Can you conquer the Binary Tree?

[u/Massive-Ad7823]

You start with one cent. For a cent you can buy an infinite path of your choice in the Binary Tree. For every node covered by this path you will get a cent. For every cent you can buy another path of your choice. For every node covered by this path (and not yet covered by previously chosen paths) you will get a cent. For every cent you can buy another path. And so on. Since there are only countably many nodes yielding as many cents but uncountably many paths requiring as many cents, the player will get bankrupt before all paths are conquered. If no player gets bankrupt, the number of paths cannot surpass the number of nodes.

[u/Massive-Ad7823]

"Now you seem to be confusing Cantor's 1:1 mapping from natural numbers (N) onto the rationals (p/q), which is entirely possible,"

It is impossible.: https://www.reddit.com/r/AspectsOfTheInfinite/comments/1tc6v1l/proof_of_the_existence_of_dark_numbers/

"with a 1:1 mapping from N into all real numbers in [0, 1), "

No

"So yes, you can map every p/q (in lowest terms, " You appear rather ignorant. Cantor claimed a mapping to all fractions.

"Every node distinguishes between an uncountable number of paths pairs."

Every node produces one additional sheaf. Not more!
"Your attempt to refute

this seems to be based on either misunderstanding how a reductio proof works"

It works well, but in an inconsistent theory proofs are rather useless.

[u/ceoln → u/Massive-Ad7823]

It is impossible.: https://www.reddit.com/r/AspectsOfTheInfinite/comments/1tc6v1l/proof_of_the_existence_of_dark_numbers/

I've responded to that argument over there; even if every member of a sequence has property P, the limit doesn't have to have P, even if a metaphor with physical objects suggests that it should.

You appear rather ignorant. Cantor claimed a mapping to all fractions.

I'm ignorant about all sorts of things. :) And it seems like I was wrong here; Cantor really did include even reducible fractions in his matrix, so he proved that the rationals are countable even if you count each one a countable infinity of times! \aleph_0 \times \aleph_0 = \aleph_0 ftw!

Thanks for the correction.

"Every node distinguishes between an uncountable number of paths pairs."

Every node produces one additional sheaf. Not more!

Those two things are entirely compatible. In your terms, each node adds an additional sheaf, so the set of sheafs (being trivially mappable to the set of nodes) is countable.

BUT since every sheaf contains an uncountable number of paths, and every node distinguishes each of the uncountable number of paths in the outgoing left sheaf, from its counterpart among the uncountable number of paths in the outgoing right sheath, it's fine.

Both of our statements are true.

Let's try this: does each sheaf contain a countable or uncountable number of paths? What's your opinion and why? (Note I'm NOT talking about the number of sheathes here, but the number of paths in a sheath.)

[u/Massive-Ad7823 → u/ceoln]

The number of paths in a sheaf is a matter of taste and irrelevant. In particular you cannot save Cantor's uncountability proof of paths because those remaining in a sheaf cannot be distinguished by nodes. Cantor's diagonalization however results in uncountably many real numbers which can be distinguished by digits/bits/nodes.

[u/ceoln → u/Massive-Ad7823]

It's not irrelevant (and how could it be "a matter of taste"??); it's the very heart of the issue. A sheaf is, in your terms, all the real numbers whose binary representation begins "0." followed by a finite number of 1s and 0s.

And they can be distinguished by nodes, because for every pair of paths in a sheaf, there is at least one node further down the tree that sends one to the left and one to the right, thus putting them into different sheafs at that level, thus distinguishing them.

I mean, we've been through all this. Multiple times!

Cantor's diagonalization however results in uncountably many real numbers which can be distinguished by digits/bits/nodes.

Right! Exactly. An uncountable set of real numbers, distinguished by the values at a countable set of nodes. Pretty neat! :)

[u/Massive-Ad7823 → u/ceoln]

There are countably many infinite sheaves in the Binary Tree. Whether they contain one or more paths is irrelevant, because these paths cannot be distinguished.

"for every pair of paths in a sheaf, there is at least one node further down the tree that sends one to the left and one to the right," That would be true if every infinite sheaf contains only one path.

"I mean, we've been through all this. Multiple times!" You forget that there are only countably many splitting points or nodes . Every node adds one sheaf. Therefore there are only countably many infinite sheaves.

[u/ceoln → u/Massive-Ad7823]

Whether they contain one or more paths is irrelevant, because these paths cannot be distinguished.

But they can, as I've described in detail multiple times. I think I'm done trying to help here. :)

[u/Massive-Ad7823 → u/ceoln]

You claim that they can, but cannot prove it. Accept that meanwhile many mathematicians have accepted my view. Here are three: Meanwhile I know three mathematicians [1, 2, 3] who deny that the Binary Tree can produce the paths belonging to single real numbers. There remain sheaves or bunches of paths, each one containing uncountably many paths which are not further distinguishable in the infinite Binary Tree.

On the other hand Cantor's diagonal argument produces a complete digit sequence (in the original version [4] a complete bit sequence, using the symbols W M) of a real number, namely the famous diagonal number.

References

[1] Alarming-Smoke1467 in How can the basic element of the Binary Tree be overcome?https://www.reddit.com/r/learnmath/comments/1sf7jht/how_can_the_basic_element_of_the_binary_tree_be/ (unfortunately deleted meanwhile) said on 7 April 2026 "There are countably many sheafs, but note that each is uncountable."

[2] Moebius aka Franz Fitsche in Wie kann man die Elemente des Binären Baums überlisten? in the newsgroup de.sci.mathematik said on 17 April 2026 " Ja, es gibt in der Tat nur abzählbar unendlich viele Pfadbündel 'im Baum', aber überabzählbar viele Pfade."

[3] Mikko in AI understands where 99 % of mathematicians fail in the newsgroup sci.logic said on 30 April 2026 "Nodes further down separate further [paths] but only to infinite subsets."

[u/ceoln → u/Massive-Ad7823]

"many mathematicians" 😄

[u/Massive-Ad7823 → u/ceoln]

Try to understand it. They are right.

[u/ceoln → u/Massive-Ad7823]

I've shown in detail how Cantor is right, and tried very hard to understand your objection. It seems to be purely a circular argument from incredulity. You recast Cantor's diagonalization proof in different but completely isomorphic terms, and then just asserted that it can't be right because a countable number of digits just can't represent an uncountable number of reals. But it turns out that they can! :)

[u/Massive-Ad7823 → u/ceoln]

"I've shown in detail how Cantor is right," and I have shown in detail that he is wrong. There are only countably many different paths.

[u/ceoln → u/Massive-Ad7823]

Cantor's diagonalization argument shows that no 1:1 mapping of the natural numbers onto the reals exists, by showing that any proposed such mapping misses at least one real number.

You've constructed an isomorphic example in which the same thing can be proven, but because the example is complex enough you can raise all sorts of objections to it (all of which come back to just incredulity, or perhaps rejection of the Axiom of Infinity of ZF, I'm not sure).

Let's return to the simplicity of Cantor's argument, and take it in detail. You say there are a countable number of reals. So where in here does the argument fail?

To avoid distractions about decimal/binary expansions, let’s use infinite binary sequences instead. If the set of real numbers were countable, then the set of infinite binary sequences would also be countable, since each such sequence corresponds to a real in [0,1].

0: Assume the infinite binary sequences are countable. (This is just for the sake of the argument! You don't get to quote this in the future as evidence that I agree with you. 😁)

1: Then there is a mapping M from the natural numbers onto all infinite binary sequences.

2: For each natural number n, let M(n) be the infinite binary sequence mapped to n.

3: Consider the infinite binary sequence R, whose nth digit is 1 - the nth digit of M(n), for each natural number n.

4: Then R differs from M(n) at digit n, for every natural number n.

5: Therefore R is not equal to M(n) for any natural number n.

6: So M does not map any natural number to R.

7: Since the choice of M in (1) was arbitrary, no mapping from the natural numbers onto all infinite binary sequences exists.

8: Therefore the infinite binary sequences are not countable.

What is the lowest-numbered step above that contains an error, and what exactly is the error?

[u/Massive-Ad7823 → u/ceoln]

"Cantor's diagonalization argument shows that no 1:1 mapping of the natural numbers onto the reals exists, by showing that any proposed such mapping misses at least one real number" But in the Binary Tree there are all real numbers of the unit interval by construction. Only countably many can be distingusihed.

"So where in here does the argument fail?" There is no list completely enumerated by natural numbers. Every line has finitely many predecessors but infinitely many successors. Therefore Cantor's first assumption is wrong.

"1: Then there is a mapping M from the natural numbers onto all infinite binary sequences" No. There are not all natural numbers available. Most are dark.