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.