# On "Every model of ZFC has an element that is a model of ZFC"

There is a theorem saying that every model of ZFC has an element that is a model of ZFC (see Joel Hamkins's post https://mathoverflow.net/questions/51754/clearing-misconceptions-defining-is-a-model-of-zfc-in-zfc/51786#51786 and Noah Schweber's answer here AST is conservative over ZF).

My question is why the theorem does not lead to a contradiction by the following argument (with the assumption that ZFC is consistent)?

Let $$\alpha$$ be the least ordinal such that there is a set $$M$$ of rank $$\alpha$$ such that $$\langle M, \in^M \rangle \models$$ ZFC with $$\in^M \subseteq M \times M$$. Fix such $$M$$. Then the theorem implies that there is some $$A \in M$$ such that $$\langle A, \in^A\rangle \models$$ ZFC for some $$\in^A \subseteq A \times A$$. But A has rank $$< \alpha$$.

The "model of ZFC in $$M$$" that exists is represented by some element $$A\in M$$ together with a relation $${\in^A}\in M$$, but this model is not literally the set $$A$$ with that relation; instead it is the set $$B=\{a:a\in^M A\}$$ together with the relation $${\in^B}=\{\langle a,b\rangle:M\models a\in^A b\}$$. This set $$B$$ can have the same rank as $$M$$, since it need not be an element of $$M$$.
(Indeed, it is easy to see that $$\alpha=\omega$$ since $$M$$ can be taken to have underlying set any countable set, say $$M=\omega$$. Then $$A$$ will be a finite ordinal, but in the structure $$\langle M,\in^M\rangle$$ it will represent an infinite set.)
• I see, thanks. But isn't this "theorem" a bit of abusing the terminology? When we say a structure models ZFC, don't we mean that the structure has a domain $D$ and an interpretation $I$ of $\in$ which is a subset of $D \times D$? Commented Feb 2 at 19:27