```Date: Dec 13, 2012 1:56 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: On the infinite binary Tree

> 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.> 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 byconstructing all nodes of the Binbary Tree by a countable set ofinfinite paths.> This is clear.> But the problem that you will face is with the INFINITE undirectional> paths beginning from 0 of the infinite binary tree, now those paths> cannot be identified with the nodes at their ends because they simply> have no ending nodes because they are infinite, so you need to> calculate another way to map those kinds of paths to nodes.I did it already.> The only> one I can think of really is to map each infinite path to the> COLLECTION of all nodes in it, which is to say we map each infinite> path to the subset of all nodes of the binary tree that simply have> all and only those nodes in that path as elements of. Now this will> bring us again to Cantor, how many subsets of nodes of the infinite> binary tree do we have?No again. The number of subsets of the set of naural numbers is ofinterest. Every subset of N is represented by one and only one singleinfinite path. It is not interesting to find out the set of subsets ofnodes - at least not for our question.> Cantor says it is uncountable, therefore the> total number of INFINITE paths in the binary tree can be uncountable!And it is countable too. Therefore countability is not well defined.> However I think you are claiming that each infinite path is to be> mapped to the Godel number of the defining formula of it (which is a> finite parameter free formula), and of course by then we'll be having> only countably many such infinite paths of the binary tree. And also> this is well known since definable sets in this way are always> countable (this is pretty much standard). Now the real point is what> if there is a path for which we cannot find a finite parameter free> formula that define it, by then those kinds of paths cannot be proved> countable in this way. However you seem to dismiss that possibility as> nonsensical.It is nonsensical because the same could be assumed for Cantor'sdiagonal. It would be undefinable and it would be impossible to provethat it differs from all lines of the list - in particular ifundefinable reals exist and are members of the list.However, this question is irrelevant. It stands that the completeBinary Tree can be constructed by countably many paths. That's enoughto have a contradiction.Regards, WM
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