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If a theory has countably many countable models (up to isomorphism) then it has at countably many types, and it follows that there exists a countable $\omega$-saturated model of such theory.

If a theory has a countable $\omega$-saturated model then it has at most countably many types, but I can't see why (if at all) it follows that the theory has then at most countably many models (modulo isomorphism). Is this converse true?

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up vote 6 down vote accepted

This is not true. The idea for producing a counterexample is to make sure that there are only countably many types, but also to make sure that countably many of them are non-isolated, so that they can be omitted and realized at will. This will give continuum many models, depending on which subset of the non-isolated types are realized.

Here's the simplest explicit counterexample that I could see:

Let $\{P_i\,|\, i\in\omega\}$ be unary predicates, and let $\{c_{i,j}\,|\,i,j\in\omega\}$ be constant symbols.

$T$ asserts that the predicates are disjoint and the constants are distinct, and it assigns countably many constants to each predicate, i.e. $T = \{\lnot\exists x\, P_i(x)\land P_j(x)\,|\,i\neq j\} \cup \{c_{i,j}\neq c_{i',j}\,|\,i\neq i'\} \cup \{P_i(c_{i,j})\,|\,i,j\in\omega\}$.

The standard arguments for toy theories like this show that $T$ has quantifier elimination and is complete. The key thing is that each type $p_i(x): \{P_i(x)\} \cup \{x\neq c_{i,j}\,|\,j\in\omega\}$, asserting that $x$ is an unnamed element satisfying $P_i$, is non-isolated.

$T$ has a countable $\omega$-saturated model $M$, which has countably many unnamed elements satisfying each $P_i$, plus countably many elements not satisfying any $P_i$.

But $T$ also has continuum many non-isomorphic models. For any $S\subseteq \omega$, let $M_S$ be a model which has one unnamed element satisfying $P_i$ for all $i\in S$ and no unnamed element satisfying $P_i$ for all $i\notin S$.

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Trivial point: to make $T$ complete you need it to assert that the constants are all different. – Chris Eagle Feb 16 '13 at 18:25
You're right - thanks! I've updated the answer. – Alex Kruckman Feb 16 '13 at 18:25
I know this is an old post, but found it while trying to understand some things, how would you prove that this thing is complete and has QE?? I would like to know that. – Sara Jun 24 '15 at 9:10
There are many tests for QE, for example the one from Marker's book quoted in this math.SE question. Try to use this test to demonstrate QE for this theory. Once you have QE, completeness is not hard: A theory with QE is complete if and only if it decides every quantifier-free sentence. And in this case it's easy to make a list of all the atomic sentences and see that they're decided by $T$. – Alex Kruckman Jun 24 '15 at 19:28

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