Why do you need the full semantics to prove the quasi categoricity of second-order ZFC

Question

Why you need the full semantics for SOL to prove Zermelo’s Quasi-Categoricity Theorem. Which step relies on it? Thanks.

Answer 1

I will follow Kanamori's proof (Theorem 1.3. of Kanamori 2013.) This theorem only shows $V_\kappa$ models second-order ZFC under full semantics iff $\kappa$ is inaccessible. However, it is not hard to prove from Kanamori's result that every model of second-order ZFC is isomorphic to $V_\kappa$ for some inaccessible $\kappa$:

Proof. By regularity, we can prove every non-empty unary relation over a model $X$ has a minimal element. Therefore, $X$ is well-founded.

Now consider its transitive collapse $M$. Since the model satisfies second-order separation, the transitive collapse is closed under subsets of elements. By induction on $\alpha<\delta$, where $\delta$ is the height of $M$, we can show $V_\alpha$ is a subset of $M$. From this, we can conclude $M=V_\delta$.

In sum, every model of second-order ZFC is isomorphic with $V_\delta$ for some $\delta$. Now apply Kanamori's Theorem 1.3.

The main feature of full semantics, unlike Henkin semantics, is it can access arbitrary subsets and function over the domain. Since Kanamori's proof uses somewhat arbitrary functions over $V_\kappa$, every part of the proof needs full semantics.

Let us examine the regularity of $\kappa$ as an example. Kanamori starts the proof with the existence of $\alpha<\kappa$ and a cofinal function $G:\alpha\to \kappa$ to derive a contradiction. We can see that $G$ can be extended to a function from $V_\kappa$ to $V_\kappa$ without changing its range (this is not what Kanamori did, but it does not harm the main outline of the proof.) Therefore, $V_\kappa$ can access the range of $G$ due to second-order replacement.

This proof breaks out if we use Henkin semantics: we do not know whether $G$ is included in a range of function variables of the given model. Therefore, we cannot ensure the second-order replacement is applicable to $G$.

We can also see that the remaining part of the proof also breaks down by a similar argument. (Even worse, we cannot even ensure every model of second-order ZFC is well-founded! In fact, there is an ill-founded model of second-order ZFC under Henkin semantics.)