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