Date: Nov 13, 2012 3:16 PM
Author: Uirgil
Subject: Re: Cantor's first proof in DETAILS

In article <k7u09t$tjd$>,
"LudovicoVan" <> wrote:

> "Zuhair" <> 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

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 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.