# Difference between revisions of "Superstrong"

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$\kappa$ is ''superstrong'' if it is 1-superstrong. 0-superstrongness is equivalent to $\kappa$-[[strong|strongness]]. | $\kappa$ is ''superstrong'' if it is 1-superstrong. 0-superstrongness is equivalent to $\kappa$-[[strong|strongness]]. | ||

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+ | The existence of a (n+1)-superstrong cardinal implies the consistency of the existence of a n-superstrong cardinal, furthermore it also implies the consistency of the existence of a [[huge|n-huge]] cardinal (for n > 0). | ||

## Revision as of 07:11, 10 September 2017

A cardinal $\kappa$ is *n-superstrong* if there is an elementary embedding $j:V\to M$ of the universe $V$ into a transitive class $M$, with critical point $\kappa$, such that $V_{j^n(\kappa)}\subset M$.

$\kappa$ is *superstrong* if it is 1-superstrong. 0-superstrongness is equivalent to $\kappa$-strongness.

The existence of a (n+1)-superstrong cardinal implies the consistency of the existence of a n-superstrong cardinal, furthermore it also implies the consistency of the existence of a n-huge cardinal (for n > 0).

### Relation to other large cardinal notions

Every superstrong cardinal is a Shelah cardinal.

If $\kappa$ is superstrong as witnessed by $j:V \to M$, then $\kappa$ is a strong cardinal in the model $M_{j(\kappa)}$. However, superstrong cardinals need not be strong in $V$. In fact, the least superstrong cardinal is less than the least strong cardinal, because superstrongness is a $\Sigma_2$ property (as can be seen using its characterization in terms of extenders) whereas every strong cardinal is $\Sigma_2$-reflecting.

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