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

On 07.11.2024 10:22, Mikko wrote:

On 2024-11-06 17:55:15 +0000, WM said:
 

On 06.11.2024 16:04, Mikko wrote:

On 2024-11-06 10:01:21 +0000, WM said:

>

I leave ε = 1. No shrinking. Every point outside of the intervals is nearer to an endpoint than to the contents.

>
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.

>
When ε approaches 0 then the measure of the real axis is, according to Cantor's results, 0. That shows that his results are wrong.

 It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.

But not between irrational points.

 

But the important question is also covered by ε = 1. The measure of the real axis is, according to Cantor's results, less than 3. That shows that his results are wrong.

 No, that is not Cantor's result,

It is Cantor's result that all rationals are countable, hence inside my intervals.
But we can use the following estimation that should convince everyone:
Use the intervals I(n) = [n - sqrt(2)/2^n, n + sqrt(2)/2^n]. Since n and q_n can be in bijection, these intervals are sufficient to cover all q_n. That means by clever reordering them you can cover the whole positive axis except "boundaries".
And an even more suggestive approximation:
Replace the I(n) by intervals J(n) = [n - 1/10, n + 1/10].
These intervals (without splitting or modifying them) can be reordered, to cover the whole positive axis except boundaries. Every rational is the midpoint of an interval.That means the real axis is covered infinitely often.
Reordering them again in an even cleverer way, they can be used to cover the whole positive and negative real axes except boundaries. And reordering them again, they can be used to cover 100 real axes in parallel.
That would be possible if Cantor was right.
Regards, WM