I have one problem and I m sure that can be solved by using compactness theorem but I cant solve it.
Let $T$ be an $L$-theory and $\{F_i (x) \mid i\in I\}$ family of $L$-formulas. Suppose further that every element of every model of theory $T$ satisfies at least one of $F_i (x)$. Prove that exist finite $J \subseteq I$ and $T \vDash \forall x (\lor F_i(x))$. Disjunction is on finite $J$.
I try to suppose that for every finite $J \subseteq I$, $T \vDash \forall x (\lor F_i(x))$ doesn't hold. So, exists some model $m \vDash T$ and $m \vDash \lnot \forall x (\lor_J F_i(x))$. So, $m \vDash \exists x\lnot (\lor_J F_i(x))$.
So, for every finite $J \subseteq I$, exists $m \vDash T$ and exists some $b\in M$ and $m \vDash \lnot (\lor_J F_i[b])$.......$m \vDash (\land_J \lnot F_i[b])$........$(\forall i\in J)$ $m \vDash \lnot F_i[b]$ .
1) We have that every model $M$ of $T$ (i.e. $M \vDash T$) satisfy at least one $F_i$.
2) Consider the set of formulae $\{ \lnot F_i \}_{i∈I}$; no model of $T$ can satisfy it, becuase for each model $M$ of $T$ there is at least one $\lnot F_i$ that is not satisfied by it.
Thus, applying Compactness theorem to the set of models of $T$, we have that exists finite $J \subseteq I$ such that $\{ \lnot F_j \}_{j∈J}$ is not satisfiable by any model of $T$.
3) Let $N$ the largest $j \in J$ and consider $\{ \lnot F_1, \lnot F_2, \ldots, \lnot F_N \}$ that is again unsatisfiable by any model of $T$. But this means that every $M$ satisfy at least one of $F_1, F_2, \ldots, F_N$ i.e. satisfy $\lor_{j∈J} F_j$ (it is a finite disjunction, and thus a well-formed formula).
4) In conclusion, we have that every model $M$ of $T$ is a model of $\lor_{j∈J} F_j$, i.e.