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

On 2024-11-10 10:56:26 +0000, WM said:

On 10.11.2024 11:20, Mikko wrote:
 > On 2024-11-09 21:30:47 +0000, WM said:
 >
 >> On 09.11.2024 15:03, Mikko wrote:
 >>> On 2024-11-08 16:30:23 +0000, WM said:
 >>
 >>>>
 >>>> If Cantors enumeration of the rationals is complete, then all rationals
 >>>> are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ... and none is outside.
 >>>
 >>> All positive rationals quite obviously are in the sequence. Non-positive
 >>> rationals are not.
 >>>
 >>>> Therefore also irrational numbers cannot be there.
 >>>
 >>> That is equally obvious.
 >>>
 >>>> Of course this is wrong.
 >>>
 >>> You may call it wrong but that's the way they are.
 >>
 >> The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
 >
 > Maybe, maybe not, depending on what is all n.
 It is, as usual, all natural numbers.
  > If all n is all reals then
 > the measure of their union is infinite.
 But n is all naturals as you could have found out yourself, by the measure < 3.

That is not something one can find out. The symbol n means whatever you
say it means, and in this case you didn't say.
--
Mikko