Montague's Reflection Principle and Compactness Theorem

Question

Here's a question I can't answer by myself: The Reflection Principle in Set Theory states for each formula $\phi(v_{1},...,v_{n})$ and for each set M there exists a set N which extends M such that the following holds

$\phi^{N} (x_{1},...,x_{n})$ iff $\phi (x_{1},...,x_{n})$ for all $x_{1},...x_{n} \in N$

Thus if $\sigma$ is a true sentence then the RFP yields a model of it and as a consequence any finite set of axioms of ZFC has a model (as a consequence ZFC is not finitely axiomatizable by Gödel's Second Incompleteness Theorem)

But why can't I just use now the Compactness Theorem (stating that each infinte set of formulas such that each finite subset has a model, has a model itself) to obtain a model of ZFC (which is actually impossible)??

Answer 1

For any finite set of axioms K of ZFC, ZFC proves "K has a model", via the reflection principle as you note. However, ZFC does not prove "for any finite set of axioms K of ZFC, K has a model". The distinction between these two is what prevents ZFC from proving that ZFC has a model.

(That is, even though, as you note, ZFC proves "if every finite set of axioms K of ZFC has a model, then ZFC has a model", as ZFC proves compactness, it does not follow that ZFC proves the consequent of this implication, as in fact ZFC does not prove the antecedent; ZFC only proves each particular instance of the antecedent, but not the universal statement itself.)

Answer 2

As Sridhar already explained, Lévy–Montague Reflection is a theorem scheme and not a single theorem which resolves the apparent contradiction, but here are a few 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's fix such a listing $\phi_0$,$\phi_1$,$\phi_2$,... If $M$ is a model of ZFC, 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 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.

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 Lévy–Montague, $M$ believes that $\{\phi_0,\ldots,\phi_n\}$ has a model for every standard $n$, but since 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 $M$ is a model of ZFC + ¬Con(ZFC). Since people in $M$ believe that their 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 $N-1$. 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 $M'$ is in fact a model of ZFC!!!

Thus, Lévy–Montague Reflection does imply that every model of ZFC contains another model of ZFC, but the models are not necessarily aware of that fact...