Date: Jan 13, 2013 7:48 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 191

On 13 Jan., 13:13, Zuhair <zaljo...@gmail.com> wrote:
> On Jan 12, 10:22 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>

> > 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.


> Read what I wrote to you. What you are saying is what I noted as
> A_CIBT (go and read my post to remember what that means), and I
> already told you that A_CIBT has a different structure from the CIBT,


And that is wrong. There are all the nodes and all the edges, hence
all the paths of the complete infinite Binary Tree - and nothing more.

> but I also noted that with the case of A_CIBT the CIBT is a SUBTREE of
> A_CIBT.


Both are improper subtrees of each other.

> And since we already have Uncountably many paths in the CIBT,
> then of course well have also Uncountably many paths in A_CIBT, simply
> because all paths of CIBT are paths of A_CIBT. So your argument fails
> with the case of A_CIBT.


No, my argument cannot fail unless there are nodes beyond any finite
level, what you called level aleph_0. But of course that is nonsense.
As you don't repeat it, you may have learned it meanwhile.
>
> The really countable tree that have all its paths indistinguishable
> from SOME paths of the CIBT is actually the A_CIBT* (review my earlier
> response to you to remember what that is). But however there are
> UNCOUNTABLY many paths of the CIBT that are distinct (up to some
> finite n position) from ALL paths of A_CIBT* You just cannot escape
> uncountability.


There are not uncountably many finite (initial segments of) paths. And
also any anti-diagonal can only differ from other paths in its (and
their) finite initial segments. Unless your silly idea of nodes at
level aleph_0 was correct (it is not) there is no chance to differ at
other places than finite (initial segments of) paths. But that is
impossible if all of them are already there. And the latter is
possible, because they form a countable set.

Stop claiming uncountability unless you can find a node that make a
finite initial segment actually infinite.

Regards, WM