Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : sci.logicDate : 22. Nov 2024, 09:42:09
Autres entêtes
Organisation : -
Message-ID : <vhpg51$13soc$1@dont-email.me>
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User-Agent : Unison/2.2
On 2024-11-21 11:03:28 +0000, WM said:
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:
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.
That is irrelevant to your question whether the whole interval becomes
black if the shifted intervals (n/2, n/2+1) are painted black.
And that question is irrelevant to the topic specified on the subject line.
-- Mikko
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