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. $\endgroup$ – Andrés E. Caicedo May 30 '15 at 12:00
  • $\begingroup$ Thanks Andres, but could you be a little more specific? $\endgroup$ – Monroe Eskew May 30 '15 at 12:14
  • $\begingroup$ And could you let me know if I am on the right track for reconstructing the details? $\endgroup$ – Monroe Eskew May 30 '15 at 12:22
  • $\begingroup$ @AndresCaicedo, I do not see how anything in the first chapter addresses my question. $\endgroup$ – Monroe Eskew May 31 '15 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. $\endgroup$ – Andrés E. Caicedo May 31 '15 at 12:46

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 '15 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. $\endgroup$ – Monroe Eskew Jun 15 '15 at 9:03
  • $\begingroup$ I removed that, sorry in advance if something got messed up. $\endgroup$ – Asaf Karagila Jun 15 '15 at 9:05

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.