# 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? Nov 14, 2014 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, 2014 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.
– user104955
Nov 14, 2014 at 21:22
• @AsafKaragila so inaccesibile is equivalent to superinaccesible(Definition 2)? Nov 25, 2014 at 14:03
• @MphLee: Maybe? Probably? Nov 25, 2014 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".