Re: Cantor and the binary tree

In article <1119540259.633206.243350@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

> Jan de Vos wrote:
> > In sci.math, mueckenh@xxxxxxxxxxxxxxxxx wrote:
> > > Each edge is mapped on that bunch which contains that edge. If this
> > > bunch splits then the resulting bunches get the respective edge (where
> > > they differ) while the former egde is dropped (no longer regarded at
> > > all - thrown away because there are enough). In this way it is fairly
> > > clear that every bunch has an edge of its own. This is not restricted
> > > to finite bunches (which do not exist in my tree) but to all bunches
> > > which represent intervals of numbers with an infinite binary
> > > representation. Hence, all non-empty subsets of numbers (ncluding
> > ^^^^^^^^^^^^^^^^^^^^^
> >
> > Not all non-empty subsets are bunches. There are no bunches that
> > contain only a single path.
>
> If they are not, then there are no single real numbers, at least they
> are lacking binary representation with a countably infinite number of
> bits.

Such a wild claim requires proof.

In fact the "bunch" of any node corresponds to the set of binary strings
that agree up to the corresponding digit position and include all
possible binary continuations beyond it. Since all such sets of strings
are uncountable, so are the corresponding sets of paths.
> >
> > I agree that you can make a bijection between the bunches (where a
> > bunch is defined as the set of all paths from a single node) and the
> > nodes of the tree. That is trivial.
> >
> > But I DIDN'T ask for such a bijection. I asked for an injection from
> > the set of paths into the set of nodes.
>
> We have a surjection from the set (better: from a subset) of edges on
> the set of bunches of paths.

Which carefully misses the point, The issue is not how many bunches, but
how big any one bunch is. Counting bunches does not answer that question.

Pick any bunch and count the paths in that bunch!

To "count" the paths in a bunch:

Since there is an easy bijection from the paths in any one bunch to the
paths in any other bunch, all bunches are of the same size, and since
there is an easy bijection from any bunch to P(N), all bunches are the
same size as P(N).

Each bunch consists of all paths passing through some node, so that it
is equivalent to a maximal binary tree with that node as root node.

For such a maximal binary tree, each path determines a suset of N ( or a
mamber of P(N)) as follows: counting branches (edges) on a path from the
root node, n is in the set if and only if the nth branch is a right
branch. An equivalent, though different, mapping results from including
n for an nth left branch and excluding n for an nth right branch.

It is easy to see that each path determines a unique subset of N (member
of P(N)) and that each subset of N (member of P(N)) is dertemined by a
unique path of the tree.
.

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