4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Any info about it in Foreman's article in the Handbook? I don't have it near me at the moment to check... $\endgroup$ – Rachid Atmai Nov 25 '20 at 17:34
  • 1
    $\begingroup$ Sorry, I don't understand the statement starting with "...; in other words, ...": $\kappa$ is not mentioned there, should not it be involved? $\endgroup$ – მამუკა ჯიბლაძე Nov 25 '20 at 17:34
  • 2
    $\begingroup$ 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 $M$. $\endgroup$ – Gabe Goldberg Nov 25 '20 at 17:56
  • 1
    $\begingroup$ Do you think you could add that as an answer? $\endgroup$ – Master Nov 25 '20 at 18:17
  • 1
    $\begingroup$ If you weaken this to, “For all $\alpha<\delta$, $j[\alpha] \in M$,” then you have the axiom I2. $\endgroup$ – Monroe Eskew Nov 25 '20 at 21:48
11
$\begingroup$

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 $M$.

$\endgroup$
  • $\begingroup$ Still doesn't mean that it's weaker than I1, though. :) $\endgroup$ – Asaf Karagila Nov 26 '20 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.