Timeline for What is the consistency strength of this kind of iterating Berkeley cardinals?
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 5, 2019 at 23:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 26, 2019 at 11:23 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
added 160 characters in body
|
| Jan 15, 2019 at 20:06 | vote | accept | Zuhair Al-Johar | ||
| Jan 26, 2019 at 11:14 | |||||
| Jan 13, 2019 at 21:02 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
added 100 characters in body
|
| Jan 13, 2019 at 20:07 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
added 2061 characters in body
|
| Jan 5, 2019 at 15:34 | comment | added | Zuhair Al-Johar | @MonroeEskew, hmm..., I'll try! Thanks | |
| Jan 5, 2019 at 15:33 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
added 106 characters in body
|
| Jan 5, 2019 at 15:31 | comment | added | Monroe Eskew | @ZuhairAl-Johar— Your question would be more interesting and motivated if you could give examples of things being Berkeley cardinals in one rank, and then failing to be so in a higher rank. | |
| Jan 5, 2019 at 15:30 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
added 237 characters in body
|
| Jan 5, 2019 at 15:20 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
deleted 48 characters in body
|
| Jan 5, 2019 at 15:07 | comment | added | Zuhair Al-Johar | @MonroeEskew, I don't know, but for the time being I don't see a clear argument against it, i.e. if the inner universe has embedding satisfying Berkeley cardinal conditions, this doesn't mean that the outer universe should obey the same embedding trends. | |
| Jan 5, 2019 at 14:54 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
added 129 characters in body
|
| Jan 5, 2019 at 14:08 | comment | added | Zuhair Al-Johar | @AndrésE.Caicedo, inaccessibility here is primarily of stages, $\kappa$ is just the mark of a stage of the cumulative hierarchy, so $inaccessible (\kappa) \leftrightarrow \kappa \text{ is a limit ordinal} \wedge V_{\kappa} \text{ is regular }$, where $V_{\kappa} \text{ is regular }$ is defined as $V_{\kappa}$ not being the union of a subset of it that is strictly subnumerous to it (strictly subnumerous here means absence of an injection from it to that subset). | |
| Jan 5, 2019 at 8:46 | comment | added | Monroe Eskew | Does a finite $S$ as above follow from the existence of some number of Berkeley cardinals? | |
| Jan 4, 2019 at 19:56 | comment | added | Andrés E. Caicedo | By strongly inaccessible do you mean that for each $\lambda<\kappa$, $2^\lambda<\kappa$? (This gives you choice and therefore a contradiction.) Maybe you should mean instead something about how the power set of $\lambda$ relates to $V_\kappa$? | |
| Jan 4, 2019 at 19:53 | comment | added | Zuhair Al-Johar | @AndrésE.Caicedo, ah I see what you mean. Well possibly if there is a clear inconsistency then this would justify my question, but if there is non, then I see what you mean. | |
| Jan 4, 2019 at 19:49 | comment | added | Andrés E. Caicedo | Yes, I am aware of all of that. There is one paper on the subject. My point is that your questions are rather premature given that state of affairs. It would be like asking whether supercompactness is consistent when people were barely coming to terms with measurability. | |
| Jan 4, 2019 at 19:47 | comment | added | Zuhair Al-Johar | @AndrésE.Caicedo, No there is some hierarchy of cardinals beyond choice, like super-Reinhardt, Club-Berkeley, and others above them. Also it might be possible to derive an inconsistency with what I'm saying for example Berkeley cardinals are closed (i.e. the limit of Berkeley cardinals is a Berkeley), and perhaps other properties that can destroy what I'm saying here and prove it inconsistent. As about inaccessibility I mean the strong inaccessibility criterion. | |
| Jan 4, 2019 at 19:32 | comment | added | Andrés E. Caicedo | Berkeley cardinals are beyond anything that has been studied. So asking for a large cardinal property that does the work really makes no sense currently. One could answer the question by saying that the consistency strength is the existence of a proper class of inaccessibles $\kappa$ such that ..., just repeating precisely what you said, and that would be a perfectly reasonable acceptable large cardinal assumption. You need to clarify what you mean. Also, precisely what version of inaccessibility do you have in mind? (We are in a ZF setting, so there are several inequivalent such versions). | |
| Jan 4, 2019 at 19:18 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
edited title
|
| Jan 4, 2019 at 18:26 | comment | added | Zuhair Al-Johar | for the first comment a proof of its inconsistency, or otherwise interpreting it in a known large cardinal property if consistent, for the second comment proving the least cardinal property that can interpret it | |
| Jan 4, 2019 at 17:33 | comment | added | Andrés E. Caicedo | Similarly, what sort of answer are you expecting for the question of its consistency strength? | |
| Jan 4, 2019 at 17:14 | comment | added | Andrés E. Caicedo | What sort of thing would count as an answer to the question of whether this is consistent? | |
| Jan 4, 2019 at 16:21 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
edited body
|
| Jan 4, 2019 at 16:14 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |