Re: Log i = 0

On 29.05.2025 01:22, 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.

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What's the point ?
It is the DEFINITION of "counting". A countable infinite set IS a set equipped with a bijection onto \N.
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This bijection does not exist because most natural numbers cannot be distinguished as a simple argument shows.

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Bijected elements need not be distinguished, it is enough to show a bijection.

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You mean it is enough to believe in a bijection?

[...]
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Either the bijection works or it doesn't. For instance, Cantor Pairing works with any unsigned integer.

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It does not.
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All positive fractions
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     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, ...
     ...
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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
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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, ... .
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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:
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XOOO...    XXOO...    XXOO...    XXXO...
XOOO...    OOOO...    XOOO...    XOOO...
XOOO...    XOOO...    OOOO...    OOOO...
XOOO...    XOOO...    XOOO...    OOOO...
M(0)       M(2)       M(3)        M(4)      ...
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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.
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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?
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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.
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It should go without saying that by rearranging the X of M(0) never a complete covering can be realized.

It sounds like you are trying to say

Don't guess. Try to understand the proof.

, even the following map does not work:
 map_to(x) return x + 1
map_from(x) return x - 1

It works for all definable numbers because they are a potentially infinite sequence: For every x there is x + 1. The above proof concerns all natural numbers. For Cantor's actually infinite set the map does not work, but that is more difficult to understand (since you can't imagine dark numbers) than the above obvious result.
Regards, WM