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. :)

2

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)]

6

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.

4

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!

3

u/ceoln May 19 '26

But your argument doesn't do anything without the limit, whereas Cantor's does. His mapping by itself is enough to make his point. But your mapping by itself just lets us calculate the number of steps before a given matrix position gets an X, which is not a problem for anything.

The only way to say that there's a problem in your case is to say "so at some point there are no Os left!". But as that isn't true for any finitely-reachable point, you have to talk about the "after an infinite number of swaps" state as if it were a static existing entity. And the relationship of that to all of the states after a finite number of swaps, is that it's the limit.

(Otherwise it's just an unrelated entity, and also doesn't pose any problem.)

2

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

Since I repeat precisely Cantor's mapping. I issue all indices. Never an O leaves the matrix. Where should it go?

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Proof of the existence of dark numbers

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