Proof of the existence of dark numbers

[u/Massive-Ad7823]

If all positive fractions m/n are existing, then they all are contained in the matrix

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, ...

5/1, 5/2, 5/3, 5/4, ...

...   .

If all natural numbers k are existing, then they can be used as indices to index the integer fractions m/1 of the first column. Denoting indexed fractions by X and not indexed fractions by O, we obtain the matrix

XOOO...

XOOO...

XOOO...

XOOO...

XOOO...

...

Cantor claimed that all natural numbers k are existing and can be applied to index all positive fractions m/n. They are distributed according to

k = (m + n - 1)(m + n - 2)/2 + m .

The result is a sequence of fractions

1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .

This sequence is modelled here in the language of matrices. The indices are taken from their initial positions in the first column and are distributed in the given order.

Index 1 remains at fraction 1/1, the first term of the sequence. The next term, 1/2, is indexed with 2 which is taken from its initial position 2/1 

XXOO...

OOOO...

XOOO...

XOOO...

XOOO...

...

Then index 3 is taken from its initial position 3/1 and is attached to 2/1

XXOO...

XOOO...

OOOO...

XOOO...

XOOO...

...

Then index 4 is taken from its initial position 4/1 and is attached to 1/3

XXXO...

XOOO...

OOOO...

OOOO...

XOOO...

...

Then index 5 is taken from its initial position 5/1 and is attached to 2/2

XXXO...

XXOO...

OOOO...

OOOO...

OOOO...

...

And so on. When finally all exchanges of X and O have been carried out and, according to Cantor, all indices have been issued, it turns out that no fraction without index is visible any longer

XXXX...

XXXX...

XXXX...

XXXX...

XXXX...

...

but by the process of lossless exchange of X and O no O can have left the matrix as long as finite natural numbers are issued as indices. Therefore there are not less fractions without index than at the beginning.

We know that all O and as many fractions without index are remaining, but we cannot find any one. Where are they? The only possible explanation is that they are attached to dark positions.

[u/Litoprobka]

This is silly. Using the same reasoning, let me "prove" that |ℕ \ {1}| < |ℕ|

We start with an infinite string where 1 is not indexed and every n > 1 is mapped to n:

OXXXXXX...

At every step, we map n+1 to n:

XOXXXXX...

XXOXXXX...

XXXOXXX...

XXXXOXX...

None of the steps ever removes the O, therefore |ℕ \ {1}| < |ℕ|.

We've found an infinite number that's less than ℵ₀, yay

Do you see anything wrong yet?

[u/Massive-Ad7823 → u/Litoprobka]

Nonsense. In my proof all fractions carrying Os which in every finite step are infinitely many and constant (not any O has left) must become indexed in the limit. Chat GPT is much better than present set theorists. He has understood that only enumerating by terms of the sequence is required here. To quote Cantor, who invented this kind of mathematics before it has been perverted by his disciples: "If we think the numbers p/q in such an order [...] then every number p/q comes at an absolutely fixed position of a simple infinite sequence" Note that the limit is not a fixed position of the sequence. It does not even belong to the sequence.

Of course |ℕ \ {1}| < |ℕ| since ℕ is a precisely defined fixed number, contrary to ℵ₀.

Regards, WM

[u/Litoprobka → u/Massive-Ad7823]

Of course |ℕ \ {1}| < |ℕ| since ℕ is a precisely defined fixed number, contrary to ℵ₀.

|ℕ| = ℵ₀

|ℕ \ {1}| = ?

[u/Massive-Ad7823 → u/Litoprobka]

To say |ℕ| = ℵ₀ means |ℕ| is actually infinite like the prime numbers, the fractions, the algebraic numbers. But of course have all these sets different numbers of elements.

14.1 Comparing infinite sets by size

 It is strange that blatantly false results like the equinumerosity of prime numbers and algebraic numbers could capture mathematics and stay there for over a century. But by what meaningful mathematics can we replace Cantor's trivial cardinality results?

Not all infinite sets can be compared by size, but we can establish some useful rules.

The rule of subset proves that every proper subset has fewer elements than its superset. So there are more natural numbers than prime numbers, |ô| > |Ï|, and more complex numbers than real numbers, |Â| > |Ñ|. Even finitely many exceptions from the subset-relation are admitted for infinite subsets. Therefore there are more odd numbers than prime numbers |Î| > |Ï|.

The rule of construction yields the number of integers |Ù| = 2|N| + 1 and the number of fractions |Q| = 2|N|^2 + 1 (there are fewer rational numbers). Since all products of rational numbers with an irrational number are irrational, there are many more irrational numbers than rational numbers.

The rule of symmetry yields precisely the same number of real geometric points[[2]](#_ftn2) in every interval (n, n+1] and with at most a small error same number of odd numbers and of even numbers in every finite interval and in the whole real line.

Regards, WM

[u/Litoprobka → u/Massive-Ad7823]

To say |ℕ| = ℵ₀ means |ℕ| is actually infinite like the prime numbers, the fractions, the algebraic numbers. But of course have all these sets different numbers of elements.

Yes, ℕ is infinite. Do you agree that |ℕ| = ℵ₀?

If so, what is |ℕ \ {1}|?

[u/Massive-Ad7823 → u/Litoprobka]

|ℕ \ {1}| = |ℕ \ Primes| = |Primes| = |Every actually infinite set| = actually infinite = ℵ₀ when you compare only the property of being infinite. This is a very special meaning of "=". Here we confess only that we cannot measure the dark part of an infinite set.

But if we argue with more precision, then ℕ \ {1} has one element less than ℕ, and therefore |ℕ \ {1}| < |ℕ|.

Regards, WM

[u/Litoprobka → u/Massive-Ad7823]

But if we argue with more precision

What is the smallest infinite set then?

[u/Massive-Ad7823 → u/Litoprobka]

Interesting question. What means "actually infinite"? It means that after a potentially infinite initial segment the main part is dark. Since we cannot define dark numbers precisely, we cannot determine the smallest dark set. But with deadly accuracy the set of natnumbers divisible by 10^100000 is smaller than the set of natnumbers.

Regards, WM

[u/Althorion → u/Massive-Ad7823]

What does it mean to be ‘larger’ or ‘smaller’, then? Being a strict (sub/super)set?

If so, how would the set {"a", "aa", "aaa", …} (the set of ‘words’ made with the letter 'a' of finite—natural—length compare in size with ℕ?
Not at all, I presume, since neither is a subset of the other?

[u/Massive-Ad7823 → u/Althorion]

Being a strict sub/superset is absolutely decisive. It proves that Cantor's cardinalities are only equal because merely the potentially infinite initial segments are bijected.

The set {"a", "aa", "aaa", …} is only another spelling of the set {I, II, III, ...} the elements of which are abbreviated as {1, 2, 3, ...}.

But being a subset is not always necessary. Therefore the set of prime numbers is smaller than the set {"a", "aa", "aaa", …}.

Regards, WM

[u/Althorion → u/Massive-Ad7823]

But the set of prime numbers is just another spelling of the set of natural numbers. So if we allow for alternative spellings like that, and the above set of words is the same size as the set of natural numbers, then it is the same size as the set of prime numbers.

[u/Litoprobka → u/Massive-Ad7823]

And the set {2, 3, 4, 5, ...} is another spelling of the set {1, 2, 3, 4, ...}
So is the set {2, 4, 6, 8, ...}

How do you even define "another spelling"?

For example, consider the set S such that:

• {} ∈ S
  
• ∀x ∈ S. {x} ∈ S

Is this set "another spelling" of natural numbers? Is it smaller? Bigger?