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Let $L=\{P_0,P_1,P_2\}$ be a first order language, and let $$T=\bigg\{\Big(\forall x\ P_i(x)\Big)\vee \Big(\forall x\ \neg P_i(x)\Big):i \in \{0,1,2\}\bigg\}\\ \qquad\qquad\qquad\qquad\qquad\bigcup\bigg\{\forall x\ P_0(x)\implies P_1(x), \forall x\ P_1(x)\implies P_2(x) \bigg\}\bigcup T_\infty,$$

where $T_\infty$ is just the theory of an infinite model.

I.e. any of the $P_i$'s is either empty or the whole set, and $P_0\subseteq P_1\subseteq P_2$.

I believe (correct me if wrong) that there are 4 countable non-isomorphic models of $T$:

$\mathfrak{A}_0=(A_0;\emptyset,\emptyset,\emptyset)$,

$\mathfrak{A}_1=(A_1;\emptyset,\emptyset,A_1)$,

$\mathfrak{A}_2=(A_2;\emptyset,A_2,A_2)$, and

$\mathfrak{A}_3=(A_3;A_3,A_3,A_3)$.

Now, my questions are:

Are any of those models the prime model of $T$? Are any of them saturated? $\omega$-homogeneous? Atomic?

I would think none of them is a prime model, since in $\mathfrak{A}_0$ (the only one that could be a prime model) the sentence $\exists x P_1(x)$ fails, but it's true in $\mathfrak{A_3}$.

Somehow my intuition tells me they are $\omega$-homogeneous. Somehow one would take advantage of the fact that any $P_i$ is either everything or nothing, hence any bijection would be elementary, right?

I'm not too sure how to go about with saturation, nor atomicity (is that the word?), though. I figure arbitrary types over finite parameteres could get really complicated. And I don't have enough intuition about atomic models to really know what to do.

Any ideas?

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All four of these models are saturated, $\omega$-homogeneous, and atomic.

How can that be, you ask? Any two countable saturated (for example) models of a theory $T$ must be isomorphic, and clearly $\mathfrak{A}_0 \not\cong \mathfrak{A}_0$.

But the result is that any two countable saturated elementarily equivalent models are isomorphic. Equivalently, any two countable saturated models of the same complete theory $T$ are isomorphic.

What's going on in your example, of course, is that $T$ is not a complete theory. There are exactly four complete theories extending $T$. Here's one, which we can call $T_0$, since $\mathfrak{A}_0\models T_0$.

$$\{\forall x\, \lnot P_0(x) \land \lnot P_1(x) \land \lnot P_2(x)\} \bigcup T_\infty$$

This is just the theory of infinite sets together with three empty predicates. This theory is countably categorical (the unique countable model up to isomorphism is $\mathfrak{A}_0$), and it follows that $\mathfrak{A}_0$ is saturated, $\omega$-homogeneous, and atomic.

The same holds for the theories of the other $\mathfrak{A}_i$.


Edit: As pointed out in the comments, none of these models are prime models of $T$. In fact, $T$ cannot have any prime models, since only complete theories have prime models (Proof: If $M_1 \models T$ and $M_2 \models T$, and $T$ has a prime model $P$, then $P\preccurlyeq M_1$ and $P\preccurlyeq M_2$, so Th($M_1$) = Th($P$) = Th($M_2$), and hence $T$ is complete).

However, each $\mathfrak{A_i}$ is a prime model for its complete theory, Th($\mathfrak{A_i}$).


Further Edit: You write "somehow one would take advantage of the fact that any $P_i$ is either everything or nothing, hence any bijection would be elementary" and "I figure arbitrary types over finite parameteres could get really complicated." The first intuition is correct, the second is incorrect.

What's really going on here is that each of the four completions of $T$ has quantifier elimination. If you've learned about quantifier elimination, it's a good exercise to (a) prove that these four theories have quantifier elimination, then use quantifier elimination to (b) figure out what all the types look like (they're not complicated at all) and (c) show that any embedding of models is an elementary embedding.

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  • $\begingroup$ None of these are prime models of $T$ -- it can have no prime models, because it's not complete. $\endgroup$ – tomasz Apr 18 '13 at 18:54
  • $\begingroup$ No, but they're all prime models of their complete theories. Since "prime model" only makes sense for complete theories, the assertion "M is prime" should mean "M is a prime model of Th(M)". $\endgroup$ – Alex Kruckman Apr 18 '13 at 19:20
  • $\begingroup$ Ah, I see now that the question asked whether they are prime models of T. I missed that the first time. $\endgroup$ – Alex Kruckman Apr 18 '13 at 19:24
  • $\begingroup$ I'm a little confused. Why are you saying "All four" of these models are saturated? $\endgroup$ – FPP Apr 22 '13 at 16:31
  • $\begingroup$ I mean that each of your four models $\mathfrak{A}_0$, $\mathfrak{A}_1$, $\mathfrak{A}_2$, and $\mathfrak{A}_3$, is saturated. Which part is confusing you? $\endgroup$ – Alex Kruckman Apr 22 '13 at 17:32

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