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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