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, WM
The binary tree has a well known structure: each node at each level is attached to TWO nodes below, while in your construction this is not the case: for example let me take the finite path 011 Now I'll append to it all definable tails from 000... to 111...., so I imagine that in the following way
Clearly we do have Aleph_0 of finitely definable tails from 0000... to 111.., All of those are appended to each node and accordingly each node would be attached to Aleph_0 of nodes at next level, while with the 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 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.
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
Notice that the paths of this tree are all undistinguishable from the paths of the above binary tree by labeling of their nodes, both trees are three level trees, both has exactly the same paths as distinguished by labels of their nodes, But yet they are DISTINCT trees, they do have different structure, here in this case the total number of distinguishable paths by labeling of nodes is equal, both are 7, however the difference in structure is OBVIOUS, the later tree is not binary in the sense that it is not the case that each node is attached to exactly two nodes below it.
I call the second tree as the UNFOLDED tree image of the three level binary tree.
Now lets take the CIBT (Complete Infinite Binary Tree), and lets construct the UNFOLDED tree image of it, lets denote the later as CIBT*. Now we know that ALL paths of CIBT* are finite! So we know for sure that the total number of paths of CIBT* is countable! BUT the condition is not the same for the CIBT, not all paths of the CIBT are finite, some of them are infinite.
Now Imagine that I took the CIBT* and to each Path of it I'll append Aleph_0 of tail paths that are finitely definable from 0-0-0-.. to 1-1-1-.... and lets call the resulting tree the A_CIBT* short for Appended CIBT*.
Now A_CIBT* is definitely COUNTABLE, i.e. has countably many paths and nodes. But however A_CIBT* has a structure that is altogether different from the CIBT which have UNCOUNTABLY many paths. The CIBT is NOT a substree of the A_CIBT*. The CIBT has MORE paths that does the A_CIBT*.
On the other hand lets take the CIBT itself and then append to each node of it an Aleph_0 of finitely definable tails from 0-0-0... to 1-1-1-.... The resulting tree would be designated as A_CIBT, short for Appended CIBT. Now this tree is also not a binary tree since each node is appended by aleph_0 of nodes below it while in the CIBT each node is attached to two nodes below it. However A_CIBT has the CIBT as a SUBTREE of it, and thus all paths of the CIBT are paths of it whether finite or not. However since the CIBT has UNCOUNTABLY Many paths, then the A_CIBT does also have UNCOUNTABLY many paths, because it is has the CIBT as a substree of it. This means that the CIBT has MORE paths that are distinguishable by labeling of their nodes than do the A_CIBT* have.