Re: The complete infinite binary tree has only countably many infinite paths.

On 26 Mrz., 11:58, "calvin.ost...@xxxxxxxxx" <calvin.ost...@xxxxxxxxx>
wrote:
On Mar 26, 5:48 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

0  [say instead (0,
Agreed. But no path ends there. So it is not important.

It provides a uniform naming scheme that makes it
much easier to make general statements about
the tree.

Can you rigorously specify the edges that are being added, in
a manner such as I have done above (instead of just with pictures
and informal words).

As well as labelling the nodes you can also label the edges leading to
the nodes by the same scheme.
(1,1), (1,2)
(2,1), (2,2), (2,3), (2,4)
  ...
 (n,1), ..., (n, 2^n)

So use just
the edges  {(n,m)| n,m in N, m<=2^n}

No...

You know, a tree consists of a set of nodes, and a set of
edges that join the nodes.   We only name the nodes so
that we can separate them out from one another in order
to specify the edges.   So, to start with, I tried to get you
to name the nodes in a way that would be convenient for
later reference when the edges were added.   So far so
good. However, those names strictly speaking said
nothing about the structure of the tree.   All the real work
comes in giving the edges.   That is what ultimately
defines the tree.

But have you done that?  No.  You merely *named*
the edges.   That is not what you must do.   You must
say what the edges are in terms of what nodes they
connect.  Edges are a subset of the cartesian product
of the set of nodes with itself.
So, please continue, and specify what your tree is by
saying what the set of edges are, as a subset of NxN,
where N is the set of nodes defined above.

Every edge leads to one and only one node. If we call the edge by E(x,
y) where (x, y) is the node, then the edge is uniquely defined. There
is not much work to do. Further if we work with edges, then we need
not work with nodes at all. Therefore even the E can be dropped. Why
do you see problems where they are not?


Then all the paths (except  p_0) can be labelled by their last node
that has value 1.

Here, I think, is your problem.  You are not noticing that when
you add a single path, you are actually adding many paths.

What many of paths do I add when I add 0.1000...?

What I said here was based on trying to understand your
account, and I did not.   It is true that if
you add paths only, you are adding only one
path at a time.    But you will not add all paths
by doing that.

Before goning on, please anwer the question: Do you agree that I
construct all the nodes respectively all the edges of the tree by my
prescription?

  The only way to add  all paths is to
add edges one at a time in the right way.  You will never get
all paths by trying to add a countable number of infinite
paths.   You cannot construct the tree by adding paths.

Why not?

You must construct it by adding edges. And in the right
way.   You will never add a single infinite
path at any time.   Only in the final tree will there be
infinite paths.  It all sounds very paradoxical, must
drive you crazy.   But the point is this:  the final tree
has infinite paths in it because these paths are sets
edges that such that each edge is added at some
*different* finite stage, not along with all its pathmates
all at once.  They are added one edge at each of stage
S(i), for some strictly increasing function S of i.  So notice:
these infinite paths do not occur in any of the finite trees made
during the construction, because no such tree has all the edges
required!  Only the final tree has all the edges required.

Therefore I do not add edges but infinite paths in my proof (A)

Infinity is tricky.

Yes. You can construct the complete tree by all path that end by
zeros. And you can deconstruct it by all paths that end by ones.
You can construct it by all paths that end by the sequence of pi. Or
you can construct it by path that end by whatever you choose from time
to time.
But you cannot construct it by using all paths that represent all real
numbers of the unit interval unless you use some paths more than once.
All paths simply do not fit into the tree. Nevertheless they are in
it, after infinity has been finished (and given they all exist).

In any case, it is probably quite possible to define the
final tree not as a limit, but directly, by clearly specifying
the set of nodes as I have asked you to do.  Once we
have all tree defined, by saying what its edges are,
we can then look to see which paths are in it, or not.

If the tree has all nodes, after the end of construction, then there
are all paths.

Regards, WM

.

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