# If $\Phi$ has only finite models, there is some $n\in \mathbb{N}$ such that each model has at most $n$ elements.

I want to prove the following claim:

Let $\Phi \subseteq \mathrm{Sent}_{\Sigma}$ be such that each model $\mathcal{M}$ of $\Phi$ is finite. Show that there is some $n\in \mathbb{N}$ such that each model of $\Phi$ has at most $n$ elements.

My try: I have proved the following statement which I can use as a fact: If $\Phi \subseteq \mathrm{Sent}_{\Sigma}$ admites finite models of arbitrarily big cardinal, the $\Phi$ has an infinite model.

So let's assume otherwise, that is, given $n\in \mathbb{N}$, suppose I can find a model $\mathcal{M}\in \mathcal{Mod}(\Phi)$ such that $|\mathcal{M}|\geq n$. This means I can find models of $\Phi$ arbitrarily big, so by the fact above, $\Phi$ possesses an infinite model $\mathcal{N}$, which contradicts the hypothesis that $\Phi$ has only finite models.

Is this prove okay? It looks like a simple proof by contradiction and that makes me doubt.

• You seem to be confusd about what you want to prove. The claim you are after says "at least $n$ elements", so yoy want to show that there are no models with less than $n$ elements. – Mariano Suárez-Álvarez Jun 4 '17 at 19:01
• As for the claim you want to prove: why don't you just take $n=1$? – Mariano Suárez-Álvarez Jun 4 '17 at 19:03
• @MarianoSuárez-Álvarez Ohh so sorry, I meant "at most", not "at least". Translation problems. I'll edit the question. Thank you! – user313212 Jun 4 '17 at 19:09

The proof is not correct. Arbitrarily big is not the same as infinite.

The standard proof of this fact makes use of the compactness theorem.

It is easy to construct a sentence $S_n$ such that $S_n$ is true in a model iff the model has more than $n$ elements (can you show this?).

Then consider $\Phi + \{S_n\}_{n=1}^{\infty}$. Clearly every finite subset of the theory is satisfiable, since $\Phi$ has arbitrarily large models.

But then by compactness the whole theory is satisfiable, and since only an infinite model can satisfy all the $S_n$, we have found an infinite model for our theory $\Phi$.

• Excuse me, but how this answers my question? You answered to "arbitrarily big finite models implies that there is an infinite one". I just said I knew this result and could actually use it as a fact. – user313212 Jun 4 '17 at 19:27
• @user313212 Then your reasoning is correct. I thought you though the result was trivial, which it is not by any means (compactness is a huge theorem) – Jsevillamol Jun 4 '17 at 19:35

Yes, that proof is correct - the statement you're trying to prove is basically the contrapositive of the statement you already know.