Deciding whether a theory $T$ is $\aleph_0$-categorical?

Question

I'm trying to solve the following exercise from my lecture notes:

1. For $n ≥ 1$, let $L_n$ be the language with $n$ unary predicate symbols $P_1, . . . , P_n$. Show there exists a theory $T_n$ asserting that each $P_i$ is infinite, the $P_i$ are disjoint ($¬(∃x)(P_1(x)\wedge P_2(x))$, etc.) and there are infinitely many elements not in any $P_i$, $i ≤ n$.

2. Show $T_n$ is $ℵ_0$-categorical. Conclude that $T_n$ is complete.

3. How many models of cardinality $ℵ_1$ (up to isomorphism) does $T_2$ have?

4. Let $T = \bigcup_n T_n$. Show that $T$ is a complete theory.

5. Is $T$ $\aleph_0$-categorical? If not how many countable models does $T$ have?

I've done everything apart from number 5.

It's easy to show that each $T_n$ is $ℵ_0$-categorical as in part 2, because if you have a countably infinite model, and each of the $P_i$ are infinite then they must also be countably infinite and you can take bijections between them. Moreover, if we take any countably infinite model to have domain $A$, then since $A\backslash(P_1\cup ...\cup P_n)$ is infinite this will also be countably infinite, and so you can take a bijection between these parts of any two countably infinite model. It's easy to check that the function defined on the domain of a whole model as the union of these bijections (which is well defined since the $P_i$ are disjoint) will be an isomorphism between any two countably infinite models, hence $T_n$ is $ℵ_0$-categorical.

I'm not sure if I can do the same in part 5 - each of the $P_i$ will still be countably infinite and so I could do the same thing for them, taking bijections between their interpretations under any two models. But I'm not sure if in this case the subset of the domain of a model not in any $P_i$ will still have to be infinite - there are infinitely many not in any finite selection of the $P_i$ because $T$ contains all the $T_n$, but I'm not convinced that means there will be infinitely many not in any of the $P_i$.

I'd really appreciate any advice you could offer - if the the subset of the domain of a model not in any $P_i$ will still have to be infinite, an explanation of why would be really helpful. If this might not be infinite, then I can see that presumably in this case there would be $\aleph_0$ countable models (up to isomorphism) as you could take a bijection between this subset of any possible size and use it to get an isomorphism as above, but I'm not sure how to show it might not be infinite - I can't think of an obvious counterexample.

Answer 1

Your intuition is absolutely correct; we can make the set $M=\bigcup_{0<i\in\omega}\{i\}\times\mathbb{Z}$ into an $L$-structure by realizing each $P_i$ as $\{i\}\times\mathbb{Z}$. This is a model of $T$; each $P_i$ is infinite, the $P_i$ are pairwise disjoint, and for any $n\in\omega$ there are infinitely many elements not in $\bigcup_{0<i\leqslant n}P_i=\bigcup_{0<i\leqslant n}\{i\}\times\mathbb{Z}$.

On the other hand, we can realize $\{\star\}\cup M$ as an $L$-superstructure of $M$ by asserting $\star\notin P_i$ for any $0<i\in\omega$, and this will again be a model of $T$ for the same reasons as $M$. However, these two structures cannot be isomorphic, as one has an element not in any $P_i$ and the other does not. In fact there is no $L$-embedding at all from $\{\star\}\cup M$ to $M$, as there is nowhere to map $\star$ to.

In this way, there are indeed $\aleph_0$ countable models of $T$ up to isomorphism, with one unique model for any possible countable size of the complement of $\bigcup_{0<i\in\omega}P_i$.