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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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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

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