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

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.
And you have not shown how the fact that there is no real number between
the intervals (n/2, n/2+1) is relevant to anything mentioned on the
subject line.
--
Mikko