```Date: Feb 17, 2013 4:24 PM
Author: Virgil
Subject: Re: Matheology � 222 Back to the roots

In article <544d47e7-f733-480a-85ab-913387d85e11@j2g2000yqj.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:> On 16 Feb., 16:26, William Hughes <wpihug...@gmail.com> wrote:> > > >      the nth FIS of l(n) is the nth FIS of d .> > >      But this does not make l(n) coFIS to d.> >> > > > > And there is not more than every n.> >> > > > > there is no line l such that d and l> > > > > are coFIS> > > > That would only be true if there was an n larger than every n> >> > > ??  The statement is yours.  Are you now withdrawing it.-> > The statement is just to the point.> > You said: there is no line l such that d and l are coFIS> I said: That (your statement) would only be true if there was an n> larger than every n (but there isn't). There is, however, a natural larger than any previously given natural. > 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.> > 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?--
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