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This is a question originally asked at MSE a few years ago; the original poster hasn't been active in a while, so I'm taking the liberty of asking it here:

Is there a complete first-order theory $T$ with exactly two countable models up to isomorphism?

If we require $T$ to be countable the answer is negative (Vaught's "Never-Two" theorem). However, the proof of this breaks down completely once we allow $T$ to be uncountable. Various cardinality-mixing results are comparatively easy - e.g. there is a countable theory with exactly two models of cardinality $\aleph_1$ up to isomorphism - but none of them (that I'm aware of anyways) bear on this question.

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    $\begingroup$ Also, fun fact: A countable complete theory cannot have precisely $4$ models (up to isomorphism) of cardinality $\aleph_1$. $\endgroup$
    – James E Hanson
    Commented Sep 3, 2023 at 2:01
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    $\begingroup$ @JamesHanson The fact in your second comment is neat, do you have a citation for that? I'm not familiar with it. $\endgroup$
    – Noah Schweber
    Commented Sep 3, 2023 at 2:12
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    $\begingroup$ @NoahSchweber This is a modern reference (with a full classification of the spectra of countable theories). I was under the impression that the result was originally due to Lachlan, but I can't find it right now. $\endgroup$
    – James E Hanson
    Commented Sep 3, 2023 at 4:55
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    $\begingroup$ After some offline discussion, James and I realized that for any finite $k$, there is a countable theory with exactly $k$ models of size $\aleph_1$ up to isomorphism: take the theory of an equivalence relation with exactly $k$ classes, each of which is infinite. Then a model of size $\aleph_1$ is determined up to isomorphism by how many of the classes are uncountable. However, there is no countable theory with exactly $4$ models of size $\aleph_2$ up to isomorphism. Other numbers which cannot be realized in $\aleph_2$ include $2$, $8$, and $9$ (assuming I've done my computations correctly). $\endgroup$
    – Alex Kruckman
    Commented Sep 5, 2023 at 16:51
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    $\begingroup$ @AlexKruckman 'Tis done. $\endgroup$
    – Noah Schweber
    Commented Sep 6, 2023 at 18:44

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I posted an example of such a theory over as math.stackexchange.com: Does any uncountable complete theory have exactly two countable models? I'll avoid copying the whole thing, in case there are corrections to be made, but here's the summary:

Yes, this is possible. I'll describe an uncountable language $\mathcal L$ and two non-isomorphic elementarily equivalent countable $\mathcal L$-structures $M_0,M_1$ such that any other elementarily equivalent countable structure is isomorphic to one of these. The idea is that $M_0$ encodes the set of all subsets of $\mathbb N$ of even order, and $M_1$ encodes the set of all subsets of $\mathbb N$ of odd order, but the language doesn't distinguish which is which.

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