What is the consistency strength of almost $\omega$-huge cardinals?

Question

What is the consistency strength of a cardinal $\kappa$, such that there is some $j: V\prec M$ such that $M^{\lt j^\omega(\kappa)}\subseteq M$; in other words, for every cardinal $\lambda\lt\delta$, $M^\lambda\subseteq M$, where $\delta$ is the least fixed point of $j$. Let's call such a cardinal almost $\omega$-huge.

It is a known theorem of $ZFC$ that there can be no $\omega$-huge cardinal; i.e. $M^\delta\nsubseteq M$. However, there seems to be no known information about almost $\omega$-huge cardinals. To my knowledge, the standard methods of deriving contradiction do not work in this scenario.

$V_{\delta+2}$ need not be a subset of $M$, so $j\restriction V_{\delta+2}$ need not be an elementary embedding. Furthermore, $|j"\delta|=\delta$, so $j"\delta$ need not be in $M$. To my knowledge, there has been no research on this front. So what is the consistency strength of an almost $\omega$-huge cardinal?

Results

• Every almost $\omega$-huge cardinal is $I2$.

• Every almost $\omega$-huge cardinal is a limit of $I2$ cardinals, because the sequence of ultrafilers witnessing $I2$ is in $M$.

• $j"\delta\notin M$ and $V_{\delta+1}\notin M$.

My hunch is that this is weaker or equivalent to $I1$ cardinals, but I have no way to prove this.

Answer 1

Almost $\omega$-huge is equivalent to $\omega$-huge, so it is inconsistent with AC. Closure under $\kappa_n$-sequences plus closure under $\omega$-sequences implies closure under $\delta$-sequences: given a $\delta$-sequence $\langle a_\alpha : \alpha < \delta\rangle\subseteq M$, $s_n = \langle a_\alpha : \alpha < \kappa_n\rangle$ is in $M$ for each $n$, so $\langle s_n : n < \omega\rangle$ is in $M$ by closure under $\omega$-sequences. Therefore $\bigcup s_n = \langle a_\alpha : \alpha < \delta\rangle$ is in $M$.