Date: Nov 13, 2012 8:12 PM
Author: Uirgil
Subject: Re: Cantor's first proof in DETAILS

In article <k7uh30\$d4p\$1@dont-email.me>,
"LudovicoVan" <julio@diegidio.name> wrote:

> "Uirgil" <uirgil@uirgil.ur> wrote in message
> news:uirgil-B3AA26.15111513112012@BIGNEWS.USENETMONSTER.COM...

> > In article <k7udtq\$np6\$1@dont-email.me>,
> > "LudovicoVan" <julio@diegidio.name> wrote:

> >> "Uirgil" <uirgil@uirgil.ur> wrote in message
> >> news:uirgil-91F13B.13165013112012@BIGNEWS.USENETMONSTER.COM...
> >>

> >> > 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.

> >>
> >> I'll have a look at Zuhair's follow-up as soon as I manage, but let me
> >> for
> >> now just point out that the above argument is obviously bogus: the
> >> rationals
> >> too are dense (have the IVP as Zuhair has called it) and, by the very
> >> same
> >> argument, we have proved that the rationals too are not countable... see?

> >
> > The difference being that a monotone but finitely bounded sequence of
> > rationals need not have a limit among the rationals but MUST have a
> > limit among the reals, a LUB or GLB.

>
> Yes, it's the *completeness* property that is required. Anyway, as
> anticipated, I'll have to come back to this when I have time: the devil is
> in the details!
>
> -LV
>

The completeness property will still be there when you come back!