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)

and others and beyond

and others and beyond

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)