On equivalent formulas Couyld someone give some hints on how to prove this : 
Claim: Let  $T$ be a  $\tau$-theory  and $\varphi$ a  $\tau$-formula, that have the same models. Then $\varphi$ is equivalent to a finite conjunction of formulas of $T$.
I was thinking about arguing that if $\mathcal{M} \models T$ then $\mathcal{M} \models \varphi$. So $T \models \varphi$. Then by compactness there must be some finite $\Delta \in T$ such that $\Delta \models \varphi$. Then I claim that $\mathcal{M} \models \varphi $ iff $\mathcal{M} \models \Delta$. Is this correct ?  
 A: Yes, your argument is correct. As long as your definition of "$\varphi$ is equivalent to $\Delta$" is that $\varphi$ and $\Delta$ have the same models. 
The last step could do with some more explanation. The reasoning is hidden below. 

 If $\mathcal{M}\models \varphi$, then $\mathcal{M}\models T$ (by assumption), so $\mathcal{M}\models \Delta$ (since $\Delta\subseteq T$). And conversely, if $\mathcal{M}\models \Delta$, then $\mathcal{M}\models \varphi$, since $\Delta\models \varphi$. 

A: You have to explicit the compactness argument.
Since any model of $\varphi$ is a model of $T$, one has $\varphi \vdash T$. In particular, for any conjunction of conjunctions of formulas of $T$, say $\Delta$, we have $\vdash \varphi \to \Delta$.
Assume for a contradiction that $\varphi$ isn't equivalent to any finite conjunction of elements of $T$. Then, the following set is finitely consistent :
$$T' := \{\Delta \wedge \neg \varphi \ \big| \ \Delta \textrm{ finite conjunction of formulas in } T \} $$
But a model of $T'$ is a model of $T$ that satisfies $\neg \varphi$, a contradiction.
