# A theory with only countably many non-isomorphic countable models can be completed

Let $L$ be a countable first order language and let $T$ be a theory in $L$ with countably many non-isomorphic countable models, then there is a $L$-sentence $\phi$ such that the theory axiomatized by $T\cup\{\phi\}$ is complete.

Here is the skeleton of my proof which may or may not work:

If $T$ is complete, let $\phi\in T$, then $T\cup\{\phi\}=T$ is complete and we're done.

Otherwise if $T$ is not complete. Let $\phi$ be such that $\phi\notin T$ and $\neg\phi\notin T$.

By Vaught's theorem, it suffices to show: 1. $Ent(T\cup\{\phi\})$ has no finite model, and 2. there exists a cardinal $\kappa\geq\aleph_0+|L|=\aleph_0$ (since $L$ is countable) such that if $M\vDash Ent(T\cup\{\phi\})$ with $|M|=\kappa$, then $M$ is unique up to isomorphism.

Since $T$ is a theory, we have $T=Ent(T)$, where $Ent(T):=\{\phi\mid$ if $M\vDash T$ then $M\vDash\phi\}$.

For 1: Suppose for a contradiction $T\cup\{\phi\}$ has a finite model $M$. Since $\phi\notin Ent(T)=T$, there is a model $M'$ such that $M'\vDash T$ and $M'\vDash\neg\phi$. How can I find a contradiction from this?

For 2, let $M_1$ and $M_2$ be models of $Ent(T\cup\{\phi\})$ such that $|M_1|=|M_2|\geq\aleph_0$. We must show that $M_1\cong M_2$. But I'm not sure how to proceed?

Where should I use the countability assumptions?

• Hint: try to prove the contrapositive. – bof Oct 1 '17 at 4:09
• If $T\cup\{\phi\}$ has a finite model $M$ then there is a sentence $\psi$ such that the theory axiomatized by $T\cup\{\phi\land\psi\}$ is the complete theory $Th(M)$ and everything is fine. – bof Oct 1 '17 at 5:28
• Wait, how do you pick a sentence $\psi$ such that $T\cup\{\phi, \psi\}$ is complete? – Sid Caroline Oct 1 '17 at 6:18
• My previous comment was somewhat silly. I deleted it. – Alex Kruckman Oct 1 '17 at 16:16
• Did you use the assumption that $T$ has only countably many countable models? – Sid Caroline Oct 1 '17 at 20:40

## 1 Answer

Hint. Show that, if no finite extension of $T$ is complete, then $T$ has $2^{\aleph_0}$ complete extensions. Each of those extensions has a countable model, and no two of those models are elementarily equivalent, much less isomorphic.

• Sorry but I still don't quite follow. Why does $T$ has $2^{\aleph_0}$ complete extensions? What does $2^{\aleph_0}$ stand for? All we know is that $||L||=\aleph_0+|L|=\aleph_0$ because $L$ is countable, and thus there are countably many formulas in $L$. Also why does each such extension have a countable model? It is not given that $T$ has a countable model since countably many could mean zero. – W.Scott Oct 1 '17 at 4:50
• If $T$ has no countable models, then (being a theory in a countable language) $T$ is an inconsistent theory, whence it has no complete extension or is already complete, depending on whether your definition of complete theory includes inconsistent theories. $2^{\aleph_0}$ is the cardinal number of the power set of $\mathbb N,$ or the set of all infinite sequences of zeroes and ones, or the set of all real numbers. – bof Oct 1 '17 at 5:14
• If inconsistent theories are deemed to be complete, then the conclusion of the statement you're trying to prove should be "the theory axiomatized by $T\cup\{\phi\}$ is complete and consistent" or "complete and satisfiable". Otherwise there is a trivial solution $\phi=\exists x(x\ne x).$ – bof Oct 1 '17 at 5:21
• But why does $T$ have $2^{\aleph_0}$ complete extensions? Is it because $L$ is countable so it can't be more than $2^{\aleph_0}$, and for the contrapositive statement, suppose $T$ has less than $2^{\aleph_0}$ complete extensions, then $T$ has at most $\aleph_0$ finite extensions, which is equal to (and thus less than or equal to) the cardinality of the language $L$, so there is a non-empty collection of $L$-sentences each of which completes $T$. – Sid Caroline Oct 1 '17 at 20:31
• Do you know how to prove that a consistent theory has a complete extension? – bof Oct 1 '17 at 20:45