Virgil
Posts:
4,482
Registered:
1/6/11
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Re: Matheology � 203
Posted:
Feb 4, 2013 4:36 PM
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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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