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

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.

• Any info about it in Foreman's article in the Handbook? I don't have it near me at the moment to check... – Rachid Atmai Nov 25 '20 at 17:34
• Sorry, I don't understand the statement starting with "...; in other words, ...": $\kappa$ is not mentioned there, should not it be involved? – მამუკა ჯიბლაძე Nov 25 '20 at 17:34
• Almost $\omega$-huge is equivalent to $\omega$-huge, so it is inconsistent with AC. Closure under $\kappa_n$-sequences for all $n$ plus closure under $\omega$-sequences implies closure under $\delta$-sequences: given a $\delta$-sequence $\langle a_\alpha :\alpha < \delta\rangle\subseteq M$, for all $n$, $s_n = \langle a_\alpha : \alpha < \kappa_n\rangle\in M$, so $\langle s_n : n <\omega\rangle\in M$, so $\bigcup s_n = \langle a_\alpha :\alpha < \delta\rangle$ is in $M$. – Gabe Goldberg Nov 25 '20 at 17:56
• Do you think you could add that as an answer? – Master Nov 25 '20 at 18:17
• If you weaken this to, “For all $\alpha<\delta$, $j[\alpha] \in M$,” then you have the axiom I2. – Monroe Eskew Nov 25 '20 at 21:48

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 $$M$$.