```Date: Feb 18, 2013 4:45 PM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 222 Back to the roots

On 17 Feb., 22:24, Virgil <vir...@ligriv.com> wrote:> In article> <544d47e7-f733-480a-85ab-913387d85...@j2g2000yqj.googlegroups.com>,>>>>>>  WM <mueck...@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.Nevertheless it is a natural number and therefore finite.>> > 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-thline. Why do you always forget that simple fact?>>>> > 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.Regards, WM
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