Please see Joel Hamkins answer: https://mathoverflow.net/q/51786.
A section from his argument:
Suppose that M is a model of ZFC. Thus, in particular, ZFC is
consistent. If it happens that M is ω-standard, meaning that it has
only the standard natural numbers, then M has all the same proofs and
axioms in ZFC that we do in the meta-theory, and so M agrees that ZFC
is consistent. In this case, by the Completeness theorem applied in M,
it follows that there is a model m which M thinks satisfies ZFC, and
so it really does.
I don't understand the following two inferences:
1) "then M has all the same proofs and axioms in ZFC that we do in the meta-theory, and so M agrees that ZFC is consistent"
2) "it follows that there is a model m which M thinks satisfies ZFC, and so it really does"
Mainly, for (1) M agrees that ZFC is conistent because its an arithmetical statement and those are absolute for $\omega$-models, so what does this have to do with having "all the same proofs and axioms in ZFC that we do in the meta-theory". And for (2) why does $M \models (m \models \text{ZFC})$ imply $m \models \text{ZFC}$? After all, we didn't assume $M$ was transitive. I only know that the $\models$-relation is absolute for transitive models of $\text{ZF-P}$ (I think?).
Best Answer
The question is what do you mean by a "model of ZFC". If you mean the axioms as we enumerate them, in the meta-theory, this may or may not be the same thing as those of the universe of sets in which we are looking. The same can be told about the inference rules of FOL.
But since we can code everything into integers, that means that if we have a model whose integers are standard (read: agree with the meta-theory), then these problems go away.
What does that mean? Well, if the model was a model of ZFC, then it proves the completeness theorem, and since the arithmetic statement "ZFC is consistent" is true, that means that we can find a model of ZFC, and that it is the same ZFC as our meta-theory's one.
You might want to argue that the relation of that model might not be "correct" in some way, but we can make this model internally countable and the relation is then coded by a subset of $\omega$. So it's all good.