In article <email@example.com>, "LudovicoVan" <firstname.lastname@example.org> wrote:
> "Uirgil" <email@example.com> wrote in message > news:uirgil-B3AA26.15111513112012@BIGNEWS.USENETMONSTER.COM... > > In article <firstname.lastname@example.org>, > > "LudovicoVan" <email@example.com> wrote: > >> "Uirgil" <firstname.lastname@example.org> 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!