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

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Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : sci.logic
Date : 25. Nov 2024, 09:43:32
Autres entêtes
Organisation : -
Message-ID : <vi1dbj$2moon$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
User-Agent : Unison/2.2
On 2024-11-24 14:01:15 +0000, WM said:

On 24.11.2024 13:38, Mikko wrote:
On 2024-11-23 08:49:18 +0000, WM said:
 
It is relevant by three reasons:
1) The limit of the sequence f(n) of relative coverings in (0, n] is 1/10, not 1. Therefore the relative covering 1 would contradict analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has left, the white must remain such that the whole real axis can never become black.
 You say that it is relevant but you don't show how that is relevant
to the fact that there is no real number between the intervals (n/2, n/2+1)
that is not a part of at least one of those intervals.
 Because that has nothing to do with the topic under discussion. See points 1, 2, and 3. They are to be discussed.
 The subject line specifies that the discussion should be about Cantor's
enumeration of the rational numbers.
 OP specifies that the discussion shall be baout the sequence of
itnrevals
 That is a mistake. Should read:
[q_n - ε*sqrt(2)/2^n, q_n + ε*sqrt(2)/2^n].
OK but the following applies to that, too:

The 1, 2, and 3 above are not relevant to the topic sepcified by the
subject line and OP.

My last example contradicts a simpler bijection, namely that between all natural numbers and all natural numbers divisible by 10: Let every unit interval on the real axis after a number 10n carry a black hat. Then it should be possible to cover all intervals with black hats.
What does "contradicts" in "contradicts a simpler bijection"?
--
Mikko

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