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

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Virgil

Posts: 7,006
Registered: 1/6/11
Re: Matheology � 190
Posted: Jan 13, 2013 4:59 PM
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In article
<fcbf94db-8b22-4a16-b837-ccd23e1766a7@k6g2000yqf.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> 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.


Since every binary path has a node at every "level" (distance from the
root), it can only be represented by an infinite set of naturals in this
way.

And there is another way of representing paths so that there is one and
only one path for each subset of |N. Each infinite path has a node at
each level n in |N, and is uniquely determined by the subset of |N of
those levels at which the path branches left rather than right.

Thus it is clear that there is a bijection between the set of paths of a
Complete Infinite Binary Tree and 2^|N, the set of all subsets of |N.
>
> Everything in this model is countable.


The power set of |N is not!



> All finite initial segments
> belong to a countable set.


True but irrelevant!


> 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.


Wm argues in effect that because every finite initial segment of an
infinite sequence is finite, the sequence itself must be finite.
--





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