8
$\begingroup$

Solovay proved that a measurable cardinal is equiconsistent with the existence of an extension of Lebesgue measure to a full measure on $P(\mathbb R)$. The inner model direction is relatively simple. Let $\kappa$ be the additivity of the $\sigma$-ideal of measure zero sets, and translate this to a $\kappa$-complete ideal $J$ on $\kappa$. $P(\kappa)/J$ is a c.c.c. boolean algebra. If $F$ is the dual filter, then one can show using Kunen’s theory that $L[F]$ is a model of $\kappa$ measurable. The key is to show $L[F] \models$ GCH.

Are there similar equiconsistencies with larger cardinals, perhaps using non-canonical models? For example, suppose $\kappa$ is weakly inaccessible there is a $\kappa$-complete uniform ideal $I$ on $\kappa^+$ such that $P(\kappa^+)/I$ is c.c.c. Is $\kappa$ $\kappa^+$-strongly compact in an inner model? What is the best we can say about the consistency strength?

Update: Related easier question. Suppose $U$ is a $\kappa$-complete uniform ultrafilter on $\kappa^+$. Consider the model $L[U]$. Is $(\kappa^+)^{L[U]}<\kappa^+$?

$\endgroup$
5
  • 3
    $\begingroup$ Foreman's paper might be relevant: "Smoke and mirrors: Combinatorial properties of small cardinals equiconsistent with huge cardinals". $\endgroup$
    – Yair Hayut
    Commented Nov 28, 2020 at 20:21
  • 2
    $\begingroup$ You probably want to demand some regularity for the ideal (or $\theta=\kappa^+$), since otherwise this is true if $\theta$ is real-valued measurable. $\endgroup$ Commented Nov 28, 2020 at 20:28
  • 3
    $\begingroup$ Having completeness exactly $\kappa$ is still not a strong enough constraint: if $\kappa$ and $\theta$ carry uniform $\kappa$-complete (resp. $\theta$-complete) ideals $I$ and $J$, you can get a countably saturated uniform ideal on $\theta$ with completeness $\kappa$ by forcing with $I$ to get $G$, then with the ideal generated by $J$ to get $H$, then taking the ideal in $V$ of sets $A$ with $[\text{id}]_G + [\text{id}]_H\notin j_H(j_G(A))$ forced by 1. (Countable saturation uses that $V[G][H]$ is a ccc extension by Jech 22.32.) $\endgroup$ Commented Nov 28, 2020 at 22:04
  • $\begingroup$ @GabeGoldberg Thanks, I will constrain the question. $\endgroup$ Commented Nov 28, 2020 at 22:56
  • $\begingroup$ @GabeGoldberg Could you send me your new email address? Your old one is bouncing. $\endgroup$ Commented Dec 17, 2020 at 12:52

0

You must log in to answer this question.

Browse other questions tagged .