# Strong cardinal

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 $\kappa$.

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