Difference between revisions of "Superstrong"

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(I think superstrong cardinals are important enough to have their own page)
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#REDIRECT [[strong#Superstrong cardinal]]
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A cardinal $\kappa$ is ''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 $V_{j(\kappa)}\subset M$.
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Since the induced [[extender| extenders]] arising from the factors below $j(\kappa)$ are all in $M$, it follows that $M_{j(\kappa)}\models\text{ZFC}+\kappa$ is strong. Thus, the existence of a superstrong cardinal implies the existence of a transitive model of ZFC with a strong cardinal. Therefore, superstrongness strictly exceeds strongness in consistency strength, if the latter is consistent.
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Meanwhile, superstrong cardinals need not themselves be strong, and the least superstrong cardinal is definitely not strong. This is simply because superstrongness is witnessed by a single object, the extender of the superstrongness embedding, and thus has complexity $\Sigma_2$, but every strong cardinal is $\Sigma_2$-[[reflecting]]. So if $\kappa$ is strong and hence $\Sigma_2$-reflecting, the existence of a superstrong cardinal must reflect below $\kappa$.
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Revision as of 13:01, 21 November 2013

A cardinal $\kappa$ is 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 $V_{j(\kappa)}\subset M$.

Since the induced extenders arising from the factors below $j(\kappa)$ are all in $M$, it follows that $M_{j(\kappa)}\models\text{ZFC}+\kappa$ is strong. Thus, the existence of a superstrong cardinal implies the existence of a transitive model of ZFC with a strong cardinal. Therefore, superstrongness strictly exceeds strongness in consistency strength, if the latter is consistent.

Meanwhile, superstrong cardinals need not themselves be strong, and the least superstrong cardinal is definitely not strong. This is simply because superstrongness is witnessed by a single object, the extender of the superstrongness embedding, and thus has complexity $\Sigma_2$, but every strong cardinal is $\Sigma_2$-reflecting. So if $\kappa$ is strong and hence $\Sigma_2$-reflecting, the existence of a superstrong cardinal must reflect below $\kappa$.


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