# Superstrong

Superstrong cardinals were first utilized by Hugh Woodin in 1981 as an upper bound of consistency strength for AD. However, Shelah had then discovered that Shelah cardinals were a weaker bound that still sufficed to imply the consistency strength for AD. After this, it was found that the existence of a proper class of Woodin cardinals was equiconsistent to AD. This is a significant weakening of superstrongness.

## Definitions

There are, like most critical point variations on measurable cardinals, multiple equivalent definitions of superstrongness. In particular, there is an elementary embedding definition and an extender definition.

### Elementary Embedding Definition

A cardinal $\kappa$ is $n$-superstrong (or $n$-fold superstrong when referring to the $n$-fold variants) iff it is the critical point of some elementary embedding $j:V\rightarrow M$ such that $M$ is a transitive class and $V_{j^n(\kappa)}\subset M$ (in this case, $j^{n+1}(\kappa):=j(j^n(\kappa))$ and $j^0(\kappa):=\kappa$).

A cardinal is superstrong iff it is $1$-superstrong.

The definition quite clearly shows that $\kappa$ is $j^n(\kappa)$-strong. However, the least superstrong cardinal is never strong. 

### Extender Definition

A cardinal $\kappa$ is $n$-superstrong (or $n$-fold superstrong when referring to the $n$-fold variants) iff there is a $(\kappa,\beta)$-extender $\mathcal{E}$ for a $\beta>\kappa$ with $V_{j^n_{\mathcal{E}}(\kappa)}\subseteq$ $Ult_{\mathcal{E}}(V)$ (where $j_{\mathcal{E}}$ is the canonical ultrapower embedding from $V$ into $Ult_{\mathcal{E}}(V)$). 

A cardinal is superstrong iff it is $1$-superstrong.

## Relation to other large cardinal notions

• The consistency strength of $n$-superstrongness follows the double helix pattern.  Specifically:
• The consistency strength of the existence of a superstrong cardinal is less than that of the existence of a $\kappa$ which is $2^\kappa$-supercompact. 
• If $\kappa$ is superstrong, then the necessary transitive class $M$ has $M\cap V_{j(\kappa)}$ satisfying $\kappa$'s strongness (although $\kappa$ may not be strong in $V$).
• There are many measurable cardinals above a superstrong cardinal. 
• Every $n$-huge cardinal is $n$-superstrong. 
• If $\kappa$ is strong but there is a superstrong cardinal at least the size of $\kappa$, then there are also $\kappa$ superstrong cardinals less than $\kappa$. 
• Every $1$-extendible cardinal is superstrong and has a normal measure containing all of the superstrongs less than said $1$-extendible. 
• Every $\kappa$ which is $2^\kappa$-supercompact is larger than a superstrong cardinal and has a normal measure containing all of the superstrongs less than it. 
• Every superstrong cardinal is Woodin and has a normal measure containing all of the Woodin cardinals less than it. 
• Superstrongness is consistency-wise stronger than Hyper-Woodinness. 
• Superstrongness is not Laver indestructible.