Date: Feb 18, 2013 8:36 PM
Subject: Re: Matheology � 222 Back to the roots
WM <firstname.lastname@example.org> wrote:
> On 17 Feb., 22:24, Virgil <vir...@ligriv.com> wrote:
> > In article
> > There is, however, a natural larger than any previously given natural.
> Nevertheless it is a natural number and therefore finite.
but for every one of them there is successor which is also one of them.
> > > There is only every d_n and for every d_n there is a line containing
> > > it. Otherwise it could not be a d_n.
> > And for EVERY line, a d_n NOT contained in it!
> > In fact, more d_n's not contained in it than are contained in it.
> Similarly, there are more lines than are contained as FISs in the n-th
> line. Why do you always forget that simple fact?
Perhaps if it were at all relevant here one might not ignore it,
but as it isn't, one doesn't.
Note that the only time a line and any FIS of that line are the same is
when the line is finite of some length n and the FIS is FISn.
But for lines that are not finite there are no FISs equal to those lines.
> > > You are, again, arguing with finished infinity, d having more than
> > > every d_n.
> > If you object to that then you must be arguing that d does not have
> > more than at least one of its d_n.
> > Which d_n would that be, WM?
> Make a proposal.
> Say n. It ius not fixed. But it is finite.
But even proposing any n proves that n is not what was asked for, since
the proposing of any 'n' implies the existence of an 'n+1'.
At least with the standard Naturals.
Does WM claim to know of a natural that does NOT have a successor
Unless WM, or someone else, can name a d_n that d cannot not exceed,
there does not exists any such d_n, and thus no such n, meaning that the
sequence of n's is endless, but nevertheless is contained in the very
name "the complete sequence of natural numbers", as there are no n's
other than in that sequence.