3 improved wording

As Sridhar already explained, Montague Lévy–Montague Reflection is a theorem scheme and not a single theorem which resolves the apparent contradiction, but here are a few more additional cool facts.

First, note that ZFC is not finitely axiomatizable (otherwise we would indeed have a contradiction) but there is a recursive listing of the axioms of ZFC. Let $\phi_0$,$\phi_1$,$\phi_2$,... be Let's fix such a listing . Suppose $\phi_0$,$\phi_1$,$\phi_2$,... If $M$ is a model of ZFC. Either , then either $M$ is an $\omega$-model (i.e. the finite ordinals of $M$ are truly finite) or it is not (i.e. $M$ has some nonstandard finite ordinals). Let's see what happens in each case.

Suppose first that $M$ is an $\omega$-model. The recursive listing $\phi_0$,$\phi_1$,$\phi_2$,... exists in $M$ and, by Montague, Lévy–Montague, people living in $M$ believe that $\{\phi_0,\ldots,\phi_n\}$ has a model for each $n < \omega$. Since people living in $M$ also believe in the Compactness Theorem, they also believe that there is a model of ZFC! . This is surprising, but note that the hypothesis that $M$ is an $\omega$-model is essential since without it we there is no reason for $M$'s notion of finite to agree with ours. This is where your initial reasoning strayed, you naturally assumed that every model of ZFC was an $\omega$-model, which is not true.\omega$-model. Suppose now that$M$is not an$\omega$-model. The recursive listing$\phi_0$,$\phi_1$,$\phi_2$,... makes sense in$M$, but since$M$has nonstandard finite ordinals this listing continues beyond the true$\omega$and people who live in$M$believe that these nonstandard$\phi_N$'s are real axioms of ZFC! By Montague, Lévy–Montague,$M$believes that $\{\phi_0,\ldots,\phi_n\}$ has a model for every standard$n$, but since Montague Lévy–Montague Reflection doesn't say anything about nonstandard axioms, there may be some nonstandard finite ordinal$N$in$M$such that people living in$M$do not believe that the nonstandard finite set $\{\phi_0,\ldots,\phi_N\}$ has a model. Now here is a funny thing that was pointed out by Joel David Hamkins in answer to another question. Suppose people in$M$is a model of ZFC + ¬Con(ZFC). Since people in$M$believe that their$\omega$is finite ordinals are wellordered, there must be a first finite ordinal$N$in$M$such that $\{\phi_0,\ldots,\phi_N\}$ has no model in$M$. This$N$must be nonstandard finite ordinal, and so must its predecessor$N-1$. Now By minimality of$N$, people in$M$believe that $\{\phi_0,\ldots,\phi_{N-1}\}$ does have a model. Let$M'$be such a model. Note that$M' \models \phi_n$for every standard axiom$\phi_n$since$n < N-1$. Therefore, although people living in$M$certainly don't believe it, this model$M'$is in fact a model of ZFC!!! Thus, Montague Lévy–Montague Reflection does imply that every model of ZFC contains another model of ZFC, but the models might are not be fully necessarily aware of that fact... 2 added link & changed some wording As Sridhar already explained, Montague Reflection is a theorem scheme and not a single theorem which resolves the apparent contradiction, but here are a few more interesting cool facts. First, note that ZFC is not finitely axiomatizable (otherwise we would indeed have a contradiction) but there is a recursive listing of the axioms of ZFC. Let$\phi_0$,$\phi_1$,$\phi_2$,... be such a listing. Suppose$M$is a model of ZFC. Either$M$is an$\omega$-model (i.e. the finite ordinals of$M$are truly finite) or it is not (i.e.$M$has some nonstandard finite ordinals). Let's see what happens in each case. Suppose first that$M$is an$\omega$-model. The recursive listing$\phi_0$,$\phi_1$,$\phi_2$,... exists in$M$and, by Montague, people living in$M$believe that $\{\phi_0,\ldots,\phi_n\}$ has a model for each$n < \omega$. Since people living in$M$also believe in the Compactness Theorem, they also believe that there is a model of ZFC! This is where your initial reasoning strayed, you naturally assumed that every model of ZFC was an$\omega$-model, which is not true. Suppose now that$M$is not an$\omega$-model. The recursive listing$\phi_0$,$\phi_1$,$\phi_2$,... makes sense in$M$, but since$M$has nonstandard finite ordinals this listing continues beyond the true$\omega$and people who live in$M$believe that these nonstandard$\phi_N$'s are real axioms of ZFC! By Montague,$M$believes that $\{\phi_0,\ldots,\phi_n\}$ has a model for every standard$n$, but since Montague doesn't say anything about nonstandard axioms, there may be some nonstandard finite ordinal$N$in$M$such that people living in$M$do not believe that the nonstandard finite set $\{\phi_0,\ldots,\phi_N\}$ has a model. Now here is a funny thing that was pointed out by Joel David Hamkins elsewhere on this sitein answer to another question. Suppose people in$M$is a model of ZFC + ¬Con(ZFC). Since people in$M$believe that their$\omega$is wellordered, there must be a first$N$such that $\{\phi_0,\ldots,\phi_N\}$ has no model . in$M$. This$N$must be nonstandard, and so must its predecessor$N-1$. Now people in$M$believe that $\{\phi_0,\ldots,\phi_{N-1}\}$ does have a model. Let$M'$be such a model. Note that$M' \models \phi_n$for every standard axiom$\phi_n$, so \phi_n$ since $n < N-1$. Therefore, although people in $M$ don't believe it, this model $M'$ is in fact a model of ZFC!!!

Thus, Montague Reflection does imply that every model of ZFC contains another model of ZFC, but the models might not be fully aware of that fact...

1

As Sridhar already explained, Montague Reflection is a theorem scheme and not a single theorem which resolves the apparent contradiction, but here are a few more interesting facts.

First, note that ZFC is not finitely axiomatizable (otherwise we would indeed have a contradiction) but there is a recursive listing of the axioms of ZFC. Let $\phi_0$,$\phi_1$,$\phi_2$,... be such a listing. Suppose $M$ is a model of ZFC. Either $M$ is an $\omega$-model (i.e. the finite ordinals of $M$ are truly finite) or it is not (i.e. $M$ has some nonstandard finite ordinals). Let's see what happens in each case.

Suppose first that $M$ is an $\omega$-model. The recursive listing $\phi_0$,$\phi_1$,$\phi_2$,... exists in $M$ and, by Montague, people living in $M$ believe that $\{\phi_0,\ldots,\phi_n\}$ has a model for each $n < \omega$. Since people living in $M$ also believe in the Compactness Theorem, they also believe that there is a model of ZFC! This is where your initial reasoning strayed, you naturally assumed that every model of ZFC was an $\omega$-model, which is not true.

Suppose now that $M$ is not an $\omega$-model. The recursive listing $\phi_0$,$\phi_1$,$\phi_2$,... makes sense in $M$, but since $M$ has nonstandard finite ordinals this listing continues beyond the true $\omega$ and people who live in $M$ believe that these nonstandard $\phi_N$'s are real axioms of ZFC! By Montague, $M$ believes that $\{\phi_0,\ldots,\phi_n\}$ has a model for every standard $n$, but since Montague doesn't say anything about nonstandard axioms, there may be some nonstandard finite ordinal $N$ in $M$ such that people living in $M$ do not believe that $\{\phi_0,\ldots,\phi_N\}$ has a model.

Now here is a funny thing that was pointed out by Joel David Hamkins elsewhere on this site. Suppose people in $M$ is a model of ZFC + ¬Con(ZFC). Since people in $M$ believe that their $\omega$ is wellordered, there must be a first $N$ such that $\{\phi_0,\ldots,\phi_N\}$ has no model. This $N$ must be nonstandard, and so must its predecessor $N-1$. Now people in $M$ believe that $\{\phi_0,\ldots,\phi_{N-1}\}$ does have a model. Let $M'$ be such a model. Note that $M' \models \phi_n$ for every standard axiom $\phi_n$, so although people in $M$ don't believe it, this model $M'$ is in fact a model of ZFC!!!

Thus, Montague Reflection does imply that every model of ZFC contains another model of ZFC, but the models might not be fully aware of that fact...