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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asked Jun 14, 2014 at 8:42
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3$\begingroup$ Related "mathoverflow.net/questions/69/…" $\endgroup$– Mohammad GolshaniJun 14, 2014 at 8:50
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1 Answer
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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.
answered Jun 14, 2014 at 9:01
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$\begingroup$ There is also a proof of Vaught's theorem in Chang & Keisler's "Model Theory", Theorem 2.3.15. $\endgroup$ Jun 14, 2014 at 14:14
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$\begingroup$ On the other hand, yes for $n\not=2$. $\endgroup$ Jun 14, 2014 at 17:23