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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.

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    $\begingroup$ @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. $\endgroup$
    – hardmath
    Aug 29, 2014 at 15:24
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    $\begingroup$ 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.) $\endgroup$
    – HTFB
    Mar 18, 2015 at 13:03
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    $\begingroup$ @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. $\endgroup$ Mar 17, 2016 at 13:37
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    $\begingroup$ @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).") $\endgroup$ Dec 19, 2021 at 23:03
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    $\begingroup$ @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) $\endgroup$ Dec 20, 2021 at 7:27

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