```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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