Date: Jan 12, 2013 2:22 PM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 191
On 12 Jan., 19:34, Zuhair <zaljo...@gmail.com> wrote:

> On Jan 12, 3:26 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

>

>

>

>

>

> > On 12 Jan., 12:45, Zuhair <zaljo...@gmail.com> wrote:

>

> > > On Jan 12, 11:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:

>

> > > > Matheology § 191

>

> > > > The complete infinite Binary Tree can be constructed by first

> > > > constructing all aleph_0 finite paths and then appending to each path

> > > > all aleph_0 finiteley definable tails from 000... to 111...

>

> > > No it cannot be constructed in that manner, simply because it would no

> > > longer be a BINARY tree.

>

> > No? What node or path would be there that is not a node or path of the

> > Binary Tree? This is again an assertion of yours that has no

> > justification, like many you have postes most recently, unfortunately.

>

> Notice also that one can have a COUNTABLE tree (i.e. a tree that has

> countably many paths and nodes) that has finite paths

> indistinguishable from the finite paths of the complete binary tree by

> labeling of their nodes.

I noticed that already some years ago.

>

> Let me show an example at finite level, take the three level binary

> tree:

>

> 0

> / \

> 0 1

> / \ | \

> 0 1 0 1

>

> I can have the following tree:

>

> 0

> / / | \ \ \

> 0 1 0 0 1 1

> | | | |

> 0 1 0 1

>

> Hope that helps.

No, you have not understood. By attaching one or more infinite tails

to every finite path the Binary Tree is not changed in any discernible

way. The path have exactly the nodes that belong to the tree. The only

difference is that infinitely many paths cross each node. But even

that is not really a difference, because it was the case in the

original Binary Tree too. Try to find out what the reason is.

Regards, WM