Theorem 2.65 in Woodin's book shows that a saturated ideal on $\omega_1$ exists after Levy-collapsing a Woodin cardinal $\delta$ to $\omega_2$. I am confused about the part of the argument where he shows that ideal he defines is a proper ideal.

Claim (2.2) on page 50 says we can find a countable structure $X \prec H_{\delta^+}$ and an $X$-generic condition $p$ such that for all inaccessible $\gamma \in X \cap \delta$ and $Col(\omega_1,<\gamma)$-names $\tau \in X$ for a semi-proper subset of $\mathcal P(\omega_1)/NS$, there is a $\sigma \in X$ such that $p \Vdash \sigma \in \tau$ and $p \Vdash X \cap \omega_1 \in \sigma$. He says we construct this pair $(X,p)$ by an elementary chain.

Suppose $X_0 \prec H_{\delta^+}$ and $p_0 \restriction \gamma \in X_0$. By definition of semi-proper, if $\tau \in X_0$ is as above, then $p_0 \Vdash_\gamma (\exists y \in \tau) Sk(X_0[G_\gamma] \cup \{ y \}) \cap \omega_1 = X_0[G_\gamma] \cap \omega_1$ and $X_0[G_\gamma] \cap \omega_1 \in y$. The problem I have is: How do we choose a name for $y$ with a similar property? There is a name $\sigma$ such that $p_0$ forces $\sigma^G$ witnesses the above property of $y$ in $V[G_\gamma]$, but could it be that $Sk(X_0 \cup \{ \sigma \}) \cap \omega_1 \not= X_0 \cap \omega_1$?

Or perhaps there is a quite different strategy for building the elementary chain. Thanks for your help!

  • $\begingroup$ The argument Hugh refers to can be seen in (more) detail at the beginning of Paul Larson's book on stationary tower forcing. $\endgroup$ May 30, 2015 at 12:00
  • $\begingroup$ Thanks Andres, but could you be a little more specific? $\endgroup$ May 30, 2015 at 12:14
  • $\begingroup$ And could you let me know if I am on the right track for reconstructing the details? $\endgroup$ May 30, 2015 at 12:22
  • $\begingroup$ @AndresCaicedo, I do not see how anything in the first chapter addresses my question. $\endgroup$ May 31, 2015 at 10:59
  • $\begingroup$ Yes, nothing there is explicitly what you need. But the construction is essentially as in 1.1.18-1.1.22, iterated. If I can think of a more explicit place, I'll post a reference. $\endgroup$ May 31, 2015 at 12:46

1 Answer 1


Woodin's argument is wrong. Let us state his definition of the ideal. Assume $G \subseteq \mathrm{Col}(\omega_1,<\delta)$ is generic over $V$.

Let $I_0 \in V[G]$ be the set of $A \subseteq \omega_1$ such that for some $f : \omega_1 \to \mathcal{P}(\omega_1) \setminus NS$,

(1.1) $A = \{ \beta < \omega_1 : \beta \notin f(\alpha)$ for all $\alpha < \beta \}$

(1.2) If $\mathcal A = \{ f(\alpha) : \alpha < \omega_1 \}$, then for some $\gamma < \delta$, $\gamma$ is strongly inaccessible in $V$, $\mathcal A \in V[G \cap V_\gamma]$ and $\mathcal A$ is semiproper in $V[G \cap V_\gamma]$.

Let $I$ be the normal ideal generated by $I_0$.

As stated earlier in the chapter, Foreman, Magidor, and Shelah proved that whenever $\delta$ is supercompact and $G \subseteq \mathrm{Col}(\omega_1,<\delta)$ is generic, then in $V[G]$, every maximal antichain in $\mathcal P(\omega_1)/NS$ is semiproper. Actually only something like $\beth_4(\delta)$-supercompactness is used.

Assume $\kappa < \delta$ are both Woodin and $\beth_4(\cdot)$-supercompact. Let $G_\delta \subseteq \mathrm{Col}(\omega_1,<\delta)$ be generic and let $G_\kappa = G_\delta \cap V_\kappa$. Let $I_\kappa \in V[G_\kappa]$ and $I_\delta \in V[G_\delta]$ be the ideals as above, and note that $I_\kappa \subseteq I_\delta$.

By a well-known forcing argument (see here), $V[G_\kappa]$ satisfies $\Diamond(S)$ for every stationary $S \subseteq \omega_1$. This easily implies that $NS \restriction S$ is not $\omega_2$-saturated for any stationary $S$. In $V[G_\kappa]$, the set of stationary $S$ which are in $I_\kappa$ is dense in $\mathcal{P}(\omega_1)/NS$. This is because otherwise there would be some $S$ such that $I_\kappa \restriction S = NS \restriction S$.

Thus there is a maximal antichain of stationary sets $\mathcal A \subseteq I_\kappa$ in $V[G_\kappa]$, and it is semiproper. In $V[G_\delta]$ there is an enumeration $\{ S_\alpha : \alpha < \omega_1 \}$ of $\mathcal A$, and the diagonal union $\nabla S_\alpha$ is put into the dual filter to $I_\delta$. But since $I_\kappa \subseteq I_\delta$ and $I_\delta$ is normal in $V[G_\delta]$, $\nabla S_\alpha \in I_\delta$. Thus $I_\delta$ is not a proper ideal.

  • $\begingroup$ There's some unicode issue in the block quote. (I see three "characters", after $A$ in the first line, and after "some" and before "is strongly" in the third line.) $\endgroup$
    – Asaf Karagila
    Jun 15, 2015 at 9:00
  • $\begingroup$ It doesn't show up for me, but I tweaked it a bit. Maybe a Mac vs PC thing. Do you still see it? Feel free to fix it. $\endgroup$ Jun 15, 2015 at 9:03
  • $\begingroup$ I removed that, sorry in advance if something got messed up. $\endgroup$
    – Asaf Karagila
    Jun 15, 2015 at 9:05

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