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

On 26 Mrz., 13:48, "calvin.ost...@xxxxxxxxx" <calvin.ost...@xxxxxxxxx>
wrote:
On Mar 26, 7:57 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

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?

It is not what you *call* the edges, it is what they *are*.  An
edge is essentially something between two nodes.  So, to
specify *what it is*, you must give the two nodes.

No. The goal node is sufficient.

You don't
do that, do you?  You just say, E(x,y) is what we call the edge
between x and y.  Fine, but what are x and y?  They are just
formal names.   You must give the set of {(x,y)}, a subset
of NxN, in order to even define the tree.  The problem I see
is that you haven't defined the tree at all here.

x is the level number. It runs from the root level 0 through the
naturl numbers.
y = 2^x. Where is the problem?

(This is assuming, again, what we should be able to do,
define the tree T in one shot, as a set of edges between
nodes, rather than building it up as a limit, which you have
been trying to do all along, but which is leading
into confusion)

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?

Above, you have said *nothing* about the tree.   Let us
look yet again at the limit method.   You start with
an empty tree, or maybe a tree containing the
edges for the path you call 0.0000..., and at each step
you add another infinite path to tree T(n) to get
tree T(n+1) (you do not say clearly
which path).   Then, I believe, you say that T =
the union of the T(i), i in N.   (Or more simply,
if the infinite set of edges added at step i is p(i),
the tree then is union p(i), in in N.

This is not a complete specification until you specify
precisely what each p(i) is.

p(x, y) = is the path that starts at the root node and goes to the yth
node in the xth level and subsequently ends with zeros.

   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?

Because there are an uncountable number of paths, and
the process has only a countable number of steps, one
for each i in N.

So all nodes will be covered.

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

Right.  You added explicitly only a countable number of
paths, but there are an uncountable number that sneak
their way in by the end,

If you believe that this can happen in mathematics, why then don't you
believe that in Cantor's list all real numbers can sneak in after it
has been completed? Of course every line number n that you search does
not contain the diagonal number. But after completion it sneaks its
way.

Concluding: If you allow legerdemain in the tree, then you must also
allow it in Cantor's list. Then uncountability is unsubstantiated
nevertheless.

Regards, WM

.

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