Re: Cantor and the binary tree

Virgil said:
> In article <MPG.1d22309d742aee97989e75@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
>
> > Martin Shobe said:
> > > On 20 Jun 2005 04:09:53 -0700, mueckenh@xxxxxxxxxxxxxxxxx wrote:
> > > >> This no more proves countability of the set of paths than the fact that
> > > >> lists of bits can represent all reals in [0,1] proves the reals
> > > >> couontable. The set of all such lists, like the set of all those maximal
> > > >> paths, is uncountable.
> > > >
> > > >By construction of the tree we see that all binary representations
> > > >which differ at a finite position are represented by bunches (=
> > > >non-empty subsets) of paths in the tree. Further we see that they can
> > > >distinguish themselves only at nodes or branches. The occasions to do
> > > >so are countable.
> > >
> > > What is it you are smoking? Can I have some? Even in finite binary
> > > trees, the number of paths is greater than the number of nodes where
> > > branching occurs.
> > >
> > > Martin
> > >
> > >
> > Excuse me Martin, but maybe you should have some of what I am smoking. Every
> > path ends in a leaf node, which are half the nodes in the tree. You start
> > with
> > one node that represents the root path. For each pair of nodes, you create a
> > new path. A finite tree with n levels (including the root) has (2^n)-1 nodes,
> > (2^n)-2 branches, and only 2^(n-1), or (2^n)/2 paths, as denoted by its leaf
> > nodes. This relationship is preserved through infinity, even in the absence
> > of
> > identifiable leaf nodes.
>
> TO is right for finite trees but wrong for maximal binary trees.
No, this property holds for all balanaced binary trees.
>
> For any finite tree there is at least one more node than there are
> paths. If every path shares the root node, eachfinite path will also
> have a leaf node, so there is at least one extra node.
Incorrect. There is one more node than branches, and twice as many branches as
paths.
>
> TO's mistake is to presume that what happens for finite trees must also
> be the case for (infinite) maximal binary trees. It is a presumption
> that can, in fact, be disproved, and has been disproved in these threads.
No it has not. Your various interpretations of the branches in the tree don't
prove things about the structure of the tree itself, and your mishmoshed
judgement of "countability" doesn't hold water for me anyway. Have you ever
actually USED a binary tree for anything? I mean, weren't you the one that told
me one can't insert a node in the middle of the tree, when computers do it
every single day? Drop the bijections, and inspect the structure of the tree.
>

--
Smiles,

Tony
.

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