Superstrong cardinals were first utilized by Hugh Woodin in 1981 as an
upper bound of consistency strength for the [](Axiom%20of%20determinacy.md).
However, Shelah had then discovered that
<a href="Shelah" class="mw-redirect" title="Shelah">Shelah</a>
cardinals were a weaker bound that still sufficed to imply the
consistency strength of $\text{(ZF+)AD}$. After this, it was found that
the existence of infinitely many
[Woodin](Woodin.md "Woodin")
cardinals was equiconsistent to $\text{AD}$. Woodin-ness is a
significant weakening of superstrongness.
*Most results in this article can be found in
{% cite Kanamori2009 %} unless indicated
otherwise.*
## Definitions
There are, like most critical point variations on
[measurable](Measurable.md "Measurable")
cardinals, multiple equivalent definitions of superstrongness. In
particular, there is an [](Elementary%20embedding.md)
definition and an
<a href="index.php?title=Extender&action=edit&redlink=1" class="new" title="Extender (page does not exist)">extender</a>
definition.
### Elementary Embedding Definition
A cardinal $\kappa$ is **$n$-superstrong** (or $n$-fold superstrong
when referring to the [](N-fold_variants.md))
iff it is the critical point of some [](Elementary%20embedding.md)
$j:V\rightarrow M$ such that $M$ is a transitive class and
$V_{j^n(\kappa)}\subset M$ (in this case,
$j^{n+1}(\kappa):=j(j^n(\kappa))$ and $j^0(\kappa):=\kappa$).
A cardinal is **superstrong** iff it is $1$-superstrong.
The definition quite clearly shows that $\kappa$ is
$j^n(\kappa)$-[strong](Strong.md "Strong").
However, the least superstrong cardinal is never strong.
### Extender Definition
A cardinal $\kappa$ is **$n$-superstrong** (or $n$-fold superstrong)
iff there is a
<a href="index.php?title=Extender&action=edit&redlink=1" class="new" title="Extender (page does not exist)">$(\kappa,\beta)$-extender</a>
$\mathcal{E}$ for a $\beta>\kappa$ with
$V_{j^n_{\mathcal{E}}(\kappa)}\subseteq$
[$Ult_{\mathcal{E}}(V)$](Ultrapower.md "Ultrapower")
(where $j_{\mathcal{E}}$ is the canonical ultrapower embedding from
$V$ into $Ult_{\mathcal{E}}(V)$).
A cardinal is **superstrong** iff it is $1$-superstrong.
## Relation to other large cardinal notions
The consistency strength of $n$-superstrongness follows the [](N-fold_variants.md)
{% cite Kentaro2007 %}. Specifically:
- [measurable](Measurable.md "Measurable")
= $0$-superstrong = [](Huge.md)
= super almost $0$-huge = $0$-huge = super $0$-huge
- $n$-superstrong
- $n$-fold supercompact
- $(n+1)$-fold strong, $n$-fold extendible
- $(n+1)$-fold Woodin, $n$-fold Vopěnka
- $(n+1)$-fold Shelah
- almost $n$-huge
- super almost $n$-huge
- $n$-huge
- super $n$-huge
- $(n+1)$-superstrong
Let $M$ be a transitive class $M$ such that there exists an elementary
embedding $j:V\to M$ with $V_{j(\kappa)}\subseteq M$, and let
$\kappa$ be its superstrong critical point. While $j(\kappa)$ need not
be an inaccessible cardinal in $V$, it is always
[worldly](Worldly.md "Worldly")
and the rank model $V_{j(\kappa)}$ satisfies $\text{ZFC+}
quot;$\kappa$
is strong" (although $\kappa$ may not be strong in $V$).
Superstrong cardinals have strong upward reflection properties, in
particular there are many
[measurable](Measurable.md "Measurable")
cardinals *above* a superstrong cardinal. Every $n$-huge cardinal is
$n$-superstrong, and so $n$-huge cardinals also have strong reflection
properties. Remark however that if $\kappa$ is
[strong](Strong.md "Strong")
or
[supercompact](Supercompact.md "Supercompact"),
then it is consistent that there is no inaccessible cardinals larger
than $\kappa$: this is because if $\lambda>\kappa$ is
inaccessible, then $V_\lambda$ satisfies $\kappa