2
$\begingroup$

Adding classes into a set theory like ${\bf ZFC}$ to get a theory like ${\bf MK}$ adds some consistency strength, but less than even a single inaccessible cardinal since $\kappa$ being inaccessible means $(V_\kappa,\in,V_{\kappa+1})$ is a model of ${\bf MK}$.

It's pointed out by Andreas Blass in the answer to this MO question that even iterating this process infinitely, with sets followed by classes followed by hyper classes followed by..., yields more consistency strength than ${\bf MK}$ but still less than a single inaccessible, which presumably means that the above construction can be tweaked to model this situation. He also mentions that the intuition for this situation is closer to an inaccessible than ${\bf MK}$, but still weaker.

I'm curious about the consistency strength of the following situation. Let us refer to sets as $0$-sets, classes as $1$-sets, hyperclasses as $2$-sets, and so on. For each pair of ordinals $\beta\leq\alpha$, let ${\bf ZFC}_{\alpha,\beta}$ be the theory obtained by adding the existence of $\gamma$-sets for all $\gamma<\alpha$ plus full comprehension on $\gamma$ sets for all $\gamma<\beta$ to ${\bf ZFC}$. The first index indicates how far up the collection hierarchy we are, and the second index indicates how far up comprehension applies; for example, $${\bf ZFC}_{2,2}={\bf MK},$$ $${\bf ZFC}_{2,1}={\bf GBC},$$ $${\bf ZFC}_{1,1}={\bf ZFC}.$$

It is unclear to me if, when Andreas says to 'continue this process indefinitely', he is talking about ${\bf ZFC}_{\omega,\omega}$ or ${\bf ZFC}_{O_n,O_n}$. Assuming the former, my question is:

Does there exist a large cardinal axiom $\phi$ and ordinals $\alpha,\beta$ such that the consistency strength of ${\bf ZFC}_{\alpha,\beta}$ is greater than or equal to the consistency strength of ${\bf ZFC}+\phi$?

If the latter is true, that ${\bf ZFC}_{\alpha,\beta}$ is weaker in consistency than a single inaccessible for all $\alpha,\beta$, my question is:

Are there other known 'consistency strength axioms' $\phi$ such that ${\bf ZFC}+\phi$ is equivalent in consistency strength to ${\bf ZFC}_{\alpha,\beta}$?

Joel Hamkins mentions in this MO answer that these iterated classes can be coded by well-founded class relations on the ordinals, and there is a PhD dissertation by Kameryn Williams exploring this encoding process and closely related topics including the consistency strength of ${\bf GBC}$ plus various transfinite recursion principles yielding theories with a consistency strength between ${\bf GBC}$ and ${\bf MK}$. The ${\bf ZFC}_{\alpha,\beta}$ hierarchy jumps from ${\bf GBC}$ to ${\bf MK}$ immediately, though, and is thusly adding more consistency strength than the transfinite recursion principles investigated in the Williams paper.

$\endgroup$
5
  • $\begingroup$ If you are treating classes as objects and forming classes of classes etc., doesn't that mean that your so-called "proper classes" are really just sets? $\endgroup$
    – bof
    Sep 17, 2020 at 7:14
  • $\begingroup$ @bof There are no classes of classes, only hyperclasses of classes. In general there are no $\alpha$-sets containing other $\alpha$-sets, only $\alpha$-sets containing $\beta$-sets for $\beta<\alpha$. $\endgroup$
    – Alec Rhea
    Sep 17, 2020 at 7:17
  • $\begingroup$ @bof Actually we do have $0$-sets (regular sets) containing other sets, but anywhere farther up the hierarchy $x$ being a member of an $\alpha$-set implies that $x$ is a $\beta$-set for some $\beta<\alpha$. $\endgroup$
    – Alec Rhea
    Sep 17, 2020 at 7:27
  • $\begingroup$ It’s not exactly clear how you formulate $\mathrm{ZFC}_{On,On}$, but a single inaccessible should be enough for that: let the collection of $\alpha$-sets be $V_{\kappa+\alpha}$. Or am I missing something? $\endgroup$ Sep 17, 2020 at 7:30
  • $\begingroup$ @EmilJeřábek That looks correct, thank you; indexing with $O_n$ would mean we use $V_{\kappa+\alpha}$ for all $\alpha\in O_n$ and allow comprehension over all of them, which would be the new full universe with an inaccessible correct? $\endgroup$
    – Alec Rhea
    Sep 17, 2020 at 7:33

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.