Re: The reality of sets, on a scale of 1 to 10 [Was: The non-existence of "dark numbers"]

On 25.03.2025 23:20, joes wrote:

Am Tue, 25 Mar 2025 20:16:10 +0100 schrieb WM:

On 25.03.2025 09:18, joes wrote:

Am Mon, 24 Mar 2025 20:40:07 +0100 schrieb WM:

On 24.03.2025 02:11, joes wrote:

Am Sun, 23 Mar 2025 18:18:15 +0100 schrieb WM:

>

The "bijection" is invalid because there are always infinitely many
elements following after every defined pair.

Which are also bijected.

How can you prove that?

How can you disprove it?

Knowing that every pair belongs to a finite initial segment.

Of course it does.
 

Upon it
follow infinitely many elements which cannot be proven to have partners
in the other set.

Neither can they be disproven.

Disproven is it for instance by the following:
X-O-Matrices
All positive fractions
1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
...
can be indexed by the Cantor function k = (m + n - 1)(m + n - 2)/2 + m which attaches the index k to the fraction m/n in Cantor's 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, ... .
Its terms can be represented by matrices. When we attach all indeXes k = 1, 2, 3, ..., for clarity represented by X, to the integer fractions m/1 and indicate missing indexes by hOles O, then we get the matrix M(0) as starting position:
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
... ... ... ... ...
  M(0)   M(2)   M(3)   M(4)   M(∞)
M(1) is the same as M(0) because index 1 remains at 1/1. In M(2) index 2 from 2/1 has been attached to 1/2. In M(3) index 3 from 3/1 has been attached to 2/1. In M(4) index 4 from 4/1 has been attached to 1/3. Successively all fractions of the sequence get indexed. In the limit, denoted by M(∞), we see no fraction without index remaining. Note that the only difference to Cantor's enumeration is that Cantor does not render account for the source of the indices.
Every X, representing the index k, when taken from its present fraction m/n, is replaced by the O taken from the fraction to be indexed by this k. Its last carrier m/n will be indexed later by another index. Important is that, when continuing, no O can leave the matrix as long as any index X blocks the only possible drain, i.e., the first column. And if leaving, where should it settle?
As long as indexes are in the drain, no O has left. The presence of all O indicates that almost all fractions are not indexed. And after all indexes have been issued and the drain has become free, no indexes are available which could index the remaining matrix elements, yet covered by O.
It should go without saying that by rearranging the X of M(0) never a complete covering can be realized. Lossless transpositions cannot suffer losses. The limit matrix M(∞) only shows what should have happened when all fractions were indexed. Logic proves that this cannot have happened by exchanges. The only explanation for finally seeing M(∞) is that there are invisible matrix positions, existing already at the start. Obviously by exchanging O and X no O can leave the matrix, but the O can disappear by moving without end, from visible to invisible positions.
Regards, WM