Assume $\kappa$ is measurable, $U$ is its unique normal measure, $V=L[U]$. We levy collapse $\kappa$ to make it become $\omega_1$.

If we don't have the inner model condition, then we only know that $\omega_1$ has a precipitous ideal. In this case if we want the nonstationary ideal to be precipitous, we need a much more complex forcing notion. So my question is about the inner model case:

Do we have that, the nonstationary ideal on $\omega_1$ is precipitous?


1 Answer 1


No. Suppose otherwise. Let $G$ be the Levy collapse generic, and $D$ be the generic for forcing with the nonstationary ideal after that. Since $\mathrm{Ult}(L[U,G],D)$ is wellfounded, letting $i_D:L[U,G]\to\mathrm{Ult}(L[U,G],D)$ be the ultrapower map, we have that $i_D(L[U])=L[i_D(U)]$ is some iterate of $L[U]$. Let $j:L[U]\to L[i_D(U)]$ be the iteration map. Let $\Gamma$ be a proper class of ordinals fixed by both $j$ and $i_D$. We have $j(U)=i_D(U)$ and $j(\alpha)=i_D(\alpha)$ for all $\alpha\in\Gamma$. But $L[U]=\mathrm{Hull}^{L[U]}_{\Sigma_1}(\Gamma)$ (using that $\Gamma$ is proper class and definable in a generic extension of $L[U]$). It follows that $j=i_D\upharpoonright L[U]$. But working back in $L[U]$, because this Levy collapse (correction) has the $\kappa$-cc, the set $X$ of all $\alpha<\kappa$ of cofinality $\omega$ remains stationary in $V[G]=L[U,G]$. So we could have taken $X\in D$, giving that $\kappa\in i_D(X)$, whereas $\kappa\notin j(X)$, but $X\in L[U]$, a contradiction.

  • $\begingroup$ You are always so good at these questions. $\endgroup$ Dec 5, 2023 at 7:50
  • $\begingroup$ One minor thing. We are using $\mathrm{Col}(\omega,<\kappa)$, which is not countably closed, but preserves stationary subsets of $\kappa$ by the $\kappa$-c.c. $\endgroup$ Dec 5, 2023 at 9:06
  • $\begingroup$ @MonroeEskew Oops, confusion, thanks for the pointer, will correct... $\endgroup$
    – Farmer S
    Dec 5, 2023 at 11:07

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