Timeline for Predicative definition and existence of ordinal numbers
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2019 at 2:37 | comment | added | Timothy | I like this question. It seems that you're looking for an answer that none of the answers to math.stackexchange.com/questions/71726/… solve your problem. I guess you were asking something more like math.stackexchange.com/questions/1212221/…. | |
| Nov 16, 2016 at 22:27 | vote | accept | Damian | ||
| Nov 16, 2016 at 22:27 | history | edited | Damian | CC BY-SA 3.0 |
make conclusion
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| Nov 5, 2016 at 21:43 | comment | added | Noah Schweber | As to the $\alpha_n$s, I am applying separation (*not* replacement this time) in the following way: given a set of well-orderings $\{S_n:n\in\mathbb{N}\}$, we can by Separation get the set of $n$s such that "There exists an ordinal $\alpha$ in order-preserving bijection with $S_n$." Now, this act of Separation on the face of it refers to the class of ordinals, so might be predicatively suspect; whether it can be made predicatively acceptable depends on what you mean by "predicative," see my previous comment. | |
| Nov 5, 2016 at 21:41 | comment | added | Noah Schweber | To the OP: I just saw your edit. The predicativity requirements you apply are going to have to be made precise before I can give a meaningful answer. Meanwhile, I have absolutely no idea what your second bullet means: what is a "Goedel-proof"? Everything I've written is a perfectly valid proof from ZFC (note that ZFC itself is not predicative). I think before I can comment more in a useful way, you need to make clearer exactly what kind of proof you are willing to accept. Otherwise I'm just chasing a moving target. | |
| S Nov 1, 2016 at 14:28 | history | suggested | Damian | CC BY-SA 3.0 |
refer to new comment
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| Nov 1, 2016 at 14:15 | review | Suggested edits | |||
| S Nov 1, 2016 at 14:28 | |||||
| Oct 31, 2016 at 19:45 | comment | added | Noah Schweber | To the OP: see my edit. You are misunderstanding the axiom of replacement. | |
| S Oct 31, 2016 at 19:30 | history | suggested | Damian | CC BY-SA 3.0 |
refined question
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| Oct 31, 2016 at 19:12 | review | Suggested edits | |||
| S Oct 31, 2016 at 19:30 | |||||
| Oct 31, 2016 at 1:01 | comment | added | Noah Schweber | @AsafKaragila Whoops, missed that. | |
| Oct 31, 2016 at 0:58 | comment | added | Asaf Karagila♦ | @Noah: As you can see below the deleted answer, I haven't forgot. :-) | |
| Oct 31, 2016 at 0:27 | comment | added | Noah Schweber | @AsafKaragila Don't forget Replacement! That's the other power player here. | |
| Oct 31, 2016 at 0:26 | answer | added | Noah Schweber | timeline score: 4 | |
| Oct 30, 2016 at 21:26 | comment | added | Asaf Karagila♦ | You don't have to prove this. Hartogs' theorem, historically, came before the von Neumann ordinals were in play. The theorem states that there is a well-ordered set which does not inject into a set. This is not the same as saying "there is a von Neumann ordinal which does not inject into a set". And indeed without the power set the collection of countable ordinals can be a proper class. But with power set and replacement we can just prove that the class of countable ordinals is in fact a set. So you have to choose, power set, replacement, both, neither. Then we can talk. | |
| Oct 30, 2016 at 21:03 | comment | added | user100477 | hello Asaf, my first objection applies before the power set argument in the proof, so let me rephrase it into a question: Why do I have to prove that the collection $\{ o | o \text{ is a countable ordinal} \}$ exists as a set, but don't have to prove that it exists as a class? | |
| Oct 30, 2016 at 20:28 | comment | added | Asaf Karagila♦ | You're missing the power set axiom. And indeed without it, we cannot prove that $\omega_1$ exists. | |
| Oct 30, 2016 at 20:23 | comment | added | Wojowu | Hartogs' lemma doesn't "take the class of ordinals as given". You can take the set of all well-orderings on subsets of $X$ (power set a few times and apply specification), and then consider the map taking a well-ordering $\prec$ to its order type $\beta$ (which is an ordinal; its existence can be proven by transfinite induction along $\prec$). Axiom of replacement then shows that the image of this function is a set, and this is precisely the Hartogs' ordinal. | |
| Oct 30, 2016 at 20:13 | review | First posts | |||
| Oct 30, 2016 at 20:23 | |||||
| Oct 30, 2016 at 20:10 | history | asked | Damian | CC BY-SA 3.0 |