Class $K^{\infty}$ is elementary provided that $K$ is.

Question

Let $K$ be an elementary class of $\Sigma$-structures. Show that the class $K^{\infty}$ formed by the structures of $K$ with infinite domain is also elementary.

I'm really clueless about this problem. I know that $K$ is elementary if there is som $\Phi \subseteq \mathrm{Sent}_{\Sigma}$ such that $K=\mathcal{Mod}(\Phi)$, that is $$K= \{ \mathcal{A} \mid \mathcal{A} \models \varphi \,\, \forall \varphi \in \Phi\}$$ so $$K^{\infty}=\{ \mathcal{A} \in K \mid |\mathcal{A}|=\infty\}$$ Obviously $K^{\infty}\subseteq K$, but I don't know what strategy to follow. Any hint would be highly appreciated. Thank you.

Answer 1

Suppose $\Phi$ is a set of sentences defining $K$ - that is, $K=\{\mathcal{A}: \mathcal{A}\models\Phi\}$. Now $K^\infty\subseteq K$ as you observe, so you want to find a bigger set of sentenecs $\Psi\supseteq \Phi$ such that $K^\infty=\{\mathcal{A}: \mathcal{A}\models\Psi\}$.

(Why bigger? Well, satisfying more sentences is a stronger constraint, and there are fewer things satisfying it.)

That means we want to add some sentences to $\Phi$. Now the obvious sentence to add is "the domain is infinite"; the structures satisfying $\Phi$ together with "the domain is infinite" are exactly the infinite models of $\Phi$, that is, the elements of $K^\infty$!

So we're done, right? Just set $\Psi=\Phi\cup\{$"the domain is infinite"$\}$.

But this doesn't work,since "the domain is infinite" isn't a first-order sentence. So instead, we need to somehow express "the domain is infinite" in a first-order way. On the plus side, we don't have to do this with a single sentence, we're allowed to add a bunch of sentences. Which leaves us with:

Can you think of a family of sentences $\Theta=\{\theta_i: i\in\mathbb{N}\}$ such that the models of $\Theta$ are exactly the infinite structures?

HINT: It's enough for the models of $\theta_i$ to be the structures with size at least $i$, for every $i\in\mathbb{N}$ ...