Timeline for Is acyclic ZF consistent?
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Feb 21, 2021 at 17:02 | comment | added | Zuhair Al-Johar | Does this argument require Replacement? I mean the existence of a Hartog number for every set requires replacement. So if we work in Zermelo only, then would this still apply? | |
| Feb 16, 2021 at 18:48 | comment | added | Zuhair Al-Johar | one last remark, the purpose of this theory was to add automorphisms on stratified versions of it, and for that purpose this might work, because simply stratfication cannot handel von Neumann's. | |
| Feb 16, 2021 at 13:58 | comment | added | Zuhair Al-Johar | hmmm... on second thought I'm realizing that the problematic factor (the one that causes ill-founded sequences) must be forbidden from comprehension axioms, so that the ill-founded sequence remain external, so here the culprit that I can see is $\mathcal R$, but if we add automorphism $j$ then that would be problematic also, since it's free use in comprehension axioms would cause internal forbidden sequences. I see. so it must be shunned from comprehension axiom. It appears that this kills my approach. Thanks | |
| Feb 16, 2021 at 13:18 | comment | added | Zuhair Al-Johar | continuation...If the website allows accepting two answers, I would have accepted BOTH, because both of them address the heart of the issue and they are the correct answers. I'm just saying that if we forbid using $\mathcal R$ then perhaps, this approach would be saved, we'd be having an external sequence of stages indexed by ill-founded ordinals, and this is pretty much standard, the real point of all of that is related to the automorphism question, this might allow us to use $j$ in separation and replacement but definitely not $\mathcal R$ | |
| Feb 16, 2021 at 13:13 | comment | added | Zuhair Al-Johar | @AsafKaragila, I don't understand your last comment really. Yes, you cannot have this sequence internally, I agree, but you can have it externally. I thought that that what happens when we don't allow $\mathcal R$ in separation and replacement. Allowing the stages being indexed by ill-founded ordinals, doesn't mean by itself that you have that sequence internally in your theory (unless you use $\mathcal R$), if somehow it can be shown that this sequence can exist internally even without using $\mathcal R$, then the whole approach fails). | |
| Feb 16, 2021 at 12:26 | comment | added | Asaf Karagila♦ | @ZuhairAl-Johar: At this point you stopped even participating in this discussion. You seem to be under the impression that you can just avoid von Neumann ordinals altogether, like they don't exist somehow. I'm not sure why you even accepted the other answer, since it also uses those. Even if you remove $\mathcal R$, you cannot have an ill-founded "sequence" of iterated power sets. Period. This is my last comment on the topic, you can do whatever you want with it. | |
| Feb 16, 2021 at 11:57 | comment | added | Zuhair Al-Johar | I just wanted to say that: all of that only means that all ordinals are von Neumann's. So the theory is still consistent but not interestng. However, I think if we forbid $\mathcal R$ from being used in separation and replacement, then possibly this can allow us to use ill-founded ordinals as ranks of stages, and even upgrad acyclic construction in the way used in Greg's answer. It's allowing the use of some ill-founded ordinals as indices of stages in an iterative hierarchy that is the whole purpose of this theory. | |
| Feb 16, 2021 at 11:32 | comment | added | Asaf Karagila♦ | @ZuhairAl-Johar: The Hartogs can jump by an infinite gap when taking power sets, so there's room to grow from the roots, even if all have the same Hartogs. | |
| Feb 16, 2021 at 11:30 | comment | added | Zuhair Al-Johar | By the way, I think all of that happend because you allow the use of $\mathcal R$ in instances of separation and replacement. I think if we shun that, we can save the approach. Or do you have a proof that the same argument would still run. | |
| Feb 16, 2021 at 11:22 | comment | added | Zuhair Al-Johar | Yes, of course, I know that. I'm speaking about the SET of the hartogs of ALL end nodes of a Specker tree with infinite rank. It appears to be a set of infinitely descending ordinals? | |
| Feb 16, 2021 at 11:19 | comment | added | Asaf Karagila♦ | @ZuhairAl-Johar: Because trees can have infinite rank without having infinite branches. | |
| Feb 16, 2021 at 11:15 | comment | added | Zuhair Al-Johar | I mean why this is not implying existence of infinitly descending hartogs coming down from a common node? I mean you say that the Specker tree cannot have any of its branches being infinite because the hartogs cannot descend down for ever, that's understood, but why it is not the same condition for the hartogs of all the end nodes of a tree, won't those be infinitely descending??? | |
| Feb 16, 2021 at 11:09 | comment | added | Asaf Karagila♦ | @ZuhairAl-Johar: Because it's a nontrivial problem? Why is it an open problem whether or not the Partition Principle implies choice? Sometimes things are just hard. | |
| Feb 16, 2021 at 11:09 | comment | added | Zuhair Al-Johar | why it is an open problem that one can have a Specker tree of infinite rank in absence of choice? | |
| Feb 16, 2021 at 11:01 | comment | added | Asaf Karagila♦ | @ZuhairAl-Johar: I'm sorry. I naturally assumed you'd understand sarcasm. No, choice is not needed here. If you look at the proof of Hartogs' theorem you'll find that it follows from it. And if you've used choice to prove Hartogs' theorem, you're doing it wrong. | |
| Feb 16, 2021 at 10:59 | comment | added | Zuhair Al-Johar | your comment is paradoxical you said it uses choice and then you said not? | |
| Feb 16, 2021 at 10:57 | comment | added | Asaf Karagila♦ | @ZuhairAl-Johar: Yes, it uses choice. That's why I go through Hartogs numbers and not just claim it is finite by Cantor's theorem. No. Of course it's not using choice. | |
| Feb 16, 2021 at 10:54 | comment | added | Zuhair Al-Johar | Does that argument use choice? | |
| Feb 16, 2021 at 9:22 | comment | added | Zuhair Al-Johar | Thanks for the answer. The definition of the Specker tree would be made inside a stage by using separation (or replacement), I think this definition would essentially use of the symbol $\mathcal R$ here. If we forbid using $\mathcal R$ in instances of separation and replacement would that objection still hold. | |
| Feb 16, 2021 at 2:17 | comment | added | Asaf Karagila♦ | (You can probably get better gaps with Lindenbaum numbers, but it doesn't matter, does it?) | |
| Feb 16, 2021 at 2:15 | comment | added | Asaf Karagila♦ | The point is, if $\aleph(\mathcal{PPP}(x))=\kappa$, then $\aleph(x)<\kappa$. Now consider the fact that a Specker tree is in a sense logarithmic when it "grows up", an infinite branch is necessarily a decreasing sequence in Hartogs numbers. | |
| Feb 16, 2021 at 2:05 | comment | added | Asaf Karagila♦ | It's not increasing, because if $x$ is amorphous $\aleph(x)=\aleph(\mathcal P(x))$. The thing is that the tree is really going downards. | |
| Feb 16, 2021 at 1:49 | comment | added | Noah Schweber | I think you want to say that $x\mapsto\aleph(x)$ is increasing with powersets, not just non-decreasing; otherwise I don't see how to get an infinite descending sequence of Hartogs numbers. | |
| Feb 15, 2021 at 23:49 | history | answered | Asaf Karagila♦ | CC BY-SA 4.0 |