# Difference between revisions of "Superstrong"

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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 $V_{j^n(\kappa)}\subset M$. | 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 | + | $\kappa$ is ''superstrong'' if it is 1-superstrong. 0-superstrongness is equivalent to [[measurable|measurability]]. |

=== Relation to other large cardinal notions === | === Relation to other large cardinal notions === |

## Revision as of 12:23, 23 October 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 measurability.

### Relation to other large cardinal notions

Every superstrong cardinal is a Shelah cardinal, and a Woodin cardinal, also there are $\kappa$ Woodin cardinals below it. Every 1-extendible cardinal is superstrong; if $\kappa$ is $2^\kappa$-supercompact or 1-extendible then there are $\kappa$ superstrong cardinals below $\kappa$.

If $\kappa$ is superstrong as witnessed by $j:V \to M$, then $\kappa$ is a strong cardinal in the model $M_{j(\kappa)}=M\cap V_{j(\kappa)}$. 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. Furthermore, if $\kappa$ is strong and there is a superstrong cardinal above it, then there are $\kappa$ superstrong cardinals below $\kappa$.

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).

### References

Kanamori, Akihiro (2003). The Higher Infinite : Large Cardinals in Set Theory from Their Beginnings (2nd ed.). Springer. ISBN 3-540-00384-3.

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