In article <email@example.com>, WM <firstname.lastname@example.org> 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.
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. --