What is "left to right" traversal? Of course you can traverse a infinite tree bottom up. For example this cantor tree can be completely traversed (its nodes):
/0 0 ... / \1 * \ /0 1 ... \1
You enumerate level 0, only one node, then you enumerate level 1, only the nodes 0 and 1, then you enumerate level 2, the nodes 00, 01, 10, 11, and so on.
Nevertheless the infinite paths in the tree are uncountable (Cantors Theorem). If you don't understand this, its you who commits a "S=Lim S" blunder.
Have a Nice day!
Am Dienstag, 3. Oktober 2017 14:09:14 UTC+2 schrieb John Gabriel: > (4) NO, I do *not* agree that you can do a ?left to right? traversal of > the infinitely-long representations. This is the problem with your whole > damned argument: you?re mixing together notions from finite representations > with infinite representations. You cannot do an ordered traversal of the > leaves of an infinite tree. It?s *meaningless*. What?s the left-neighbor of > 1/3 in your tree? You cannot specify it ? it doesn?t really exist: there > simply is no real number which is ?closest? to 1/3 without being 1/3. But a > left-to-right traversal supposes that there *is*. And that?s the problem. > You?re trying to get a result using a property of a finite tree, when that > property doesn?t exist on a tree extended to infinity.