Talk:Ordinal collapsing function

wall street

I think this is a joke: http://science.slashdot.org/story/10/05/27/0258245/Sudden-Demand-For-Logicians-On-Wall-Street

"In an unexpected development for the depressed market for mathematical logicians, Wall Street has begun quietly and aggressively recruiting proof theorists and recursion theorists for their expertise in applying ordinal notations and ordinal collapsing functions to high-frequency algorithmic trading...."

69.228.170.24 (talk) 06:04, 27 May 2010 (UTC)

Yes, all my sources assure me it's a joke, and nobody has stepped up to provide evidence that infinite ordinals are used in high-frequency trading. John Baez (talk) 21:04, 24 November 2012 (UTC)

Greetings. I am the author. FLeℵgyel (ta|k) —Preceding undated comment added 23:29, 17 November 2013 (UTC)

question on the section: predicative part

It is written

"Now $\psi (\zeta _{0})=\zeta _{0}$ but $\psi (\zeta _{0}+1)$ is no larger, since $\zeta _{0}$ cannot be constructed using finite applications of $\phi _{1}\colon \alpha \mapsto \varepsilon _{\alpha }$ and thus never belongs to a $C(\alpha )$ set for $\alpha \leq \Omega$, and the function $\psi$ remains “stuck” at $\zeta _{0}$"

but should not $\zeta _{0}$ be an element of $C(\zeta _{0}+1)$ since we have $\psi (\zeta _{0})=\zeta _{0}$ and $\zeta _{0}<\zeta _{0}+1$? If this is correct $\psi (\zeta _{0}+1)$ should be larger than $\zeta _{0}$ — Preceding unsigned comment added by 88.131.62.36 (talk) 11:14, 15 June 2013 (UTC)

If you check the definition of C(ζ₀+1), you will see that you would have to show that ζ₀ belongs to it (for some other reason) in addition to ζ₀ < ζ₀+1 before you can conclude that ζ₀ belongs to it on account of being ψ(ζ₀). JRSpriggs (talk) 11:01, 16 June 2013 (UTC)

I was also initially confused about the values of ψ, but I understand it now. "ψ(α) is the smallest ordinal which cannot be expressed from 0, 1, ω and Ω using sums, products, exponentials, and the ψ function itself (to previously constructed ordinals less than α)." The key part is "previously constructed"; I need to be able to create the ordinal number in a finite number of steps from {0, 1, ω, Ω} before I can apply the ψ function to it. Since ζ₀ cannot be constructed in a finite number of steps from {0, 1, ω}, the only way it can be generated is as ψ(Ω); and by definition, ψ(Ω) is not a member of any constructed sets before C(Ω+1). - Mike Rosoft (talk) 05:43, 22 August 2014 (UTC)

Beyond $\Gamma _{0}$

There is a statement that: $\psi (\Omega ^{\Omega }+\Omega ^{\alpha })=\phi _{\Gamma _{0}+\alpha }(0)$

...but I guess this would need a reference where someone has worked out the correspondence between the collapsing function and the Veblen functions to this level, because I'm trying to work out the same thing and getting a different answer.

Before you get to $\phi _{\Gamma _{0}+\alpha }(0)$, or even $\phi _{\Gamma _{0}+1}(0)$, you have to take the limit of $0,\ \phi _{\Gamma _{0}}(0)=\Gamma _{0},\ \phi _{\Gamma _{0}}(\Gamma _{0}),\ \phi _{\Gamma _{0}}(\phi _{\Gamma _{0}}(\Gamma _{0})),\ ...$, which should equal $\phi _{\Gamma _{0}+1}(0)$.

Before even that you have to get to $\phi _{\Gamma _{0}}(1)$, which is presumably the supremum of $\phi _{\alpha }(\Gamma _{0}+1)$ for $\alpha <\Gamma _{0}$.

To my reasoning, it should be something like

$\psi (\Omega ^{\Omega }+\Omega ^{\alpha })=\phi _{1+\alpha }(\Gamma _{0}+1)$

and specifically,

$\psi (\Omega ^{\Omega }+1)=\epsilon _{\Gamma _{0}+1}$

and

$\psi (\Omega ^{\Omega }+\Omega )=\phi _{2}(\Gamma _{0}+1)=\eta _{\Gamma _{0}+1}.$

--Stephen J. Brooks (talk) 16:10, 30 July 2021 (UTC)