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Let $\mathcal{L}$ be a countable first order language. For a natural number n, can we find a complete $\mathcal{L}$-Theory $T$ which has exactly n non-isomorphic countable models ?

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No, not for $n=2$. A theorem of Vaught says that a complete theory cannot have exactly two nonisomorphic models. A proof of this can be found in Shawn Hedman's "A First Course in Logic", for example.

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  • $\begingroup$ There is also a proof of Vaught's theorem in Chang & Keisler's "Model Theory", Theorem 2.3.15. $\endgroup$ – user50948 Jun 14 '14 at 14:14
  • $\begingroup$ On the other hand, yes for $n\not=2$. $\endgroup$ – Noah Schweber Jun 14 '14 at 17:23

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