1
$\begingroup$

In Asaf's answer (https://math.stackexchange.com/q/2442035) to this question:

Proper-class-many categorical extensions of ZFC2

he explains the impossibility of internally characterizing proper class-many categorical extensions of ZFC2, by adding an axioms stating that "There are exactly $\alpha$ inaccessibles". As I understand it, the problem only arises when $\alpha$ is the height of the model being characterized. In this situation, since $\alpha$ is not a parameter in the model, the axiom "There are exactly $\alpha$ inaccessibles" is not a formula of the (object) language and must be replaced by "There are proper class many inaccessibles". The latter axiom is not specific enough to categorically determine a domain.

My question is where the following "proof" would break down. Suppose we try to prove by induction on the ordinals that for every ordinal $\alpha$, the theory $T_{\alpha}$ = ZFC2 + "There are exactly $\alpha$ inaccessibles" is categorical.

It seems that the height of the models determined at the $\alpha$th stage vastly surpass $\alpha$, so it is not immediately evident that one must reach a point in the induction where the parameters $\alpha$ was too big to be in the model. Is there an argument involving fixed points of normal functions that could be used to prove formally that this induction will fail?

$\endgroup$
2
$\begingroup$

As I understand it, you are asking for a proof that the function taking an ordinal $\alpha$ to the $\alpha$th inaccessible cardinal $\kappa_\alpha$ has a fixed point. However, no such proof exists in ZFC. For instance, there may not exist any inaccessible cardinals at all, in which case the function is never defined. In that case all your theories are categorical because they have no models at all.

Even assuming that $\kappa_\alpha$ is defined for all $\alpha$ (i.e., there exists a proper class of inaccessibles), there is no such proof. Indeed, if $\kappa$ is the least inaccessible limit of inaccessible cardinals, then in $V_\kappa$, there is a proper class of inaccessible cardinals but $\alpha\mapsto\kappa_\alpha$ has no fixed points.

So the fixed point you ask about is not guaranteed to exist. It just could exist, depending on what inaccessible cardinals you have in your base model, so you can't prove in ZFC that it doesn't exist (assuming the consistency of sufficiently large cardinals).

$\endgroup$
7
  • $\begingroup$ Are there assumptions that could be added in the metatheory (short of "there is a proper class of inaccessibles") that would make it provable that the induction does not work? $\endgroup$ – Mallik Nov 17 '17 at 16:51
  • 2
    $\begingroup$ The induction does not work iff there exists an inaccessible cardinal that is a limit of inaccessible cardinals. But that is rather tautological, since $\alpha$ is a fixed point iff it is an inaccessible limit of inaccessibles. It is implied by lots of large cardinal axioms, though (including pretty much every major large cardinal axiom stronger than the existence of an inaccessible). $\endgroup$ – Eric Wofsey Nov 17 '17 at 16:59
  • $\begingroup$ But adding "There is no inaccessible limit of inaccessibles" in the metatheory would still not enable one to prove that the induction does work, I take it. What about in conjunction with the axiom "Every categorical extension has a model whose domain is a set in a higher model of ZFC"? $\endgroup$ – Mallik Nov 17 '17 at 17:28
  • $\begingroup$ If you assume there is no inaccessible limit of inaccessibles, that does let you prove the induction works. The only way the induction could fail is if there did exist an inaccessible limit of inaccessibles, and you can prove that in the metatheory (assuming the metatheory includes ZFC). $\endgroup$ – Eric Wofsey Nov 17 '17 at 17:47
  • $\begingroup$ What would that proof that the induction works look like? I have no sense of how the assumption is sufficient. $\endgroup$ – Mallik Nov 17 '17 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.