
Re: Cantor's first proof in DETAILS
Posted:
Nov 13, 2012 1:58 AM


On Nov 12, 9:18 pm, "LudovicoVan" <ju...@diegidio.name> wrote: > "Zuhair" <zaljo...@gmail.com> wrote in message > > news:86a85cce2a844c9fb860527958274b50@o8g2000yqh.googlegroups.com... > > > Let a_0 = x_0 > > Let b_0 be the first entry in (x_n) such that b_0 > a_0. > > Let a_i+1 be the first entry in (x_n) such that a_i < a_i+1 < b_i. > > Let b_i+1 be the first entry in (x_n) such that a_i+1 < b_i+1 < b_i. > > In Cantor's proof a_{i+1} and b_{i+1} are the two first entries encountered > (in any order) in (x_n) *after* the entries corresponding to a_i and b_i. > This does not seem to be the case with your proof, where it instead seems > that entries are just picked every time restarting from the beginning of > (x_n). > > Could you clarify? I'd like to be sure before I proceed reading it... > > LV
Yes in this proof the entries are as you said will be picked every time restarting from the beginning of (x_n) provided you follow the rules stipulated in definition of the a's and b's. BUT you'll see as you go down the proof that this will also lead to every a_i+1 and b_i +1 coming after a_i , b_i in (x_n). (See Result 6 of this proof).
As regards the "order" of a_i and b_i, then if you meant by order the quantitative comparison of their "values" then clearly in this proof we MUST have a_i < b_i by definition. But if you meant by "order" the places of them in (x_n), then the mere definition of the a's and b's do not mention itself any place order restriction, but as I said still by following those rules this will eventually lead to each b_i coming after a_i in (x_n) [Result 4].
Also I agree with your notation that a_i+1 is better written as a_{i +1}. But I guess it is understood like that anyway.
Zuhair

