In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> 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).
The union of every FIS of a line is not a FIS unless the line of which those FIS's are FIS's is a FIS of itself. > > > > 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.
Actually, one can leave out as many of them as one keeps.
> Again you drift astray > with actual infinity.
Actual infinity woks better than WMytheology almost everywhere. > > 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.
Given any FIS one can always look at the next FIS, without end. At least until WM can name a FIS for which there is no successor FIS. --