# Difference between revisions of "Strong"

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A cardinal $\kappa$ is ''strong'' if it is $\theta$-strong for every ordinal $\theta$, meaning that there is an elementary embedding $j:V\to M$ of the universe $V$ into a transitive proper class $M$, having critical point $\kappa$ and with $V_\theta\subset M$. One may without loss also suppose that $j(\kappa)\gt\theta$. | A cardinal $\kappa$ is ''strong'' if it is $\theta$-strong for every ordinal $\theta$, meaning that there is an elementary embedding $j:V\to M$ of the universe $V$ into a transitive proper class $M$, having critical point $\kappa$ and with $V_\theta\subset M$. One may without loss also suppose that $j(\kappa)\gt\theta$. | ||

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+ | Every strong cardinal is $\Sigma_2$-[[reflecting]], because if any $V_\theta$ satisfies some sentence $\sigma$, then if $j:V\to M$ witnesses $\theta$-strongness, then $M$ satisfies that some ordinal $\theta$ below $j(\kappa)$ has $V_\theta\models\sigma$, and so by elementarity $V$ satisifies that some $\theta\lt\kappa$ has $V_\theta\models\sigma$, as desired. | ||

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* In particular, $\kappa$ is $P^2(\kappa)$-hypermeasurable if and only if it is $\kappa+2$-strong. This hypothesis appears in many theorems. | * In particular, $\kappa$ is $P^2(\kappa)$-hypermeasurable if and only if it is $\kappa+2$-strong. This hypothesis appears in many theorems. | ||

* A cardinal $\kappa$ is [[measurable]] if and only if it is $\kappa^+$-hypermeasurable, since $P(\kappa)\subset M$ for any $j:V\to M$ with critical point $\kappa$. | * A cardinal $\kappa$ is [[measurable]] if and only if it is $\kappa^+$-hypermeasurable, since $P(\kappa)\subset M$ for any $j:V\to M$ with critical point $\kappa$. | ||

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+ | == 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 07:07, 6 January 2012

A cardinal $\kappa$ is *strong* if it is $\theta$-strong for every ordinal $\theta$, meaning that there is an elementary embedding $j:V\to M$ of the universe $V$ into a transitive proper class $M$, having critical point $\kappa$ and with $V_\theta\subset M$. One may without loss also suppose that $j(\kappa)\gt\theta$.

Every strong cardinal is $\Sigma_2$-reflecting, because if any $V_\theta$ satisfies some sentence $\sigma$, then if $j:V\to M$ witnesses $\theta$-strongness, then $M$ satisfies that some ordinal $\theta$ below $j(\kappa)$ has $V_\theta\models\sigma$, and so by elementarity $V$ satisifies that some $\theta\lt\kappa$ has $V_\theta\models\sigma$, as desired.

## Hypermeasurability

To give a more refined hiearachy, a cardinal $\kappa$ is *$A$-strong* or equivalently *$A$-hypermeasurable*, if there is an elementary embedding $j:V\to M$ of the universe $V$ into a transtive class $M$, with critical point $\kappa$, for which $A\in M$.

When $\delta$ is a cardinal, we say that $\kappa$ is *$\delta$-hypermeasurable* to mean really that $\kappa$ is $H_\delta$-strong. (Note, this terminology is technically ambiguous, since one could also interpret $\delta$-hypermeasurability as asserting merely that $\delta\in M$, but this reading is not the intended reading, and the ambiguity is avoided by realizing that $M$ contains all ordinals, and so this latter notion is not needed.)

Note that the concept of $\theta$-strong is indexed by ordinals $\theta$ and the concept of $\delta$-hypermeasurable is indexed by cardinals. The latter concept provides a more refined hierarchy, particularly when the GCH fails.

- A cardinal $\kappa$ is $\theta$-strong if and only if $\kappa$ is $\beth_\theta$-hypermeasurable.
- In particular, $\kappa$ is $P^2(\kappa)$-hypermeasurable if and only if it is $\kappa+2$-strong. This hypothesis appears in many theorems.
- A cardinal $\kappa$ is measurable if and only if it is $\kappa^+$-hypermeasurable, since $P(\kappa)\subset M$ for any $j:V\to M$ with critical point $\kappa$.

## Superstrong cardinal

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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