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Topic: Matheology � 233
Replies: 37   Last Post: May 12, 2014 10:24 AM

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Posts: 8,833
Registered: 1/6/11
Re: Matheology � 233
Posted: Mar 27, 2013 9:54 PM
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In article
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
> >
> > 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.


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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