Virgil
Posts:
4,482
Registered:
1/6/11
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Re: Matheology � 198
Posted:
Jan 25, 2013 7:46 PM
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In article
<3e2a64f8-3cf0-478a-8411-2db261dbc8d5@u7g2000yqg.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 25 Jan., 09:20, William Hughes <wpihug...@gmail.com> wrote:
> > On Jan 25, 9:11 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 25 Jan., 09:07, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Jan 25, 8:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > "Write a sequence like xxxxxxxxxx... and do never stop"
> >
> > > > Perfectly acceptable finite description of an infinite sequence.
> >
> > > Of course, nothing however that could be diagonalized in a Cantor-list
> > > or could be the result of diagonalization.
> >
> > Why not? The diagonalization of a list with finite description is a
> > finite
> > desciption.
>
> The diagonalization of finite descriptions does not necessarily supply
> a description at all.
Maybe not in Wolkenmuekenheim, but then no one but WM ever works there.
> >
> Of interest is this: If the same set of
> nodes has to describe both, the Binary Tree with finite paths and that
> with infinite paths, then it is impossible to discern, alone by nodes,
> whether we work in the former or the latter.
There is no such thing as a Complete Infinite Binary Tree with finite
paths.
At least not according to any standard definition of Complete Infinite
Binary Trees.
A path in any binary tree of the sort being considered, a Complete
Infinite Binary Tree, is, by definition, necessarily maximal, in that no
node can be appended to any path and still have a path in that tree. and
so none of WM's finite chains of nodes in a CIBT can be a maximal
chain, thus also never a path, as there is always a child of its last
node which can be added to it making a longer but still finite chain of
nodes.
> Hence I can assert
No you cannot, since your basic assumptions about Complete Infinite
Binary Trees are proved false.
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