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-th
line. 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