In order to prove the completeness theorem we obviously need a framework such as ZFC (I'm aware that ZFC isn't the only possibility) so that we can talk about a language $\mathcal{L}$ and also about models of $\mathcal{L}$.
Now the completeness theorem makes perfect sense to me in so far as the language which we study has a model like first order logic ($\mathcal{L_=}$), ordered fields, groups, etc.
But then I hand a thought. By the completeness theorem if $\mathbf{ZFC} \models \varphi$ then $\mathbf{ZFC} \vdash \varphi$ (by $\mathbf{ZFC}$ I mean the axioms of ZFC). However $\mathbf{ZFC} \models \varphi$ means that any model $\mathfrak{U}$ for which $\mathfrak{U} \models \mathbf{ZFC}$ it is true that $\mathfrak{U} \models \varphi$. But how do we know that any model of ZFC exists? Since we can't construct a model of ZFC within ZFC, how can we interpret what the completeness theorem is saying?
So does the completeness theorem just not say anything about ZFC itself? Or is it the case that even though we don't know anything about models of ZFC inside ZFC we can still talk about them and because of this the completeness theorem does hold.
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The completeness theorem is a theorem of ZFC Set Theory. It says that if a first-order theory $T$ is consistent, then $T$ has a model (and vice versa).
This is a conditional statement: you can use the completeness theorem to get a model of the theory only if you know that said theory is consistent. Similarly, if you know that a theory has a model, you can be sure that theory is consistent.
The following is a particular instance of the theorem, obtained by taking ZFC as the theory $T$:
This theorem of ZFC leaves open two possibilities,
while ruling out the following others:
If ZFC really is free of contradictions, then it can't prove its own consistency or come up with its own model, so the axioms of ZFC do not pin down which of I or II hold.
Nevertheless, certain ZFC extensions might lean towards one outcome. E.g. the theory ZFC+$\neg \mathrm{Con}(ZFC)$ proves that ZFC is inconsistent, and by the completeness theorem also proves that ZFC has no model (I). Similarly, Tarski-Grothendieck set theory proves both that ZFC has a model and that ZFC is consistent (II).
But ZFC has no consistent extension which proves that ZFC is consistent but not that it has a model (III), nor a consistent extension which proves that ZFC has a model but not that it is consistent (IV): this is the meaning of the completeness theorem when applied to ZFC itself.