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

On 11/15/2024 7:00 AM, WM wrote:

On 15.11.2024 11:43, Mikko wrote:

Translated intervals are not
the same as the original ones.
Not only their order
but also their positions can be different
as demonstrated by your example and mine, too.

>
If the do not cover the whole figure
 in their initial order,
then they cannot do so in any other order.

Sets for which that is true
are finite sets.
Insisting on "for all sets"
does not change infinite sets into finite sets.
Insisting on "for all sets"
changes what.you.are.talking.about.
You will no longer be talking about infinite sets.

Since according to Cantor's formula
the smaller parts of ℝ+ are frequently covered,
in the larger parts much gets uncovered.

ℝ⁺ holds points.between.splits of ℚ⁺
ℚ⁺ holds ratios of numbers in ℕ⁺
ℕ⁺ holds numbers countable.to from.1
ℝ⁺ ℚ⁺ ℕ⁺ are infinite sets.
You (WM) are both talking.about and not.talking.about
infinite sets.
(Multi.tasking, I suppose.)

Every definable rational is covered.

countable.to from.1 ⟨i,j⟩ ↦ kᵢⱼ countable.to from.1
kᵢⱼ = (i+j-1)⋅(i+j-2)/2+i
For each countable.to from.1 k
⟨iₖ,jₖ⟩ is a pair of countable.to from.1
countable.to from.1 k ↦ ⟨iₖ,jₖ⟩ countable.to from.1
(iₖ+jₖ) = ⌈(2⋅k+¼)¹ᐟ²+½⌉
iₖ = k-((iₖ+jₖ)-1)⋅((iₖ+jₖ)-2)/2
jₖ := (iₖ+jₖ)-iₖ
(iₖ+jₖ-1)⋅(iₖ+jₖ-2)/2+iₖ = k

That is called potential infinity.

⎛ A finite set A can be ordered so that,
⎜ for each subset B of A,
⎜ either B holds first.in.B and last.in.B
⎜ or B is empty.
⎜
⎜ An infinite set is not finite.
⎜ An infinite set C can _only_ be ordered so that
⎜ there is a non.empty subset D of C, such that
⎜ either first.in.D or last.in.D or both don't exist,
⎜ not visibly and not darkly.
⎜
⎝ And our sets do not change.