Timeline for Strategic vs. tactical closure
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
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| Jan 24 at 19:51 | vote | accept | Monroe Eskew | ||
| Jan 24 at 16:11 | answer | added | Andreas Lietz | timeline score: 3 | |
| Dec 5, 2022 at 13:52 | answer | added | Will Brian | timeline score: 5 | |
| Dec 5, 2022 at 6:39 | answer | added | bof | timeline score: 6 | |
| Dec 5, 2022 at 4:52 | answer | added | Steven Clontz | timeline score: 1 | |
| Dec 4, 2022 at 17:24 | comment | added | Joel David Hamkins | Yes, I was going to suggest descending for $\tau(p)\leq p$. | |
| Dec 4, 2022 at 16:31 | history | edited | Monroe Eskew | CC BY-SA 4.0 |
edited body
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| Dec 4, 2022 at 16:17 | comment | added | Joel David Hamkins | Not that I am aware. | |
| Dec 4, 2022 at 16:16 | comment | added | Monroe Eskew | @JoelDavidHamkins You’re right that they’re equivalent, but I’m not sure a strictly decreasing strategy plays much role in arguments. Does it? | |
| Dec 4, 2022 at 14:52 | comment | added | Joel David Hamkins | Ah, but I see that you actually wanted $\tau(p)\leq p$, which isn't quite the same as what is usually called regressive. Why do you allow the reflexive play? I guess it is to handle the case that the poset has atoms. But that case is basically trivial since player II would go immediately for the atoms. Since I think of the atomless case as the main case, and player I can gain no advantage with reflexive play, I usually think about the game as having strictly descending play. | |
| Dec 4, 2022 at 14:16 | history | edited | Monroe Eskew | CC BY-SA 4.0 |
added 84 characters in body
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| Dec 4, 2022 at 14:14 | comment | added | Monroe Eskew | @JoelDavidHamkins Thanks, I agree. | |
| Dec 4, 2022 at 14:13 | comment | added | Joel David Hamkins | Monroe, you call a tactic a decreasing function, but I would want to call it a regressive function, since what you want is $\tau(p)<p$, rather than $q<p\to\tau(q)<\tau(p)$. | |
| Dec 4, 2022 at 13:57 | answer | added | Joel David Hamkins | timeline score: 7 | |
| Dec 2, 2022 at 22:30 | history | edited | Aurel | CC BY-SA 4.0 |
typo sigma -> tau
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| Dec 2, 2022 at 18:53 | vote | accept | Monroe Eskew | ||
| Dec 2, 2022 at 22:28 | |||||
| Dec 2, 2022 at 16:49 | history | became hot network question | |||
| Dec 2, 2022 at 15:15 | answer | added | Joel David Hamkins | timeline score: 11 | |
| Dec 2, 2022 at 14:31 | comment | added | Asaf Karagila♦ | @Joel: Interesting you say that. Every time I try to count the types of chocolate, there seem to be new ones. I can only conclude that there are uncountably many types, and therefore, you seem to be saying, gluttony is a bad way to experience chocolate... | |
| Dec 2, 2022 at 14:04 | comment | added | Joel David Hamkins | Sorry the relevant post for the Chocolatier's game is mathoverflow.net/a/401136/1946, since that is where I prove there is no winning tactic for the Glutton, if there are uncountably many chocolate types. | |
| Dec 2, 2022 at 13:37 | comment | added | Monroe Eskew | @AsafKaragila Oh sorry, my eyes skipped over "club shooting." This one is distributive, but it is not $\sigma$-strategically closed, because one can use the strategy to build a generic condition over any sufficiently elementary countable model. | |
| Dec 2, 2022 at 13:36 | comment | added | Joel David Hamkins | A similar distinction appears in the Chocolatier's game, where the Glutton has a winning strategy, but no winning tactic. mathoverflow.net/questions/401151/… But I don't see immediately how to transform this to your setting, so I'm not sure whether this is helpful. | |
| Dec 2, 2022 at 13:29 | comment | added | Asaf Karagila♦ | Club shooting cannot be $\sigma$-closed, since it would imply it being proper and therefore SSP, and the whole point is to destroy the stationarity of a set. There is a classical theorem that it is $\sigma$-distributive, but I think you can also show that it is in fact strategically closed. | |
| Dec 2, 2022 at 13:14 | comment | added | Monroe Eskew | @AsafKaragila The examples like that are all countably closed, as far as I know, but they serve to show for example that being $\omega_1+1$-strategically closed is weaker than some other closure notions. | |
| Dec 2, 2022 at 13:09 | comment | added | Asaf Karagila♦ | Isn't something with add or threading squares/club shooting/something like that an example? | |
| Dec 2, 2022 at 12:41 | comment | added | Monroe Eskew | @AsafKaragila A bit. It is hard to find examples that are not just forcing-equivalent to countably closed. Jech and Shelah show that consistently such an example exists, but it is somewhat complicated and I don't know if it is tacitcally closed. A google search suggests that my question may be equivalent to a long-standing open problem in topology. | |
| Dec 2, 2022 at 10:18 | comment | added | Asaf Karagila♦ | Have you looked at the standard examples of strategically closed forcings? | |
| Dec 2, 2022 at 10:12 | history | edited | Monroe Eskew | CC BY-SA 4.0 |
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| Dec 2, 2022 at 7:55 | history | asked | Monroe Eskew | CC BY-SA 4.0 |