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

On 2024-11-03 16:40:01 +0000, WM said:

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.

There are no adjacent intervals. Between any two non-overlapping intervals
there are rational numbers. Around one of those to numbers is an interval
that does not touch the first mentioned intervals.
Between any two intevals there is space and that space contains other
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.

Some intervals overlaps because there are infintely many rational numbers
in every interval and each of these rationals has its own inteval.
There is space that is not part of any interval but all numbers in that
space are irrational.

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.

There are no adjacent intervals. Non-existent intervals don't prevent
anything.
--
Mikko