I am referring to the proof of (4) implies (1) in Theorem 3.16 of Woodin's The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal. His proof leverages on the fact that if the sharp of every real exists, then $\delta^1_2 = u_2$, where $u_2$ is the second uniform indiscernible (the least ordinal above $\omega_1$ which is an $x$-indiscernible for every real $x$).

This is roughly how it goes:

  • Fix any ordinal $\alpha$ strictly between $\omega_1$ and $\omega_2$, and a well-ordering $<_{\alpha}$ of $\omega_1$ of ordertype $\alpha$.
  • Find a club $C$ of $\omega_1$ such that for every $\gamma \in C$, the rank of $<_{\alpha}$ restricted to $\gamma$ is less than the least ordinal in C greater than $\gamma$.
  • By hypothesis, there is a $D \subset C$, $D$ a club of $\omega_1$, such that $D$ is constructible from a real $z$. We can assume $D$ is definable from $z$ and $\omega_1$ in $L[z]$ since $z^\sharp$ exists.
  • By reflecting the definition of $D$ downwards, we have that for every $\gamma \in D$, $rank(<_{\alpha} \restriction \gamma) < rank(\mathcal{M}(z^{\sharp}, \gamma + 1))$, where $\mathcal{M}(z^{\sharp}, \gamma + 1)$ stands for (quoting Woodin) "the $\gamma$[$+1$] model of $z^{\sharp}$". I am not exactly sure what that quote means, so I assumed $rank(\mathcal{M}(z^{\sharp}, \gamma + 1))$ is the least $z$-indiscernible above $\gamma$, which works for the arguments hitherto. Did I make a mistake in my interpretation here?
  • Now, following the previous inequality of ranks, he immediately concluded that $\alpha < rank(\mathcal{M}(z^{\sharp}, \omega_1 + 1))$, which finishes the proof.

It is at the last step (the final bullet point) that I got lost. How does $\alpha < rank(\mathcal{M}(z^{\sharp}, \omega_1 + 1))$ follow from the previous steps?

EDIT: Alternatively, I would appreciate it if anyone can point me to another proof of the statement "if the sharp of every real exists and every club contains a club constructible from a real, then $\delta^1_2 = \omega_2$"

  • $\begingroup$ I think what's involved in the last step is indiscernibility. Since the Silver indiscernibles w.r.t. $z$ form a club, you have the next-to-last bullet point for some indiscernibles $\gamma$, and you can replace such a $\gamma$ with $\omega_1$ in any statement that $L[z]$ understands. This replacement will turn $<_\alpha\upharpoonright\gamma$ into $<_\alpha$. $\endgroup$ May 26, 2020 at 13:52
  • $\begingroup$ But we do not know if $<_{\alpha}$ is constructible from any real, so the set of inequalities in the penultimate bullet point may not be expressible in L[z]. Or am I mistaken here? $\endgroup$
    – Zoorado
    May 26, 2020 at 14:13

1 Answer 1


Let $\kappa$ be a sufficiently large regular cardinal. Take a countable elementary substructure $H$ of $H(\kappa)$ containing $z$ and $<_\alpha$ with $\omega_1\cap H\in D$. Let $\pi : M\to H(\kappa)$ be the inverse of the transitive collapse of $H$. To see the fifth bullet point, it suffices to show that in $M$, $\pi^{-1}(<_\alpha)$ has rank less than $(\text{rk}(\mathcal M(z^\#,\omega_1^M+1)))^M$. But this follows from the fourth bullet point since $\omega_1^M = \omega_1\cap H \in D$, $\pi^{-1}({<}_\alpha) = {<}_\alpha \restriction \omega_1^M$, and $M$ correctly computes $\text{rk}(\mathcal M(z^\#,\omega_1^M+1))$ since $z^\#$ is in $M$.

  • $\begingroup$ Thanks for the clear explanation! $\endgroup$
    – Zoorado
    May 27, 2020 at 2:56

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