An excercise about omitting types theorem

I am studying the omitting types theorem, which I know in the following form (where we implicitly work inside a monster model of a complete theory without finite models)

Theorem (Omitting types) Assume $$L(A)$$ is countable. Then for every consistent type $$p(x) ⊆ L(A)$$ the following are equivalent:

1. All models containing $$A$$ realize $$p(x)$$ $$\quad\quad\quad$$
2. $$A$$ isolates $$p(x)$$

Now, I am asked to proove the following:

Let $$p(x) ⊆ L(B)$$ and $$p_n (x) ⊆ L(A)$$, for $$n < ω$$, be consistent types such that: $$p(x) →\lor_{n<\omega}\ p_n(x)$$ Prove that there is an $$n<ω$$ and a formula $$φ(x) ∈ L (A)$$ consistent with $$p(x)$$ such that $$p(x) ∧ φ(x) → p_n (x)$$

Now, I know that it must hold that $$p(x)\rightarrow p_m(x)$$ for some $$m$$. And intuitively I would like to find some $$\phi(x)\in L(A)$$ which isolates something like $$p(x)\rightarrow p_m(x)$$, but I don't know how to go further, since I cannot translate this "implication" between types into a type itself, which moreover should be something over $$A\cup B$$

I am also asked to prove the following:

Let $$A\subset B$$ and $$p(x)\subset L(A)$$ suppose that for any solution $$a$$ of $$p(x)$$ the complete type over $$B$$, $$tp(a/B)$$ is isolated. Show that $$p(x)$$ is isolated.

Now, applying the theorem (with $$B$$) instead of $$A$$, I am able to conclude that any model containing $$B$$, contains also, for each solution $$a$$ of $$p(x)$$ an element which is in the same orbit (under automorphisms fixing $$B$$) of $$a$$. Then this will also be a solution of $$p(x)$$, since it is the image of $$a$$ through an automorphism which fixes $$B$$ and hence also $$A$$. But how to extend to models which contain $$A$$ but maybe not $$B$$ in order to apply the theorem again proving the claim?

Thanks in advance for any help

• Does "consistent type" mean complete type (i.e. $\varphi(x)\in p(x)$, or $\lnot \varphi(x)\in p(x)$, for all formulas $\varphi(x)$), or is a consistent type just a consistent set of formulas with free variables from $x$? – Alex Kruckman Nov 15 '20 at 18:31
• The second: it simply means a consistent set of formulas. Let me know if any other nomenclature sounds strange to you (my professor's is usually nonstandard). Thanks! – Francesco Bilotta Nov 15 '20 at 18:39
• It is not true that we must have $p\rightarrow p_m$ for some $m$ in general. For example, letting $A = \emptyset$ and $B = \{b_n\mid n\in \omega\}$ a countable set, the empty type over $A$ implies $\{x\neq b_n\mid n\in \omega\}\lor \bigvee_{n\in \omega} \{x = b_n\}$, but the empty type certainly does not imply any of these types. – Alex Kruckman Nov 15 '20 at 20:09

First question:

Let's prove the contrapositive. Assume that for all $$n$$ and all $$L(A)$$-formulas $$\varphi(x)$$, $$p(x)\cup \{\varphi(x)\}\not\models p_n(x)$$. We want to prove that $$p(x)\not\models \bigvee_{n\in\omega} p_n(x)$$, i.e. we want to realize $$p(x)$$ by an element which does not satisfy any of the $$p_n(x)$$.

One way to go is to introduce a new constant symbol $$c$$, consider the theory $$T\cup p(c)$$, and use the omitting types theorem to omit all the $$p_n(x)$$. But this is a little messy, what with the changing of languages, and it's also overkill: you don't really have to omit all the types $$p_n(x)$$, just make sure that $$c$$ doesn't realize them.

So I would prefer to prove this result by redoing a small part of the proof of the omitting types theorem. Note that we don't have to assume the language is countable in this proof, which is a big advantage over using omitting types!

We build a sequence of $$L(A)$$-formulas $$(\varphi_n(x))_{n\in \omega}$$ by induction, ensuring as we go that for all $$n$$, $$p(x)\cup \{\varphi_k(x)\mid k\leq n\}$$ is consistent. To pick $$\varphi_n(x)$$, we let $$\psi_n(x) = \bigwedge_{k. By induction, $$\psi_n(x)$$ is consistent with $$p(x)$$. So by our assumption, $$p(x)\cup \{\psi_n(x)\}\not\models p_n(x)$$. Thus we can find some element $$b$$ satisfying $$p(x)\cup \{\psi_n(x)\}$$ but not satisfying $$p_n(x)$$. Let $$\varphi_n(x)$$ be some formula which is true of $$b$$, but whose negation is in $$p_n(x)$$. Then $$p(x)\cup \{\varphi_k(x)\mid k\leq n\}$$ is consistent, since it is realized by $$b$$.

Now let $$q(x) = p(x)\cup \{\varphi_n(x)\mid n\in \omega\}$$. By compactness, $$q(x)$$ is consistent. Let $$c$$ be some realization. Then $$c$$ realizes $$p$$, but $$c$$ does not realize any $$p_n(x)$$, since for any $$n$$, $$c$$ satisfies $$\varphi_n(x)$$, and $$\lnot\varphi_n(x)\in p_n(x)$$.

Second question:

To me, this question seems totally unrelated to the first question and the omitting types theorem... I would prove it like this:

Let $$Q$$ be the set of complete types $$q(x)$$ over $$B$$ with $$p(x)\subseteq q(x)$$. For each $$q(x)\in Q$$, $$q(x)$$ is isolated by an $$L(B)$$-formula $$\varphi_q(x)$$. So $$p(x)\cup \{\lnot \varphi_q(x)\mid q\in Q\}$$ is inconsistent. By compactness, there are finitely many types $$q_1,\dots,q_n$$ such that $$p(x)\models \bigvee_{i=1}^n \varphi_{q_i}(x)$$. Let's call this disjunction $$\psi(x)$$. Note that also $$\psi(x)\models p(x)$$, since if $$\psi(a)$$, then $$a$$ realizes $$q_i(x)$$ for some $$i$$, and $$p(x)\subseteq q_i(x)$$. Thus we have shown that $$p(x)$$ is isolated by a formula over $$B$$.

The question as stated is a bit ambiguous, but I suppose the goal is to show that $$p(x)$$ is isolated by a formula over $$A$$. But this follows from the above, since if a set definable over $$B$$ is invariant over $$A\subseteq B$$ (i.e. fixed by all automorphisms of the monster model which fix $$A$$ pointwise), then the set is actually definable over $$A$$. I'll leave that as an exercise for you.

• Thanks, exceptionally clear as usual. Just one thing: in the answer to the first question, aren't you assuming that $p_n(x)$ is complete, to find the appropriate $\varphi_n(x)$? – Francesco Bilotta Nov 17 '20 at 11:56
• @FrancescoBilotta No - if $b$ fails to satisfy $p_n$, there is some formula in $p_n$ that is not true of $b$. The negation of such a formula is $\varphi_n$. – Alex Kruckman Nov 17 '20 at 13:14
• of course, thanks again – Francesco Bilotta Nov 17 '20 at 13:50