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

On 2024-11-23 08:49:18 +0000, WM said:

On 23.11.2024 09:07, Mikko wrote:

On 2024-11-22 10:53:32 +0000, WM said:
 

On 22.11.2024 09:42, Mikko wrote:

On 2024-11-21 11:03:28 +0000, WM said:

 

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.

 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
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n]
without specifying what it means to mutiply an interval with ε;
where q_n is Cantor's enumeration of rationals.
The 1, 2, and 3 above are not relevant to the topic sepcified by the
subject line and OP.
--
Mikko