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

On 11/21/2024 5:21 AM, WM wrote:

On 21.11.2024 10:16, Mikko wrote:

On 2024-11-20 11:42:15 +0000, WM said:

On 19.11.2024 17:26, Jim Burns wrote:

On 11/19/2024 6:01 AM, WM wrote:

That implies that
our well-known intervals

>
Sets with different intervals are different.
Our sets do not change.

>
The intervals before and after shifting are not different.
Only their positions are.

Intervals with different points are different.
Our sets do not change.

The intervals are different.
A shifted interval contains a different set of numbers.
>

The intervals before and after shifting are not different.
Only their positions are.

>
The intervals are different.
A shifted interval contains a different set of numbers.

>
Consider this simplified argument.
Let every unit interval after
a natural number n which is divisible by 10
be coloured black: (10n, 10n+1].
All others are white.
Is it possible to shift the black intervals

No,
it is not possible to shift the black intervals.
Because our sets do not change.
  Is it possible to match the black intervals
  to the proper superset of all the intervals?
Yes.
10⋅n ↦ n ↦ 10⋅n
The two sets, of black and of all, are infinite.

Is it possible to shift the black intervals
so that the whole real axis becomes black?
>
No.
Although there are infinitely many black intervals,
the white intervals will remain in the majority.

Here, you have used 'majority' in a way which
includes 10⋅n ↦ n ↦ 10⋅n
That makes your answer irrelevant to the question.

For every finite distance (0, 10n)

Each, unlike the whole, finite.

 the relative covering is precisely 1/10,

The measure of black and the measure of all are
unbounded by any countable.to number, thus are +∞
+∞/+∞ is undefined.

whether or not the intervals have been moved or
remain at their original sites.
That means the function decribing this,
1/10, 1/10, 1/10, ...
has limit 1/10.
That is the quotient of
the infinity of black intervals and
the infinity of all intervals.

The Paradox of the Discontinuous Function
(not a paradox):
lim.⟨ rc(1), rc(2), rc(3), ... ⟩  ≠
rc( lim.⟨ 1, 2, 3, ... ⟩ )
You (WM) do not "believe in"
proper.superset.matching sets
discontinuous functions
similar triangles in proportion
And you (WM) feel that
  those feelings should be enough.