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

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 nubers, whether rational or
irrational, there are both rational and irrational numbers. There are
also intervals from the above specified set.

(Even the existence of neighbouring intervals is problematic.)

Not at all. Between any two non-interlapping intervals there is another
interval so there are not neighbouring intervals. Consequentely, all
these interval are lonely, and so are the rational numbers in their center.

Therefore there is nothing between the intervals, and the complete real axis has measure 0.

As long as ε > 0 the intervals overlap so "between" is not well defined.
Anyway, there are real numbers that are not in any interval.

This result is wrong but implied by the premise that Cantor's enumeration is complete.

Your result is wrong.
Cantor's enumeration is complete. Numbers not enumerated are not rational.
--
Mikko