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

On 2024-11-27 11:10:51 +0000, WM said:

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:

 

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

 What does "contradicts a bijection" mean?
 

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.

No, it is not. You merely deny it, disregarding obvious facts. The function
f(x) = 10 * f obviously maps every element of ℕ to a different element of
D and there is no element of D that is not 10 * f for some f so this f is
a bijection between ℕ and D.
--
Mikko