Standard Complete Model of ZFC and Reflection

Question

I am reading Levy's Axiom Schemata of Strong Infinity in Axiomatic Set Theory. In theorem 6, which says that this Reflection principle $N_0$ is equivalent to Replacement + Infinity in $ZF$ (in $S$ which is $ZF$ minus Replacement + Infinity, to be more precise), he skips the step from $N_0$ to Infinity & Replacement, which is the interesting part for me. Instead, Levy refers to a book by Montague and Feferman, The method of arithmetization and some of it's applications, that was never finished, as far as I know. (see bottom part of page 8)

This $N_0$ says the following: $$ \exists u(Scm^S(u) \& \forall x_1, \ldots, x_n(x_1, \ldots, x_n \in u \& \varphi \iff Rel(u, \varphi))) $$ Definition 1 from said paper says that $\alpha$ is called inaccessible iff $R(\alpha)$ ($V_\alpha$ in today's notation) is a standard complete model of ZF.

My question is: How can this possibly hold? Does is not violate Gödel's incompleteness?

Levy says he uses formulation of ZF axioms in non-simple applied first-order functional calculus, which sounds like second-order logic to me, but that still shouldn't allow him to prove the existence of inaccessibles, right?

What am I missing? Any help or further reference would be very appreciated, I'm stuck for several weeks already.

Answer 1

First of all, note that the reflection theorem doesn't provide an $\alpha$ such that $V_\alpha$ satisfies all of ZFC; rather, the reflection theorem provides, for each theorem $\varphi$ of ZFC, an $\alpha$ such that $V_\alpha$ satisfies $\varphi$. So we're not getting inaccessibles from reflection.

Second, note that while each instance of the reflection theorem is provable in ZFC - that is, for each theorem $\varphi$ of $ZFC$, $ZFC$ proves "There is some $\alpha$ such that $V_\alpha\models\varphi$" - ZFC does not prove "Every instance of the reflection theorem holds" (indeed, that's not even expressible!).

Finally, the proof of the above claim - that ZFC proves each instance of reflection - takes place in a metatheory slightly stronger than ZFC. So there is no contradiction.

A more modern source, like Jech's book, might be clearer about these things.

Answer 2

Regarding inaccessibles, the usual (modern) def'n of a weakly inaccessible cardinal is an uncountable regular limit cardinal, and a strongly inaccessible cardinal is an uncountable regular strong-limit cardinal.

If ZFC is consistent then the existence of an inaccessible is not a theorem of ZFC, because if $k$ is inaccessible then $(V_k)^L$ is a model of ZFC (where $L$ is Godel's constructible class). Adding an axiom that there is an inaccessible is analogous to the axiom of Infinity, which implies the consistency of Peano Arithmetic.

See "Set Theory: An Introduction To Independence Proofs", by Kenneth Kunen. There are also 2 books by Jech on set theory that I am not familiar with, but have a good reputation.