Timeline for Is stratified acyclic ZF consistent with non-trivial automorphisms over $V$?
Current License: CC BY-SA 4.0
39 events
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| Feb 16, 2021 at 14:23 | history | undeleted | Zuhair Al-Johar | ||
| Feb 16, 2021 at 13:59 | history | deleted | Zuhair Al-Johar | via Vote | |
| Feb 16, 2021 at 11:53 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Feb 15, 2021 at 12:29 | comment | added | Zuhair Al-Johar | @AsafKaragila, see acyclic ZF, the second of axioms about ranks, explicitly states that for every ordinal there is a stage that it indexes, just define $V_\alpha = x \iff \mathcal R(\alpha,x)$, and then you get stages indexed after every ordinal. That's the formula that I can show relative to that system. | |
| Feb 15, 2021 at 12:18 | comment | added | Zuhair Al-Johar | yes it is different, I know, there is some resemblance. Anyhow, what I've put forwards is an axiomatic system and I'm asking if there is an obvious inconsistency with it, and if not, what would be its consistency strength? is it so powerful like for example the wholeness axiom? or is it within ZFC strength or near to it? I didn't claim I have a proof of it. By the way my system forbids $x=\{x\}$, it only allow acyclic membership ill founded kind of sets. | |
| Feb 15, 2021 at 12:14 | comment | added | Asaf Karagila♦ | I've seen Randall's proof back in 2016 when he explained it to Thomas and me in person over a few days. It's a very clever proof, and it is based on an iteration of permutation models (which I reckon can be retrofitted as an iteration of symmetric extensions). It has no business with ill-founded "ordinals" and "iterations" over these. | |
| Feb 15, 2021 at 12:13 | comment | added | Asaf Karagila♦ | If you're not going to listen, I'm not going to speak. You keep telling me "you can, you can", but I am saying that I can't. So the onus is on you, as the one making the mathematical claim, to provide me with a complete proof. What your comment is telling me is that you're merely postulating an axiomatic system. In which case, we can just postulate Con(NF), it gives a much much simpler proof of Con(NF). | |
| Feb 15, 2021 at 12:13 | comment | added | Zuhair Al-Johar | @AsafKaragila Con(NF) does follow from ZF. Holmes had proved that, its just a matter of time to see it, but his approach shares some similarities with that. | |
| Feb 15, 2021 at 12:11 | comment | added | Zuhair Al-Johar | @AsafKaragila, I've already referred the system, see "acyclic ZF", this is done by stipulation and not through definitions, the axioms gives it the meaning. The iteration occuring here is different, you can index it with ill-founded ordinals (or actually pre-ordinals). | |
| Feb 15, 2021 at 12:08 | comment | added | Asaf Karagila♦ | Look, the proof that iteration and recursion require well-foundedness are not "convention". It is exactly what prevents circularity and ill-founded definitions. Writing an equation (e.g. $x=\{x\}$) is meaningless without an axiomatic system which can prove that the equation has a solution. What is your equation and what is the proof that it has a solution? (And you'd think that something that implies Con(NF) in a very simple manner would be very very hard to justify, given the long-standing (and hopefully soon to be closed) open problem of whether or not Con(NF) follows from ZF. | |
| Feb 15, 2021 at 12:05 | comment | added | Zuhair Al-Johar | @AsafKaragila, I understand that for the conventional treatment. But this is not the same, here those are defined in such a manner as to violate well-foundedness, that's the point of all of that, so yes this theory goes against what is known traditionally, but I'm not seeing an obvious inconsistency with it. Here you can have stages inside stages inside stages,... and so on iteratively, all being isomorphic to each other. If that is consistent, then a proof of Con(NF) is almost straightforward. | |
| Feb 15, 2021 at 11:58 | comment | added | Asaf Karagila♦ | Do you understand that "iteration" and $V_\alpha$ are inherently well-founded? The whole point of well-foundedness is that you can iterate (recursively) over something if and only if it is well-founded. So your entire premise is flawed here. | |
| Feb 14, 2021 at 23:36 | comment | added | Zuhair Al-Johar | @NoahSchweber, yea, I'd agree with that, possibly they'd be more appropriately called as Pre-ordinals. | |
| Feb 14, 2021 at 23:30 | comment | added | Noah Schweber | @ZuhairAl-Johar I think that's worth explicitly mentioning - total order is such a core property of ordinals that, once dropped, it's not clear that the resulting objects deserve the name anymore. | |
| Feb 14, 2021 at 23:26 | comment | added | Zuhair Al-Johar | @NoahSchweber, well the membership relation $\in$ constitute a strict pre-order on ordinals. There is no assumption of it being a total order on them. | |
| Feb 14, 2021 at 23:16 | comment | added | Noah Schweber | Are the generalized ordinals in this context linearly ordered for some reason? | |
| Feb 14, 2021 at 23:12 | comment | added | Zuhair Al-Johar | @NoahSchweber, all of these details are shown in the referred theory "acyclic ZF". That said, then $x=V_\alpha \iff \mathcal R(\alpha,x)$, and ordinals are the ill-founded ordinals of that posting (transitive sets of transitive sets). $V$ when put like that (un-indexed) means the whole universe, it is not an object in this theory. | |
| Feb 14, 2021 at 22:34 | comment | added | Noah Schweber | What exactly are "ordinals" here, and what exactly is "$V$" here? | |
| Feb 14, 2021 at 11:42 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Feb 14, 2021 at 11:36 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Feb 14, 2021 at 11:29 | history | undeleted | Zuhair Al-Johar | ||
| Feb 11, 2021 at 22:45 | history | deleted | Zuhair Al-Johar | via Vote | |
| Feb 11, 2021 at 19:02 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Feb 11, 2021 at 18:48 | comment | added | Zuhair Al-Johar | @AsafKaragila, I'll correct the axiom to make it more explicit. | |
| Feb 11, 2021 at 18:48 | comment | added | Asaf Karagila♦ | That's a non-standard term, to say the least. So I have no idea what you're talking about. | |
| Feb 11, 2021 at 18:47 | comment | added | Zuhair Al-Johar | @AsafKaragila, agreed! the critical point would be a non-standard ordinal. | |
| Feb 11, 2021 at 18:43 | comment | added | Asaf Karagila♦ | That is not mentioned in the axiom. Also, since $j$ must map well-founded sets to well-founded sets, its restriction to the von Neumann universe would be an automorphism. Extensional and well-founded objects don't like automorphisms. | |
| Feb 11, 2021 at 18:42 | comment | added | Zuhair Al-Johar | @AsafKaragila, $<$ is ordinal strict smaller than. | |
| Feb 11, 2021 at 18:41 | comment | added | Zuhair Al-Johar | @NoahSchweber, well I didn't think of that really. But no problem you can shun those formulas, the most important is that $j$ can be used in separation and replacement. | |
| Feb 11, 2021 at 18:40 | comment | added | Asaf Karagila♦ | Ah, I didn't notice that. What's < here? | |
| Feb 11, 2021 at 18:40 | comment | added | Zuhair Al-Johar | @AsafKaragila, what about the last axiom? where is the critical point? | |
| Feb 11, 2021 at 18:15 | comment | added | Asaf Karagila♦ | You've omitted regularity now? So you can take ZFA with a non-empty set of atoms. Say, two atoms. Then the function that switches the atoms is a definable automorphism. And that's before we even stratified the axioms. | |
| Feb 11, 2021 at 15:07 | comment | added | Noah Schweber | Quick comment: $j$ cannot appear in the $\phi$s of the isomorphism axiom. | |
| Feb 11, 2021 at 12:23 | history | undeleted | Zuhair Al-Johar | ||
| Feb 11, 2021 at 12:21 | history | deleted | Zuhair Al-Johar | via Vote | |
| Feb 11, 2021 at 12:19 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Feb 11, 2021 at 8:03 | comment | added | Zuhair Al-Johar | @Hanul Jeon, what is known is that "stratified ZF + every set is the same size as a set of singletons", is equivalent to ZF. I'll try to find a source on that. However, I'm not sure of the strength of str ZF, perhaps it would be as strong as Zermelo or slightly stronger? | |
| Feb 10, 2021 at 23:29 | comment | added | Hanul Jeon | Side question: is there anything known about the consistency strength of stratified $\mathsf{ZF}$? | |
| Feb 10, 2021 at 12:22 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |