```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:>>>>>>>>>> > 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, WMRead what I wrote to you. What you are saying is what I noted asA_CIBT (go and read my post to remember what that means), and Ialready 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 ofA_CIBT. And since we already have Uncountably many paths in the CIBT,then of course well have also Uncountably many paths in A_CIBT, simplybecause all paths of CIBT are paths of A_CIBT. So your argument failswith the case of A_CIBT.The really countable tree that have all its paths indistinguishablefrom SOME paths of the CIBT is actually the A_CIBT* (review my earlierresponse to you to remember what that is). But however there areUNCOUNTABLY many paths of the CIBT that are distinct (up to somefinite n position) from ALL paths of A_CIBT* You just cannot escapeuncountability.Zuhair
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