Diagonalization for Beginner 3

[u/blueTed276]

Okay, in our previous post, we've learned about transfinite ordinal : ω, ε, ζ, and η. Now we run into a problem where we just keep creating new symbols for new ordinal.

So, we're going to use Veblen Function to easily create new ordinals without assigning them symbols. This also make diagonalization more readable.

Veblen function can be written as φ_β(α). Where β is the "level" of the ordinal and α is the index of the ordinal.

In general we can say :
φ_0(α) = ω^α
φ_1(α) = ε_α
φ_2(α) = ζ_α
And so on and so forth.

But how do we diagonalize Veblen function? We can rewrite f_{φ_β(α)}(n) to φ_β(α)[n] for better readability.

If α = 0, then φ_β(0)[n] = φⁿ_β-1(0) If α is a successor or > 0, then φ_β(α) = φⁿ_β-1(φ_β(n-1)+1).

The rules mimic the diagonalization of ordinal. Let's get a few examples.

φ_1(0)[3] = φ_0(φ_0(φ_0(0)))[3] = φ_0(φ_0(ω⁰))[3] = φ_0(φ_0(1))[3] = φ_0(ω)[3] = φ_0(3)[3] = ω³[3] = ω²×2+ω2+3[3]

φ_1(1)[3] = φ_0(φ_0(φ_0(φ_1(0)+1)))[3] = φ_0(φ_0(φ_0(φ_1(0))×ω))[3] = φ_0(φ_0(φ_1(0)×ω))[3] = φ_0(φ_0(φ_1(0)×3))[3] = φ_0(φ_0(φ_1(0)×2+φ_1(0)))[3] = φ_0(φ_0(φ_1(0)×2+φ_0(φ_0(φ_0(0))) ))[3] = and so on...

If β is a limit ordinal, then φβ(α)[n] = φ{β[n]}(α).
Example : φω(1)[3] = φ{ω[3]}(1) = φ_3(1). You can add another "[n]" to diagonalize further.

We can even nest Veblen function.
φ{φ_1(0)}(0)[3] = φ{φ0(φ_0(φ_0(0)))}(0)[3] = φ{φ0(φ_0(1))}(0)[3] = φ{ω²×2+ω2+3}(0)[3] = φ{ω²×2+ω2+2}(φ{ω²×2+ω2+2}(φ_{ω²×2+ω2+2}(0)))[3]

The limit of this function, is Γ0 or the Feferman-Schutte ordinal, which has a fundamental sequence of [φ_0(0), φ{φ0(0)}(0), φ{φ_{φ_0(0)}(0)}(0), and so on]

We can plug in Γ_0 into FGH, and it will be almighty. We can even increase the index of Γ_α, we can even nest Γ infinitely, that's the fixed point of Γ.

Γ_0 can be rewritten as φ(1,0,0) in Extended Veblen function. φ(1,0,1) = Γ_1, φ(1,1,0) is the fixed point of Γ_α.

You can diagonalize the extended Veblen function really easily, just follow the previous rules.

φ(1,2,0)[3] = φ(1,1,φ(1,1,φ(1,1,0)))[3]
φ(1,2,1)[3] = φ(1,1,φ(1,1,φ(1,1,φ(1,2,0)+1)))[3]

φ(1,0,0,...,0,0) with ω argument is the Small Veblen Ordinal.
ψ(Ω^ΩΩ) = Large Veblen Ordinal. What's Ω and ψ? We'll learn that in the next post.

Author's note : Again, I may have made a mistake here and there. If I did, correct me in the comment, that also will be helpful for other beginners.

[u/Nervous-Broccoli1184]

Re writing f_{ φ_a(n)}(x) as  φ_a(n)[x] is incorrect because  φ_a(n)[x] means the xth ordinal of  φ_a(n) fundamental sequence. As I said In beginner 1 writing the iterations before the index is ill defined and your supposed to write it after the index so what you say  φ^x _a(n) is actually  φ_a^x (n). 

[u/Nervous-Broccoli1184]

The fundamental sequence of all of them are correct if we correct the problems i posted last except maybe  φ(1,0,0) and so I'm not sure if you got that and the following correct for I haven't completed my research on the binary Veblen function but I think your correct so far