User blog comment:Rgetar/Higher weakly inaccessible and weakly Mahlo cardinals/@comment-28606698-20200106214759/@comment-35470197-20200112002246

> Normal function on α is normal function f(x) such as for any δ < α there is x such as δ < f(x) < α

I did not know such a terminology of "on α". Is it correct? It is just a normal function such that α is not a successor ordinal and belongs to the image of f. (Of course, you can define the terminology in that way. Maybe just I have not known the terminology, but I am afraid that the definition includes a typo.)

> 1

The proof includes circular logic, unless you show that β+1 is the least ordinal above β, because Theorem 1 is often used to show that β+1 is the least ordinal above β. If you want to show such a statement, you need to show it only using the definition of the notion of ordinals in order to avoid circular logic. (Since such arguments using the definition are usually very long, it might be good to skip writing the proof here if you understand it.)

> 2

The proof is good, but I note that you do not have to divide the cases because the statement holds even if we only require f to be increasing.

> 10

The proof is good. I note that the theorem follows from Kleene's fixed point theorem, which can be verified in the same method. Since Kleene's fixed point theorem is useful, it is good to know the name.

> 16

You need to assume that the given set of fixed points is not empty. Otherwise, the supremum of the set is 0, which is not necessarily a fixed point of f. Please show "sup{α_i} = f(α) for some α" in a precise manner. Then you will find that {α_i} should not be empty in order to apply the continuity to obtain the statement.

> 17

It has counterexamples. For example, cosider f(x) = ω+x. Then f is a normal function on ω+ω by the definition of your terminology of "on α", but {n|f(n)<ω+ω} = {n|n<ω} is not a fundamental sequence of ω+ω.

> 18～

I think that your definition of "on α" includes a typo.