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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