# Superextendibles defined analogously to Superstrong cardinals: Where are they Consistency Strength-wise?

A common trend in large cardinal axioms dealing with critical points of elementary embeddings from $V$ into a transitive class $M$ is to make some large cardinal axiom with an ordinal parameter, similar to "$M$ is closed under $\theta$-sequences" or "$V_\theta$ is a subset of $M$". The resulting cardinals from these definitions are the $\theta$-Supercompacts and the $\theta$-Strongs.

After this, expansions could be made to make the cardinal even larger; specifically, by taking that parameter of $M$ and making it $j^n(\kappa)$. When doing this to $\theta$-Strong cardinals, one achieves $n$-Superstrong cardinals. When doing this to Supercompact cardinals, one achieves $n$-Huge cardinals.

Of course, a natural question at this point is to see what happens when one does it to $\eta$-Extendible cardinals. Here are the results:

Let $\kappa$ be $n$-Superextendible if there is some ordinal $\eta$ such that $\kappa$ is the critical point of some nontrivial elementary embedding $j:V_\theta\prec V_\eta$ such that $\theta>j^n(\kappa)$.

So far, here are the results:

• The $0$-Superextendible cardinals are precisely the $0$-Extendible cardinals
• I3 cardinals are $n$-Superextendible for every $n$ (this property could be called $\omega$-Superextendible)
• The $1$-Superextendible cardinals, if one lets an arbitrary such be $\kappa$, are $\kappa$-Extendible. Of course, every $1$-Superextendible cardinal is a Supercompact cardinal and is, in fact, a stationary limit of such due to this property.
• If a cardinal is $n+1$-Superextendible it is $n$-Superextendible

These properties are, yes, very obvious. Are there any other possibly less obvious properties that I am missing? Where do these cardinals fall under consistency strength?

• The strength of these $n$-fold large cardinal axioms have been calibrated in the paper ac.els-cdn.com/S0168007207000127/… – Joseph Van Name Oct 21 '17 at 23:51
• @Joseph_Van_Name That definition is slightly stronger than this one I believe. Correct me if I'm wrong though – Zetapology Oct 24 '17 at 1:46

These large cardinals actually already have a name. They are called $$n$$-huge* cardinals.
Theorem: If $$\kappa$$ is $$n$$-huge*, then $$\kappa$$ is $$n$$-huge and a limit of $$n$$-huge cardinals.
Proof. Let $$j: V_\alpha\rightarrow V_\beta$$ be an $$n$$-hugeness* embedding, with $$\lambda=j^n(\kappa)$$. Furthermore, let $$Y=\{j(\alpha)|\alpha\lt\lambda\}$$ and $$D=\{X\subseteq\lambda|Y\in j(X)\}$$. It is immediately a normal measure. Now note that $$ot(Y\cap j^{i+1}(\kappa))=j^i(\kappa)$$. For the rest, note that $$D\in V_\beta$$.■
Theorem: If $$\kappa$$ is $$n+1$$-huge, then $$\kappa$$ is $$n$$-huge* and a limit of $$n$$-huge* cardinals.
Proof. Let $$j: V\rightarrow M$$ be an $$n+1$$-hugeness embedding, with $$\lambda=j^n(\kappa)$$, and let $$k=j\restriction V_{\lambda+1}$$. Then $$k\in M$$ and so $$M\vDash\kappa\text{ is }n\text{-huge*}$$ and so $$M\vDash V_{j(\lambda)}\vDash(\kappa\text{ is }n\text{-huge*})$$, but $$V_{j(\lambda)}\subseteq M$$ and so $$\kappa$$ is actually $$n$$-huge*. A similar argument goes for the rest .■