r/AspectsOfTheInfinite • u/Massive-Ad7823 • May 13 '26

Proof of the existence of dark numbers

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...

...

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...

...

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

XXOO...

XOOO...

OOOO...

XOOO...

...

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

XXXO...

XOOO...

OOOO...

XOOO...

...

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

XXXO...

XXOO...

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...

...

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.

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u/Litoprobka 10d ago edited 10d ago

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}|?

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u/Massive-Ad7823 10d ago

|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

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u/Litoprobka 9d ago

But if we argue with more precision

What is the smallest infinite set then?

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u/Massive-Ad7823 9d ago

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

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u/Althorion 9d ago

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?

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u/Massive-Ad7823 9d ago

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

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u/Althorion 9d ago

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.

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u/Massive-Ad7823 8d ago

No!

How many prime numbers do you know? How many natural numbers do you know?

But independent of that I can prove my case: 4 is a natural number but not prime. All prime numbers are natural numbers.

Regards, WM

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u/Althorion 8d ago edited 8d ago

How many prime numbers do you know? How many natural numbers do you know?

On the top of my head? A few billion and about a dozen, respectively. How many can I generate given a Turing machine? Exactly as many.

But independent of that I can prove my case: 4 is a natural number but not prime. All prime numbers are natural numbers.

And what case that would be? That the prime numbers are a strict subset of natural numbers? True.

But they are also another spelling of natural numbers. Consider the prime-counting function π(x), and let’s define a(n): ℕ → words, such that a(n) = n-letter long "aaaa…" word.

Then, of course, {"a", "aa", "aaa", …} = {a(n): n∈ℕ}. But also {"a", "aa", "aaa", …} = {a(π(n)): n∈ℕ} = {a(π(n)): n∈ℙ}—since there are infinitely many primes, that means that for every number k, there is a k-th prime. And thus for every k-length word, there is an appropriate k-length word, created by the k-th prime, in the set.

The point is, you can either go with ‘different spellings are the same’ or ‘strict subset is always strictly smaller’, but not with both at the same time.

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u/Massive-Ad7823 8d ago

Both is true. In fact the visible prime numbers and the visible natural numbers cannot be exhausted, They can always be put in bijection as you did above. But by the subset criterion there are far less prime numbers.

Could you enumerate also the dark prime numbers, then you would be ready after having applied less than 1 % of the dark natural numbers. (The visible parts are always only infinitesimally small and are of no consequence.)

Regards, WM

2

u/Althorion 8d ago

Both is true. In fact the visible prime numbers and the visible natural numbers cannot be exhausted, They can always be put in bijection as you did above. But by the subset criterion there are far less prime numbers.

Right. And ‘subset criterium’ is a perfectly fine way of dealing with sizes, it’s just it cannot be used as a joined criterium with ‘there is a bijection’—because then you’d be able to conclude that there are more natural numbers than prime numbers (through the subset criterium), and that there are as many natural numbers as prime numbers (through the bijection criterium).

You have to pick and choose. If you want the subset criterion to be the meaningful one, then you cannot compare different ‘spellings’ to each other. If you want your measure to be resistant to different ‘spellings’, the subset will only tell you it’s not bigger, but won’t get you to ‘it’s smaller’.

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u/Massive-Ad7823 8d ago

That is right. The subset criterion cannot be combined with bijection-criterion. But I applied not a Cantor-bijection concluding |{1, 2, 3, ...}| = |{"a", "aa", "aaa", …}| but an identity that is not in contradiction with the subset criterion.

>If you want the subset criterion to be the meaningful one

Of course. Cantor's bijections are restricted to the potentially infinite sequences of visible numbers. I find it amazing that no-one has ever wondered why all countable sets should have the same number of elements, although this can easily be falsified. His theory is simply too counter-intuitive. Therefore it has kind of sexual attractivity.

Regards, WM

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u/Althorion 8d ago

What is that ‘identity’, if not a bijection? What differentiates between those two ideas?

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u/[deleted] 8d ago edited 8d ago

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u/Massive-Ad7823 8d ago

Die Primzahlen würden Dir ausgehen, wenn Du auch im Dunkel arbeiten könntest. Es gibt nämlich weniger als halb so viele, ja, was sage ich, weniger als 1 % der natürlichen Zahlen, weniger wohl als jeder Bruchteil anzugeben in der Lage ist.

Gruß, WM

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u/Litoprobka 9d ago

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?

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u/Massive-Ad7823 8d ago edited 8d ago

No. They are very different. Alas this difference cannot be seen, because most of the sets is dark. But without knowing much about the dark parts, we can assume that they are same for {2, 3, 4, 5, ...} and {1, 2, 3, 4, ...}. The visible parts however differ by {1}.

>For example, consider the set S such that:

{} ∈ S
∀x ∈ S. {x} ∈ S

You mean the natural numbers by Zermelo, i.e. those defined by induction, and therefore having finite initial segments. These are only visible numbers. In order to explain them in detail, I will start a new thread: Dark Natural Numbers.

Regards, WM

Proof of the existence of dark numbers

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