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

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 Virgil Posts: 8,833 Registered: 1/6/11
Re: Matheology � 190
Posted: Jan 12, 2013 4:11 PM

In article
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 12 Jan., 09:45, Zuhair <zaljo...@gmail.com> wrote:
> > On Jan 12, 11:07 am, 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, ...

> >
> > > But wait! The Binary Tree has aleph_0 levels. At each level the number
> > > of nodes doubles. We start with the (empty) finite path at level 0 and
> > > get 2^(n+1) - 1 finite paths within the first n levels. The number of
> > > all levels of the Binary Tree is called aleph_0. That results in
> > > 2^(aleph_0 + 1) - 1 = 2^aleph_0 finite paths.

> >
> > NO. We get (2^(n+1)) -1 PATHS within the first n levels, This of
> > course INCLUDES paths at level n itself.
> >
> > At level 0 the number of PATHS is 2^(1) -1 which is 1 that is the
> > empty path
> > At level 1 the number of PATHS is 2^2 -1 which is 3 this INCLUDES
> > paths at level 0 and those at level 1 ITSELF! which are the empty path
> > (at level 0), and 0-1, 0-1 paths at level 1.
> >
> > Now the number of ALL levels of the Binary Tree is called aleph_0,
> > correct.
> >
> > But the result is The number of ALL *PATHS* up to Aleph_0

>
> aleph_0 is not a natural number. Therefore I carefully excluded any
> reference to last levels. And this is not required in mathematics.

The set of all levels is |N.

The number of paths will be equal to the number of SUBSETS of the set of
levels as each path is uniquely determined by the set of such levels at
which it branches left.

>
> > (and
> > INCLUDING PATHS at level Aleph_0 OF COURSE: The later ones are the
> > INFINITE paths)

>
> You are in error. There are, according to set theory, aleph_0 finite
> natural numbers. aleph_0 is not a natural number.

Zuhair should probably have said "paths having Aleph_0 levels", or
equivalently, "paths with no last level".
>
> Instead of paths you could also say nodes, because there is a
> bijection between finite paths and nodes. Do you defend the opinion
> that there must be nodes at an infinite level in order to collect
> infinitely many nodes? (This would be a sober position, but I doubt
> that it is your position.)

His position should be that paths have no last level, since that is the
truth for Complete Infinite Binary Trees .
>
> > is (2^(Aleph_0 +1)) -1 which is 2^Aleph_0, this
> > includes both FINITE as WELL as INFINITE paths of the complete
> > infinite binary tree and NOT just finite paths as you are deceived by.

>
> And you think that in the proof of Nicole d'Oresme also infinite
> natural numbers play a role? You are in error.

WM seems to be the only one who thinks that.
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