In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 4 Feb., 11:19, William Hughes <wpihug...@gmail.com> wrote: > > > Of course, every FIS is in a line. > > > > True but irrelevant. We can use induction to > > show that there is no natural number n, such > > that the nth line of L contains every FIS > > of 0.111.... > > We can use induction to show that there is no natural number which > would allow us to draw any final conclusion, i.e., there are > infinitely many lines remaining beyond line number n. > > > > The question is now > > > > Can a potentially infinite list > > of potentially infinite 0/1 > > sequences have the property that > > if s is a potentially infinite 0/1 > > sequence, then there is a line, g, of L > > with the property that every > > initial segment of s is contained in g > > ? > > > > Yes or No please > > No. There cannot be a line g where the FIS are complete, because the > FISs cannot be complete at all. But I already showed you the list > which satisfies the possible claim: Every FIS is in a line. > > 0.1 > 0.11 > 0.111 > ... > > (Because there is no FIS of the diagonal that is missing in every > line.)
But a "diagonal" whose nth digit is taken from the nth FIS, will have for every n in |N a successor position, n+1. > > By the way, the FISs are isomorphic to the natural numbers. There > cannot be a line g where the natural numbers are complete.
But such a "diagonal", not being a line in your list, is not constrained by any such limitation. --