
Re: Matheology § 233
Posted:
Apr 3, 2013 6:01 AM


On 3 Apr., 00:03, Virgil <vir...@ligriv.com> wrote: > In article > <1b631b4421cc4e3a80281bcfb1cc5...@m1g2000vbe.googlegroups.com>, > > > > > > WM <mueck...@rz.fhaugsburg.de> wrote: > > On 2 Apr., 01:54, Virgil <vir...@ligriv.com> wrote: > > > > I am not at all sure that it is even possible to build any binary tree > > > this way but it is clearly impossible to built a COMPLETE INFINITE > > > BINARY TREE, this way. > > > Given the foundations of matheology, it is possible to construct every > > node / every finite path of a Binary Tree that is complete with > > respect to its nodes. > > > > For one thing, in a CIBT, every path is by definition maximal in the > > > sense that no additional node can be added to a path without making the > > > result not a path, and is also minimal in the sense that no node can be > > > removed from it without making the result not a path. > > > > In WM's "trees", every FISON (Finite Initial Sequncee Of Nodes) appears > > > to be a path, which is quite differnt notion of path. > > > Call it as you like. I call it finite path as an abbreviation of FIS > > of an infinite path. > > > > > It is impossible to write them out. Yes. But they are constructed like > > > > the finite paths. It is impossible to prohibit infinite paths (of > > > > rationals and of irrationals) to be constructed when the complete set > > > > of nodes of the Binary Tree is constructed by means of all finite > > > > paths. > > Therefore it is impossible to distinguish infinite paths by > > > > > nodes other than be naming infinite sets of nodes. Alas there are only > > > > countably many names available. > > > > Thus not all CIBTpaths are nameable, just like not all real numbers are > > > nameable. > > > They are not even distinguishable by nodes. They are purest belief. > > What in mathematics is not a matter of belief?
Given that an enumeration of all rational numbers is believed to form a Cantorlist, then the following is not a matter of belief. Beyond the nth line there are f(n) rational numbers the first n digits of which are the same as the first n digits d_1, d_2, ..., d_n of the antidiagonal. For every n in N, f(n) > k for every k in N. Define for every n in N the function g(n) = 1/f(n) = 0. In analysis the limit of this function is lim[n>oo] g(n) = 0.
Regards, WM

