Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 26.11.2024 14:07, Richard Damon wrote:

On 11/26/24 6:07 AM, WM wrote:

On 26.11.2024 10:09, Mikko wrote:

On 2024-11-25 14:38:13 +0000, WM said:

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The simple example contradicts a bijection between the two sets described above.

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What does "contradicts a bijection" mean?
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It shows that the mapping claimed to be a bijection is not a bijection.

 Where did you do that with the ACTUAL bijection, and not just your strawman "equivalent".

An actual bijection is assumed: Consider the black hats at every 10 n and white hats at all other numbers n. It is possible to shift the black hats such that every interval (0, n] is completely covered by black hats. There is no first n discernible that cannot be covered by  black hat. But the origin of each used black hat larger than n is now covered by a white hat. Without deleting all white hats it is not possible to cover all n by black hats. But deleting white hats is prohibited by logic. Exchanging can never delete one of the exchanged elements.

 Which element of which infinite set did not participate in the bijection?

That is the crucial point! The white hats remain as long as logic is valid. But their carriers cannot be found. They are dark.
Regards, WM