Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : wolfgang.mueckenheim (at) *nospam* tha.de (WM)
Groupes : sci.logicDate : 21. Nov 2024, 12:03:28
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vhn420$jf6v$3@dont-email.me>
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User-Agent : Mozilla Thunderbird
On 21.11.2024 11:59, Mikko wrote:
On 2024-11-21 10:21:40 +0000, WM said:
On 21.11.2024 10:16, Mikko wrote:
On 2024-11-20 11:42:15 +0000, WM said:
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The intervals before and after shifting are not different. Only their positions are.
>
The intervals are different. A shifted interval contains a different
set of numbers.
>
Consider this simplified argument. Let every unit interval after a natural number n which is divisible by 10 be coloured black: (10n, 10n+1]. All others are white. Is it possible to shift the black intervals so that the whole real axis becomes black?
Yes. Shift the interval (10n, 10n+1) to (n/2, n/2+1).
For every finite (0, n] the relative covering remains f(n) = 1/10, independent of shifting. The constant sequence has limit 1/10.
Regards, WM