and one containing $P_i(x)\land \lnot P_{i+1}(x)$ for each $i\in \omega$ (Morley rank $1$), and one non-realized type containing all the $P_i(x)$ (Morley rank $2$).