# An $\omega$-categorical theory $T$ with no finite models is complete.

I'm trying to understand the following proof of this result, but I don't understand:

1. where the assumption that $$T$$ has no finite models is used;

2. why the final step is valid.

Take two models $$\mathcal{M},\mathcal{N}$$ of $$T$$, and suppose $$\mathcal{M}\models\phi$$. We need to show $$\mathcal{N}\models\phi$$ (the proof works exactly the same if $$\mathcal{M}\not\models \phi$$).

By the definition I have of $$\omega$$-categorical, $$T$$ must be in a countable language with an infinite model. So, by the Downward Lowenheim Skolem theorem, $$\mathcal{M},\mathcal{N}$$ both have countable elementary substructures, $$\mathcal{M}',\mathcal{N}'$$. Since $$T$$ is $$\omega$$-categorical, $$\mathcal{M}'$$ is isomorphic to $$\mathcal{N}'$$.

Thus far I am fine, but the proof then concludes "by elementarily, $$\mathcal{N}\models\phi$$". This is the step that I don't follow. What I have thought of is: "by the Tarski-Vaught test, $$\mathcal{M}\models \phi$$ precisely when $$\mathcal{M}' \models \phi$$, which is precisely when $$\mathcal{N}'\models\phi$$, and if $$\mathcal{N}'\models\phi$$ then $$\mathcal{N}$$ models $$\phi$$ since $$\mathcal{N}'$$ is a substructure", but I'm not comfortable with this.

• The fact that a substructure is elementary says exactly that it satisfies the same formulas. May 7, 2019 at 14:56

## 1 Answer

You use the fact that $$T$$ has no finite models because you apply Löwenheim-Skolem to $$\mathcal{M}$$ and $$\mathcal{N}$$, which you can only do if they are infinite structures (Löwenheim-Skolem does not apply to finite structures).

For the second question: this follows directly from the definition of an elementary embedding. An embedding $$f: \mathcal{M}' \to \mathcal{M}$$ is elementary if for all tuples $$a \in \mathcal{M}'$$ and every formula $$\varphi(x)$$ we have that $$\mathcal{M}' \models \varphi(a)$$ if and only if $$\mathcal{M} \models \varphi(f(a))$$.

In this case we do not even need any parameters. That is, we are only interested in sentences. So particularly, having an elementary embedding $$f: \mathcal{M}' \to \mathcal{M}$$ means that for every sentence $$\varphi$$ we have $$\mathcal{M}' \models \varphi$$ if and only if $$\mathcal{M} \models \varphi$$.

The proof of Löwenheim-Skolem uses the Tarski-Vaught test to show that the embeddings obtained in that proof are elementary. After that, you need not concern yourself with the Tarski-Vaught test, it is the fact that we have elementary embeddings that is useful.

So if $$f: \mathcal{M'} \to \mathcal{M}$$ and $$g: \mathcal{N'} \to \mathcal{N}$$ are the elementary embeddings obtained from Löwenheim-Skolem ($$\mathcal{M'}$$ and $$\mathcal{N'}$$ both countable), then the argument should go as follows. Suppose that $$\mathcal{M} \models \varphi$$, then $$\mathcal{M'} \models \varphi$$ because $$f$$ is an elementary embedding. So by isomorphism of $$\mathcal{M'}$$ and $$\mathcal{N'}$$ (which we have by $$\omega$$-categoricity) we must have $$\mathcal{N'} \models \varphi$$. Then because $$g$$ is an elementary embedding, we find $$\mathcal{N} \models \varphi$$.

• This is extremely clear. It was partially the lack of parameters confusing me, because all the definitions I have use parameters, so that explanation is useful. May 7, 2019 at 15:03