# Complete, Finitely Axiomatizable, Theory with 3 Countable Models

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.

• 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
Commented Mar 18, 2015 at 13:03
• @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. Commented Mar 17, 2016 at 13:37
• @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).") Commented Dec 19, 2021 at 23:03
• @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) Commented Dec 20, 2021 at 7:27
• 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. Commented Dec 20, 2021 at 7:28