On 20 Feb., 17:44, William Hughes <wpihug...@gmail.com> wrote: > On Feb 20, 5:29 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 20 Feb., 13:31, William Hughes <wpihug...@gmail.com> wrote: > > > > On Feb 20, 12:37 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 20 Feb., 11:30, William Hughes <wpihug...@gmail.com> wrote: > > > <snip> > > > > > A statement you can make is that there > > > > > is no line of the list with the property > > > > > that it is coFIS to d. > > > > > (you do not need every line or every FIS to "actually > > > > > exist" to make this statement) > > > > > You need all FISs of d to make this statement. > > > > No, for each line of L you only need some > > > of the FISs. For every line you need every > > > not all. > > > Then you have a statement for finitely many lines and none for > > infinitely many lines. > > > > <snip> > > > > > > Let z be a potentially infinite sequence such that > > > > > for some natural number m, the mth FIS of > > > > > z contains a zero. > > > > > > Are z and x coFIS? > > > > > No. > > > > Is the following statement true > > > > For every natural number n we have > > > > the (n+1)st FIS of the nth line > > > of L contains a 0. > > > Of course. > > Is the statement > > For every natural number n we have > the nth line of L and x > are not coFIS > > true?-
True but irrelevant.
Relevant is only this: For every natural number n we have a line that is identical with d_1, ..., d_n.
You try and try to have a one-sided look onto the infinite.
For every natural number n we have a FIS 1,2,3,...,n, ..., m that is longer and contains n. And for every FIS 1, 2,3 ,..., m we have a natural number, for instance 2m, that is not contained in FIS 1, 2,3 ,..., m.
