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Topic: Matheology § 233
Replies: 20   Last Post: Apr 4, 2013 10:16 AM

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fom

Posts: 1,968
Registered: 12/4/12
Re: Matheology § 233
Posted: Apr 3, 2013 11:37 AM
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On 4/3/2013 5:01 AM, WM wrote:
> On 3 Apr., 00:03, Virgil <vir...@ligriv.com> wrote:
>> In article
>> <1b631b44-21cc-4e3a-8028-1bcfb1cc5...@m1g2000vbe.googlegroups.com>,
>>
>>
>>
>>
>>
>> WM <mueck...@rz.fh-augsburg.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 CIBT-paths 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 Cantor-list, then the following is not a matter of belief. Beyond
> the n-th 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
> anti-diagonal. 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?








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