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? > > -LV >
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.
Density is not enough distinguish between Q and R, but the GLB/ LUB property is enough.
Any densely ordered interval of positive length having the GLB/LUB property is uncountable.