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Topic: Matheology § 190
Replies: 15   Last Post: Jan 14, 2013 3:35 PM

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mueckenh@rz.fh-augsburg.de

Posts: 16,036
Registered: 1/29/05
Re: Matheology § 190
Posted: Jan 13, 2013 10:16 AM
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On 13 Jan., 00:26, Virgil <vir...@ligriv.com> wrote:
> In article
> <4bffb7f3-9bfa-4dae-9108-da5e24389...@f4g2000yqh.googlegroups.com>,
>
>
>
>
>
>  WM <mueck...@rz.fh-augsburg.de> wrote:

> > On 12 Jan., 22:00, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <c0615860-6190-4c10-9185-78ed2f6a2...@x10g2000yqx.googlegroups.com>,

>
> > >  WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > Matheology 190
>
> > > > The Binary Tree can be constructed by aleph_0 finite paths.
>
> > > >         0
> > > >       1, 2
> > > >   3, 4, 5, 6
> > > > 7, ...

>
> > > Finite trees can be built having finitely many finite paths.
> > > A Complete Infinite Binary Tree cannot be built with only finite paths,
> > > as none of its paths can be finite.

>
> > Then the complete infinite set |N cannot be built with only finite
> > initial segments {1, 2, 3, ..., n} and not with ony finite numbers 1,
> > 2, 3, ...? Like Zuhair you are claiming infinite naturals!

>
> A finite initial segment of |N is not a path in the unary tree |N.
>
> And neither |N  as a unary tree nor any Complete Infinite Binary Tree
> has any finite paths.
>
>    "A Complete Infinite Binary Tree cannot be built with only
>    finite paths, as none of its paths can be finite."
>
> Means the same as
>
>    "A Complete Infinite Binary Tree cannot be built HAVING only
>    finite paths, as none of its paths can be finite."
>
> WM has this CRAZY notion that a path in a COMPLETE INFINITE BINARY TREE
> can refer to certain finite sets of nodes.
>
> And no one is claiming any infinite naturals, only infinitely many
> finite naturals.


So each n belongs to a finite initial segment (1,2,3,...,n).
Same is valid for the nodes of the Binary Tree: Each node belongs to a
finite initial segment of a path, the natural numbers (1,2,3,...,n)
denoting the levels which the nodes belong to.

Everything in this model is countable. All finite initial segments
belong to a countable set. Everything that possibly differs from an
entry of a Cantor-list belongs to a finite initial segments of the
anti-diagonal. There is nothing infinite that could explain or justify
uncountability.

Regards, WM



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