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In his 1965 paper Omitting Classes of Elements (found in The Theory of Models: Proceedings of the 1963 International Symposium at Berkeley, published by North-Holland Publ. Co., Amsterdam (1965)), Morley proved the following omitting types theorem:

For a complete countable theory $T$ and a 1-type $\Sigma$, if for every $\alpha<\omega_1$ there exists a model of $T$ with cardinality $>\beth_\alpha$ which omits $\Sigma$ then there is a model of $T$ omitting $\Sigma$ in every infinite cardinality.

Morley then claims without proof that this theorem still holds if $T$ is of cardinality $\lambda>\aleph_0$ by replacing $\omega_1$ in the statement by $(2^\lambda)^+$, but the proof given in the paper appears to fail for uncountable languages. Specifically, even after replacing $\omega_1$ with $(2^\lambda)^+$, the inductive step of the proof (Theorem 3.1 in the paper) fails at the limit cases.

So is there a way to prove that this statement indeed holds for uncountable languages? Or is the statement actually false when $T$ is uncountable?

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The key step in the proof is to show that given a sequence $(a_i : i < \beth_{(2^{|T|})^+})$, there is a sequence $(b_i : i < \omega)$ of indiscernibles such that for every $m$ there are $i_0 < i_1 < ... < i_{m-1}$ with $$tp(b_0...b_{m-1})=tp(a_{i_0}...a_{i_{m-1}}).$$

The proof of this assertion (in this generality) can be found in say Tent and Ziegler's book "A Course in Model Theory", or in Casanovas' book "Simple Theories and Hyperimaginaries".

The full proof of Morley's omitting types theorem can be found in the appendix of Baldwin's book Categoricity (I think it is even available for free from Baldwin's webpage.)

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  • $\begingroup$ Thanks for your reply, but could you explain how this fact can be used in the proof? Specifically, I am unsure of what the sequence $(a_i:i<(2^{|T|})^+)$ is supposed to correspond to in the context of Morley's proof for the countable case. $\endgroup$
    – Hanif Cheung
    Nov 5, 2015 at 22:57
  • $\begingroup$ More details can be found in Baldwin's book. (I've edited the answer to add this.) $\endgroup$
    – Levon Haykazyan
    Nov 6, 2015 at 13:01
  • $\begingroup$ Baldwin's categoricity is actually what got me lost: the proof started out trying to construct $\Phi_i$ for $i<(2^{|T|})^+$, but the actual induction given only has length $\omega$ (in fact, in his errata Baldwin corrects himself, claiming the induction should be of length $\omega$). But since the induction only deals with one term (or formula, for the indiscernibility part) per step, in the uncountable case you are still left with terms which have not been guaranteed to omit the type. $\endgroup$
    – Hanif Cheung
    Nov 7, 2015 at 10:25
  • $\begingroup$ How about this approach. Assume the theory has Skolem function (if not, add them). Now let $(a_i : i < (2^{|T|})^{+})$ enumerate a model $M$. Now given a cardinal $\kappa$ construct $(b_\alpha : \alpha < \kappa)$ as above. Let $N$ be the Skolem hull of $b_\alpha$'s. I think it should be easy to prove that every type realised in $N$ is already realised in $M$. $\endgroup$
    – Levon Haykazyan
    Nov 7, 2015 at 11:27
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    $\begingroup$ It seems to me that for the above outline to work you should start with a sequence indexed by $\beth_{(2^{|T|})^+}$ not just ${(2^{|T|})^+}$, to get indiscernible you want to have enough space to apply the Erdos-Rado theorem. $\endgroup$
    – Rami Grossberg
    Dec 10, 2015 at 17:56
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The most general theorems, covering also uncountable languages and uncountable sets of types can be found in Shelah's Classification Theory in section 5 of Chapter VII. Theorems VII.5.3 together with Theorem VII.5.5(2) prove not only the result of Morley you are asking about but at the same time extend two-cardinal ("wide apart") theorems named after Vaught and Morley.

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