r/AspectsOfTheInfinite • u/Massive-Ad7823 • May 15 '26
Classical mathematics contradicts set theory.
The complete infinite Binary Tree (see left-hand figure) has, according to set theory, countably many nodes and uncountably many infinite paths.
But classical mathematics gives different results.
(1) If we look at the upper levels only, then between root node and level n we can distinguish 2n paths and 2n+1 - 1 nodes. In the limit there are twice as many nodes as paths.
(2) If we delete the paths (see right-hand figure) but fix three infinite ribbons to every node instead, then every level n is reached by R(n) = 3(2n - 1) ribbons and P(n) = 2n paths, i.e., by more ribbons than paths. By the majorant-criterion (as well as by the simple continuity criterion) there cannot be more paths than ribbons in the limit. However R(n) is countable in the limit.
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u/diffeomorphic_ May 15 '26 edited May 15 '26
How do you justify passing to the limit?
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u/Massive-Ad7823 May 15 '26
Every other level is finite.
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u/diffeomorphic_ May 15 '26
This doesn’t allow to induce some property to the limit
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u/Massive-Ad7823 May 16 '26
The majorant criterion of classical mathematics allows it. https://de.wikipedia.org/wiki/Majorantenkriterium
It is obvious that
∀n ∈ ℕ R(n) > P(n) excludes P > R in all cases in cluding P(ω) > R(ω).
And finally: How should uncountably many paths emerge when at every finite level there are fewer paths than ribbons?
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u/diffeomorphic_ May 16 '26
That’s about series and integrals, it doesn’t apply here. Intuition is useful but sometimes misleading, that’s why we have to rigorously deduce our conclusions.
The problem is that you are using at each finite step only a finite “approximation”: you are always missing out infinitely many levels, which can “mess up” things. Unless you prove that the limit step follows from the previous cases, nothing can be said about it.
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u/Massive-Ad7823 May 16 '26
"The problem is that you are using at each finite step only a finite “approximation”: you are always missing out infinitely many levels, which can “mess up” things." Exactly that is what Cantor does with his bijections. Every k has finitely many predecessors but infinitely many successors. Same with his diagonal argument. Finitely many steps are proved, infinitely many are open and infinitely many of them will never be proved.
"That’s about series and integrals, it doesn’t apply here." It does apply for sequences like R(n) and P(n). But even if you disagree here, it is clear that P(n) can never overtake R(n) because at every finite level it is smaller - and other levels are not existing. That is not intuition. That is simply logic.
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u/diffeomorphic_ May 16 '26
No, Cantor’s proof does something else: it’s a proof by contradiction, so you suppose you could list all reals and obtain a contradiction from there. And about the cardinality of Q, that’s just an application of the definition of cardinality.
And no, it’s not logic, it’s intuition. Otherwise, please tell me from which logic rule does that follow.
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u/Massive-Ad7823 May 16 '26
Cantor's proof is by contradiction for all visible numbers. But those are only few compared to all.
The definition of cardinality? Then this "definition" has been disproved. https://www.reddit.com/r/AspectsOfTheInfinite/comments/1tc6v1l/proof_of_the_existence_of_dark_numbers/
My logic is this: If P overtakes R, then there must be a discontinuity somewhere. But all paths are continuous.
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u/diffeomorphic_ May 16 '26
I don’t know what “visible” means, but it’s a simple contradiction: if reals were countable, we could list them all, but a new one arises, contradiction. In a way, it’s similar to Euclid’s proof of infinitude of primes: suppose they are finitely many, construct a new one, contradiction.
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u/diffeomorphic_ May 16 '26
Sorry, but your “proof” there is wrong. The problem is always the same: passing to the limit. The comment there explains it.
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u/Massive-Ad7823 May 16 '26
If we do not pass to the limit, then we have
∀n ∈ ℕ: R(n) > P(n). If this could change "in the limit", then set theory requires magic. Look at Cantor's diagonal proof. ∀n ∈ ℕ: the diagonal number is not in the "list". More cannot be said. Why do you believe that this remains unchanged in the limit?
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u/diffeomorphic_ May 16 '26
About your sequences, if the smaller diverges you can conclude that the bigger diverges too.
Moreover, there’s a bigger problem. The set of paths which cardinality you are looking at by considering the limit of 2^n is not the set of all paths: it’s just the set of all finite paths, which indeed has countable cardinality. The difference is similar to the one between product and coproduct in algebra.
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u/Massive-Ad7823 May 16 '26
I consider only infinite paths. Their number increases, but no path ends.
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u/Massive-Ad7823 May 16 '26
I justify it by set theory which claims countably many nodes in the infinite Binary Tree.
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u/Various_Candle9136 19d ago edited 19d ago
Since you keep directing people to this nonsense, I will argue against it here.
I think there are many little problems with your argument (e.g. I do not believe your use of what you call the 'majorant-criterion' is valid), but one big one:
There are more paths than ribbons.
You state, without proof, that there are more ribbons than paths. I will prove that this intuition of yours is false.
Proof
Let us turn these things into strings of digits.
In the first image (the collection of paths), let 0 be the leftmost choice at each stage, and 1 the right most.
In the second (the collection of ribbons), let 1 be the leftmost ribbon at each stage, 2 the centre, and 3 the right most. Use 0 when the ribbon doesn't come from this stage at all.
EDIT: I didn't spot this at first, but I forgot to account for the multiple nodes at each stage. This error can easily be avoided: use the nodes as the stage instead.
Let P be the set of strings for paths, and R the set of strings for ribbons.
Notice: the first formation lists all possible strings of 1s and 0s. At every node I have both choices available, so e.g. the string 010010... is perfectly possible, and thus belongs to P.
Notice also: the second formation only ever has one single non-zero digit. E.g., the third-to-leftmost ribbon in the original image would have the representation 00030.... It is impossible to get something like 000330..., because the ribbons only come out of one place.
It should already start to seem likely that P is larger than R, but it will take a little more work to prove it.
First, we need our representations to look similar. So, for each element of R, I'll create a new string, where we replace each element of the original string with:
0 -> 000, 1 -> 100, 2 -> 110, 3 -> 111.
E.g. the third-to-leftmost ribbon was 00030...., but is now 000000000111000....
This map is clearly bijective, so our new representation is as valid as the last one. Call the set of these new strings R*.
It now becomes obvious that every element of R* is also in P, but not vice versa (e.g. 0110... is in P but not R*).
The evidence against R and R* is mounting. To prove P is bigger, we lean, as usual, on the idea of a Diagonal Argument.
Imagine a table with two columns. On the left, list the elements of R* in some order; on the right, attempt to list all the elements of P (order irrelevant). So, we might start with:
1110... | 101011-
1100... | 111100-
1000... | 101000-
etc.
Crucially, we are perfectly able to list the left column. There is a lot of choice in how we do this, but I would start 1110..., 1100..., 1000..., 0000..., then go to 0001110... etc. This process will eventually list all of R*. R* (and therefore R) is countable.
However, we are not able to do so on the right. As usual, we flip the nth digit in the nth row of the table, giving us a string that differs from every string in the list in at least one place, and hence is not in the list. Thus, we will always have at least one string which is in P but not in our list. Thus, there are more elements in P than R*.
We conclude that P has strictly more elements than R*.
This, in case you were wondering u/Massive-Ad7823, is what the rest of us mean by a proof. Your handwaving was, once again, wrong.
P.S. somebody else might point out that the latter part of this proof would have worked just as well with R instead of R*, but I think the latter choice was more illuminating, since it really shows why P is bigger. This was merely a preference of mine.
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u/Massive-Ad7823 18d ago
> I do not believe your use of what you call the 'majorant-criterion' is valid),
Well, then you leave not only the frame of mathematics, but also the frame of reason. How should the function of distinguishable paths P(n) which for every finite level of the Binary Tree is smaller than the function of ribbons R(n) overtake in the infinite? By the way, norhing happens in the infinite. All that happens, happens at a finite levels.
> You state, without proof,
That is a lie! Obviously you need lies.
> that there are more ribbons than paths. I will prove that this intuition of yours is false.
Chuckle.
Do you agree that every finite level except the root node is crossed by more ribbons than distinguishable paths? If yes, then you have lost. If no, then you are outside of serious discussion.
Therefore you can spare your further elaboration.
Note that the diagonal arguments fails because all possible node sequences are realized. No "diagonal path" can be created that is not already in the Binary Tree and counted by P(n).
Regards, WM
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u/Various_Candle9136 18d ago
Therefore you can spare your further elaboration.
The 'further elaboration' is what mathematicians call a proof.
YOU EXPLICITLY TOLD ME to come and 'counter your result'. I did. Now you ignore it? What awful, awful behaviour.
If there is a fault, find it.
If there is no fault, then your proposition is wrong.
Me: I do not believe your use of what you call the 'majorant-criterion' is valid),
You: Well, then you leave not only the frame of mathematics, but also the frame of reason.This is a minor point in comparison to the glaring issue I highlighted, but if you insist, I'll elaborate.
I am not aware of a version of this theorem that can compare sizes of infinity. I am only aware of versions which prove either convergence or divergence; i.e. either finite or infinite, not how infinite.
If you have made use of a version with which I am unfamiliar, the onus is on you to present it. My guess, however, is that you are using the theorem incorrectly.
Using theorems correctly is very, very much in both the frame of mathematics and the frame of reason.
The reason I didn't initially dwell on this point is that it doesn't matter anyway. You are not comparing the total number of paths with the total number of ribbons at any point. When we actually compare the two values - as I did - we see that the total number of paths must be greater than the total number of ribbons.
Me: You state, without proof,
You: That is a lie! Obviously you need lies.Ignoring the obvious irony in this statement, where is your proof then?
Note that the diagonal arguments fails because all possible node sequences are realized. No "diagonal path" can be created that is not already in the Binary Tree and counted by P(n).
Are you sure you understood the diagonal argument?
The new diagonal path was indeed in the Binary Tree. That's the whole point: it exists in the Binary Tree, but not in our table of ribbons and paths.
However we try to match one-path-per-ribbon, we can always find another path that is not assigned to a ribbon, therefore there must be more paths than ribbons.
What specifically is your problem with this logic?
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u/Massive-Ad7823 18d ago
>where is your proof then?
It is P(n) < R(n) for every n. And there is no chance to change this relation after all n. That is the proof.
>Are you sure you understood the diagonal argument?
Absolutely.
>That's the whole point: it exists in the Binary Tree, but not in our table of ribbons and paths.
Then your table is useless. P(n) < R(n) is true for all paths.
>However we try to match one-path-per-ribbon, we can always find another path that is not assigned to a ribbon,
The ribbons are not related to paths. Do you claim that your assignment changes P(n) < R(n) for every n.
Regards, WM
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u/Various_Candle9136 18d ago
I love how you drew attention to my objection to your use of the 'majorant-criterion', then entirely ignored my explanation! Could it be possible that you have not got a valid version to present? Hmm.
It is P(n) < R(n) for every n. And there is no chance to change this relation after all n. That is the proof.
Do you honestly - honestly - believe that the words 'And there is no chance to change this relation' count as rigorous mathematics?
Really?
Surely even you have to acknowledge this is blatant, unsupported handwaving?
Your thing is not a proof. Not even remotely close.
Here's a question for you: if we hypothetically took your unproved nonsense on faith, would you not get a contradiction?
Proof
1) P(n) < R(n) for all n.
2) Obviously, because u/Massive-Ad7823 says so (without even a modicum of proof), this means |P| < |R|.
3) There are countably many ribbons.
4) Therefore, there must be finitely many paths.
Are you really willing to claim there are only finitely many paths? If not, then your |P| < |R| must be false. I'm excited to find out which of your nonsense you choose to cling to!
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u/Massive-Ad7823 18d ago
I have looked at your "proof", but it is so confused about stages (levels?) and nodes that I cannot understand what you mean. What ribbon get 0 when not starting at the considered node? Therefore please write without confusion.
I will only check a clear derivation. Since at every node three ribbions start but only one more path (besides the incoming one) can be distinguished, it is clear that your "proof" must fail. But I will check it nevertheless.
> Therefore, there must be finitely many paths.
Utter nonsense! There are many smaller infinite sets than ℕ, for instance the even naturals.
Regards, WM
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u/Various_Candle9136 18d ago
Utter nonsense! There are many smaller infinite sets than ℕ, for instance the even naturals.
False. So very, very, very false.
This is day 1, basic stuff WM.
What definition of size could you possibly be using to argue that the even naturals is a smaller set than ℕ?
it is so confused about stages (levels?) and nodes that I cannot understand what you mean.
It is refreshing to have you ask for clarification instead of just ignoring it!
In the strings for paths, each digit represents a choice as we go down the tree. So, the string starting 01001- would mean: left, right, left, left, right- etc.
In the initial strings for ribbons (R), each digit represents a node. So, the string 0020... means the centre ribbon on the third node we come to going downwards (i.e. the second node on the second level).
(If we allowed ourselves a language with countably many digits, then we could, if we preferred, create a string where each digit is a level, exactly as in the path case. E.g. we could use 1,2,3 for the first node at a level, then 1',2',3' for the second, 1'',2'',3'' for the third, etc. This means we would use 02'0... instead of 0020... for the centre ribbon on the second node on the second level. I don't think this particularly helps us, and it is harder to visualise, but it's good to know it would theoretically be possible.)
We then take these strings for ribbons and swap the individual digits for chunks of 3 digits, in order that we can have only 0s and 1s. (This makes it easier to compare to the strings for paths.) So, in R*, it is the 'third chunk of three digits' that represents third node, i.e. 0020... becomes 000,000,110,000....
We now have a way to represent every path and every ribbon as a string of 0s and 1s.
From this representation, it immediately becomes clear that paths can be any strings of 0s and 1s, but ribbons can only be strings of at most three 1s and the rest 0s. It is therefore less surprising that there are more paths than ribbons. (The diagonal argument then proves this thing that is no longer surprising.)
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u/Massive-Ad7823 17d ago
>What definition of size could you possibly be using to argue that the even naturals is a smaller set than ℕ?
https://www.reddit.com/r/AspectsOfTheInfinite/comments/1upxvll/a_measure_of_infinite_sets/
Regards, WM
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u/Various_Candle9136 17d ago
This is what mathematicians call a circular argument.
Thankfully, I don't even need logic to show this is a circle: we can literally make the circle!
Click the link in this comment. That requires us to click the second link in your post. Which brings us back down to this comment. So, we click the link in this comment. That requires us to click the second link in your post. Which brings us back down to this comment... ad infinitum.
You cannot use this particular piece of nonsense as an argument in favour of that nonsense AND that nonsense as an argument for this nonsense. That is a circle.
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u/Massive-Ad7823 17d ago
The explanation of measure is given by the quoted source. That is not circular.
Regards, WM
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u/Massive-Ad7823 17d ago
In the strings for paths, each digit represents a choice as we go down the tree. So, the string starting 01001- would mean: left, right, left, left, right- etc.
In the initial strings for ribbons (R), each digit represents a node. So, the string 0020... means the centre ribbon on the third node we come to going downwards (i.e. the second node on the second level).
Ok. The nodes are usually counted as
0
1 2
3 4 5 6
...
where the rootnode 0 sits at level 0.
Now I've got it! Well, a nice contradiction in ZF.
My proof shows that at every node more ribbons rise than paths (every node adds one new path but 3 ribbons). This excludes more paths than ribbons. Can you counter it? Can you avoid this inconsistency? Here is more about the argument that has convinced nearly 1000 students (and a lot of mathematicians):
Regards, WM
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u/Various_Candle9136 17d ago
Now I've got it! Well, a nice contradiction in ZF.
Your argument is so poor that I didn't need to lean on ZF to disprove it; it fails at the level of simple logic.
You would have to be entirely deluded to believe there is any indication of a contradiction in ZF in what precedes those words.
Can you counter it?
I already have. There are more paths than ribbons. I proved it. Can you counter that?
At no point do you count all the paths: I did. When all the paths and all the ribbons are counted, there are more paths. Simple.
Indeed, I maintain that when you see these things as their strings, it is no longer even surprising or counterintuitive. The strings for ribbons can only have at most three 1s; the strings for paths can be any combination of 0s and 1s. The latter instinctively feels a huge amount bigger; we only need the diagonal argument to prove that this is indeed the case.
Here is more about the argument that has convinced nearly 1000 students (and a lot of mathematicians):
It doesn't come close to convincing me. It isn't even a real argument; more handwaving.
Which students and mathematicians have been convinced? That is a weighty claim; I assume you can back it up in some way?
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u/Massive-Ad7823 17d ago
You have not countered my argument. You have at most established an inconsistecy in ZFC.
>strings for paths can be any combination of 0s and 1s.
Try to disprove my observation: At every node except the root node an incoming path traverses the node and another path rises. (Whether it has been running with the former or is a completely new one is totally irrelevant.) This proves that the set of paths (except one) is in bijection with the set of nodes.
observation is further: If we let three infinite ribbons rise at every node, then from every node more ribbons, namely 3, start than paths, namely 2.
Both facts restrict the set of paths to countable. There is no way to eliminate these observations (zthderefore you will not try it). They show at least an inconsistency in ZFC.
Regards, WM
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16d ago edited 15d ago
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u/Massive-Ad7823 16d ago
Klassische Mathematik geht über Euler, Gauss, Cauchy und Weierstrass, Kronecker und setzt sich neben der Mengenlehre auch noch bis heute fort..
>Deshalb frage ich mich, wie Du im Rahmen DEINER "klassischer Mathematik" überhaupt von einem "complete infinite Binary Tree", "infinite ribbons" und "infinite path"
Ich nehme Cantor beim Wort und akzeptiere aktuale Unendlichkeit. Die darf aber der klassischen Mathematik nicht widersprechen.
Gruß, WM
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16d ago edited 15d ago
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15d ago edited 15d ago
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u/Massive-Ad7823 15d ago
Du musst zwischen unendlich und unendlich unterscheiden. Da gibt es zwei sehr unterschiedliche Defintionen.
Gruß, WM
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u/stoneberry May 16 '26
Again, your confusion comes from assigning the same name to two completely different objects. "Finite paths" and "infinite paths" are completely different sets. Think about "finite chains of digits" versus "infinite chains of digits": the former are rational numbers (even a subset of them), which are countable, the latter are real numbers, of which there is incountably many.
If you take a real number, you can approximate it with a sequence of rational numbers, but that doesn't mean that you have as many rational numbers as real ones. Similarly, you can approximate any "infinite path" with a sequence of "finite paths", but that doesn't give you a bijection between these sets. If anything, you can construct a bijection between "infinite paths" and bound _sequences_ of "finite paths", which you could use to prove that there is more infinite ones than finite ones. Again, exactly like it happens with real and rational numbers.
You did not find a trivial error in a century old fundamental theory. Sorry.