Date: Nov 13, 2012 4:28 PM Author: Zaljohar@gmail.com Subject: Re: Cantor's first proof in DETAILS On Nov 13, 11:16 pm, Uirgil <uir...@uirgil.ur> wrote:

> In article <k7u09t$tj...@dont-email.me>,

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> "LudovicoVan" <ju...@diegidio.name> wrote:

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

> >news:86a85cce-2a84-4c9f-b860-527958274b50@o8g2000yqh.googlegroups.com...

> > <snip>

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> > > Theorem 4. for all i. x_i =/= L

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> > > <...>

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> > > Let J=L

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> > > QED

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> > 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".

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> The limit of a strictly increasing sequence (the a_i) is NOT EVER a

> member of the sequence.

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> The limit of a strictly decreasing sequence (The b_i) is NOT EVER a

> member of the sequence.

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> No values which are bounded below by a strictly increasing sequence and

> bounded above by a strictly decreasing sequence are members of either

> seequence.

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> Thus proving that, given any sequence of values in R, there must be

> values in R not appearing in that sequence.

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> > You should rather try and show the mistake in my objection instead of

> > proposing the same argument again and again.

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> Until you can falsify the original argument, which you have not done,

> there is no need to falsify your false argument against it.

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> As I had put it there:

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> > << an omega-th end-point, a_oo, would necessarily be drawn from an omega-th

> > entry of the sequence! Formally, we have the following property:

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> > A m : a_m e (x_n) & b_m e (x_n)

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> > That works not only for n and m in N, but also for n and m in N*. >>

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> That assumes that one is forced to work with N*, whereas sensible people

> work with N and have no problems.

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> > Objection to Cantor's First Proof

> > <https://groups.google.com/d/msg/sci.math/T2V4Jh7zzD8/wDM_wsyQZ0QJ>

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> > (Note that mine is against Cantor's First Proof of which yours remains a

> > paraphrase, not a faithful reproduction.)

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> 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