# $\Sigma\subset Th(\mathbb{N})$ is finite, so $\Sigma$ has $2^{\aleph_0}$ non-isomorphic models

I want to prove the statement in the title using Godel's theorems. Without Godel's theorem, I may use the theorem about infinite model from cardinality $\kappa$ $\Rightarrow$ infinite models from any cardinality $>\kappa$ to get as many non-isomorphic models as I wish (cause $\mathbb{N} \models \Sigma$)

I was trying to use Godel's theorem in the next manner, every recursive $\Sigma \subset Th(\mathbb{N})$ is not complete (and so for finite subsets as they are recursive), therefore one can find sentence $\alpha$ such that $\Sigma\nvdash \alpha , \neg\alpha$ so $\Sigma\cup\alpha$ , $\Sigma\cup\neg\alpha$ has models $M_1 , M_2$ because there are consistent ($\Sigma\subset Th(\mathbb{N})$ and therefore consistent ) . $M_1,M_2$ are not isomorphic cause they are not elementary equivalent.

I don't find an Idea to generalize this approach to get $2^{\aleph_0}$ models, any other approach will be appreciated, as long as it is using Godel's theorems

• For yet another approach, Andrés' comment on this answer suggests a compactness-based argument that $\mathbb N$ (or any other model of Q) has continuum many nonisomorphic countable elementary extensions. – Henning Makholm Jun 23 '18 at 18:37

At each node in the tree you have a finite set of sentences $B$ such that $PA+\Sigma+B$ is consistent. Apply the Gödel-Rosser construction to this theory to get an undetermined $\alpha$. The two successors of the node will arise by appending either $\alpha$ or $\neg\alpha$ to $B$.
The tree has countably many nodes, but $2^{\aleph_0}$ many infinite branches. Each branch has a model thanks to compactness. The models of two different infinite branches are non-isomorphic because they disagree about the $\alpha$ where their branches diverge from each other.
• I am having some trouble with "models of two different branched are non-isomorphic" - when branching - the descendants of $M\models \Sigma$ , $M_1\models \Sigma \cup \alpha, M_2\models \Sigma\cup\neg\alpha$ are non isomorphic but potentially one of them might be isomorphic to $M$ from which they were branched, so maybe this process doesn't result with $2^\aleph_0$ models which are not isomorphic in pairs (every 2 are non-isomorphic to each other) – dan Jun 23 '18 at 13:03
• @dan: Sorry, sloppy phrasing on my part. I'm really only considering models for the infinite branches -- I don't count models of the finite intermediate steps at all. So in your example, $M$ is not counted in its own right among the $2^{\aleph_0}$ models. – Henning Makholm Jun 23 '18 at 13:06
• In Gödel's original work the condition was that $\Sigma$ is "$\omega$-consistent", which is a rather technical property that he could argue is "obviously true" for a theory that is true about the actual integers. Later Rosser improved the construction so it now works for any $\Sigma$ that is merely consistent and contains sufficiently much arithmetic, no matter if it is true in $\mathbb N$. That's what I actually meant when I said "Gödel's construction". – Henning Makholm Jun 25 '18 at 9:58