Timeline for A question about 0# and truth in levels of the L hierarchy
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 21, 2010 at 17:29 | vote | accept | Amit Kumar Gupta | ||
| Dec 20, 2010 at 3:03 | comment | added | Andrés E. Caicedo | Glad to help. There are some nice ideas in that paper. In the matter of fragments of PFA, you may also want to look at the recent paper by Neeman and Schimmerling. | |
| Dec 20, 2010 at 3:00 | comment | added | Amit Kumar Gupta | Bah! I started writing that comment before you posted yours but didn't finish til after, so just ignore it! Thanks again. | |
| Dec 20, 2010 at 2:56 | comment | added | Amit Kumar Gupta | 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 $L_{\theta +1}$. | |
| Dec 20, 2010 at 2:51 | comment | added | Andrés E. Caicedo | @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 $\varphi'(a)$. | |
| Dec 20, 2010 at 2:32 | comment | added | Amit Kumar Gupta | 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 $a \notin L_{\omega}$. | |
| Dec 20, 2010 at 1:25 | comment | added | Andrés E. Caicedo | 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 $\omega_2$. | |
| Dec 20, 2010 at 1:20 | history | answered | Andrés E. Caicedo | CC BY-SA 2.5 |