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So, I'm not sure to what extent this is a thing. John Baez mentions in this blog post that common large countable ordinals beyond the large Veblen ordinal can also "be defined as fixed points". He doesn't expand on that but I take it that means fixed points of a further Veblen-like / klammersymbol-like construction.

But, I haven't been able to find any good account of this. Is this a known/standard thing? I want to know just how far the Veblen construction can be pushed -- like, beyond the large Veblen ordinal, is there an "ultimate Veblen ordinal" somewhere; and might it be equal to an already-named / well-known one, such as Bachmann-Howard? (I suppose a negative answer here would be if arbitrarily large computable ordinals could be obtained this way, making the "ultimate Veblen ordinal" just $\omega_1^{\mathrm{CK}}$.)

Now I can certainly think of ways of continuing beyond the large Veblen ordinal myself... but I don't really want to reinvent the wheel here when I expect others have likely already done it better. So, is there a good account of this somewhere?

Thank you all!

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    $\begingroup$ If one takes Veblen-like construction as an informal concept, then it would be hard to justify some $\alpha<\omega_1^{\mathsf{CK}}$ to be the ultimate Veblen ordinal (the suprema of all ordinals obtained by Veblen-style constructions). The reason for this is that it wouldn't be clear why one shouldn't be allowed to make the next step of diagonalization and overcome this particular ordinal. $\endgroup$
    – Fedor Pakhomov
    Commented Sep 8, 2020 at 9:51
  • $\begingroup$ However, I think that it is possible to give Bachmann-Howard ordinal as a (non-least) upper bound for reasonable constructions of this kind. The motivation here is that it is reasonable to expect that this constructions should be formalizable in Kripke-Platek set theory with infinity $\mathsf{KP}\omega$ and the proof-theoretic ordinal of $\mathsf{KP}\omega$ is Bachmann-Howard ordinal. $\endgroup$
    – Fedor Pakhomov
    Commented Sep 8, 2020 at 9:52
  • $\begingroup$ I think a link to original paper (of howard I think?) was posted on fom (if I am re-calling correctly). I don't think it was that long ago (maybe around an year or bit more). I will try to find and post the link (in few days perhaps). Other than that, one could probably find a number of relevant topics by using a search or following the links in some of the topics under "Related". $\endgroup$
    – SSequence
    Commented Sep 8, 2020 at 14:35
  • $\begingroup$ Here is the link I was referring to in comment above: arxiv.org/abs/1903.04609 $\endgroup$
    – SSequence
    Commented Sep 9, 2020 at 4:54
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    $\begingroup$ @HarryAltman I think that it is very unlikely that there is some concrete point of failure. But in my opinion the reasoning that either there is a concrete point of failure or we could reach unlimitely large ordinals isn't valid. This is because we are dealing with an informal notion here. Furthermore, it is rather unlikely that any particular formalization would be completely satisfactory (precisely because then it would lead to a concrete point of failure, which we would be able to diagonalize against). $\endgroup$
    – Fedor Pakhomov
    Commented Sep 10, 2020 at 10:23

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Here is some relevant information by a grandmaster on the subject: http://www.mathematik.uni-muenchen.de/~buchholz/articles/jaegerfestschr_buchholz3.pdf

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  • $\begingroup$ OK, taking a look at this I'm afraid I don't understand this. He explains how to construct $\Gamma_\alpha$ in the preliminaries, but I'm having trouble following it beyond that. His $\phi$ only has one subscript, but (going by the equation before Lemma 1.6) it acts like multiple (i.e., acts like Klammersymbols) if you go into uncountables? I don't really understand how to get beyond that though, especially if merely getting beyond (the equivalent of) a single subscript requires getting to $\Omega$! I'd hope for each step to be somehow expressible in terms of things constructed so far... $\endgroup$
    – Harry Altman
    Commented Oct 19, 2020 at 2:12

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