User blog comment:Edwin Shade/Rank-on-Rank Turing Ordinals and Beyond/@comment-1605058-20180119225409/@comment-1605058-20180120101519


 * \(\alpha-1\) only makes sense if \(\alpha\) is a successor ordinal, but in other cases we can use, for example, \(f_\beta<^*f\) for all \(\beta<\alpha\). With this definition, it's not clear that every function has its ordinal well-defined (even if we fix some fundamental sequences for all countable ordinals) - why is it not possible that \(f_\alpha<^*f\) for all countable ordinals \(\alpha\)?


 * I should have pointed it out, I apologize that I didn't - the point here is that what was done in the Math.SE post for \(\varepsilon_0\) can be done for many other ordinals, in particular for \(\omega_1^{CK}\) (generally, for any ordinal which is a limit of limit ordinals). However, if you put the assumption that we need to have \(f_\beta<^*f_\alpha\) for \(\beta<\alpha\), this example will break. Nevertheless, I do think that even then we can set up fundamental sequences cleverly in such a way that \(f_{\omega_1^{CK}}\) is computable, however I don't have a reference for that. If you can disprove it, I'd be glad to hear the argument.


 * My point here is the same as in Upquark's, please also see my reply there.


 * On the busy beaver article things are intentionally explained in somewhat vague way, and importantly they do not touch on an important issue - dependence on the choice of enumeration of machines. While for finite levels this is not too much of a problem as long as machines' enumeration is recursive, to proceed in a manner described there for transfinite ordinals, we need to arrange all the machines of level \(\beta<\alpha\) in a two-dimensional array of size \(\omega\times\omega\), which essentially requires us to fix a bijection between \(\alpha\) and \(\omega\), and the problem here is that for different choice of this bijection we will get different busy beaver functions which need not to grow in the same way.