Ordinal Notation[]
I have never seen an ordinal notation associated to Madore's OCF. Is there an ordinal one? Or is it just defined up to an OCF?
p-adic 01:37, February 15, 2020 (UTC)
- Yes, it seems, there not exists an ordinal notation associated to the Madore's function. Only OCF was defined.--Denis Maksudov (talk) 12:10, February 15, 2020 (UTC)
- Thank you. Then I will add an explanation of the non-existence of a canonical one.
- p-adic 12:15, February 15, 2020 (UTC)
- If an ordinal notation system \((OT,<)\) is created associated to Madore's function in the traditional way, it's likely impossible for the function assigning order type to be an order-preserving isomorphism. To my knowledge, such order-preserving isomorphism has to be a bijection, however not all members of \(C(\varepsilon_{\Omega+1})\) (e.g. \(\varepsilon_0\times\omega=\omega^{\varepsilon_0+1}\)) are uniquely representable in terms of a suitable normal form for Madore's function. We can define a normal form forbidding terms including \(``\omega^{\varepsilon_0+1}\!"\) (or \(``\varepsilon_0\times\omega\!"\), taking the other approach), however I don't consider such a normal form to be suitable for an interpretation of Madore's function. The only other idea I have for an associated ordinal notation with an order-preserving bijection would be letting \(OT\) be an equivalence class under a relation \(\equiv\) (e.g. \(``\omega^{\varepsilon_0+1}\!"\equiv``\varepsilon_0\times\omega\!"\)), then proving order-preservingness of \(<\) up to \(\equiv\) C7X (talk) 03:01, 7 April 2021 (UTC)
- Using ≡ does not solve the issue, because then you need to show the recursive ness of <. (I do not know the issue on normal form, because I have never heard Madore's intention on the definition of normal form.)
- p-adic 14:16, 7 April 2021 (UTC)
Alternate link[]
Are there objections on changing the source to this page? The main problem that I could think of would be if there had been changes to the draft since its copying to the mainspace C7X (talk) 23:33, 3 April 2021 (UTC)
- How about putting both links?
- p-adic 23:35, 3 April 2021 (UTC)
Value of \(\psi(0)\)[]
Hello! In line 2 of the definition I read \(C_{n + 1}(\alpha) = \{\gamma + \delta , \gamma \delta , \gamma^{\delta} , \psi(\eta) \mid \gamma , \delta , \eta \in C_n(\alpha) ; \eta < \alpha\}\). Substituting \(0\) for \(\alpha\) gives \(C_{n + 1}(0) = \{\gamma + \delta , \gamma \delta , \gamma^{\delta} , \psi(\eta) \mid \gamma , \delta , \eta \in C_n(0) ; \eta < 0\}\). There are no ordinals \(<0\), this'd mean that \(\eta < 0\) is always false and thus would mean that \(C_{n+1}(0)\) is empty. As \(C(0) = \bigcup_{n < \omega}C_n(\alpha)\) and \(C_0(0) = \{0 , 1 , \omega , \Omega\}\), this'd mean that \(C(0) = \{0 , 1 , \omega , \Omega\}\) and \(\psi(0)\), the least countable ordinal that's not in the set \(C(0)\), would be \(2\). Is this correct? DaVinci103 (talk) 07:26, 13 September 2023 (UTC)
- For \(C_{n + 1}(0) = \{\gamma + \delta , \gamma \delta , \gamma^{\delta} , \psi(\eta) \mid \gamma , \delta , \eta \in C_n(0) ; \eta < 0\}\), there are no ordinals \(<0\), this'd mean that \(\eta < 0\) is always false and thus would mean that \(\psi(\eta)\) is not in \(C_{n+1}(0)\). Still we have \(\gamma + \delta , \gamma \delta , \gamma^{\delta}\). So \(C_{n + 1}(0) = \{\gamma + \delta , \gamma \delta , \gamma^{\delta} \mid \gamma , \delta \in C_n(0)\}\). 🐟 Fish fish fish ... 🐠 08:35, 13 September 2023 (UTC)
- Okay, thanks! Does the definition of \(;\) in set-builder notation then differ from the definition of \(\land\)? If that isn't the case, then there would there be no \(\gamma\) and \(\delta\) for which \(\eta < 0\) and thus there would be no \(\gamma + \delta\), \(\gamma\delta\) and \(\gamma^{\delta}\). DaVinci103 (talk) 11:31, 13 September 2023 (UTC)
- The original Wikipedia page included two separate sets for \(\gamma + \delta , \gamma\delta , \gamma^{\delta}\) and \(\psi(\eta)\). Shall I change it here to the union of two separate sets too? DaVinci103 (talk) 13:20, 13 September 2023 (UTC)
- Okay! I've edited the page: https://googology.fandom.com/wiki/Madore%27s_function?oldid=398165 . DaVinci103 (talk) 14:07, 13 September 2023 (UTC)