```Date: Feb 4, 2013 4:36 PM
Author: Virgil
Subject: Re: Matheology � 203

In article <2a1f9bd0-4845-4503-ad23-c892c0fc95c6@y4g2000yqa.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> 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.--
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