
Re: Construction of the Binary Tree
Posted:
Oct 6, 2017 6:11 PM


On Friday, October 6, 2017 at 4:22:13 PM UTC4, WM wrote: > Am Freitag, 6. Oktober 2017 15:57:54 UTC+2 schrieb Dan Christensen: > > On Friday, October 6, 2017 at 4:35:58 AM UTC4, WM wrote: > > > A countable set can be constructed by using always half of the remaining time for the next step. An uncountable set cannot be constructed such that uncountably many elements can be distinguished. So it is possible to construct N and with it all its subsets. But these subsets cannot be distinguished unless it is indicated which elements are to combine. Therefore we find: > > > > > >  The Binary Tree can be constructed because it consists of countably many nodes and edges. > > > > > >  The Binary Tree cannot be constructed because it consists of uncountably many distinct paths. > > > > > > > Should be easy enough to a construct binary tree on N. > > > > Construct: L: N > N such that L(n)=2n+1 (the left node of n) > > Construct: R: N > N such that R(n)=2n+2 (the right node of n) > > > > So, we have L(0)=1, R(0)=2, L(1)=3, R(1)=4, L(2)=5, R(2)=6, .... > > > > Does this result in a contradiction? If so, please demonstrate exactly how using ONLY the rules of arithmetic on N, the axioms of set theory and the rules of logic. > > There are aleph_0 steps completing all paths,
You will need a formal proof of this, Mucke. But first, you will have to formalize the notion of "steps completing all paths."
> all are distinct, each one is completed by its own step. If so, then there are not more than aleph_0 paths. >
This all looks rather far removed from set theory or formal logic, Mucke. And it makes no mention of the functions L and R as I have them defined above. Both seem perfectly reasonable, don't you think? And they do seem to implement a binary tree as you requested with no apparent contradictions arising from them  none that you have been able to demonstrate anyway.
Dan
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