Date: Jan 12, 2013 1:34 PM
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, 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
/ \ \ \ .. \
0 1 1 1
| | | |
0 0 1 1
. | | |
. 0 0 1
. . | .
. . 0 .
. . . .
. . . .
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
/ \ | \
0 1 0 1
I can have the following tree:
/ / | \ \ \
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
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
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
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
Hope that helps.