On 12 Dez., 11:26, Zuhair <zaljo...@gmail.com> wrote: > WM has presented the idea that the infinite binary tree must have > countably many paths. It seems that he thinks that the total number of > paths in a binary tree is always smaller than or equal to the total > number of nodes. So since obviously the number of nodes in an infinite > binary tree is countable, then the number of paths must be so. But > since each real can be represented by a path in a 0,1 labeled infinite > binary tree, then we must have countably many reals thus defying > Cantor. > > The 0,1 labeled infinite binary tree he is speaking of is the > following: > > 0 > / \ > 0 1 > / \ / \ > 0 1 0 1 > . . . . > . . . . > > Answer: in order to justify this claim one must first see what is > happening at the finite level. And make a simple check of the number > of nodes and the number of paths through those nodes. If that > confirmed that the total number of nodes is more or equal to paths > then there would be some justification for that claim. > > Let's see: > > Lets take the first degree binary tree which is the following > > 0. One node, no paths > > The second degree binary tree is: > > 0 > / \ > 0 1 > > This has three nodes and two paths. > > Lets take the third degree binary tree > > 0 > / \ > 0 1 > / \ / \ > 0 1 0 1 > > Now this has 7 nodes BUT 8 paths, those are > > 0-0 > 0-1 > 1-0 > 1-1 > 0-0-0 > 0-0-1 > 0-1-0 > 0-1-1
Wrong. Every path starts at the root node. Every real of the interval (0, 1) starts with "0."
