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.