Timeline for What is the strength of this strict constructible iterative hierarchy?
Current License: CC BY-SA 4.0
16 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Nov 26, 2018 at 5:43 | vote | accept | Zuhair Al-Johar | ||
| Nov 24, 2018 at 17:16 | comment | added | Nik Weaver | @AndreasBlass: I agree, I misunderstood the question. | |
| Nov 24, 2018 at 16:46 | comment | added | Andreas Blass | @NikWeaver I agree with you about getting all the recursive ordinals very soon, but that seems to show only that the construction lasts for $\omega_1^{CK}$ steps, not that it lasts strictly longer. In fact, it seems to end at $\omega_1^{CK}$; see my answer below. | |
| Nov 24, 2018 at 16:44 | answer | added | Andreas Blass | timeline score: 4 | |
| Nov 24, 2018 at 16:02 | comment | added | Zuhair Al-Johar | To be more precise there is a well ordering class $R$ on the class $\{L_0,L_1,..,L_{\alpha}\}$ and that this $R$ is isomorphic to $s$ | |
| Nov 24, 2018 at 15:57 | comment | added | Zuhair Al-Johar | .. continuation, ...to the class $\{L_{\alpha}| \alpha \text{ is a countable}\}$ | |
| Nov 24, 2018 at 15:56 | comment | added | Zuhair Al-Johar | @AndrésE.Caicedo by a level I mean a stage $L_{\alpha}$ of the constructible hierarchy, the well ordering is on the class $\{L_0, L_1,....,L_{\alpha}\}$, in other words suppose you have a well ordering $s$ that is a set (i.e. an element of some level $L_{\alpha}$, then there is a subclass $\{L_0,L_1...,L_{\kappa}\}$ (that is a hierarchy) of the class of all levels of the hierarchy that is isomorphic to $s$. For example the stage $L_{\omega_1}$ is not reachable from below, because no stage $L_{\alpha}$ for a countable $\alpha$ would contain a set that is a well ordering that is isomorphic to | |
| Nov 24, 2018 at 15:11 | comment | added | Nik Weaver | You get (orderings of $\omega$ isomorphic to) all the recursive ordinals at $L_{\omega + 1}$, so the answer is greater than $\Omega_1^{CK}$. Probably it's consistent both that the answer is $\omega_1$ and that it is not. | |
| Nov 24, 2018 at 14:45 | review | Close votes | |||
| Dec 2, 2018 at 3:05 | |||||
| Nov 24, 2018 at 14:20 | comment | added | Andrés E. Caicedo | I still don't understand. At stage $\alpha$, by "a well ordering on the class of all levels from the empty set till $L_\alpha$" you may mean either "a well-ordering of $\alpha$", "a well-ordering of $L_\alpha$", or something else. Which one do you mean? (If you mean something else, please clarify.) Thanks. | |
| Nov 24, 2018 at 14:08 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
deleted 8 characters in body
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| Nov 24, 2018 at 14:02 | comment | added | Zuhair Al-Johar | Ok, I've deleted 'the', it is some well ordering, not a specific one. | |
| Nov 24, 2018 at 14:02 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
added 38 characters in body
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| Nov 24, 2018 at 13:54 | comment | added | Andrés E. Caicedo | What do you mean by "the well-ordering"? Both when you talk of the well-ordering class and the well-ordering of some set. I'm afraid I don't understand yet what you are trying to say. | |
| Nov 24, 2018 at 13:23 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |