Re: Two results of set geometry

On Sep 28, 7:14 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:
On 27 Sep., 16:47, "*** T. Winter" <***.Win...@xxxxxx> wrote:

In article <1190633128.867363.104...@xxxxxxxxxxxxxxxxxxxxxxxxxxx> WM <mueck...@xxxxxxxxxxxxxxxxx> writes:
> On 20 Sep., 15:14, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> > Each and every node in the mapping of WM
> > maps to either paths terminating in a continuous stream of 0's or in
> > a continuous stream of 1's. 2/3 is not such a number. The 11-th node
> > on the path from the root of 0.1000... is mapped to:
> > 0.100000000000111...
> > if I understand WM's mapping correctly.

> You do. And it is easy to understand that every node of 2/3 is
> included in the mapping. Therefore there is nothing of 2/3 which is
> not included.

Except that it is not in the range of the mapping because there is no
node that maps to that number.

There is a bijection between initial segments of the column and the
rows in

No. There is a bijection between the *finite* initial
segments of the column and the rows in

1
11
111
...,
not with the full column, because the full column has no element which
was not in a finite segment. For the numnber 2/3 in the tree this
situation is different?


In the matrix case we have

There is a bijection between the set of
finite initial segments,
elements are finite initial segments, and the set of
rows, elements are rows.

In the matrix case the finite initial segments
are elements of the domain.

In the binary tree case, your putative bijection
is between the set of nodes, elements are nodes
and the set of paths, elements are paths.

In the binary tree case, your putative bijection
deos not have finite initial segments as elements of the domain
or as elements of the range,

- William Hughes


.

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