$L$ models set theory and so does $V_\kappa$ for $\kappa$ inaccessible

Question

Background: I have been reading the 1980 edition of Kunen. Theorem VI.2.1 states it is provable from ZF that $\mathbf L$ (Kunen writes classes in bold) is a model of ZF. Also, it is a well-known argument that since $V_\kappa$ (Kunen would write this as $R(\kappa)$) models ZFC for $\kappa$ inaccessible, ZFC does not prove the existence of strong inaccessibles, by Goedel's Second Incompleteness Theorem.

Question 1: What is the difference between these two situations? Why is it that in the former the existence of a model of ZF is supposedly harmless while in the latter $R(\kappa)$ modeling ZFC is problematic?

Attempted answer to Question 1: Of course that is because $\mathbf L$ is a proper class while $R(\kappa)$ is a set if $\kappa$ exists. Thus in the latter situation one can formalize soundness theorem to get the formalized consistency sentence Con(ZFC) provable from ZFC. In the former, one only obtains the consistency of ZF (which is modeled by $\mathbf L$) in the metatheory under the condition of ZF being consistent in the metatheory (Theorem IV.8.2). Proper classes do not really exist.

Question 2: But people often treat proper classes in the same way as sets. Why can you not do the same thing to $\mathbf L$ as $R(\kappa)$ here? What is the general principle regarding this kind of issue?

Answer 1

There are two notions of "model" here. One is truly model-theoretic, using the relation of satisfaction which holds between structures and formulas. The other is merely proof-theoretic, using only the notion of provability of relativizations within a theory.

In the first sense of "model", it is nontrivial that $ZF$ proves that a subtheory has a model. For that's to say that $ZF$ proves a single statement "there is a set which satisfies all axioms of $T$". As you note, $ZF$ proves the soundness theorem, i.e., it proves "if some set satisfies all axioms of $T$, then $T$ is consistent". So if, in this first sense, $ZF$ proves that $T$ has a model, then $ZF$ proves "$T$ is consistent".

On the other hand, according to the merely proof-theoretic notion of "inner" model, the existence of models of $ZF$ is trivial. For, to say that $\mathbf C$ is an inner model of $T$ (according to $ZF$) is just to say that $ZF$ proves the relativization to $\mathbf C$ of every axiom of $T$. So, for example, $ZF$ has a one-line proof of the relativization to $\mathbf V$ of every instance of an axiom of $ZF$. So in this sense it might be said that $\textbf V$ is a model of $ZF$ according to $ZF$. Yet nothing follows about the provability in $ZF$ of $ZF$'s consistency. The application of soundness would require that the derived formulas have been relativized to a set, not just a class.

The relationship between the two notions of model and the set-class distinction is roughly as follows. $ZF$ can ask about "true" modelhood only with respect to what it considers to be sets. For Tarski's theorem implies that a concept of satisfaction is not in general definable with respect to classes. But conversely, Kunen makes clear (e.g. in the discussion around IV.8.2) that an inner model can have a set-sized domain: relative to some parameter $v$, the formula $x\in v$ may define an inner model of this or that theory.

Answer 2

The thing about "It is provable from $\sf ZF$ that $L$ is a model of $\sf ZF$" is that this is a bit misleading.

For every axiom of $\sf ZF$, it is provable from $\sf ZF$ that $L$ satisfies that axiom. Therefore it is a meta-theorem that $L$ satisfies $\sf ZF$. But that's fine for relative consistency argument, since those are meta-theory arguments anyway.

In the case of $V_\kappa$, this is not a meta-theorem, since $V_\kappa$ is a set, and therefore we have a truth predicate. In fact it means much more. Consider the case where $V$ has non-standard integers. This means that $\sf ZFC$ of the meta-theory and $\tt ZFC$ as internalized by $V$ are different. Now we can prove that $L\models\sf ZFC$, but there is no way we can prove that $L\models\tt ZFC$. Since those axioms do not live in the meta-theory and satisfaction of proper classes is not defined internally at all. On the other hand, $V_\kappa\models\tt ZFC$ since we can prove internally that it satisfies all the axioms.

And this, in a nutshell, is the heart of the issue for your second question. Since proper classes are not objects, they do not have a truth predicate, and therefore the statement $L\models\sf ZFC$ is a statement in the meta-theory, and the meta-theory's interpretation of the $\sf ZFC$ predicate. Whereas $V_\kappa$ is a set, so it has a truth predicate, and it can satisfy the internal interpretation of $\sf ZFC$.

Let me finish by presenting a theorem that I learned from this wonderful anwer by Joel Hamkins.

If $M$ is a model of $\sf ZF$, then there is some $N\in M$ such that $N$ is a model of $\sf ZF$.

Proof. If $M$ is an $\omega$-model, then it must agree with $V$ on the truth value of $\operatorname{Con}\sf (ZF)$. So the consequence is true. If $M$ is not an $\omega$-model, then in $M$ the axioms of $\sf ZFC$ from $V$ make a bounded part of the internal version of $\sf ZFC$ as interpreted by $M$.

Since bounded part means "internally finite", by the reflection theorem there is some $M$-ordinal $\theta$ such that $M_\theta\models\sf ZF$. $\square$

The point here is that:

1. $M$ thinks that $\sf ZFC$ as $V$ interprets it, is just a finite part of $\sf ZFC$ as it interprets it. In fact, it cannot recognize the "true part" of the axioms, but we know that we can use any non-standard integer as an upper bound.

2. Since $M$ is not well-founded, the membership of $M$ is not the real membership relation. So this does not contradict the regularity axiom.