Re: Simple enough for every reader?

On 13.06.2025 11:53, Mikko wrote:

On 2025-06-12 09:41:51 +0000, WM said:
 

On 12.06.2025 09:56, Mikko wrote:

On 2025-06-11 11:30:26 +0000, WM said:

>

For all natural numbers that can be chosen as individuals:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
The "..." can be removed by the inductive proof
for all definable n ∈ ℕ:
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
==> |ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.

>
That the set difference of an infinite set and a finite set is infinite
is well understood and therefore an uninteresting result.

>
That is not the only result.

 That is the result of your proof. Other results can be discussed in
other contexts.

The following result cannot be circumvented: After mankind will have ceased, there is a largest natural number ever named. It belongs to a finite initial segment. Almost all numbers are greater and have never defined. In our system they have remained undefined or dark.
Same is true for the past at every time.
Analogously: There are only finitely many prime numbers known up to every time. The others are dark. Almost all will remain dark.

 

Interesting is that all defined numbers belong to a finite initial segment.

 About natural numbers that is obvious. One interesting thing is that
rational numbers can be ordered so that every one of them belongs
to a finite initial segment.

That has been disproved by X-O-Matrices. No-one has ever identified a mistake. Can you???
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...
XOOO...    OOOO...    XOOO...    XOOO...
XOOO...    XOOO...    OOOO...    OOOO...
XOOO...    XOOO...    XOOO...    OOOO...
M(0)       M(2)       M(3)        M(4)      ...
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.

Another intresting thing is that real
numbers cannot be ordered so.

The reason is that most are dark too.

 Note that in order to dscuss infinities you must use second or higher
order logic. In first order logic you cannot say that an initial
segment is finite.

Both is nonsense. Simply use logic.

 

Therefore most numbers are dark.

 That does not follow.
 

It does.
Regards, WM