"Fix some signature $L$ and let $T$ be an $L$-theory. Consider some family of formulae $\{ \varphi_n(x) ~|~n \in \mathbb{N} \}$. Let $P(x)$ denote the infinite conjunction $\bigwedge_{n \in \mathbb{N}}\varphi_n(x)$. Because this is an infinite conjunction, $P(x)$ is not defined in first order logic, but nevertheless for some $L$-models $\mathcal{A}$ and elements $a \in A$ (where $A$ is the universe of $\mathcal{A}$) it is still possible that $P(a)$ is true. Show that the following are equivalent:
- There is a first-order $L$-formula $\varphi(x)$ such that for all $\mathcal{A} \models T$ and $a \in A$ we have: $P(a)$ holds if and only if $\varphi(a)$ holds.
- There is some finite subset $F \subseteq \mathbb{N}$ such that for all $\mathcal{A} \models T$ and $a \in A$ we have: $P(a)$ holds if and only if $\bigwedge_{n \in F} \varphi_n(a)$ holds."
Thoughts: Showing that (2) implies (1) seems simple enough, just take $\varphi(x) = \bigwedge_{n \in F} \varphi_n(x)$.
I'm not sure how to go about proving that (1) implies (2), the infinite conjunction is throwing me off. We were given the hint that a compactness or an ultraproduct argument will work, but I'm having trouble applying their respective theorems to this question.
Any hints/answers are appreciated!
Hint : if for no finite $F\subset \mathbb{N}$, $\models \displaystyle\bigwedge_{n\in F} \phi_n(x) \implies \phi(x)$, then for all such $F$, the theory $\{\phi_n(x) \mid n\in F\}\cup\{\neg \phi(x)\}$ is consistent. What about the theory $\{\phi_n(x)\mid n\in \mathbb{N}\}\cup\{\neg\phi(x)\}$ ?