|
|
Re: Matheology § 191
Posted:
Jan 13, 2013 7:13 AM
|
|
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.
>
> Regards, WM
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,
but I also noted that with the case of A_CIBT the CIBT is a SUBTREE of
A_CIBT. 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.
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.
Zuhair
|
|
