# How large is an uncountable regular cardinal which is closed under arbitrary fast operators?

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?

• 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. – user180918 Nov 14 '14 at 20:21
• 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. – GME Nov 14 '14 at 21:22
• @AsafKaragila so inaccesibile is equivalent to superinaccesible(Definition 2)? – MphLee Nov 25 '14 at 14:03
• @MphLee: Maybe? Probably? – Asaf Karagila Nov 25 '14 at 14:58

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".