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Let $Card$ be the proper class of all cardinals, define an infinite set of operators like $\otimes_{n}:(Card\setminus \omega)\times (Card\setminus\{0\})\longrightarrow Card$ which are defined for each natural number $n\geq 0$ recursively:

Definition 1: For $n=0$ define $\otimes_{0}$ as follows:

$\forall \kappa\geq \aleph_0~\forall \lambda>0~~~~~\kappa\otimes_{0}\lambda:=\kappa^\lambda$

If $\otimes_{n}$ is defined, consider $\otimes_{n+1}$ as follows:

Fix $\kappa\geq \aleph_0$, then define:

$\kappa\otimes_{n+1}1:=\kappa$

$\forall \lambda>0~~~~~\kappa\otimes_{n+1}\lambda^{+}:=(\kappa\otimes_{n+1}\lambda)\otimes_{n}\kappa$

$\forall \lambda>0~~~~$ if $\lambda$ is a limit cardinal then $\kappa\otimes_{n+1}\lambda:=\sup(\{\kappa\otimes_{n+1}\delta~|~\delta<\lambda\})$

Definition 2: An uncountable regular cardinal $\kappa$ is super inaccessible if for all $n\in \omega$ and for all $\lambda,\theta<\kappa$ we have $\lambda\otimes_{n}\theta<\kappa$ (i.e. $\kappa$ is closed under all operators $\otimes_{n}$)

Question 1: What is the consistency strength of the existence of a super inaccessible cardinal?

Question 2: Is every strongly inaccessible cardinal super inaccessible?

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    Do you have a reason to believe that inaccessible cardinals are not superinaccessible? – Asaf Karagila Nov 14 '14 at 19:40
  • @AsafKaragila This seems so possible but I'm not sure. Maybe assuming GCH simplifies the operators and clarifies the situation. –  Nov 14 '14 at 20:21
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    I agree with Asaf. When $\kappa$ is inaccessible, $V_\kappa$ models ZFC and so these operations will be well-defined. They'll also be absolute. So the superinaccessibles are just the inaccessibles. –  Nov 14 '14 at 21:22
  • @AsafKaragila so inaccesibile is equivalent to superinaccesible(Definition 2)? – m_l Nov 25 '14 at 14:03
  • @MphLee: Maybe? Probably? – Asaf Karagila Nov 25 '14 at 14:58
  • @AsafKaragila this does not help xDD ... maybe is too obvious for set theretic monsters...is it? – m_l Nov 25 '14 at 18:12

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When I calculate these operations using your recursion equations, I get $\kappa\otimes_n\lambda=2^\kappa$ whenever $1\leq n<\omega$ and $\lambda\geq 2$. Am I misreading something, or is your definition not what you intended? Of course, if the definition is what you intended and I'm computing correctly, then "superinaccessible" is trivially equivalent to "inaccessible".

Andreas Blass
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  • My definitions are a generalization of arithmetical hyper-operators for infinite cardinals. Of course as there is no clear intuition for the "correct" definition of these operators at limit stages, I used the weakest possible operator at limits, "sup". It seems this reduces all definitions to trivial version, as you mentioned. So I think your computation should be correct. –  Jan 01 '15 at 05:07
  • This MO post and this question of mine are also related. –  Jan 01 '15 at 05:11