Re: Incompleteness of Cantor's enumeration of the rational numbers (doubling-spaces)

On 12.11.2024 21:03, Jim Burns wrote:

On 11/12/2024 1:06 PM, WM wrote:

On 12.11.2024 17:47, Jim Burns wrote:

 

[4-⅒,4+⅒] ≠ [1/3-⅒,1/3+⅒]

>
|[4-⅒,4+⅒]| = |[1/3-⅒,1/3+⅒]|
and only that is important for my argument.

 Yes,
μ[4-⅒,4+⅒] = μ[1/3-⅒,1/3+⅒]
 ⎛ Also, |[0,1]| = |[0,2]|
⎝ so I think you mean 'measure', not 'cardinality'.

Yes.

Your point is that
μ⋃{ [n-⅒,n+⅒]:n∈ℕ⁺ }
isn't in the extended reals.
 I get it.

Here is the final solution of the problem:
1) If Cantor was right, then the initial intervals would be sufficient to cover all rationals in the order 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, ... (these rationals are then centres of intervals).
2) If all rationals could be covered, then this could be accomplished in arbitrary order, not only along Cantors sequence. Because only the measure is important, not the identity of an interval.
3) Then we could first cover all naturals and then all halves and then all quarters and so on. But we know that already after covering all naturals no further intervals are available.
Regards, WM