User blog comment:Rgetar/Higher weakly inaccessible and weakly Mahlo cardinals/@comment-28606698-20200106214759/@comment-35470197-20200121081819

> What is "it"? Normal function on α according another definition?

"It" meant a normal function in your sense, but I made a typo. My explanation "α is not a successor ordinal and belongs to the image of f" is a typo of "α is not the image of a successor ordinal and belongs to the image of f". For example, f(x) = ω^x and α = ω^ω. Then f is a normal function on α, because sup_{n<ω}f(n) = α. It is what you intended? (If it is waht you intended, then it is ok. I just felt strange because it is not a usual use of the word "on" in a property of a map unlike the original use of "on" in the book which you quoted.)

> Can empty set be considered as a "set of fixed points"?

Of course. A set of sets satisfying a specific property, e.g. fixed points can be empty. For example, an ordinal is a set of ordinals. The statement does not mean that an ordinal is non-empty.

> 17 states that {f(n)|f(n) < α} forms a fundamental sequence of α,

It makes sense. Sorry for the misunderstanding of what you meant.

> We get a normal function f(n) on α, n < cof(α),

I guess that it is a normal function on α in the convention in the book which you quoted, but not a normal function on α in your convention. Or are you assuming that a normal function on α is small, i.e. its domain is an ordinal? Then your proof using the composition does not work, because in that case composition will be ill-defined. Usually, we do not assume the smallness, i.e. we also consider the case where the domain is the proper class "On" of ordinals.

In order to fix the problem, you need to extend f so that it is defined on "On".