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This is a clarificatory question regarding Asaf's answer to this question:

What axioms need to be added to second-order ZFC before it has a unique model (up to isomorphism)?

He describes the technique of defining a categorical extension of ZFC2 by adding an axiom asserting "There are exactly $\kappa$-many inaccessibles", where the extension is satisfied by a model of ZFC2 of the form $V_{\kappa_1}$. (Where $\kappa$ is a cardinal and $\kappa_1$ is the next inaccessible after $\kappa$)

(a) How many categorical extensions is it possible to obtain by this technique?

(b) If we wanted to describe proper-class-many categorical extensions, and we allow ourselves proper-class-many ordinals as parameters, would that make the technique described above work for proper-class-many inaccessibles? (I.e. if we use inaccessible ordinals in place of $\kappa$ in "$\kappa$-many"...)

(c) What sort of problems is he referring to if K $\cap V_{\kappa}$ is "really complicated" or $\kappa$ is really large, or if "crazy reflections" occur?

Mallik
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In general, there are countably many formulas in the language of set theory. Even the second-order language. So there are at most countably many extensions.

To your second question, if we allow ordinal parameters, then we can always add "There are exactly $\alpha$ inaccessible cardinals". This will characterize your model completely. But the problem is when $\alpha$ is internally the height of your model. This would cause a problem, since that axiom would be "there is a proper class of inaccessible cardinals". To overcome this, we can assume that there a proper class of inaccessible cardinals, but no inaccessible limit of inaccessible cardinals.

The third question is the inherent problem I mention above. If $\kappa$ is very large, then it is possible that any second-order properties of $\kappa$ are reflected to a proper class below it. And so there is no way to internally describe $V_\kappa$ in a unique way. So we cannot characterize such models internally with a single (or a schema of) second-order axioms. For example measurable cardinals have this property.

Asaf Karagila
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  • Clarification for the OP: the first paragraph of course only works given that we are looking at extensions by a single formula (which we are). – Noah Schweber Sep 23 '17 at 18:00
  • Sometimes I feel that we need to form something like the Justice League of America. – Asaf Karagila Sep 23 '17 at 18:01
  • In your second paragraph, is the problem that we can't express, in the language of second-order ZFC, that there are proper-class-many whatever, or is the problem that the axiom "there is a proper class of inaccessible cardinals" would not render the theory categorical? – Mallik Sep 23 '17 at 21:15
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    The latter. We can express "proper class of inaccessible" using a first-order statement, of course: every ordinal has a large inaccessible cardinal. – Asaf Karagila Sep 23 '17 at 21:15
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    Of course, you can say something about "how many models have proper class of inaccessible cardinals", but that runs out at some point. And yes, there are ways to strengthen this. But these will run out eventually too. When we reach a measurable cardinal, then there is no second-property which holds for the first time at the measurable cardinal. In fact, any second-order property holds for "a lot" of inaccessible cardinals below it. So eventually, this is going to fail. – Asaf Karagila Sep 23 '17 at 21:18
  • Thanks, that's really helpful. How does assuming that there is no inaccessible limit of inaccessibles solve that problem? – Mallik Sep 23 '17 at 21:23
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    @Mallik: It means that in any model of $\sf ZFC_2$ there will only be a set of inaccessible cardinals. So using parameters from the model we can still characterize the order type of that set. – Asaf Karagila Sep 23 '17 at 21:24
  • @AsafKaragila If one were trying to prove by induction on $\alpha$ that the theory $T_{\alpha}$ = ZFC2 + "There are exactly $\alpha$ inaccessibles" is categorical for all $\alpha \in ON$, where would that proof break down in the suggested way, i.e., how do we know there must come a point in the induction at which $\alpha$ is the height of a model of T$_{\alpha}$? (I have in mind the fixed point of a normal function $f$ but not sure what that $f$ would be.) – Mallik Nov 17 '17 at 03:12