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

In article
<a1bebea8-801c-4ca5-9d5f-2b87f28f3a74@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
WM <mueckenh@xxxxxxxxxxxxxxxxx> wrote:

On 26 Mrz., 23:17, "calvin.ost...@xxxxxxxxx" <calvin.ost...@xxxxxxxxx>
wrote:
On Mar 26, 12:45 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

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.

No, a tree is a set of edges. To specify the tree, you must
specify the edges. An edge is essentially a set of two nodes.
You must specify both nodes. To specify a tree, you must
give a set of nodes N and a set of edges E in NxN. That is
just what a tree is.


Look here: To easen understanding enumerate he nodes by numbers:

0
/ \
1 2
/ \ / \
3 4 5 6
...
The edge leading to node number 3 is completely determined by node
number 3. This edge starts at node number 1. There is no other
possibility.

More mathematically, the left child of node n is node numbered 2*n+1 and
the right child is numbered 2*n+2.
So that 'n == 1 mod 2' indicates a left child node and 'n == 0 mod 2' a
right child node.

But if you had any understanding
of what you were doing you would have understood that.

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.

Why not just p(n), using the numbering shown above? Since the nodes are
countable, assign each one a natural as indicated above and simplify
everything.

No, p(i) is the path you add at stage i. What is this path,
precisely, you have never said and still don't say and just
say nonsense instead.

If you want to call that number, belonging to (x, y), i then you can
do that. It does not change anything. The path runs from the root node
to node number i and then always turns left.

Which omits the vast majority of 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.

NO. Geez. Because the digital strings in each number in Cantor's
list do not interact in the way your path's do.

All you do in Cantor's list is to search lines 1 to n whether the
diagonal number is in. It is not. What a surprise! You search a finite
initial segment of the list, leave out the infinite rest, and then
claim that the number could not be found. Perfect!

If that is how WM misreads the Cantor proof, no wonder he is so screwed
up.

The diagonal number
is the only one that is made this way, and it is not in the list. But
in your tree, most of the paths are made by selecting one edge
from each of a number of the paths you have added. In effect
the process itself automatically creates an uncountable number of
something like diagonal numbers, and sticks them in the tree
without you being able to do anything about it.

No.

Wm is in denial again.

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

You omitted an account of how the path you denote
as .11111.... gets in the tree

After the infinite tree including all its nodes has been completed,
the nodes making up the path 0.111... are obviously in the tree.

But it appeared nowhere in WM's construction so must be an artifact of
the completion and not of the construction itself. as are all the others
with infinitely many 1's in them.

And
all path ending by a sequence of 111... are also there. You can even
deconstruct the tree by removing the countable set of all path that
end by 111...


Therefore there are two explanations possible:
Either, after finishing infinity breathtaking things happen that are
out of our control.

It is in the invisible gap between unfinished and finished that these
things happen.

Or, there is no finished infinity.

If WM continues to believe this, then he is in no position to argue the
details of a complete infinite binary tree or anything else involving
completed infinities.



No path is in the tree unless it has been added.

But most of them are "added" by the completion, which, by his own
choice, is out of WM's control.
.

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