# Proving the categoricity of an extension of ZFC2

In Asaf's answer to this question:

Proper-class-many categorical extensions of ZFC2

he confirmed that it is possible to categorically characterize a model of second-order ZFC by adding an axiom stating that There are exactly $\alpha$ inaccessible cardinals'', where $\alpha$ is an ordinal parameter.

Since $\alpha$ is an ordinal not a cardinal, presumably the axiom would really be asserting existence of a bijection from $\alpha$ to the inaccessibles in the model.

I have been trying to prove that second order ZFC + There's a bijection from $\alpha$ to the inaccessibles'' is categorical for all $\alpha$. Intuitively, it seems like it should be easy; we know from Zermelo that for any two models of ZFC2, either they are isomorphic, or one is isomorphic to a proper initial segment of the other. Moreover, any model of ZFC has height equal to an inaccessible.

I have been trying to prove this by showing that you get a contradiction by supposing that the heights of the models differ; if I can show that they have the same height, isomorphism follows by another theorem of Zermelo.

But I just can't get the right contradiction - I've tried violations of injectivity and surjectivity of the functions in each model, but since the bijections don't guarantee any sort of order on the range, I can't use the standard methods to show that the "shorter" model wouldn't have enough inaccessibles to satisfy the bijection.

Am I missing something obvious here?

Note that $\{\kappa\mid\kappa\text{ is an inaccessible cardinal}\}$ is a definable class of ordinals. So it has a uniquely determined order isomorphism with either an ordinal $\alpha$ or $\sf Ord$ itself.

What you're looking for is not a bijection, but rather an order isomorphism. So the fact that $\alpha$ is not a cardinal plays no role here.

• Am I understanding you correctly as follows: The proper spelling out of "There are exactly $\alpha$ inaccessibles" is rather "There is an order isopmorphism between $\alpha$ and the set $\{\kappa\mid\kappa\text{ is an inaccessible cardinal}\}$". In which case I should be able to prove that adding this to ZFC2 is categorical by some variation of the theorem that no well-ordering is order isomorphic to a proper initial segment of itself, right? – Mallik Jun 13 '18 at 21:17
• That is correct. – Asaf Karagila Jun 13 '18 at 21:22

First of all, the assertion "there are exactly $\alpha$ many inaccessibles" is a statement about the order type of the inaccessibles in the model; e.g. "there are exactly $\omega+1$-many inaccessibles" is different from "there are exactly $\omega$-many inaccessibles, since the former implies that there is a biggest inaccessible while the latter implies that there is no biggest inaccessible.

The desired categoricity result is a consequence of the following characterization of models of ZFC$_2$:

The set models of ZFC$_2$ are exactly (up to isomorphism) the sets $V_\beta$ for $\beta$ (strongly) inaccessible.

(For example, if in the "real world" there are exactly $17$ inaccessibles, then in the "real world" there are exactly $17$ isomorphism types of set models of ZFC$_2$.)

The proof of the characterization above proceeds broadly as follows. One direction - that $V_\alpha\models$ ZFC$_2$ if $\alpha$ is inaccessible - is trivial. The other direction is the interesting one. First, we establish that any set model $M\models$ ZFC$_2$ is well-founded, hence isomorphic to a transitive set. Assuming WLOG that $M$ literally is a transitive set, we next show that $M$ "computes powersets correctly" - that is, for each $x\in M$ we have $\mathcal{P}(x)\in M$ (alternatively, for each $x\in M$ we have $\mathcal{P}(x)^M=\mathcal{P}(x)$).

This means that $V_\alpha^M=V_\alpha$ for each $\alpha\in Ord^M$, and hence $M$ is $V_\beta$ for some ordinal $\beta$. Since $M$ computes powersets correctly, $\beta$ must be a strong limit ordinal, and so all that's left is to show that $\beta$ is regular (and this follows from the second-order Replacement axiom). So we've shown that every model of ZFC$_2$ is (isomorphic to) some $V_\beta$ with $\beta$ inaccessible, and we're done.

At this point it's clear that two set models of ZFC$_2$ are isomorphic iff they have the same ordertype of inaccessibles, since for each ordinal $\gamma$ there is at most one inaccessible $\beta$ such that $\{\alpha<\beta: \alpha$ is inaccessible$\}$ has ordertype $\gamma$, and so we're done.