Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : sci.logicDate : 01. Dec 2024, 11:17:25
Autres entêtes
Organisation : -
Message-ID : <vihd3l$2d9fk$1@dont-email.me>
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User-Agent : Unison/2.2
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
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