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Recall Vaught conjecture: the number of countable models of a first-order complete theory in a countable language is finite or $\aleph_0$ or $2^{\aleph_0}.$

Now let $\lambda$ be an uncountable cardinal. Is the following version of Vaught conjecture true:

Vaught conjecture for uncountable languages. If $T$ is a complete theory in a language of size $\lambda,$ and if $T$ has more than $\lambda-$many non-isomorphic models of size $\lambda$, then $T$ has $2^\lambda$ many non-isomorphic models models of size $\lambda.$

Also, do we have the following version of Moreley theorem:

Morley theorem for uncountable languages. If $T$ is a complete theory in a language of size $\lambda,$ and if $T$ has more than $\lambda^+-$many non-isomorphic models of size $\lambda$, then $T$ has $2^\lambda$ many non-isomorphic models models of size $\lambda.$

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I am wondering about your formulation of "Morley's theorem for uncountable languages". In fact, there is a theorem with this name and it has been proven by Shelah. It states: If a theory in a language with cardinality $\kappa \geq \omega$ is categorical in some cardinality greater than $\kappa$, then it is categorical in every cardinality greater than $\kappa$. However, I cannot immediately see how this fits to your variant of Morley's theorem. –  user42061 Oct 30 '13 at 13:29
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tobi, I think he's referring to the following theorem of Morley (which can be found in chapter 4 of Marker's book): Let $T$ be a complete theory in a countable language. If $I(T,\aleph_0)>\aleph_1$ then $I(T,\aleph_0)=2^{\aleph_0}$. –  Haim Oct 30 '13 at 18:52
    
Thanks Haim, now it makes more sense to me! I did not knew this theorem till now. –  user42061 Oct 30 '13 at 20:31
    
By the way, it's interesting to note that the proof of Morley's theorem relies on the fact that if $X$ is countable and $Y\subseteq 2^{X}$ is analytic, then either $|Y|\leq \aleph_0$ or $|Y|=2^{\aleph_0}$. So, towards a proof of a generalized Morley theorem, it would be natural to expect a generalization of this classical result from descriptive set theory. Perhaps something like that was already done by Sy Friedman and his collaborators? I don't know. –  Haim Oct 31 '13 at 0:20

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