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Topic: Matheology � 233
Replies: 36   Last Post: Apr 2, 2013 5:56 PM

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mueckenh@rz.fh-augsburg.de

Posts: 13,478
Registered: 1/29/05
Re: Matheology § 233
Posted: Mar 28, 2013 8:48 AM
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On 28 Mrz., 02:54, Virgil <vir...@ligriv.com> wrote:
> In article
> <32f47105-a873-44a0-bbcc-f744d1e71...@k1g2000yqf.googlegroups.com>,
>
>  WM <mueck...@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.


Matheology is the teaching of unnameable names and of actually
infinite decimal farctions that nobody can apply (opposite to
countably many names which define potentially infinite decimal
expansions). That religion is not what I adhere to.
>
>

> > > 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?
And the foundation of matheology is that such a node-complete tree
exists since the actually infinite set of all finite decimal
expansions exist, namely the set of all rationals that end in a
decimal period 000...

> > 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.


For a matheologian, that is inconsequent, because the result is a
contradiction of uncountably sets.
>
>
>

> > > > 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!


That means irrationals cannot be distinguished from terminating
rationals by nodes. They appear and disappear by the form of writing.
So you agree that irrational numbers as decimal fractions cannot be
distinguished from the terminating rationals by nodes. But this is the
principle of Cantor's argument, which, therefore, must fail.

> 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.


And why should the set of all rationals not contain that limit?
Does the sequence of all natural numbers, when written in the form 1,
2, 3, ... contain its limit, but when written as 7, 5, 11, ... not
contain it? Amazing.
>
> > 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 rationals
0.11000...
0.10000...
when written in the Binary Tree have the first node 1 in common, so
one of the first nodes disappears in the Binary Tree.
>
>
>

> > > > 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?


In an inclusion monotonic set, written in the Binary Tree, all
included elements belong to the element including them. Therefore the
set can only shrink. It is not possible in the Binary Tree to combine
elements destroying the tree structure.
>
> That inequality holds for ALL sets anywhere.


No. There are aleph_0 elements
0.1
0.11
0.111
...
which in the Binary tree are combined to only one path 0.111...

Since the set of all rationals contains aleph_0 elements, which only
lose nodes when combined to the tree, the Binary tree cannnot contain
more paths because no path is added.
>
> > Here only special subsets can be formed,
>
> But enough of them.


A statement of unjustified faith. See above. Always two rationals are
combined. And there cobination belongs to not more than countably many
paths.
>
>
>

> > > > 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.
>

If the set of all rationals exists, then that limit exists already in
that set. Combining paths with loss of nodes is not useful to increase
the number of paths.

Regards, WM




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