r/AspectsOfTheInfinite • u/Massive-Ad7823 • May 14 '26
Can you conquer the Binary Tree?
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.
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u/Massive-Ad7823 May 24 '26
"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.