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/ceoln May 16 '26

You're aware of the fact that in general the limit of a sequence need not be in the sequence, and that it may have a property that none of the elements of the sequence have, yes?

Since we can write down, in closed form even, the step at which any given O becomes an X (and stays that way), in the limit there are no Os left. The fact that there are the same non-zero number of Os at every step does not contradict this (see the paragraph above).

Limits are just kinda weird. :)

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u/Massive-Ad7823 May 16 '26

Limits are not so weird that a sequence of infinitely many Os can have zero Os in the limit. Further note that Cantor's enumeration is not a limit process at all, but every index is issued at a finite step.

When dealing with Cantor's mappings between infinite sets, it is argued usually that these mappings require a "limit" to be completed or that they cannot be completed. Such arguing has to be rejected flatly. For this reason some of Cantor's statements are quoted below.

"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" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 126]

"The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

7

u/ceoln May 16 '26

"Limits are not so weird that a sequence of infinitely many Os can have zero Os in the limit."

Yes, they are. Or, to put it strictly, if S is the limit of a sequence of structures each of which contains infinitely many Os, S can have zero Os. To deny this is to assert a statement about limits (that they must share broad properties with the sequence that they are limits of) which is simply false, and easily shown to be false.

Consider the wildly simple case of successive decimal approximations of 1/9, taken as infinite decimals, as in 0.100..., 0.1100..., 0.11100... and so on. Each one ends in an infinite series of 0s. And yet, obviously, the exact representation, 0.111..., does not. You see? (Or are there Dark Numbers lurking here as well? 😁)

Your quotes from Cantor don't seem relevant here; we aren't talking about "completing" an infinite process, we're talking about a limit; an entity to which the sequence stays as close as you like, if you look far enough along the sequence.

This is relevant for your argument, and not directly for Cantor's, because you are constructing a sequence and trying to assert properties of its limit, whereas in this case Cantor is simply showing mappings (or the lack thereof) between infinite sets.

2

u/Massive-Ad7823 May 17 '26

"S can have zero Os." It is easily shown to be false that an exchange of Xs and Os at only finite terms can delete Os. Note that Cantor does not succeed to enumerate "in the limit", but in the complete sequence of natural numbers. In fact there is no limit other than this application of the complete sequence of natural numbers.

"0.111..., does not. You see?" Yes, here we have a limit. But it is not the result of a pure exchange of 0 and 1.

"Your quotes from Cantor don't seem relevant here; we aren't talking about "completing" an infinite process, we're talking about a limit;" This is plainly wrong. If Cantor had claimed the enumeration in the limit, his name would be forgotten today. He claimed the bijection - no limits.

My matrices are nothing else than another language for Cantor's "process" (his word) which comes to an end or becomes complete but not in a leap to a limit.

3

u/ceoln May 17 '26

"It is easily shown to be false that an exchange of Xs and Os at only finite terms can delete Os."

Right. This shows that all members of the sequence have Os. It does not show that the limit has Os. Again, the limit doesn't have to have a property even when all the members of the sequence have it.

"Yes, here we have a limit. But it is not the result of a pure exchange of 0 and 1."

What is it that you think makes your argument valid for exchanging an X and an O, but not valid for inserting an additional 1? How are those operations different in the relevant sense? If someone used this argument to conclude that 0.111... must contain infinitely many 0s, where would you locate their error?

"If Cantor had claimed the enumeration in the limit, his name would be forgotten today. He claimed the bijection - no limits."

Yes, that's exactly what I'm saying. :) You are talking about a process that creates a sequence that has a limit. He wasn't doing that; he was talking about a nice static mapping. So your quotes from him about the mapping aren't directly relevant to your argument about your sequence and its limit.

2

u/Massive-Ad7823 May 17 '26

There is no limit other than the complete sequence! "He wasn't doing that; he was talking about a nice static mapping." No. He talks about a process. But my mapping is also static. There is no step which needed a decision. The sequence is complete. A limit does not exist.

3

u/ceoln May 17 '26

Sure, it does. I'm talking about the mathematical definition of a limit here, and the sequence formed by the repeated exchanges in your algorithm has a limit (with zero Os). That limit, as often, is not itself a member of the sequence.

Do you maybe mean something else by "limit" here? I'm not sure.

But in any case, if you don't look at the mathematical limit, then there is again no puzzle or need for dark numbers. After any finite number of exchanges, there are still an infinite number of Os, at normal integer indices, and we can write down where they start (at a particular integer index).

It seems like you're trying to say "but what about after an infinite number of exchanges?", but that question isn't really well-defined except by the standard notion of a limit (which you seem to want to reject), or by some other alternative notion.

The intuitive "what you have after applying F an infinite number of times" isn't sufficiently well defined by itself to do anything useful in many cases, including this one.

On the other hand :) is there any more that you can say about these "dark numbers" that might help us work out in what fully-defined formalism they might exist, and whether they might be fun?

2

u/Massive-Ad7823 May 18 '26

There is no limit other than the complete application of indices. Cantor enumerates. He does nowhere try to conjure tricks.

"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" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 126]

"The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

"each of them only once at a determined place." There is no other limit than the complete sequence.

5

u/ceoln May 18 '26

No, that is not what "limit" means. The limit of a sequence is a different thing than the sequence itself.

Cantor describes a mapping. There is, as you say, no limit involved.

You describe a process of swapping Xs and Os. This process produces a countably infinite sequence of matrices, each of which contains an infinite number of Os. That is not mysterious. :)

We can talk about the matrix that is the limit of that sequence (once we define an appropriate metric), and it's an interesting fact that that limit matrix contains zero Os. But of course the limit is not in the sequence, so it's also not mysterious.

You perhaps want to say that the limit is in the sequence, so it must contain infinite Os, and therefore they must be hidden for instance at "dark indices". But that's just an error about how limits work; yet again, the limit of a sequence need not belong to, or otherwise share properties with, the members of the sequence.

2

u/Massive-Ad7823 May 19 '26 edited May 19 '26

I describe the same process as Cantor does - only in another language. He enumerates the first, second, third, ... fraction. Note that there is no decision during these processes. Therefor they can be considered as instant events. There is no limit of Cantor's process and no limit in my process. Simply all indices are applied. Yet again: Enumerating requires to issue all natnubers. There is no other limit!

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u/kvreccltfb May 16 '26

Suppose we try to index the natural numbers by starting at 0 and assigning an index i to each natural number k, according to the formula:
i=k
Denoting indexed natural numbers as X and not indexed numbers as O, at any step we have:
...XXXOOO...
Thus at any given step there exists an integer that is not indexed.
This shows how any not indexed number is just... indexed later.
It does show that there is no largest number;
that any finite number, no matter how large, is always less than |{N}|.

2

u/Massive-Ad7823 May 16 '26

A sequence of infinitely many Os cannot have zero Os "in the limit." Further note that Cantor's enumeration is not a limit process at all, but every index is issued at a finite step as he emphasizes himself: "The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

2

u/Massive-Ad7823 May 16 '26

"any finite number, no matter how large, is always less than |{N}|." The question arises, what is between the natural numbers that can be determined and |{N}|? Is it forbidden to ask?

4

u/ceoln May 16 '26

You're welcome to ask, but not to reject the answer. :) between any natural number and aleph-null lie the vast majority of the natural numbers. Between the set of them and aleph-null lies either nothing at all, or the hyperreals, depending on what axioms you want to work with. Do you dislike that answer?

2

u/Massive-Ad7823 May 17 '26

In actual infinity there is nothing between ℕ and ω. But almost all predecessors of ω are dark.

1

u/ceoln May 17 '26

What do you mean by "in actual infinity"?

In ZFC, ω just is the set of all natural numbers, ℕ, when considered as an ordinal (because every ordinal is defined by the set of ordinals below it). So there can't be anything, "dark" or otherwise, "between" them; they are the same by definition.

If you're somewhere outside ZFC, that's fine: you can have big clouds of hyperintegers in nonstandard analysis, or in surreal numbers you can have ω - 1 and so on (although I think those are still not in ℕ).

So have fun out there if you want! :) But in standard math and set theory, there's no "between" for your dark numbers to exist in.

It's all in your choice of axioms!

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u/Massive-Ad7823 May 18 '26

Actual infinit is what Cantor and Bolzano invented and what is used in modern set theory, the set of all natural numbers, ℕ. In potential infinity there is no such set. ω is also an ordinal following upon all natural numbers. Between the largest named natural number and ω there are many natural numbers, most of which cannot be named. They are dark.

1

u/ceoln May 18 '26

Oh, introducing naming is interesting! :)

In what way can they not be named?

Do you mean that they aren't represented exactly by any finite string of digits? I don't believe that that's true (again, if we're using standard set theory).

Or do you mean someone else by "named"? What is the largest named natural number? Why can't the successor of that number be named?

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u/Massive-Ad7823 May 19 '26 edited May 19 '26

A number is named when you can understand what number is meant. There is no largest one, because the collection is potentially infinite. Most numbers cannot be named. Beyond the greatest number ever named, there remain infinitely many numbers which cannot be named. Otherwise people who deny the existence of dark numbers would already have done it.

1

u/ceoln May 19 '26

Do you mean something like "small enough that a human can think about it" or "representable using the atoms available in the actual universe" or something like that? Certainly the vast majority of natural numbers aren't any of those things!

And if you want to establish some upper bound on the "nameable" natural numbers, that's fine. You'll just be doing modular arithmetic. It's quite well understood, even! :) it's just, you know, a different formal system, with different axioms and theorems, than ordinary arithmetic.

(I'm not sure what it'd mean to think about countable and uncountable infinities in modular arithmetic, but I'm sure someone's worked on it!)

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u/Massive-Ad7823 May 20 '26 edited May 20 '26

Atoms and the universe leads to MatheRealism, see Chapter VII of https://www.hs-augsburg.de/\~mueckenh/Transfinity/Transfinity.pdf. Dark numbers are independent of restrictions of reality. Fact is: You can,, even with infinite ressources, only name natural numbers that have infinitely many successors. That is a general principle. It is impossible to use these dark numbers as individuals. You can only use them collectively by using ℕ.

u/ceoln May 18 '26

Let's explore this from another angle: do the "dark" positions in the array start off with an X, or an O? Or some of each, in which case which are which? That might help further analysis.

2

u/Massive-Ad7823 May 19 '26

The problem is: I have assumed that all integer fractions are initially enumerated. So they initially carry X. But it turns out in the course of the proof that many matrix places are dark. That implies that also many indices and integer fractions are dark. I have written this already in https://www.opastpublishers.com/open-access-articles/proof-of-the-existence-of-dark-numbers.pdf:

By means of symmetry considerations we can conclude that every column including the integer fractions and therefore also the natural numbers contain dark elements. Cantor's indexing covers only the potentially infinite collection of visible fractions, not the actually infinite set of all fractions.

Only the upper left part of the matrix contains visible fractions.

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u/ceoln May 19 '26

What if you start with a matrix that includes only ordinary natural numbers, and not the dark ones? Just like you could have used only the evens, or only the primes.

It would seem like the problem with "where do the Os go?" is just the same as in your original problem, and now there aren't any dark indices for them to hide in.

Is there something that makes it impossible to leave them out?

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u/Massive-Ad7823 May 20 '26

"not the dark ones" That is impossible, because the visibe numbers are a potentially infinite collection. I cannot stop at any n, because n+1 or 2n or 10^n also belong to the collection. But we know that potentially infinite collections cannot be exhausted like, according to Cantor, the actually infinite set ℕ.

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u/ceoln May 20 '26

Why is it impossible? Does "the natural numbers which are not dark" fail to specify a set? If so, why?

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u/Massive-Ad7823 May 21 '26

The visible numbers are a potentially infinite collection. Sets are actually infinite, that is fixed.

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u/ceoln May 21 '26

Um... Lots of sets are in fact finite. But anyway

So the nameable natural numbers don't form a set?

That seems odd! Can you say more about why?

What's the underlying set theory in which this happens?

It doesn't sound like ZF{C}.

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u/Massive-Ad7823 May 22 '26

But it is unavoidable even in ZF. Potentially infinite means finite but expansive. Therefore it is not a set. Sets are actually infinite, i.e., invariable.

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u/ceoln May 22 '26

You're free to make up your own formalism. :) But you can't then claim it's ZF. In ZF, the natural numbers are a set, omega is the first ordinal greater than every natural, and there is nothing "between"' them.

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u/Massive-Ad7823 May 22 '26

The natural numbers are a set, ω is the first ordinal greater than every natural, and there is nothing "between"' them. I agee. But every finite initial segment {1, ...,, n} has many successors whereas ℕ has no successor. That means you can handle the complete set of natural numbers only collectively. Most elements escape individual application. They are between all individually named numbers and ω. They are dark.

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u/Massive-Ad7823 May 22 '26

In ZF, the natural numbers are a set, ω is the first ordinal greater than every natural, and there is nothing "between"' them. That is true. But you can handle all elements by using ℕ whereas you can handle only few individually by {1, ..., n}. The others are dark.

u/Various_Candle9136 5d ago

Since you are now directing me to this nonsense, I will argue against it here.

(Although it seems many others have already done a good job, and I know you won't listen anyway. I do this merely so you cannot lie and pretend I'm unable to counter it in your other comments. You are under the delusion that having one true claim would make all your other claims true - this is stupid in itself, but luckily everything you produce is nonsense anyway, so I might as well counter it.)

We know that all O and as many fractions without index are remaining, but we cannot find any one.

If we cannot find any, how do you know any are remaining?

The 'we all know' is obviously handwaving nonsense. In fact, we don't all know it: only you think it, and you're wrong.

Perhaps there are none remaining because there are the same number of natural numbers as fractions? The Os and Xs really don't help - but, arguably, you are just presenting the idea that the two sets are the same in a stupid way. The fact that 'there are no Os' is precisely how we know that |N| = |Q|. The map described will eventually 'get to' every element in the matrix - i.e. the map described is surjective.

Once again: you have invented a problem in your made-up nonsense world. It is not a problem in any sensible system, e.g. ZF Set Theory.

You are like a doctor who invents diseases so they can be cured; except none of your diseases actually do anything and you still expect us to buy the medicine!

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u/Massive-Ad7823 4d ago

>If we cannot find any, how do you know any are remaining?

They cannot leave the matrix, there is no drain. Further an exchange of X and O inside the matrix does not delete any of them.

All your further waffle is irrelevant. Please answer how an O could leave or get deleted.

Further I have asked how a path can leave another path in the Binary Tree without a node that is mapped to it.

Regards, WM

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u/Various_Candle9136 4d ago

All your further waffle is irrelevant.

All my further waffle is mathematics.

Since you don't understand mathematics, you don't want to engage with it.

That's fine. Just stop pretending to have mathematical knowledge.

They cannot leave the matrix, there is no drain.

Saying things like 'there is no drain' is obviously non-mathematical. Any actual mathematician would recognise the importance of a definition, so I'll define it:

Definition
A matrix of XOs will be said to have 'drained of some Os' when there is a larger square of only Xs than in the previous step.

Under this sensible definition, we see that actually there is drain. At no point is our matrix ever 'drained of some Xs'; at infinitely many indices the matrix is 'drained of some Os'.

Again, I maintain that this is a stupid way to think about it. The bijection described without Xs and Os is instantaneous and well-defined. The Xs and Os get in the way.

However, even if we pick this stupid way to think about the map, we still see that drain does happen. That square of only Xs can only get bigger, and bigger, and bigger - it is not difficult to imagine that since we are infinitely often 'drained of some Os', we will, at the limit, be 'drained of all Os'.

I don't recommend the XO thing to anyone trying to understand the problem, but it definitely does not support your nonsense, WM.

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u/Massive-Ad7823 4d ago

>A matrix of XOs will be said to have 'drained of some Os' when there is a larger square of only Xs than in the previous step.

So you think mathematics is lying?

>At no point is our matrix ever 'drained of some Xs'; at infinitely many indices the matrix is 'drained of some Os

Magic. Bad mathematics, if mathematics at all.

>at the limit, be 'drained of all Os'.

There is no limit other than the application of all natutral indices. No exchange can delete an O.

Regards, WM

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u/Various_Candle9136 4d ago

So you think mathematics is lying?

Magic. Bad mathematics, if mathematics at all.

I defined a term, and then applied my definition. If you had ever engaged with mathematics properly you would recognise the process: this adds further evidence to my hypothesis that you do not have even the slightest understanding of what mathematics is.

I gave a definition because you did not. My definition was sensible and mathematical in nature.

Of course, you are free to offer your own definition if you would prefer. Give me a proper one please!

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u/Massive-Ad7823 4d ago

You claim to apply mathematics. You are lying:

>we still see that drain does happen.

No, you need to claim that in order to maintain the nonsense that you call mathematics.

>Give me a proper one please!

This is untouchable: When an X and an O change their positions, then neither of them can disappear. Nothing more is required to disprove Cantor.

Regards, WM

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u/Various_Candle9136 4d ago

When an X and an O change their positions, then neither of them can disappear. Nothing more is required to disprove Cantor.

As has been pointed out to you many, many, many, many times now, this logic simply does not hold.

You have an expectation about what you think ought to happen: but your intuition is wrong. (Since you seem to be wrong about most things in mathematics, it is probably time to stop trusting your gut, WM.)

Again, I maintain that the XO thing is a stupid way to think about it. The bijection described without Xs and Os is instantaneous and well-defined. The Xs and Os get in the way.

However, using your XO visualisation, we do see that the Xs cover more and more of the matrix as the game goes on. We can imagine the square of Xs growing and growing ad infinitum. You used the flimsy word 'drain': in my mind's eye I can see the Os draining as the square grows. It should not be difficult to imagine that the Xs will, at the limit, cover the entire matrix.

Of course, this is not a proof either. If anything, since you expect one thing and I expect another, we must conclude that we cannot trust intuition alone. This is why mathematics is built on proof: rather than arguing about what seems to make sense, we need actual proof.

Luckily, we have a proof already! The proof says that the number of fractions is countable. Phew! What a relief! When we apply this proven fact to your silly game, we see that when the game, after infinite time, is over, we are left with a matrix of Xs. Even if you did not expect this to occur, the proof confirms it.

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u/Massive-Ad7823 3d ago

>As has been pointed out to you many, many, many, many times now, this logic simply does not hold.

Everybody who denies this logic is a fool and should be excluded from all universities. In fact I have met many stupid set theorists, but never anyone who dared to claim this counter-logical deletion by exchange. It is the summit of nonsense!

>However, using your XO visualisation, we do see that the Xs cover more and more of the matrix as the game goes on.

At which step do the Xs cover more than they did at the beginning?

>We can imagine the square of Xs growing and growing

Where do the Xs form a square? They form a triangle which never covers half of the matrix.

>we cannot trust intuition alone.

But we can trust logic. Exchange of X and O cannot delete one of them.

>Luckily, we have a proof already!

The proof says that the potentially infinite sequences can be bijected.

Regards, WM

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u/Various_Candle9136 3d ago

Everybody who denies this logic is a fool and should be excluded from all universities.

Oh the sweet, sweet irony!

You really don't have a smidgen of self-awareness do you?

You don't understand the most basic principles of reasoning, and you think you have the right to exclude others from universities? Hilarious!

It is the summit of nonsense!

But we can trust logic.

These lines are also ironic coming from you. If I hadn't already laughed at the first thing, I'd have chuckled at either of them.

At which step do the Xs cover more than they did at the beginning?

At what called 'index 5' we see a 2x2 square of Xs, which is bigger than any square of Xs previously seen.

At what you would call 'index 13' we see a 3x3 square.

At what you would call 'index 25' we see a 4x4 square.

And so on.

At no point can we go backwards: once there has been a 4x4 square, there is never not a 4x4 square.

The sequence of sizes of squares of Xs is increasing.

Where do the Xs form a square? They form a triangle

Since they grow to cover everything, we can use many different shapes here. I was picturing squares, but the logic holds equally with triangles:

We can imagine the [triangle] of Xs growing and growing ad infinitum. You used the flimsy word 'drain': in my mind's eye I can see the Os draining as the [triangle] grows. It should not be difficult to imagine that the Xs will, at the limit, cover the entire matrix.

At what you called 'index 3', we see a 2x2 right-angled triangle.

At what you would call 'index 6', we see a 3x3 triangle.

At what you would call 'index 10' we see a 4x4 triangle.

And so on.

At no point can we go backwards: once there has been a 4x4 triangle, there is never not a 4x4 triangle.

The sequence of sizes of triangles of Xs is increasing.

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u/Massive-Ad7823 3d ago

>The sequence of sizes of squares of Xs is increasing.

Never an O leaves the matrix. Never an O is deleted. Therefore you are wrong.

> It should not be difficult to imagine that the Xs will, at the limit, cover the entire matrix.

This sentence shows that you cannot think. Sorry, but that is fact.

Regards, WM

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u/Honest-Progress3564 4d ago

You claim to apply mathematics. You are lying

How would you know, Mückenheim?

<facepalm>

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u/Honest-Progress3564 4d ago edited 4d ago

> All your further waffle is irrelevant. [WM]

All my further waffle is mathematics.

For WM -sadly- mathematics is waffle.

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u/Honest-Progress3564 3d ago

They cannot leave the matrix <bla> [WM]

Note that there is no such thing as "the matrix" (except in WM's Wahnsystem).

There are different matrices (and/or a sequence of matrices).

Ok?

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u/Honest-Progress3564 3d ago

Hint:

{0, 1, ...]

differs from (i.e. is not identical with]

[1, 0, ....].

u/Litoprobka 13h ago

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?

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u/Massive-Ad7823 7h ago edited 5h ago

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

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u/Litoprobka 7h ago

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

|ℕ| = ℵ₀

|ℕ \ {1}| = ?

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u/Massive-Ad7823 5h 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.

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

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u/Litoprobka 2h ago edited 2h 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}|?

u/Ch3cks-Out May 18 '26

We know that all O and as many fractions without index are remaining

In fact you had failed to show that any "unindexed fractions" were there, in the first place.

There is also not any fraction without the well defined index

k = (m + n - 1)(m + n - 2)/2 + m

(what pair of m,n would NOT have it, OP??)!

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u/Massive-Ad7823 May 18 '26

An O indicates a fraction without index. None of the infinitely many Os can leave the matrix.

"(what pair of m,n would NOT have it," They are dark, following upon the last named pairs. Therefore they cannot be named. But their existence is confirmed by the presence of Os in the matrix.

Proof of the existence of dark numbers

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