Talk:Ordinal collapsing function: Difference between revisions

Content deleted Content added
Assessment: Mathematics: class=C (assisted)
adjust spacing of lines (placement of blank lines) to improve readability
Line 11:
 
Greetings. I am the author. [[User:FLengyel|FLe&#8501;gyel]] ([[User talk:FLengyel|ta&#124;k]]) <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|undated]] comment added 23:29, 17 November 2013 (UTC)</span><!--Template:Undated--> <!--Autosigned by SineBot-->
 
 
 
 
 
== question on the section: predicative part ==
Line 28 ⟶ 24:
 
:If you check the definition of ''C''(&zeta;<sub>0</sub>+1), you will see that you would have to show that &zeta;<sub>0</sub> belongs to it (for some other reason) in addition to &zeta;<sub>0</sub> < &zeta;<sub>0</sub>+1 before you can conclude that &zeta;<sub>0</sub> belongs to it on account of being &psi;(&zeta;<sub>0</sub>). [[User:JRSpriggs|JRSpriggs]] ([[User talk: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 ζ<sub>0</sub> 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). - [[User:Mike Rosoft|Mike Rosoft]] ([[User talk:Mike Rosoft|talk]]) 05:43, 22 August 2014 (UTC)