Virgil
Posts:
7,016
Registered:
1/6/11


Re: Matheology � 233
Posted:
Mar 27, 2013 9:54 PM


In article <32f47105a87344a0bbccf744d1e71da0@k1g2000yqf.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.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 nonexistence 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 nary tree contains all (eventually constant) paths paths corresponding to nary rationals between 0 and 1, it must contain the path for any limit of a sequence of such rational nary (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 nonfinite length. At least when not incarcerated in Wolkenmuekenheim. 

