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Topic: Cantor's first proof in DETAILS
Replies: 34   Last Post: Dec 1, 2012 10:56 AM

 Messages: [ Previous | Next ]
 Zaljohar@gmail.com Posts: 2,665 Registered: 6/29/07
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...@dont-email.me>,
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>  "LudovicoVan" <ju...@diegidio.name> wrote:

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

>
> >     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

>
> > (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

Date Subject Author
11/12/12 Zaljohar@gmail.com
11/12/12 Zaljohar@gmail.com
11/12/12 Charlie-Boo
11/12/12 Zaljohar@gmail.com
11/13/12 Charlie-Boo
11/15/12 Zaljohar@gmail.com
12/1/12 Frederick Williams
11/12/12 LudovicoVan
11/12/12 LudovicoVan
11/12/12 Uirgil
11/12/12 Shmuel (Seymour J.) Metz
11/12/12 Uirgil
11/13/12 Zaljohar@gmail.com
11/13/12 LudovicoVan
11/13/12 Zaljohar@gmail.com
11/13/12 Zaljohar@gmail.com
11/13/12 Uirgil
11/13/12 LudovicoVan
11/13/12 Uirgil
11/13/12 LudovicoVan
11/13/12 Uirgil
11/13/12 Shmuel (Seymour J.) Metz
11/13/12 Zaljohar@gmail.com
11/13/12 LudovicoVan
11/13/12 Uirgil
11/14/12 Zaljohar@gmail.com
11/14/12 Uirgil
11/14/12 Zaljohar@gmail.com
11/14/12 Zaljohar@gmail.com
11/14/12 Uirgil
11/16/12 LudovicoVan
11/16/12 Uirgil
11/16/12 Zaljohar@gmail.com
11/16/12 LudovicoVan
11/13/12 Shmuel (Seymour J.) Metz