Date: Apr 1, 2013 6:14 PM
Author: William Hughes
Subject: Re: Matheology § 224
On Apr 1, 10:54 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

> On 1 Apr., 15:19, William Hughes <wpihug...@gmail.com> wrote:

>

>

>

>

>

>

>

>

>

> > On Mar 30, 3:36 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

>

> > > On 30 Mrz., 10:17, William Hughes <wpihug...@gmail.com> wrote:

>

> > > > On 24 Mrz., 18:09, WM <mueck...@rz.fh-augsburg.de> wrote:

> > > > <snip>

>

> > > > > > The only difference is that in the second case you consider

> > > > > > some subsets of the nodes to be paths, that are not considered

> > > > > > to be paths in the first case.

>

> > > > > Well, that is a correct description. It implies that these additional

> > > > > subsets cannot be distinguished by nodes from the finite subsets

>

> > > > Piffle. It is trivial to distinguish a subset that has a node

> > > > at a last level from a subset that does not have a node

> > > > at a last level.

>

> > > No, that is impossible if an infinite path consists of infinitely many

> > > finite subsets.

>

> > Let the subset of nodes in the infinitely many finite subsets

> > be Q.

>

> > Q is contained in both trees, is not a path

> > in the Binary Tree that contains only all

> > finite paths (Q does not have a node at

> > a last level)

>

> The path to which all finite paths

> 0.1

> 0.11

> 0.111

> ...

> contribute is a path too in the Binary Tree that contains all finite

> paths.

Nope. This set of nodes has no node at a last level.

Every path in the Binary tree that contains all finite paths

has a node at a last level.

The difference between the trees is not which

subsets of nodes exist, but which subsets are

considered to be paths. Only in one of the trees

can a subset of nodes with no node at a last

level be considered a path.