Date: Nov 13, 2012 1:58 AM
Author: Zaljohar@gmail.com
Subject: Re: Cantor's first proof in DETAILS
On Nov 12, 9:18 pm, "LudovicoVan" <ju...@diegidio.name> wrote:

> "Zuhair" <zaljo...@gmail.com> wrote in message

>

> news:86a85cce-2a84-4c9f-b860-527958274b50@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