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

  • 3
    $\begingroup$ In order to answer the question "which step of the proof relies on using the full semantics?", we need to know what proof you have in mind. Do you have a reference? $\endgroup$ Mar 9, 2020 at 0:11
  • 1
    $\begingroup$ Button and Walsh 2019 offer a more elaborate version of the one in Konamori 2013. That is the one I am looking at. But I’d be happy to look elsewhere for a different proof that makes answering my question easier. $\endgroup$
    – Carlo Lori
    Mar 9, 2020 at 6:34

1 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.)

  • $\begingroup$ My favorite (I forget where I saw this) is "it is hard not to prove." $\endgroup$ Mar 9, 2020 at 19:01
  • $\begingroup$ @NoahSchweber The main role of this sentence is to omit proofs and make beginners perplexing. $\endgroup$
    – Hanul Jeon
    Mar 9, 2020 at 19:13
  • $\begingroup$ Thank you, that is really helpful. $\endgroup$
    – Carlo Lori
    Mar 10, 2020 at 6:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .