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

On 11/16/2024 2:54 PM, WM wrote:

On 16.11.2024 20:30, Jim Burns wrote:

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 they 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.

>
Sets for which that is true are
sets which obey geometrical rules.

'Geometry' joins 'mathematics' and 'logic' among
words you (WM) invoke _in place of_ an argument.
Do you (WM) know whether, in your geometry,
triangles with equal angles (AKA, similar)
have corresponding sides in the same ratio?
If your geometry is like our geometry,
then they do,
  and
there is a procedure in your geometry
(with similar triangles) which,
any one integral point determines
two integral points,
  and
another procedure in your geometry
(with similar triangles) which,
from any two integral points determines
one integral point:
  and
they are inverses of each other:
  k ↦ ⟨i,j⟩ ↦ k

⎝ And our sets do not change.

>
Therefore
the set of intervals cannot grow.

An infinite set can match a proper superset
without growing.
Because it is infinite.