Example of first order logic,equivalence class,categoricity,abstract elementary classes

Question

I have problems in a paper about AEC,with an example. In fact,I need to explain most of the details in that example.
Let $\tau$ contain infinitely many unary predicates $P_n$ and one binary predicate $E$.Define a first order theory $T$ such that $P_{n+1}(x) \implies P_n(x)$,$E$ is an equivalence relation with two classes, which are each represented by exactly one point in $P_{n+1}(x) - P_n(x)$, for each $n$.
Question.Does this precisely mean that each $P_{n+1}(x) - P_n(x)$ has at least 2 points,one in each equivalence class?
Next,let $K$ be a class of models in $T$,that omit the type of two inequivalent points that satisfy all the $P_n$.
Question what does this mean in words?
Then a model $M$ in $K$ is determined by isomorphism by $\mu(M):=|\{x\in M:(\forall n)P_n(x)\}|$.
Question why?
The paper now says, so $K$ is categorical in every uncountable powers,but has $\aleph_0$ countable models (non of them is finite).
Question why?
..........
Questions about AEC side of the example.
Let $M_0, M_1 , M_2 \in K$ be such that $\mu(M_0)=0,\mu(M_1)=\mu(M_2)=1$ and $M_1,M_2$ are not isomorphic over $M_0$.Then there is no amalgamation of $M_1,M_2$ over $M_0$.
Question why?
Now if $\lambda > \aleph_0$ then every model $M\in K_{\lambda^+}$ is saturated (over $\lambda$) but it is not saturated over $\aleph_0$.
Question why?

Answer 1

I’m not familiar with the AEC material, but I can answer the basic model theory questions.

Question $\mathbf{1}$: Suppose that $\mathfrak{M}\vDash T$, and $M$ is the universe of $\mathfrak{M}$. I’ll write $\hat P_n$ for the interpretation of $P_n$ in $\mathfrak{M}$. Then $P_{n+1}(x)\implies P_n(x)$ means that $\hat P_{n+1}\subseteq\hat P_n$. Thus, $\hat P_{n+1}\setminus\hat P_n=\varnothing$, and you clearly have the difference backwards: the statement should be that each class is represented by exactly one point in $P_n(x)-P_{n+1}(x)$ for each $n$. In other words, we must have $|\hat P_n\setminus\hat P_{n+1}|=2$ for each $n$, and moreover, if $x$ and $y$ are the two points in $P_n\setminus P_{n+1}$, then $\neg\hat E(x,y)$: $x$ and $y$ are not related by $\hat E$, the interpretation of $E$ in $\mathfrak{M}$. At this point we know that $\mathfrak{M}$ must look something like this:

$M$ is the disjoint union of two sets $X$ and $Y$, the equivalence classes of $\hat E$. There are sets $X_0=\{x_n:n\in\omega\}\subseteq X$ and $Y_0=\{y_n:n\in\omega\}\subseteq Y$, where the $x_n$ and $y_n$ are all distinct, such that $X\cap\hat P_n=\{x_k:k\ge n\}$ and $Y\cap\hat P_n=\{y_k:k\ge n\}$ for each $n\in\omega$. Thus, $\hat P_n\setminus\hat P_{n+1}=\{x_n,y_n\}$ for each $n\in\omega$. Points in $X\setminus X_0$ and $Y\setminus Y_0$, if any, must be in $\bigcap_{n\in\omega}\hat P_n$ or in $M\setminus\hat P_0$. (At this point I would make a sketch.)

Question $\mathbf{2}$: The problem is to determine what it means for $\mathfrak{M}$ to omit the type of two inequivalent points that satisfy all of the $P_n$. A point that satisfies all of the $P_n$ is in all of the sets $\hat P_n$, so it’s in $\bigcap_{n\in\omega}\hat P_n$. The equivalence relation $\hat E$ has only two classes, $X$ and $Y$, so two points are inequivalent iff one of them is in $X$ and the other in $Y$. In other words, at least one of the sets $X\cap\bigcap_{n\in\omega}\hat P_n$ and $Y\cap\bigcap_{n\in\omega}\hat P_n$ must be empty.

The type itself is a set of formulas in two free variables: $$\{P_n(x)\land P_n(y):n\in\omega\}\cup\{\neg E(x,y)\}\;.$$ Saying that a model omits it is saying that no two points of the model satisfy all of the formulas in the type.

Question $\mathbf{3}$: You mean that a model $\mathfrak{M}$ in $K$ is determined up to isomorphism, not by isomorphism, by $\mu(\mathfrak{M})$. Note that $\mu(\mathfrak{M})=\left|\bigcap_{n\in\omega}\hat P_n\right|$ in the notation that I’ve been using. Since $\mathfrak{M}$ omits the above type, the members of $\bigcap_{n\in\omega}\hat P_n$ are all in the same $\hat E$-equivalence class. (The two classes are not actually labelled: my use of $X$ and $Y$ above was merely for convenience, and the labels $X$ and $Y$ could be interchanged without affecting anything.)

Now suppose that $\mathfrak{M}_0$ and $\mathfrak{M}_1$ are two models in $K$. I’ll use $\hat P_n^i$ and $\hat E^i$ for the interpretations of $P_n$ and $E$ in $\mathfrak{M}_i$. Suppose further that $\left|\bigcap_{n\in\omega}\hat P_n^0\right|=\left|\bigcap_{n\in\omega}\hat P_n^1\right|$ and that $M_i=\hat P_0^i$ for $i\in\{0,1\}$.

If $\mu(\mathfrak{M}_i)>0$, let $f:\bigcap_{n\in\omega}\hat P_n^0\to\bigcap_{n\in\omega}\hat P_n^1$ be a bijection, fix $x_\omega^0\in\bigcap_{n\in\omega}\hat P_n^0$, and for $n\in\omega$ let $x_n^0$ be the unique member of $\hat P_n^0\setminus\hat P_{n+1}^0$ such that $\hat E^0(x_n^0,x_\omega^0)$ and $x_n^1$ the unique member of $\hat P_n^1\setminus\hat P_{n+1}^1$ such that $\hat E^1(x_n^1,f(x_\omega^1)$. For $n\in\omega$ and $i\in\{0,1\}$ let $y_n^i$ be the unique member of $\hat P_n^i\setminus\hat P_{n+1}^i$ that is not $x_n^i$. Finally, extend $f$ to $P_0^0$ by letting $f(x_n^0)=x_n^1$ and $f(y_n^0)=y_n^1$ for each $n\in\omega$.

If $\mu(\mathfrak{M}_i)=0$, let $x_0^i$ and $y_0^i$ be the two points of $\hat P_0^i\setminus\hat P_1^i$ for $i\in\{0,1\}$. For $n>0$ let $x_n^i$ and $y_n^i$ be the unique points of $\hat P_n^i\setminus\hat P_{n+1}^i$ such that $\hat E_i(x_0^i,x_n^i)$ and $\hat E^i(y_0^i,y_n^i)$, respectively, and define $f:M_0\to M_1$ by $f(x_n^0)=x_n^1$ and $f(y_n^0)=y_n^1$ for $n\in\omega$.

In each case $f$ is an isomorphism of $\mathfrak{M}_0$ and $\mathfrak{M}_1$. Note that I made the additional assumption that $P_0^i=M_i$ for $i\in\{0,1\}$; this appears to be necessary, and its omission from the paper appears to be a small oversight. In terms of the theory $T$ this amounts to adding the sentence $\forall x\,P_0(x)$.

Question $\mathbf{4}$: We just saw that the isomorphism type of a model $\mathfrak{M}$ in $K$ is completely determined by the cardinal number $\mu(\mathfrak{M})$: for each cardinal $\kappa$ there is up to isomorphism just one $\mathfrak{M}$ in $K$ such that $\mu(\mathfrak{M})=\kappa$. If $\kappa>\omega$, clearly $|M|=\kappa$, while if $\kappa\le\omega$, then $|M|=\omega$. Thus, up to isomorphism there is just one such model of each uncountable cardinality, and $K$ is therefore categorical in every uncountable power. $K$ has $\omega$ (or $\aleph_0$, if you prefer) distinguishable countable models, one for each countable cardinal, and they’re all countably infinite.