Timeline for Countable choice in $L(\mathbb{R}^*_G)$
Current License: CC BY-SA 3.0
12 events
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| Oct 28, 2014 at 21:13 | comment | added | Trevor Wilson | @Asaf I agree; it does seem plausible that $\omega_1$ itself would be measurable in some inner model; it's true in the example obtained by Prikry forcing and also in the example where $\lambda$ is a limit of Woodins (because $\mathsf{AD}$ implies $\omega_1$ is measurable in $\text{HOD}$.) So maybe the hypothesis of the question implies that $\lambda$ is measurable in $\text{HOD}^{L(\mathbb{R}^*_G)}$. | |
| Oct 28, 2014 at 21:09 | comment | added | Trevor Wilson | @Yair I will have to think some more about why there would be no partial extenders on $K$ between $\lambda$ and $\lambda^+$. The part after that seems right to me, although some people get nervous about constructing core models without AC, so I guess there is something to check. | |
| Oct 26, 2014 at 18:01 | comment | added | Asaf Karagila♦ | Trevor, any luck on proving that if $L(\Bbb R^*_G)$ satisfies countable choice then $\omega_1$ is measurable in some inner model? Perhaps you could also show directly that $\omega_1$ is measurable in that model, it might work. | |
| Oct 24, 2014 at 4:29 | comment | added | Asaf Karagila♦ | Trevor, we already know that $\lambda$ is regular in $L$ (and so in other inner models). Maybe some sort of a Mathias theorem would work here. Consider a sequence witnessing the singularity of $\lambda$, and construct a measure in an inner model. | |
| Oct 24, 2014 at 3:45 | comment | added | Trevor Wilson | @Asaf I think you are right, because the core model below a measurable satisfies the strong form of covering if I recall correctly. But now, unlike with $L$, one has to consider where the core model is constructed. I will think about this. | |
| Oct 24, 2014 at 3:42 | comment | added | Asaf Karagila♦ | That sounds like a reasonable claim. My guess is that a measurable is right on the money. | |
| Oct 24, 2014 at 3:38 | comment | added | Trevor Wilson | @Asaf Aha, that does seem to show that it follows from a measurable. Prikry forcing doesn't change $\mathbb{R}^*_G$, so it doesn't change $L(\mathbb{R}^*_G)$. And because $\lambda$ was measurable (hence inaccessible) to begin with, $L(\mathbb{R}^*_G) \models \mathsf{DC}$. So together with Monroe's comment this shows that the consistency strength is between a measurable and "every real has a sharp." | |
| Oct 24, 2014 at 2:56 | comment | added | Asaf Karagila♦ | Also, the first thing I'd try is to shoot a Prikry to a measurable and collapse. | |
| Oct 24, 2014 at 2:53 | comment | added | Trevor Wilson | @Monroe Oh, quite right. And it looks like every bounded subset of $\lambda$, and hence every real in $\mathbb{R}^*_G$, has a sharp for the same reason. To go further and get an inner model with a Woodin cardinal, we could try the same thing with $K$ instead of $L$. But I'm not sure how that would work, because we used (full) covering for $L$ to get sharps and $K$ may only satisfy weak covering (as far as I know.) | |
| Oct 24, 2014 at 2:16 | comment | added | Asaf Karagila♦ | I always thought that the trick was to collapse the Woodin cardinals, not everything. You might think there is no difference, but there is. Compare the Feferman-Levy model to Truss models. | |
| Oct 24, 2014 at 2:06 | comment | added | Monroe Eskew | It must be at least $0^\sharp$ because if $\lambda$ is a singular cardinal and Jensen's covering holds, then $\lambda$ is singular in $L$. | |
| Oct 23, 2014 at 23:40 | history | asked | Trevor Wilson | CC BY-SA 3.0 |