User blog comment:Rgetar/Higher weakly inaccessible and weakly Mahlo cardinals/@comment-28606698-20200106214759/@comment-32213734-20200111214912

The proof referred to some other parts of the book that were not displayed in the preview so I had to find the whole book "Frank R. Drake. Set Theory. An Introduction to Large Cardinals." I understood the proof.

Definitions.
 * Increasing function f(x): if x > y then f(x) > f(y).
 * Continuous function f(x): if x is limit ordinal then f(x) = sup{f(y)|y < x}.
 * Normal function is increasing and continuous function.
 * Normal function on α is normal function f(x) such as for any δ < α there is x such as δ < f(x) < α
 * Fixed point of function f(x) is α such as α = f(α)
 * Functional power fn(x):
 * f0(x) = x
 * fn + 1(x) = f(fn(x))
 * for limit n fn(x) = sup{fm(x)|m < n}

Theorems.
 * 1. If α > β then α ≥ β + 1. Proof: β + 1 is least ordinal larger than β so α ≥ β + 1.
 * 2. If f is normal function then for any x f(x) ≥ x. Proof: by transfinite induction. We need to consider cases for 0, successors and limits:
 * For 0. f(0) ≥ 0 since 0 is least ordinal.
 * For successors. Let f(x) ≥ x. Then
 * f(x + 1) > f(x) since f is increasing,
 * f(x + 1) > f(x) ≥ x
 * f(x + 1) > x
 * f(x + 1) ≥ x + 1 by 1
 * For limits. Let x is limit ordinal. Then
 * f(x) = sup{f(y)|y < x}
 * x = sup{y|y < x}
 * Let f(y) ≥ y for any y < x
 * Let f(x) < x. Then
 * f(x) > f(y) ≥ y for any y < x
 * f(x) > y for any y < x
 * y < f(x) for any y < x
 * y < f(x) < x for any y < x so x is not supremum of y < x
 * So if f(y) ≥ y then f(x) ≥ x
 * 3. If f is normal function then if f(x) > f(y) then x > y. Proof: if x = y then f(x) = f(y); if x < y then f(x) < f(y). So x > y.
 * 4. If f is normal function then if f(x) = f(y) then x = y. Proof: if x > y then f(x) > f(y); if x < y then f(x) < f(y). So x = y.
 * 5. If f is normal function, α is not a fixed point of f, then f(α) > α. Proof: by 2 f(α) ≥ α. If f(α) = α then α is fixed point. So f(α) > α.
 * 6. If f is normal function, α is not a fixed point of f, then f(α) is not a fixed point of f(x). Proof: by 5 f(α) > α. Since f is increasing, f(f(α)) > f(α), that is f(f(α)) ≠ f(α), so f(α) is not a fixed point of f.
 * 7. If f is normal function, α is not a fixed point of f, then fn(α) is not a fixed point of f for any n < ω. Proof: by induction
 * f0(α) = α ≠ f(α)
 * Let fn(α) ≠ f(fn(α)). Then by 6
 * f(fn(α)) ≠ f(f(fn(α)))
 * fn + 1(α) ≠ f(fn + 1(α))
 * 8. If f is normal function, α is a fixed point of f, then fn(α) = α for any n. Proof: by transfinite induction:
 * f0(α) = α
 * Let fn(α) = α. Then fn + 1(α) = f(fn(α)) = f(α) = α
 * Let n is limit ordinal. Let fm(α) = α for any m < n. Then sup{fm(α)|m < n} = sup{α|m < n} = sup{α} = α
 * 9. If f is normal function, α is not a fixed point of f, then if n > m then fn(α) > fm(α) for any n < ω. Proof: by 7 fn(α) is not a fixed point of f for any n < ω. By 5
 * f(fn(α)) > fn(α)
 * fn + 1(α) > fn(α)
 * f0(α) < f1(α) < f2(α) < f3(α) < ...
 * So if n > m then fn(α) > fm(α)
 * 10. If f is normal function, then for any α fω(α) is its fixed point. Proof: if α is fixed point of f then by 8
 * fω(α) = α = f(α) = f(fω(α))
 * fω(α) = f(fω(α))
 * Let α is not a fixed point of f. Since fω(α) = sup{fn(α)|n < ω}, it is a limit ordinal. Since f is continuous, f(fω(α)) = sup{f(fn(α))|n < ω} = sup{fn + 1(α)|n < ω}
 * sup{fn(α)|n < ω} = sup{f0(α), f1(α), f2(α), f3(α), ...}
 * sup{fn + 1(α)|n < ω} = sup{f1(α), f2(α), f3(α), f4(α), ...}
 * By 9 f0(α) < fn(α) for 0 < n < ω so
 * sup{fn(α)|n < ω} = sup{fn + 1(α)|n < ω}
 * fω(α) = f(fω(α))
 * 11. If f is normal function, α is not a fixed point of f, then fω(α) is the next fixed point above α. Proof: by 10 fω(α) is fixed point of f. Let β < fω(α) is fixed point of f above α:
 * α < β < fω(α)
 * By 7 fn(α) are not fixed points, so β ≠ fn(α) for all n < ω. Since fω(α) = sup{fn(α)|n < ω}, that is least ordinal ≥ any fn(α) such as n < ω, there exists m < ω such as
 * β < fm(α)
 * Since β > α = f0(α), 0 < m < ω. Let k + 1 = m. So
 * fk(α) < β < fk + 1(α), k < ω
 * f is increasing function, so
 * f(fk(α)) < f(β) = β
 * fk + 1(α) < β
 * but should be
 * fk + 1(α) > β
 * So least fixed point above α is fω(α).
 * 12. If f is normal function, α is not a fixed point of f, then α + 1 is not a fixed point of f. Proof: by 11 fω(α) is the next fixed point above α, and fω(α) is limit ordinal, but α + 1 is successor ordinal. So α + 1 is not a fixed point of f.
 * 13. If f is normal function, α is not a fixed point of f, then fω(α) = fω(α + 1). Proof: by 11 fω(α) is the next fixed point of f above α, and fω(α + 1) is the next fixed point of f above α + 1. So there is no fixed point β such as
 * α < β < fω(α)
 * By 12 α + 1 is not a fixed point of f, that is α + 1 ≠ β. Since there are no ordinals between α and α + 1, then there is no fixed point β such as
 * α + 1 < β < fω(α)
 * that is fω(α) is the next fixed point of f above α + 1, so
 * fω(α) = fω(α + 1)
 * 14. If f is normal function, then fω(α + 1) is the next fixed point of f above α for any α. Proof: let α not a fixed point of f. Then by 11 fω(α) is the next fixed point of f above α, and by 13 fω(α + 1) = fω(α), so fω(α + 1) is the next fixed point of f above α. Let α is a fixed point of f. Let α + 1 is a fixed point of f. Then by 8 fω(α + 1) = α + 1, so fω(α + 1) is a fixed point of f. Let α + 1 is not a fixed point of f. Then by 11 fω(α + 1) is the next fixed point of f above α + 1. Since there are no ordinals between α and α + 1, and α + 1 is not a fixed point of f, fω(α + 1) is the next fixed point of f above α.
 * 15. If f is normal function, then for any α there is fixed point of f larger than α. Proof: by 14 fω(α + 1) is the next fixed point of f above α, that is fω(α + 1) is fixed point of f and fω(α + 1) > α.
 * 16. Supremum of any set of fixed points of normal function f is fixed point of f. Proof: if a set of fixed points of f has maximal element, then it is its supremum, and it is a fixed point of f. If a set of fixed points of f {αi} has no maximal element, then, since f is continuos function
 * sup{αi} = f(α) for some α
 * Let α is not a fixed point of f. Then α < f(α). Since f(α) = sup{αi}, that is least ordinal ≥ any αi, there exists j such as
 * αj > α
 * f is increasing, so
 * f(αj) > f(α)
 * αj > f(α)
 * but should be
 * αj ≤ f(α)
 * Hence α = f(α)
 * 17. If f is normal function on α then f(n) for n such as f(n) < α forms a fundamental sequence of α. Proof: since for any δ < α there is n such as δ < f(n) < α
 * α = sup{f(n)|f(n) < α}
 * f is increasing so if n > m then f(n) > f(m). So
 * f(n) = α[n] for f(n) < α
 * 18. If f is normal function on α and α is regular then α is a fixed point of f. Proof: from 17 f(n) = α[n] for f(n) < α, so
 * α = sup{f(n)|f(n) < α}
 * α is limit ordinal and f is continuous, so
 * α = f(δ) = sup{f(n)|n < δ}
 * By 2
 * δ ≤ f(δ)
 * δ ≤ α
 * f(n) = α[n], n < δ, so
 * δ ≥ cof(α) = α
 * δ ≥ α
 * δ = α
 * δ = f(δ)
 * α = f(α)
 * 19. If f is a normal function, α is a fixed point of f, then for if β < α then f(β) < α. Proof: f is increasing so
 * f(β) < f(α) = α
 * f(β) < α
 * 20. If f is a normal function on regular α > ω, then f has a fixed point less than α. Proof: let
 * β < α
 * β = f0(β)
 * f0(β) < α
 * By 19
 * if fn(β) < α then fn + 1(β) < α
 * fn(β) < α, n < ω
 * fω(β) is least ordinal larger than all fn(β) < α for n < ω, so
 * fω(β) ≤ α
 * By 8 fω(β) is a fixed point of f. Since fω(β) = sup{fn(β)|n < ω} fn(β), where n < ω, is a fundamental sequence of fω(β). So
 * cof(fω(β)) ≤ ω
 * but
 * cof(α) = α ≥ ω
 * cof(α) ≥ ω
 * So
 * fω(β) ≠ α
 * fω(β) < α
 * 21. If for any normal function f on α f has a fixed point less than α, then α is regular and α > ω. Proof: let α = ω. Then f(x) = 1 + x is a normal function on ω, but for f(x) < ω f(x) > x. If α is not a limit ordinal then there is no normal function on α. If α is a limit ordinal and cof(α) < α then it has a fundamental sequence of length cof(α). Take a fundamental sequence of length cof(α), then add all its limits less than α (adding limits does not change length of a fundamental sequence, since it just increases n in α[n] by 1 after inserted limits till next limit), remove all elements less than cof(α) and enumerate elements. We get a normal function f(n) on α, n < cof(α), which has no fixed points less than α, since n < cof(α), but f(n) ≥ cof(α), so f(n) > n.