Using relativization to prove relative consistency Suppose I want to prove that, for some sentence $\phi$, $Con(ZFC) \to Con(ZFC + \phi)$. And to do this, I prove that for some classe $C = \{ x : \psi(x)\}$, $C \vDash ZFC$ and $C \vDash \phi$ - using the concept of relativization of formulas. Is this enough to prove the desired result?
I thought it would go along those lines: Supose $(\mathbf{M}, E)$ is a model for ZFC and consider the set $\mathbf{N} = \{ a \in \mathbf{M} : M \vDash \psi(a) \} \subseteq \mathbf{M} $. Does the relativization argument prove that $(\mathbf{N}, E)$ is a model of $ZFC + \phi$?
 A: Yup, this is correct - that would do the job. (EDIT: although there is a serious issue here, as pointed out by Eric Wofsey's answer; this is often swept under the rug because its treatment is fairly standard, but it's important to observe at first.)
It's worth noting however that this isn't the only way to prove consistency, though. In particular, by far the most important technique in set theory for establishing consistency results, forcing, does the opposite: given a countable$^1$ model $M$ of ZFC (which exists if ZFC is consistent, by the downward Lowenheim-Skolem theorem), forcing gives us a way to try to produce models of ZFC + [desired principle] which are bigger than $M$, rather than looking for a substructure inside $M$.

$^1$OK, a bit of clarification is necessary since otherwise you might very reasonably be confused if you look at the literature on forcing:


*

*In most presentations you'll see the hypothesis that a countable well-founded (or transitive) model of ZFC exists, which is strictly stronger than the mere consistency of ZFC. However, this is unnecessary: forcing can be developed over arbitrary countable models of ZFC without difficulty.

*A variant of forcing can be developed even over uncountable models of ZFC, where instead of building a literally larger model we produce a more general object - a "Boolean-valued model" - which is still enough for consistency results. This approach actually has a lot of virtues, but it's definitely more technically involved, and one should understand the usual approach to forcing first.

*Obnoxiously, actual research in set theory tends to be glib about the actual behavior of forcing. Most egregiously, we often talk about the existence of a forcing extension of $V$, the "true universe of all sets," which is clearly nonsense! What's going on here is that  we're really using a shorthand for either the Boolean-valued-model or the countable-model approach to forcing, since these are sufficiently well-understood that the translation between them (and the translation of the abuse of language used) is left to the reader. But this is absolutely a huge stumbling block to anyone learning this for the first time. That said, it's also totally addictive: it makes it so much easier to think up forcing arguments, even if the reasoning it produces looks a priori like gibberish!
A: Yes, that's correct.  Note that strictly speaking, you can't prove (or even state) something like $C \vDash ZFC$, since $ZFC$ has infinitely many axioms and you can only relativize one of them at a time.  What you can prove is a metatheorem which says that for each individual axiom $\psi$ of $ZFC$, $C\vDash\psi$ is a theorem of $ZFC$.  This is then enough to conclude that $(\mathbf{N},E)\vDash ZFC$, by interpreting each of these theorems $C\vDash\psi$ in the model $(\mathbf{M},E)$.
