Date: Jan 13, 2013 7:13 AM Author: Zaljohar@gmail.com Subject: Re: Matheology § 191 On Jan 12, 10:22 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

> On 12 Jan., 19:34, Zuhair <zaljo...@gmail.com> wrote:

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

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

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

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> > > > > Matheology § 191

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> > > > > 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...

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

> > > > longer be a BINARY tree.

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> > > 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.

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> > 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.

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> I noticed that already some years ago.

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

> > tree:

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> > 0

> > / \

> > 0 1

> > / \ | \

> > 0 1 0 1

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> > I can have the following tree:

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> > 0

> > / / | \ \ \

> > 0 1 0 0 1 1

> > | | | |

> > 0 1 0 1

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> > Hope that helps.

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> 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.

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> 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