Timeline for Axiomatic strength of the cumulative hierarchy
Current License: CC BY-SA 4.0
23 events
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| Jul 7 at 21:03 | comment | added | Tim Button | @ZuhairAl-Johar: there are several options for how to define "level", certainly. In fact, my definition does entail that levels are transitive and potent (/supertransitive) without any axioms (see my 2021, Lemma 3.4). I do see the virtue of defining a level as a member of a history. But I wanted my basic version of LT to carry no information at all about the height of the hierarchy, so I didn't want to assume that every level was in some set. (Of course our aims here have been a bit different!) | |
| Jul 7 at 20:51 | history | edited | Tim Button | CC BY-SA 4.0 |
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| Jul 7 at 20:47 | comment | added | Tim Button | @HarryWilson yup, clearly I need to come up with a better nom de plume | |
| Jul 7 at 20:40 | history | edited | Tim Button | CC BY-SA 4.0 |
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| Jul 7 at 19:38 | comment | added | Zuhair Al-Johar | @AlecRhea, Your welcome! Thank you for this rather nice question! | |
| Jul 7 at 18:34 | comment | added | Zuhair Al-Johar | @TimButton see: mathoverflow.net/q/474492/95347, and mathoverflow.net/a/474522/95347. I personally believe your definition of level works even if it posteriorly depends on separation. But, for simplicity I prefer the accumulation function in defining histories rather than the potentiation, because it settles transitivity (and supertransitivity) of levels before the axioms. | |
| Jul 7 at 18:31 | comment | added | Zuhair Al-Johar | @TimButton, very nice! I've posted this question before two day, see Hamkins answer to it also. There is a point I like to raise about the definition of level, I personally prefer the definition to settle matters before the axioms, so I see Scott's definition of history to be an easier one to work with, also levels can be defined as set unions of histories, or as elements of histories, or as accumulation of histories. But, the simplest is as elements of histories. Your definition of a level a potentiation of a history, is nice, but not simpler since it needs axioms to prove transitivity. | |
| Jul 7 at 17:48 | comment | added | Harry Wilson | You ARE Button 2021 !! | |
| Jul 7 at 17:48 | comment | added | Harry Wilson | Heeey wait a minute! | |
| Jul 7 at 17:41 | comment | added | Alec Rhea | @ZuhairAl-Johar Thank you for the input here Zuhair, I've enjoyed reading this exchange very much. (and learned something, two birds!) | |
| Jul 7 at 17:38 | vote | accept | Alec Rhea | ||
| Jul 7 at 17:06 | history | edited | Tim Button | CC BY-SA 4.0 |
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| Jul 7 at 13:49 | comment | added | Tim Button | @ZuhairAl-Johar Thanks! I shall update the answer :) | |
| Jul 7 at 13:24 | comment | added | Zuhair Al-Johar | @TimButton. That's Ok. Of course you can edit your answer, since this comment is basically an offshoot of your comment. I only made a small modification, the argument is basically yours! | |
| Jul 7 at 12:21 | comment | added | Tim Button | @ZuhairAl-Johar: excellent! Yes, you're right. Well I think this fully answers the OP's question. Shall I edit my answer to incorporate your suggestion, or would you like to post a separate answer? | |
| Jul 7 at 12:13 | comment | added | Tim Button | I'll just quickly explain (for anyone else reading this) why @ZuhairAl-Johar's scheme gives Extensionality. Fix $c$, and let $f$ be this partial function: $f(x) = x$ if $x \in c$, but $f(x)$ is undefined otherwise. By the scheme: there is a level, $s$, with $c \in s$, and nothing else in $s$ is coextensive with $c$. But we must also check that there is no $d \notin s$ such that $d$ is coextensive with $c$. This holds because $s$, being a level, is closed under subsets, so that if $d$ is co-extensive with $c \in s$ then $d \in s$. | |
| Jul 7 at 12:07 | comment | added | Zuhair Al-Johar | @TimButton, the singleton function will give you infinity. | |
| Jul 7 at 12:05 | comment | added | Tim Button | @ZuhairAl-Johar: I think that's a very neat way to combine all of: Extensionality, Separation, Unbounded, Endless and Stratification. But I don't think it gives you Infinity; indeed, I think your scheme holds in the hereditarily finite sets. | |
| Jul 4 at 16:38 | comment | added | Zuhair Al-Johar | Actually we can get matters down to ONE axiom scheme for all of $\sf ZF$. That is: if $f$ is a partial function symbol, then: $$\forall a \exists s \ni a: Lev(s) \land \forall x \in s \exists! y \in s: y=\{f(z)\mid z \in x\}$$. | |
| Jul 4 at 10:11 | comment | added | Zuhair Al-Johar | I think we can get matters to one axiom scheme on top of Extensionality. Given your definition of $Lev$ we can add that for any partial function symbol $f$, we have: $$ \forall a \exists s \ni a: Lev(s) \land \forall x \in s \exists y \in s: y=\{f(z)\mid z \in x\}$$. There is no need for Separation here. | |
| Jul 3 at 21:51 | comment | added | Alec Rhea | Very interesting Tim, thank you. (and cool to see you here on MO!) | |
| S Jul 3 at 14:19 | review | First answers | |||
| Jul 3 at 16:46 | |||||
| S Jul 3 at 14:19 | history | answered | Tim Button | CC BY-SA 4.0 |