Re: Incompleteness of Cantor's enumeration of the rational numbers

On 03.11.2024 14:57, Mikko wrote:

On 2024-11-03 08:38:01 +0000, WM said:
 

Apply Cantor's enumeration of the rational numbers q_n, n = 1, 2, 3, ... Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].
Let ε --> 0.
Then all intervals together have a measure m < 2ε*sqrt(2) --> 0.
>
By construction there are no rational numbers outside of the intervals. Further there are never two irrational numbers without a rational number between them. This however would be the case if an irrational number existed between two intervals with irrational ends.

 No, it would not. Between any two distinct numbers, whether rational or
irrational, there are both rational and irrational numbers.

Not between two adjacent intervals. Such intervals must exist because space between intervals must exist. Choose a point of this space and go in both directions, find the adjacent intervals.

As long as ε > 0 the intervals overlap

Let ε = 1. If all intervals overlap and there is no space "between", then the measure of the real line is less than 2*sqrt(2). Therefore not all intervals overlap.

Anyway, there are real numbers that are not in any interval.

That is not possible because between two adjacent intervals there is no rational number and hence no irrational number.
Regards, WM