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Does there exist a complete, finitely axiomatizable, first-order theory $T$ with exactly 3 countable non-isomorphic models?

A few relevant comments:

There is a classical example of a complete theory with exacly $3$ models. This theory is not finitely axiomatizable (For the trivial reason that the language is infinite).

In this post, Javier Moreno explains how to rephrase this example in a finite language. Still, the theory is not finitely axiomatizable.

Some less relevant comments:

I would like to know if finite axiomability has ever be asked in this context.

There have been research in connection with stability. Lachlan has proved that a superstable theory with finitely many countable models is $\omega$-categorical. And it is still open if this can be extended to all stable theories.

bittahProfessional
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Primo Petri
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    @James pointed out my misreading of the Question that these 3 countable nonisomorphic models are the only models, interpreting instead that among its countable models, there are exactly three isomorphism classes. – hardmath Aug 29 '14 at 15:24
  • Thanks for the comments, which make the question much more accessible to someone who hasn't thought about such things before. – Carl Mummert Aug 29 '14 at 18:01
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    Wilfred Hodges' Model Theory theorem 12.2.18 states that a totally categorical theory $T$ (a) is not finitely axiomatisable and (b) is quasi-finitely axiomatisable; where the latter property means that it is definitionally equivalent to a single sentence together with all the sentences "there are at least $n$ elements" for each $n$. And further that an $\omega$-stable and $\omega$-categorical $T$ is not finitely axiomatisable either. (So your second question--has anybody thought about these questions before?--has answer "yes". Hodges says the theorem answers a conjecture of Vaught.) – HTFB Mar 18 '15 at 13:03
  • Note though that the usual Ehrenfeucht example of a 3-model theory is unstable. – HTFB Mar 22 '15 at 17:06
  • This post might give you a lead, but I don't have David Marker's model theory book -- http://mathoverflow.net/questions/143394/are-all-complete-finitely-axiomatizable-first-order-theories-aleph-0-categori – Jonny May 29 '15 at 01:14
  • Do you mean that there are $n\ge 3$ models, but only $3$ isomorphically distinct models, or that there are $3$ models, and they are not isomorphic? – DanielV Mar 11 '16 at 23:48
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    @DanielV There are 3 models up to isomorphism. "Up to isomorphism" is usually left implicit. If you do not count "up to isomorphism" there is always a proper class of models. – Primo Petri Mar 17 '16 at 13:37
  • Let ∆ be the theory of dense linear orders without endpoints (finitely axiomatizable), then the classic Ehrenfeucht example theory T is axiomatized by $∆∪{c_k<c_{k+1} | k ∈ ω}$ which has 3 countable models up to isomorphism. If you want it to be finitely axiomatized, we can simply syntactically change above set of countably infinite number of axioms to a single $\Pi_1$ sentence: $\forall k. c_k<c_{k+1}$ where $k$ ranges over another sort of natural number type. This theory's alphabet is infinite, but it by no means we cannot have a finitely axiomatized theory using this language. – cinch Dec 19 '21 at 04:17
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    @mohottnad That doesn't help at all. "$\forall k.c_k<c_{k+1}$" is not a first-order sentence: first-order logic doesn't let you quantify over symbol indices like that. And re: "If you want it to be finitely axiomatized," finite axiomatizability is the whole point of the question (see e.g. the passage "There is a classical example of a complete theory with exacly 3 3 models. This theory is not finitely axiomatizable (For the trivial reason that the language is infinite).") – Noah Schweber Dec 19 '21 at 23:03
  • @NoahSchweber thanks for your critique and I agree my above sentence is not well formed. However, similarly here's some post suggests "every recursively axiomatizable theory in first-order logic with identity that has only infinite models, has a finitely axiomatized conservative extension... simply endow the given theory with a new sort.." which seems applicable here for above theory with 3 countable models. Another way for a single sentence using $L_{w1,w}$ logic? – cinch Dec 20 '21 at 07:24
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    @mohottnad It is not true that if $T$ is a conservative extension (even in the "strong" sense) of $S$ then $T$ has the same number of countable models (up to isomorphism) as $S$; the "additional structure" of $T$ may introduce additional models. All we can say is that there will be the same number of reducts to the language of $S$ of countable models of $T$, up to isomorphism, as there are countable models of $S$ up to isomorphism. But that's much weaker. As to changing to $\mathcal{L}_{\omega_1,\omega}$, note that changing the logic also changes the meaning of "complete," (cont'd) – Noah Schweber Dec 20 '21 at 07:27
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    and by Scott's theorem a complete $\mathcal{L}{\omega_1,\omega}$-theory has at most one countable model up to isomorphism (as long as the language is countable, anyways); actually, Scott showed that for every countable structure $\mathcal{A}$ (in a countable language), there is a single $\mathcal{L}{\omega_1,\omega}$-sentence $\varphi$ such that for all countable $\mathcal{B}$ we have $\mathcal{B}\models\varphi\iff\mathcal{B}\cong\mathcal{A}$ (google "Scott sentence"). So none of this is relevant. – Noah Schweber Dec 20 '21 at 07:28

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