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NNTP-Posting-Date: Thu, 29 May 2025 15:36:53 +0000
Subject: Re: Log i = 0
Newsgroups: sci.math
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From: Ross Finlayson <ross.a.finlayson@gmail.com>
Date: Thu, 29 May 2025 08:37:00 -0700
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On 05/28/2025 04:22 PM, Chris M. Thomasson wrote:
> On 5/27/2025 1:13 PM, WM wrote:
>> On 27.05.2025 22:07, Chris M. Thomasson wrote:
>>> On 5/27/2025 12:34 PM, WM wrote:
>>>> On 27.05.2025 18:05, FromTheRafters wrote:
>>>>> WM wrote :
>>>>>> On 26.05.2025 22:25, efji wrote:
>>>>>>> Le 26/05/2025 à 16:36, WM a écrit :
>>>>>>>> That is wrong. Present mathematics simply assumes that all
>>>>>>>> natural numbers can be used for counting. But that is wrong.
>>>>>>>
>>>>>>> What's the point ?
>>>>>>> It is the DEFINITION of "counting". A countable infinite set IS a
>>>>>>> set equipped with a bijection onto \N.
>>>>>>>
>>>>>> This bijection does not exist because most natural numbers cannot
>>>>>> be distinguished as a simple argument shows.
>>>>>
>>>>> Bijected elements need not be distinguished, it is enough to show a
>>>>> bijection.
>>>>
>>>> You mean it is enough to believe in a bijection?
>>> [...]
>>>
>>> Either the bijection works or it doesn't. For instance, Cantor
>>> Pairing works with any unsigned integer.
>>
>> It does not.
>>
>> All positive fractions
>>
>>      1/1, 1/2, 1/3, 1/4, ...
>>      2/1, 2/2, 2/3, 2/4, ...
>>      3/1, 3/2, 3/3, 3/4, ...
>>      4/1, 4/2, 4/3, 4/4, ...
>>      ...
>>
>> can be indexed by the Cantor function k = (m + n - 1)(m + n - 2)/2 + m
>> which attaches the index k to the fraction m/n in Cantor's sequence
>>
>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
>> 5/1, 1/6, 2/5, 3/4, ... .
>>
>> Its terms can be represented by matrices. When we attach all indeXes k
>> = 1, 2, 3, ..., for clarity represented by X, to the integer fractions
>> m/1 and indicate missing indexes by hOles O, then we get the matrix
>> M(0) as starting position:
>>
>> XOOO...    XXOO...    XXOO...    XXXO...
>> XOOO...    OOOO...    XOOO...    XOOO...
>> XOOO...    XOOO...    OOOO...    OOOO...
>> XOOO...    XOOO...    XOOO...    OOOO...
>> M(0)       M(2)       M(3)        M(4)      ...
>>
>> M(1) is the same as M(0) because index 1 remains at 1/1. In M(2) index
>> 2 from 2/1 has been attached to 1/2. In M(3) index 3 from 3/1 has been
>> attached to 2/1. In M(4) index 4 from 4/1 has been attached to 1/3.
>> Successively all fractions of the sequence get indexed. In the limit
>> we see no fraction without index remaining. Note that the only
>> difference to Cantor's enumeration is that Cantor does not render
>> account for the source of the indices.
>>
>> Every X, representing the index k, when taken from its present
>> fraction m/n, is replaced by the O taken from the fraction to be
>> indexed by this k. Its last carrier m/n will be indexed later by
>> another index. Important is that, when continuing, no O can leave the
>> matrix as long as any index X blocks the only possible drain, i.e.,
>> the first column. And if leaving, where should it settle?
>>
>> As long as indexes are in the drain, no O has left. The presence of
>> all O indicates that almost all fractions are not indexed. And after
>> all indexes have been issued and the drain has become free, no indexes
>> are available which could index the remaining matrix elements, yet
>> covered by O.
>>
>> It should go without saying that by rearranging the X of M(0) never a
>> complete covering can be realized.
>>
>> Regards, WM
>>
>>
>
> It sounds like you are trying to say, even the following map does not work:
>
> map_to(x) return x + 1
> map_from(x) return x - 1
>
> map_to(0) = 1
> map_to(1) = 2
> ...
>
> map_from(1) = 0
> map_from(2) = 1
> ...
>
> ?


Didn't you already have this conversation, and why are you having it here?

It seems you're describing a simple book-keeping of an integer continuum
in areal terms.

Also called a geometrization sometimes.