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