User blog comment:Rgetar/Higher weakly inaccessible and weakly Mahlo cardinals/@comment-28606698-20200106214759/@comment-28606698-20200107085328

> π is an uncountable regular cardinal iff every normal function f:π → π has a fixed point. Why? Proof see on the page 115

Note that uncountable regular cardinals and weakly Mahlo cardinals have rather similar definitions:

π is uncountable regular iff every normal function on it has a fixed point,

π is weakly Mahlo iff every normal function on it has a regular fixed point.

Let I_ρ is a function enumerating ρ-weakly inaccessible cardinals and their limits. We can obtain larger cardinals using regular fixed points of functions κ↦I_ρ(κ) or κ↦I_κ(0) or κ↦I(κ,0,0), but we can’t obtain a weakly Mahlo cardinal π from below using this approach, because π is so large that every such function on π has a regular fixed point less than π.

Thus, in some kind, the first weakly Mahlo cardinal plays the same role for the functions I_ρ as the first uncountable cardinal for the Veblen functions ϕ_α (note that Veblen function ϕ_α is a function enumerating ordinals γ such that ϕ_β (γ)=γ for all β<α, meanwhile I_ρ is a function enumerating regular cardinals π, such that I_σ(π)=π for all σ<ρ, and their limits).

So we can’t reach M_1 starting with 0 and using κ↦I_ρ(κ), κ↦I_κ(0) , κ↦I(κ,0,0) , …

and we can’t reach M_2 starting with M_1 and using κ↦I_ρ(κ), κ↦I_κ(0) , κ↦I(κ,0,0) , …

…

>Isn't that already in the definition of a normal function?

Yes, it is. I just added some properties (which are already in the definition) trying to make the explanation more clear.