In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 6 Mrz., 21:54, Virgil <vir...@ligriv.com> wrote: > > > > The list is nothing but the union of all lines. The union of all lines > > > is the same as the sequence of all lines - without its limit. > > > > Actually, it is the the union of all lines that will contain any limit > > and the sequence of all lines which will not contain any limit. > > > > That sequence will HAVE a limit, but will not CONTAIN it, as every > > member of it is a FIS, and no FIS contains all other FISs. > > You see that this sequence of lines does not contain its limit |N. > You assert, however, that the union of all lines contains the limit | > N. > > You cannot explain that difference, because in this special example > union and sequence are the same infinite entity. > In order for the sequence of lines to contain |N, there would have to be a line equal to |N, but there is no such line.
In order for the union of the lines, considered merely as sets, to contain |N, all it needs is to have every member of |N in at least one line, which is trivially the case.
Thus again, common sense is right and WM is wrong!.
And where is WM's long overdue proof that some mapping from the set of all binary sequences to the set of all paths of a CIBT is a linear mapping?
Several time WM claimed this, and even presented a claimed but phony argument a couple of times, but never a real proof.
He should either present a real proof or withdraw his claim. --