Talk:Basis Representation Theorem for Ordinals

Length?
What is meant by "length" in the missinglinks? --Andrew Salmon 08:13, 18 August 2012 (UTC)


 * It means there is no link to a definition of the word "length" as it appears in the body of the text. --prime mover 12:56, 18 August 2012 (UTC)


 * Issue resolved. --Lord_Farin 13:31, 18 August 2012 (UTC)

what this is
I think I get it now. This is just the Basis Representation Theorem, am I right? In that case, is it appropriate to merge this with that, or that with this? Is it necessary that we prove this from the standpoint of pure ordinals, when in fact as it is shown that an ordinal is only a post word for a number and we've proved it for numbers? I'm looking in trepidation for every existing result in number theory being stated and proved separately in the context of ordinals. --prime mover 13:40, 18 August 2012 (UTC)


 * It is a result in that direction, yes. However, the stuff goes the other way around, in fact: The theorem for numbers is an instance of that for ordinals. --Lord_Farin 14:02, 18 August 2012 (UTC)


 * But the point is: it's proved for numbers, and then exactly the same result is proved for ordinals, with no indication that they are effectively the same thing. I believe that an "also see" won't be adequate to reflect this instance of the same work having been done twice. From what I understand, numbers are completely isomorphic to the finite ordinals so the proof for finite ordinals and numbers can be merged and transcluded as Proof 1, Proof 2 etc. (they are effectively different proofs as they are posted), then the extra stuff that is ordinal-only can be added to indicate what happens when the ordinals are not finite. --prime mover 14:20, 18 August 2012 (UTC)


 * I agree there is value in such an approach. But there are some intricacies to the compatibility of natural number addition and ordinal addition, multiplication, exponentiation to be dealt with before this is viable. I'd suggest placing an extra proof on the numbers side referring to Natural Numbers are Elements of Minimal Infinite Successor Set and then invoking the ordinal result. That's easier in the long run, I think. --Lord_Farin 14:27, 18 August 2012 (UTC)


 * Prime.Mover's suggestion sounds very good. This could even be renamed.  An isomorphism theorem should (eventually) be added showing that $( \omega, + , \times , \in )$ for ordinal addition and multiplication is isomorphic to $( \N , + , \times , < )$ for natural number addition and multiplication.  In effect, this is already done because we have already shown that the minimal infinite successor set is a Peano system under the successor set function.  Then, this theorem could simply transfer to a second proof for natural numbers. --Andrew Salmon 18:54, 18 August 2012 (UTC)


 * Exactly. --prime mover 19:39, 18 August 2012 (UTC)