Why can't we prove the consistency of ZFC by proving its axioms is satisfiable?

Question

We can prove the consistency of Peano Arithmetics by given a model of natural numbers, within this model, PA's axioms are satisfiable，thus PA is consistent. Why can't we do the same thing to ZFC

Answer 1

I think it may help to think about the Soundness Theorem as saying the following:

Let $\Sigma$ be a collection of sentences over some language $\mathcal{L}$, and let $\mathsf{S}$ be a collection of axioms of set theory. Suppose $\mathsf{S}$ proves that there is a model for $\Sigma$. Then $\mathsf{S}$ proves that $\Sigma$ is consistent. In particular, if $\mathsf{S}$ itself is consistent then so is $\Sigma$.

So if you prove the "consistency" of $\mathsf{PA}$ by constructing a model for it, using the axioms of $\mathsf{ZFC}$, then what you have really proved is relative consistency: if $\mathsf{ZFC}$ is consistent then so is $\mathsf{PA}$. Of course, you do not actually know whether $\mathsf{ZFC}$ is consistent, though it seems very likely.

You can attempt to take $\Sigma = \mathsf{ZFC}$ in the Soundness Theorem, but the question then is what do you take for $\mathsf{S}$? Gödel's second incompleteness theorem implies that you cannot take $\mathsf{S} = \mathsf{ZFC}$. Usually to construct a model of $\mathsf{ZFC}$, your set theory $\mathsf{S}$ is all the axioms of $\mathsf{ZFC}$ plus some extra ones, e.g. $\mathsf{I}=$ "there exists an inaccessible cardinal". But this is not very satisfying as a way to decide whether $\mathsf{ZFC}$ itself is consistent, since it's already trivial that $\operatorname{Con}(\mathsf{ZFC}+\mathsf{I}) \to \operatorname{Con}(\mathsf{ZFC})$.

It is very rare in logic to actually prove that a collection of axioms is consistent; it's usually only possible for very trivial examples.