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I'm re-reading a paper of Stevo Todorcevic's entitled "Localized Reflection and Fragments of PFA" and there's a claim in the proof of one of the lemmas that I thought I understood but now I'm not so sure. The claim is this:

Suppose $0^{\sharp}$ does not exist, and let $a \in L_{\omega_2}$, $\varphi$ a formula, $\theta$ a regular cardinal in $L$ such that $L_{\theta} \vDash \varphi (a)$. Then there exists $\lambda$, a cardinal in $V$, such that $\theta \leq \lambda ^+ = (\lambda ^+)^L$ and $L_{\lambda ^+} \vDash \varphi (a)$.

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Amit, I do not think this is exactly what Stevo is claiming. He writes:

Let $\varphi$ be a given formula of set theory and suppose that for some $a\in L_{\omega_2}$ there is $\theta$, a regular cardinal in $L$ such that $L_\theta\models\varphi(a)$. We need to find $\theta'\lt\omega_2$, a regular cardinal in $L$, such that $L_{\theta'}\models\varphi(a)$. Clearly we may assume that $0^\sharp$ does not exist, and therefore (by increasing $\theta$), that there is a cardinal $\lambda$ in $V$ such that $\theta=\lambda^+=(\lambda^+)^L$.

The point is that he is effectively changing the formula when passing to a larger $\theta$. Say, given $\theta$ and $\varphi$, he finds a (large) singular cardinal $\lambda$ in $V$ whose successor is computed correctly in $L$ (this is the key use of the assumption that $0^\sharp$ does not exist, so we have covering), and notes that $L_{\lambda^+}$ models, say, the statement $\varphi'(a)$ that "the length of the universe is the successor of a cardinal $\lambda$ such that there is a regular cardinal $\theta<\lambda$ such that $L_\theta\models\varphi(a)$." He then proceeds to reflect this statement down, and therefore finds a $\gamma'\lt\omega_2$ regular and successor in $L$ that models $\varphi'(a)$. But then there is a $\theta'\lt\gamma'$ regular in $L$ such that $L_{\theta'}\models\varphi(a)$, and we are done.

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  • $\begingroup$ Of course, if $0^\sharp$ exists, then the existence of $\theta'$ is easy: Let $\theta$ be the least $L$-regular such that $L_\theta\models\varphi(a)$. Then $\theta$ is definable from the finite set $A\subset\omega_2$ of indiscernibles that define $a$, and therefore below the first indiscernible larger than the elements of $A$. But this is still below $\omega_2$, since the indiscernibles form an unbounded subset of $\omega_2$. $\endgroup$
    – Andrés E. Caicedo
    Commented Dec 20, 2010 at 1:25
  • $\begingroup$ Thanks Andres, I didn't realize the formula was changing. So first off, I don't think we need a large $\lambda$, any singular $\lambda \geq \theta$ should do, no? Also I don't think you can just hope to reflect down. Suppose $V=L$, then $\gamma '$ would have to be $\omega _1$, and $\theta '$ would have to be $\omega$, but generally $a \notin L_{\omega}$. $\endgroup$
    – Amit Kumar Gupta
    Commented Dec 20, 2010 at 2:32
  • $\begingroup$ @Amit: 1. Yes, that's all I meant by "large $\lambda$": Larger than $\theta$. 2. No, of course not. I meant: Now Stevo runs the argument that is the content of the lemma, and obtains at the end a $\gamma'<\omega_2$ with $L_{\gamma'}$ a model of $\varphi'(a)$. $\endgroup$
    – Andrés E. Caicedo
    Commented Dec 20, 2010 at 2:51
  • $\begingroup$ I assume it's more than a matter of mere reflection, otherwise he would have no need for the canonical square sequence he uses in the proof. The issue is, if $\theta <\gamma$ with $\theta$ regular in $L$ and $\gamma$ successor in $L$, then $L_{\gamma}\vDash \mathrm{Ord}$ is a successor cardinal and $\theta$ regular, but the converse needn't hold. By reflection there are plenty of $\gamma <(\theta ^+)^L$ where $L_{\gamma}$ thinks $\mathrm{Ord}$ has successor length; and if $\theta$ is singular in $L$, a subset of $\theta$ witnessing this need not show up immediately at stage $L_{\theta +1}$. $\endgroup$
    – Amit Kumar Gupta
    Commented Dec 20, 2010 at 2:56
  • $\begingroup$ Bah! I started writing that comment before you posted yours but didn't finish til after, so just ignore it! Thanks again. $\endgroup$
    – Amit Kumar Gupta
    Commented Dec 20, 2010 at 3:00

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