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Topic: Matheology � 233
Replies: 37   Last Post: May 12, 2014 10:24 AM

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fom

Posts: 1,968
Registered: 12/4/12
Re: Matheology § 233
Posted: Mar 31, 2013 9:34 AM
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On 3/31/2013 3:31 AM, WM wrote:
> On 31 Mrz., 02:56, Virgil <vir...@ligriv.com> wrote:
>> In article
>> <2bc13fff-5cbb-43dd-a06a-218c68c99...@m9g2000vbc.googlegroups.com>,
>>
>>
>>
>>
>>
>> WM <mueck...@rz.fh-augsburg.de> wrote:

>>> On 30 Mrz., 22:11, Virgil <vir...@ligriv.com> wrote:
>>
>>>>>> Thus there is always at least one bit of any listed entry disagreeing
>>>>>> with the antidiagonanl, just as the Cantor proof requires.

>>
>>>>> In a list containing every rational: Is there always, i.e., up to
>>>>> every digit, an infinite set of paths identical with the anti-
>>>>> diagonal? Yes or no?

>>
>>>> The set of paths in any Complete Infinite Binary Tree which agree with
>>>> any particular path up to its nth node is equinumerous with the set of
>>>> all paths in the entire tree i.e., is uncountably infinite.

>>
>>> This was the question: In a list containing every rational: Is there
>>> always, i.e., up to every digit, an infinite set of paths identical
>>> with the anti-diagonal? Yes or no?

>>
>> Lists and trees are different. And anti-diagonals derive from lists, not
>> trees.
>> The entries in list are well ordered.
>> The entries in a Complete Infinite Binary Tree are densely ordered.

>
> The entries in form of nodes are well ordered. Anything else is not
> added to the list.
>

>> Those order types are incompatible.
>
> This was the question: In a list containing every rational: Is there
> always, i.e., up to every digit, an infinite set of paths (rational
> numbers) identical with the anti-diagonal? Yes or no?


As noted before, WM's questions often make no sense.

As explained before:

> But note that the question also demonstrates WM's
> complete lack of understanding of the diagonal
> argument.
>
> He has been told time and time again that it is
> an argument scheme which only has application
> under certain assumptions.
>
> He chooses to believe otherwise for the agenda
> of his fanaticism.
>
> Suppose one is given a countable listing of
> the rationals (with the appropriate restriction
> on double representation) according to the
> infinite listing of an expansion.
>
> Suppose one performs a diagonalization on
> that listing.
>
> Is the resultant a rational number? No.
>
> What may be concluded? That the rational numbers
> do not exhaust the capacity of the algorithm
> to generate representations if that algorithm
> is to generate a representation for every
> rational number.
>
> Although the burden of proof lies with WM
> concerning the nature of the diagonal, it
> is a simple matter to understand if one
> uses a Baire space representation instead.
>
> In the Baire space, rationals are in correspondence
> with eventually constant sequences. Since,
> by construction, a list of Baire space rationals
> would exhaust all of the eventually constant
> sequences, the resultant of a diagonal argument
> could not have an eventually constant sequence
> unless the original premise had been false.








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