
Re: Cantor's first proof in DETAILS
Posted:
Nov 13, 2012 4:28 PM


On Nov 13, 11:16 pm, Uirgil <uir...@uirgil.ur> wrote: > In article <k7u09t$tj...@dontemail.me>, > > > > > > > > > > "LudovicoVan" <ju...@diegidio.name> wrote: > > "Zuhair" <zaljo...@gmail.com> wrote in message > >news:86a85cce2a844c9fb860527958274b50@o8g2000yqh.googlegroups.com... > > <snip> > > > > Theorem 4. for all i. x_i =/= L > > > > <...> > > > > Let J=L > > > > QED > > > Same argument, same objection: as easily proven, the limit interval here > > must be degenerate, that is it is the singleton (in interval notation) > > [L;L]. So what you claim amounts to saying that the limit value L is not in > > (x_n), but that is just incorrect in that if you consider the limit value, > > then of course it does belong to (x_n), in the limit! More formally L = > > lim_{n>oo} (a_n) = lim_{n>oo} (b_n) , then just consider an injection from > > N* instead of N and you can even talk meaningfully about that "last value". > > The limit of a strictly increasing sequence (the a_i) is NOT EVER a > member of the sequence. > > The limit of a strictly decreasing sequence (The b_i) is NOT EVER a > member of the sequence. > > No values which are bounded below by a strictly increasing sequence and > bounded above by a strictly decreasing sequence are members of either > seequence. > > Thus proving that, given any sequence of values in R, there must be > values in R not appearing in that sequence. > > > > > You should rather try and show the mistake in my objection instead of > > proposing the same argument again and again. > > Until you can falsify the original argument, which you have not done, > there is no need to falsify your false argument against it. > > As I had put it there: > > > > > << an omegath endpoint, a_oo, would necessarily be drawn from an omegath > > entry of the sequence! Formally, we have the following property: > > > A m : a_m e (x_n) & b_m e (x_n) > > > That works not only for n and m in N, but also for n and m in N*. >> > > That assumes that one is forced to work with N*, whereas sensible people > work with N and have no problems. > > > > > Objection to Cantor's First Proof > > <https://groups.google.com/d/msg/sci.math/T2V4Jh7zzD8/wDM_wsyQZ0QJ> > > > (Note that mine is against Cantor's First Proof of which yours remains a > > paraphrase, not a faithful reproduction.) > > Your alleged argument against the Cantor proof does not work against > either Cantor's proof, nor Zuhair's proof, nor my proof for that matter, > since your N* is irrelevant for all of them.
I showed in the Corollary that even if he use N* as the domain of (x_n), still we can prove there is a missing real from the range of (x_n). So Cantor's argument or my rephrasing of it both can easily be shown to be applicable to N* (any set having a bijection with N) as well as N.
Zuhair

