# $n$-fold variants and $M^{(n)}$

The $n$-fold variants of large cardinals were given in a very large paper by Sato Kentaro. Most of the definitions involve giving large closure properties to the $M$ used in the original large cardinal in an elementary embedding $j:V\rightarrow M$. They are very large, but $I3$ cardinals are stronger than most $n$-fold variants of large cardinals. The critical piece of information required in most $n$-fold variants was $M^{(n)}$, a modification of a transitive class $M$.

Given a transitive class $M$ and an elementary embedding $j:V\rightarrow M$, $M^{(n)}$ and $j^{(n)}$ defined as follows: $$M^{(n)}=\bigcup_{\alpha\in\mathrm{Ord}}j^n(V_\alpha)$$ $$j^{(n)}=\bigcup_{\alpha\in\mathrm{Ord}}j^{n-1}(j\upharpoonright V_\alpha)$$

These can be thought of as a new version of $V$ using the $n$-th $j$-iteration on $V_\alpha$ rather than $V_\alpha$ itself; this of course makes $M^{(0)}=V$ for any transitive class $M$ and elementary embedding $j:V\rightarrow M$. Because of particularly this reason, the $0$-fold variants are particularly weak relative to $1$-fold variants. However, because most $n$-fold variants are more easily defined without $M^{(n)}$, although it is easier to prove things about them with $M^{(n)}$, the variants are defined in this page without them.

## $n$-fold variants which are simply the original large cardinal

There were many $n$-fold variants which were simply renaming the original large cardinal. This was due to the fact that some $n$-fold variants, if only named $n$-variants instead, would be confusing to the reader (for example the $n$-fold extendibles rather than the $n$-extendibles). Here are a list of such cardinals:

• The $n$-fold superstrong cardinals are precisely the $n$-superstrong cardinals
• The $n$-fold almost huge cardinals are precisely the almost $n$-huge cardinals
• The $n$-fold huge cardinals are precisely the $n$-huge cardinals
• The $n$-fold superhuge cardinals are precisely the $n$-superhuge cardinals

## $n$-fold supercompact cardinals

A cardinal $\kappa$ is $n$-fold $\lambda$-supercompact iff it is the critical point of some nontrivial elementary embedding $j:V\rightarrow M$ such that $\lambda<j(\kappa)$ and $M^{j^{n-1}(\lambda)}\subset M$ (i.e. $M$ is closed under all of its sequences of length $j^{n-1}(\lambda)$). This definition is very similar to that of the $n$-huge cardinals.

A cardinal $\kappa$ is $n$-fold supercompact iff it is $n$-fold $\lambda$-supercompact for every $\lambda$. Consistency-wise, the $n$-fold supercompact cardinals are stronger than the $n$-superstrong cardinals and weaker than the $n+1$-fold strong cardinals (see later in the page for such cardinals). In fact, if an $n$-fold supercompact cardinal exists, then it is consistent for there to be a proper class of $n$-superstrong cardinals.

The $1$-fold supercompact cardinals are precisely the supercompact cardinals. The $0$-fold supercompact cardinals are precisely the measurable cardinals.

## $n$-fold strong cardinals

A cardinal $\kappa$ is $n$-fold $\lambda$-strong iff it is the critical point of some nontrivial elementary embedding $j:V\rightarrow M$ such that $\kappa+\lambda<j(\kappa)$ and:

$$V_{j^{n-1}(\kappa+\lambda)}\subset M$$

A cardinal $\kappa$ is $n$-fold strong iff it is $n$-fold $\lambda$-strong for every $\lambda$. Consistency-wise, the $n+1$-fold strong cardinals are stronger than the $n$-fold supercompact cardinals, equivalent to the $n$-fold extendible cardinals, and weaker than the $n+1$-fold Woodin cardinals. More specifically, in the rank of an $n+1$-fold Woodin cardinal there is an $n+1$-fold strong cardinal.

The $1$-fold Strong cardinals are precisely the strong cardinals. The $0$-fold strong cardinals are precisely the measurable cardinals.

## $n$-fold Woodin cardinals
A cardinal $\kappa$ is $n$-fold Woodin iff for every function $f:\kappa\rightarrow\kappa$ there is some ordinal $\alpha<\kappa$ such that $\{f(\beta):\beta<\alpha\}\subseteq\alpha$ and $V_{j^{n}(f)(j^{n-1}(\alpha))}\subset M$. Consistency-wise, the $n+1$-fold Woodin cardinals are stronger than the $n+1$-fold strong cardinals, and weaker than the $n+1$-fold Shelah cardinals. Specifically, in the rank of an $n+1$-fold Shelah cardinal there is an $n+1$-fold Woodin cardinal, and every $n+1$-fold Shelah cardinal is also an $n+1$-fold Woodin cardinal.
The $1$-fold Woodin cardinals are precisely the Woodin cardinals. The $0$-fold Woodin cardinals are precisely the measurable cardinals.