Re: Cantor Confusion

In article <1166361248.951652.84540@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Virgil schrieb:

In article <1166185295.562736.84050@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Virgil schrieb:



No I simply state that the diagonal proof fails in case of the tree.

AS the diagonal proof is not allied to trees, so what?

Its contradiction is.

Show me!

If you believe in a diagonal longer than any line of the matrix and in
paths which have no beginning, then I can't show you anything.

You have things backwards, as usual. The reason that you cannot show me
things is that you have not presented any axiom system on which to base
them. When you want to prove something, you merely pull assumptions out
of the air to serve as justifications.

You cannot show me things within in ZFC or NBG that require you to
assume things that contradict the XFC or NBG axioms because in an axim
system you are not allowed to make assumptions that contradict the
axioms already assumed.

A minimum of logic is required for any arguing.

And WM does not have that minumum.

Every finite beginning of a set of paths, however long, is the beginning
of as many paths as the set of all paths.

So there are as many paths following any such beginning as there are
paths altogether.

And there are as many beginnings following such a beginning.

if a beginning is either a node or an edge, there are fewer beginnings
than unending paths from such a beginning in an infinite binary tree.

So, most of the paths have no beginning?

Uncountably many paths have the same beginning, at least for any finite
beginning to any paths.

Or where, do you think, they
begin as separated entities?

The single root node is the beginning of all paths.

The finite sub-path from the root node to any other node is the
beginning of just as many paths.

Therefore the beginnings of separated parts of the paths are
countable.


It is the endings that are not countable.

It is the endings, which do not exist.

Each path has a head node and a tail consisting of everything else.

The number of heads in an infinite binary tree is countable, the number
of endless tails from any node is equal to the number of endless binary
strings which is not countable.

So how many tails has one head?

Uncountably many tails to each head, where a head is the finite sub-path
from the root node to any node.

And where begin these mysterious uncountably many tails?

At the end node of their head.

For you it is not a problem that there are more tails than heads and
simultaneously that every head has one and only one tail?


Who said a head could only have one tail?

By my definitions, a "head" is a sequence of nodes and branches from the
root node to any end node and a "tail" is the endless subpath starting
at the end node of some head.

Thus every path is the tail for the root node as head.
For every head, there is a natural bijection of its tails onto the set
of all paths.

So that the set of tails for any head is always as uncountable as the
set of all paths.
.

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