On 20 Feb., 11:30, William Hughes <wpihug...@gmail.com> wrote: > On Feb 20, 11:15 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 18 Feb., 14:11, William Hughes <wpihug...@gmail.com> wrote: > > > > > Please answer as politely: Is there any n with no line containing > > > > d_1, ..., d_n > > > > No. > > > Please answer without using phrases like "all" or "totality" or > > "complete collection" or "whole class" or "finished infinite" or "the > > set": Can you name a union of FISs of the diagonal that is missing in > > every line of the list? > > No (The union of *every* FIS does not mean > anything)
The union of every FIS means that for every n FIS(1) to FIS(n) are united. Of course this union does not differ from FIS(n). > > 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. Again you drift astray with actual infinity.
Correct is: For every FIS of d there is an identical line. Look, you cannot, within this framework, that you agreed to, prove anything for "all" FIS of d. You can only look for FIS 1 to FIS n. Nothing further is possible. > > Clearly any FIS that "actually exists" > is a line of the list.
Therefore we have identity with a line although we cannot name the last line or last FIS of d. That's infinity. > > 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?
