```Date: Jan 12, 2013 1:34 PM
Author: Zaljohar@gmail.com
Subject: Re: Matheology § 191

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.>> Regards, WMThe binary tree has a well known structure: each node at each level isattached to TWO nodes below, while in your construction this is notthe case: for example let me take the finite path 011 Now I'll appendto it all definable tails from 000... to 111...., so I imagine that inthe following way0|1 \  1 / \ \ \ .. \0 1 1     1|  |  |      |0 0 1     1.  |  |      |.  0 0     1.  .  |      ..  .  0     ..  .  .      ..  .  .      .Clearly we do have Aleph_0 of finitely definable tails from 0000... to111.., All of those are appended to each node and accordingly eachnode would be attached to Aleph_0 of nodes at next level, while withthe binary tree each node is only attached to TWO nodes below it.Notice also that one can have a COUNTABLE tree (i.e. a tree that hascountably many paths and nodes) that has finite pathsindistinguishable from the finite paths of the complete binary tree bylabeling of their nodes.Let me show an example at finite level, take the three level binarytree:    0   /  \  0   1 / \   | \0 1  0 1I can have the following tree:       0 /  /  | \ \  \0 1  0 0 1 1       |  |  |  |       0 1 0 1Notice that the paths of this tree are all undistinguishable from thepaths of the above binary tree by labeling of their nodes, both treesare three level trees, both has exactly the same paths asdistinguished by labels of their nodes, But yet they are DISTINCTtrees, they do have different structure, here in this case the totalnumber of distinguishable paths by labeling of nodes is equal, bothare 7, however the difference in structure is OBVIOUS, the later treeis not binary in the sense that it is not the case that each node isattached to exactly two nodes below it.I call the second tree as the UNFOLDED tree image of the three levelbinary tree.Now lets take the CIBT (Complete Infinite Binary Tree), and letsconstruct the UNFOLDED tree image of it, lets denote the later asCIBT*. Now we know that ALL paths of CIBT* are finite! So we know forsure that the total number of paths of CIBT* is countable! BUT thecondition is not the same for the CIBT, not all paths of the CIBT arefinite, some of them are infinite.Now Imagine that I took the CIBT* and to each Path of it I'll appendAleph_0 of tail paths that are finitely definable from 0-0-0-.. to1-1-1-.... and lets call the resulting tree the  A_CIBT*  short forAppended CIBT*.Now A_CIBT* is definitely COUNTABLE, i.e. has countably many paths andnodes. But however A_CIBT* has a structure that is altogetherdifferent from the CIBT which have UNCOUNTABLY many paths. The CIBT isNOT a substree of the A_CIBT*. The CIBT has MORE paths that does theA_CIBT*.On the other hand lets take the CIBT itself and then append to eachnode of it an Aleph_0 of finitely definable tails from 0-0-0... to1-1-1-.... The resulting tree would be designated as A_CIBT, short forAppended CIBT. Now this tree is also not a binary tree since each nodeis appended by aleph_0 of nodes below it while in the CIBT each nodeis attached to two nodes below it. However A_CIBT has the CIBT as aSUBTREE of it, and thus all paths of the CIBT are paths of it whetherfinite or not. However since the CIBT has UNCOUNTABLY Many paths, thenthe A_CIBT does also have UNCOUNTABLY many paths, because it is hasthe CIBT as a substree of it. This means that the CIBT has MORE pathsthat are distinguishable by labeling of their nodes than do theA_CIBT* have.Hope that helps.ZuhairZuhair
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