fom
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Registered:
12/4/12


Re: Matheology § 233
Posted:
Apr 3, 2013 11:37 AM


On 4/3/2013 5:01 AM, WM wrote: > 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.
So, when did 'oo' become a natural number to which a function on the natural numbers could be mapped?

