Timeline for Stationary sets in HOD
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 8, 2023 at 16:17 | comment | added | Joel David Hamkins | @Holo Sorry, I hadn't ever formed a counterexample... | |
| Jan 8, 2023 at 15:43 | comment | added | Holo | @JoelDavidHamkins Did you find a counterexample yet? You had plenty enough time to think | |
| Jun 10, 2014 at 10:30 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 10, 2014 at 10:29 | comment | added | Joel David Hamkins | Trevor, thanks; I have now edited. As for your further point, I am inclined to agree in thinking that HOD cannot necessarily define that class, even though it is amenable, but I'll have to think about it to make a counterexample. | |
| Jun 10, 2014 at 3:58 | comment | added | Trevor Wilson | A typo: the thing you say is a set in your last paragraph is a proper class, because you didn't say "$S \subseteq \lambda$". By the way, this makes me wonder: can $\text{HOD}$ can define the class of $V$-stationary sets? At a glance this seems to me like it might be stronger than the fact that this class is amenable to $\text{HOD}$, and I'm not sure if you (and Woodin) meant to make this stronger claim. | |
| Jun 10, 2014 at 3:03 | vote | accept | Everett Piper | ||
| Jun 10, 2014 at 3:02 | comment | added | Everett Piper | Right. That was sloppy of me. I'll need to think about your great answer some more. In the meantime, I guess I'm trying to articulate a question about the definability (not necessarily using only ordinal parameters) of clubs and stationary sets. | |
| Jun 10, 2014 at 2:54 | comment | added | Joel David Hamkins | Meanwhile, yes, the only way for stationarity to be non-absolute is for the smaller model to have a set that it thinks is stationary, when the larger model has a club disjoint from it. So the larger model has a club that is not in the smaller model, which is disjoint from the stationary set. The club-shooting forcing is a way of making this kind of situation happen. | |
| Jun 10, 2014 at 2:52 | comment | added | Joel David Hamkins | No, not every stationary set is definable. But the collection of sets in HOD that are stationary in V is definable, and this is what he is using in connection with (3). | |
| Jun 10, 2014 at 2:51 | comment | added | Everett Piper | @JDH. Thank you for your answer. One interesting fact that I glean from it is that stationary sets are definable, whereas there are club sets that are not not (at least from ordinal parameters). Is this correct? Since you mention the forcing which kills a stationary/co-stationary set, is this the only way to conclude that there are club sets which are not definable, or is there an easier argument to see this? | |
| Jun 10, 2014 at 2:49 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 10, 2014 at 2:39 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 10, 2014 at 2:25 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 10, 2014 at 2:18 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |