Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : wolfgang.mueckenheim (at) *nospam* tha.de (WM)
Groupes : sci.logicDate : 01. Dec 2024, 11:55:15
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vihfai$2cnof$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
User-Agent : Mozilla Thunderbird
On 01.12.2024 11:17, Mikko wrote:
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:
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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.
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If so, no bijection is contradicted.
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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.
Obvious is that for every interval (0, n] the relative covering is 1/10, and that there are no further black hats beyond all natnumbers n.
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.
It appears so. I have shown by a different example that it is wrong. The relative covering for every interval is 1/10, independent of the configuration of the hats available inside. The limit of this sequence is 1/10.
Regards, WM
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