Date: Mar 27, 2013 9:54 PM Author: Virgil Subject: Re: Matheology � 233 In article

<32f47105-a873-44a0-bbcc-f744d1e71da0@k1g2000yqf.googlegroups.com>,

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

> On 27 Mrz., 18:26, Virgil <vir...@ligriv.com> wrote:

>

> > IF a decimal tree containing a different path for every possible finite

> > decimal from 0 to 1 exists, it also must contain a path for every real

> > from 0 to 1.

>

> And what says matheology about the existence of the set of every

> possible finite decimal path of the unit interval without any tree

> structure?

Since WM is the only one speaking for matheology, he must answer his own

question.

> >

> > So that if WM denies existence of paths for those reals, he

> > automatically also denies the existence of any such trees.

>

> I ask: What can be concluded, IF such a tree exists?

> >

> > Standard mathematics has no problem with that.

> >

> >

> >

> > > A view without faith is this: There is no irrational path at all.

> >

> > Then the view, either with or without faith, but using standard logic,

> > must be that there is no such tree at all.

>

> That is correct. But from the assumption of the actually infinite tree

> without irrationals, the proof of its non-existence follows easily.

WM makes my case for me.

> >

> > > But

> > > that would destroy the pet dogma of matheology, namely uncountability.

> >

> > Standard mathematics says that if certain trees exist, then they

> > necessarily have uncountably many paths.

> >

> > WM is the one claiming such trees exist.

> >

> I assume that the countable set of all rationals exists. If written in

> form of a tree, all uncountably many irrationals are created.

RIGHT!

If an n-ary tree contains all (eventually constant) paths paths

corresponding to n-ary rationals between 0 and 1, it must contain the

path for any limit of a sequence of such rational n-ary (eventually

constant) paths, thus also for all reals between 0 and 1.

> You say

> that certain subsets are considered. But that is definitely wrong,

> because only the existing paths are written such that everyone remains

> - only some nodes are united. This does not create new sets but only

> deletes some nodes.

Which nodes get deleted when one has each binary rational other than 0

and 1 in the infinite binary tree being represented by two infinite

paths, one ending with infinitely many 0's the other ending with

infinitely many 1's.

> >

> >

> > > The question is: How come uncountably many irrational paths into being

> > > during the countable process of constructing the complete decimal tree

> > > by constructing all its countably many nodes.

> >

> > Because all sets have more subsets than they have members.

> >

> Not in case of the tree.

How does the existence of a tree structure prevent any set from having

more subsets than members?

That inequality holds for ALL sets anywhere.

> Here only special subsets can be formed,

But enough of them.

> >

> >

> > > And an additional question for skilled matheologians: If we delete all

> > > paths containing digits 2, 3, 4, 5, 6, 7, 8, and 9 from the decimal

> > > tree of finite paths: Do all irrationals remain?

> >

> > Since that will delete everything from 0.12 to 0.21, inclusive, some

> > rationals and some irrationals vanish,

> > but the irrational 0.10110111011110... won't.

> >

> First it must become existing.

If the tree exists, then that path exists within it.

>

> > There are binary trees with only finite paths, but only when having some

> > finite maximum path length, but there are no binary trees containing

> > paths of all finite path lengths that do not contain at least one

> > infinite path as well.

>

> So the structure of the tree is something very special. Uncountability

> seems to be a matter of how numbers are written.

In a binary tree, either the entire set of path lengths is bounded by

some maximum finite number of nodes or the tree has at least one path of

non-finite length.

At least when not incarcerated in Wolkenmuekenheim.

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