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Topic: Matheology § 233
Replies: 20   Last Post: Apr 4, 2013 10:16 AM

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mueckenh@rz.fh-augsburg.de

Posts: 16,032
Registered: 1/29/05
Re: Matheology § 233
Posted: Apr 3, 2013 6:01 AM
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On 3 Apr., 00:03, Virgil <vir...@ligriv.com> wrote:
> In article
> <1b631b44-21cc-4e3a-8028-1bcfb1cc5...@m1g2000vbe.googlegroups.com>,
>
>
>
>
>
>  WM <mueck...@rz.fh-augsburg.de> wrote:

> > On 2 Apr., 01:54, Virgil <vir...@ligriv.com> wrote:
>
> > > I am not at all sure that it is even possible to build  any binary tree
> > > this way but it is clearly impossible to built a COMPLETE INFINITE
> > > BINARY TREE, this way.

>
> > Given the foundations of matheology, it is possible to construct every
> > node / every finite path of a Binary Tree that is complete with
> > respect to its nodes.

>
> > > For one thing, in a CIBT, every path is by definition maximal in the
> > > sense that no additional node can be added to a path without making the
> > > result not a path, and is also minimal in the sense that no node can be
> > > removed from it without making the result not a path.

>
> > > In WM's "trees", every FISON (Finite Initial Sequncee Of Nodes) appears
> > > to be a path, which is quite differnt notion of path.

>
> > Call it as you like. I call it finite path as an abbreviation of FIS
> > of an infinite path.

>
> > > > It is impossible to write them out. Yes. But they are constructed like
> > > > the finite paths. It is impossible to prohibit infinite paths (of
> > > > rationals and of irrationals) to be constructed when the complete set
> > > > of nodes of the Binary Tree is constructed by means of all finite
> > > > paths.

>
>  Therefore it is impossible to distinguish infinite paths by
>

> > > > nodes other than be naming infinite sets of nodes. Alas there are only
> > > > countably many names available.

>
> > > Thus not all CIBT-paths are nameable, just like not all real numbers are
> > > nameable.

>
> > They are not even distinguishable by nodes. They are purest belief.
>
> What in mathematics is not a matter of belief?


Given that an enumeration of all rational numbers is believed to form
a Cantor-list, then the following is not a matter of belief. Beyond
the n-th line there are f(n) rational numbers the first n digits of
which are the same as the first n digits d_1, d_2, ..., d_n of the
anti-diagonal. For every n in |N, f(n) > k for every k in |N. Define
for every n in |N the function g(n) = 1/f(n) = 0. In analysis the
limit of this function is lim[n-->oo] g(n) = 0.

Regards, WM



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