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.
