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

On 27.11.2024 10:33, Mikko wrote:

On 2024-11-26 11:07:57 +0000, WM said:
 

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.

 If so, no bijection is contradicted.

The possibility of a bijection between the sets  ℕ = {1, 2, 3, ...} and D = {10n | n ∈ ℕ} is contradicted.
Assume that a bijection between natural numbers divisible by 10 and all natural numbers is possible. First establish a bijection between the numbers 10n ∈ ℕ and the numbers 10n ∈ D. This can be visualized by attaching black hats to the natural numbers of the form 10n ∈ ℕ and white hats to the remaining natural numbers. The black hats indicate that a number n ∈ ℕ has a partner in D. Now it should be possible to shift the black hats such that all natural numbers are covered by black hats. The mapping f(10n) is started by exchanging white hats and black hats precisely as would have been defined without the intermediate step: 1 gets it from 10, 2 gets it from 20, 3 gets it from 30, and so on, precisely as would be done without the intermediate first step.
However when all numbers of an interval (1, 2, 3, ..., n) are equipped with black hats, then some black hats have been taken from outside of the interval, from larger 10n which in turn have received white hats. If all natural numbers are equipped with black hats, then all white hats have disappeared. But hats cannot disappear by exchanging them.
Regards, WM