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

Lemme just point out to a few things which are addressed in every rant on the topic.

"The ordinal associated with the growth rate of f(n) is be regarded as the ordinal that would be in the subscript of the fast-growing hierarchy to match f(n)." What does it mean that \(f_\alpha\) "matches f(n)"?

"any function with a growth-rate equivalent to \(f_{\omega_1^{CK}}\) must be an uncomputable function." Fun fact, there are choices of fundamental sequences for which \(f_{\omega_1^{CK}}\) is computable, see here.

"This is because no matter what means you use to define the fundamental sequence of \(\alpha\), f(n) will always have the same growth rate in relation to neighboring functions" wrong. For any function \(F\), there is a choice of a fundamental sequence for \(\omega\) such that already \(f_\omega\)

The above remarks should point out that no, the notion of an ordinal assigned to a function is not well-defined.

"level-\(\alpha\) Turing machine" What are those? Even if we agree that for finite \(\alpha\) those are Turing machines with iterated halting oracle, for infinite \(\alpha\) it's not so clear, since diagonalizing the halting oracle can lead to many inequivalent things, especially after \(\omega_1^{CK}\).