On Dec 13, 9:56 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > Ah I see, so you are imposing another condition on the definition of a > > path, > > No, that is *the* definition of a path in a Binary Tree. > Actually it is not. There is no need at all to stipulate that a path must begin by 0. It is a fixed definition.
I already showed you that the number of paths in a second degree binary tree (or third degree if you want to adop the empty path) IS larger than the total number of nodes. And what I mean by paths those that can start with 1 or with 0, but with the condition that it must be unidirectional. And showed it clearly and I've illustrated each path. You have 9 paths (inclusive of the empty path) and only 7 nodes.
> > that it must begin with 0 since it is representing a number in > > the interval [0,1], that's OK, then according to that I can see that > > you are correct since the number of paths do corresponds to the number > > of ending node, and with my calculation it would be less than the > > total number of nodes by one. I'll need to check this again, but let > > me agree with you on that for the current moment. > > > But still you have a problem. What you really managed to prove is that > > the total number of FINITE undirectional paths beginning from 0 in the > > infinite binary tree is countable, since we can simply draw an > > injective map from those paths to their ending nodes. > > No. I proved that the number of infinite paths is countable by > constructing all nodes of the Binbary Tree by a countable set of > infinite paths. >
This only means that you can have a bijective function from a countable subset of infinite paths of the binary tree to the set of all nodes, which everyone already know that this is possible, because we all agree that the total number of nodes of the infinite binary tree is countable.
What would be a proof is if you manage to define an injection from the set of ALL infinite paths of the binary tree to the set of all nodes of the binary tree.