I am trying to understand the following consequence of Löwenheim-Skolems theorem:
Let ∑ be a set of sentences in a countable language. If ∑ has any model, then it has a countable model.
I do not really understand if it means that ∑ has a countable model but also can have non-countable models or if it only can have a countable model/models? I appreciate if anyone would help me clarify this, thanks in advance!
This has already been answered in the comments, but to make things complete let me explain why. I think Brian's statement phrases this very well:
Here we take countable to mean "finite or countably infinite".
If $\Sigma$ has a finite model, then that is directly the countable model we were looking for. If $\Sigma$ has an infinite model $M$ we will need to produce a countable model. In fact, we will show that we can produce a model of cardinality $\kappa$ for any infinite $\kappa$, just as in Brian's improved claim. By compactness can find an elementary extension $M'$, so $M \preceq M'$, such that $|M'| \geq \kappa$. This is sometimes also called "upwards Löwenheim-Skolem", but it really is a simple argument where you just add $\kappa$ many new constant symbols and declare them to be distinct. Now pick any subset $A \subseteq M'$ with $|A| = \kappa$, then by the downward Löwenheim-Skolem theorem there is some $N$ with $A \subseteq N \preceq M'$ such that $|N| = |A| + |\Sigma| + \aleph_0$. Since $|A| = \kappa \geq \aleph_0$ and $|\Sigma| = \aleph_0$ we get $|N| = \kappa$. So $N$ is our model of cardinality $\kappa$.
So in fact this shows that the only way that $\Sigma$ can only have countable models is if it has only finite models. Any theory with infinite models will have arbitrarily large infinite models.