Re: Cantor Confusion

In article <1175104726.767746.12720@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
On 26 Mrz., 16:58, "*** T. Winter" <***.Win...@xxxxxx> wrote:
....
But that is not precisely the same. You want me to reverse a series that
does not terminate. I ask you how I do that, and you refrain to answer.

> Cantor, unless working every digit simultaneously, will never finish.
> In fact we need not know how to proceed, but it is sufficient to know
> hat the sum remains 2 what ever we do.

No, the sum is undefined. If you think it is defined, *prove* it and
*prove* that the sum is 2. What is the sum of the "sequence":
"..., 1/4, 1/2, 1"?

Well, what is it? If any infinite series ever had a value, then the
sum of this sequence is 2.

But it is not a sequence according to mathematical definitions.

Proof: Sum the first n terms from the right hand side and find that
the difference of their sum and 2 is not more than 1/2^(n-1). Further
find that the sum is always less than 2. Voila.

Ok, what node contributes '1' to that sum? I ask this because you want it
to apply to nodes contributing things to a path.

> However, you may proceed as follows (but only after having read my
> book): By transpositions (as we know them from Cantor's work) bring
> the n-th term (with n = 2, 3, 4, ...) into the first position and
> after that bring the n+1-th term into first position. If you are
> really fast, then you reverse the whole sequence of terms (because you
> can determine, for *every* term number n, the number of transpositions
> required to have it at the first position).

Yes, and when are we done?

When is Cantor done?

Well, at each step in the reversal process you have a sequence with a first
element and no last element. I do not know of a way to define what the
result is after infinitely many steps. What is the first element after
infinitely many steps? And how do you define that at all? Or is your
definition simply that it is {..., 3-rd term, 2-nd term, 1-st term}? In
that case the result is not a sequence and you have to redo quite a bit
of mathematics before you can draw any conclusions about it.

The problem with this is that there is a
dependency in the order of transpositions, so they can not be performed
simultaneously.

Cantor's work cannot be done simultaneously. You are in error. In
order to exchange a_n you have to find n. That is done by counting.
You cannot point to n unless you have pointed to n-1.

The *exchanges* can be done simultaneously. This can *not* be done in
your process.

Ever tried to count the students attending a lesson or the peas in a
cup full of peas?

Why should I? To perform the n-th exchange in Cantor's process you do
*not* have to do the first n-1 preceding exchanges first. Or do you think
you can do your exchanges all at once?

> I am arguing within the tree. There is never an uncountable number of
> separated paths. You say all paths were uncountable and separable.
> Conclusion. there is no "all paths" within the tree.

You have to first *prove* your statement that there is not an uncountable
number of separated paths. Until now your proofs depend crucially on the
assertion that there is not an uncountable number of separated paths, and
so are circular.

Wrong. The proof depends on the fact that the set of nodes is
undisputedly countable and only nodes are points of separation. A very
simple logical conclusion, in principle.

So, please, prove it, using simple logic.

> Therefore, if they existed isolated and were so many as you say, they
> would populate a level with uncountably many nodes.

Why? Each two paths are isolated from each other (by your definition), so
all paths are isolated from all other paths. *But* there is *no* level
where a single path is isolated from all other paths.

There is no *finite* level where all paths are separated. But the tree
is *infinite* and it contains all levels where something could happen.
A path which is not separated form another one within the tree is
never separated.

Yes, you are not contradicting what I state. Each path is separated from
each other path.

> > Well, within set theory it can be proven.
>
> Within the tree it can be disproven.

Until now you did not succeed, because your proofs either contain a denial
of the axiom of infinity or other logical flaws.

Wrong.

Right.

> That you do believe in set theory is not a crime. But that you do
> believe in uncountaly many separations without uncountably many
> separations, that is hard to believe.

I do not. You think I do, but that is due to a lack of logical reasoning.

The lack of logical reason is as follows. You say:
For every pair of path, there is a node, where they separate. There is
no node where all have separated.

I do *not* state that because that would be nonsensical. If you change
"node" in the last part to "level" I would agree. There is *no* node
where more than two paths do separate.

You should say, however:
For every pair of path, there is a node in finite distance from the
root node, where they separate. There is no node in finite distance
from the root node where all have separated.

The same problem here. But if you correct again the "no node" to "no
level" I would agree, and I have stated such already quite some time.

And you should recognize
that the binary tree is infinite. Therefore, everything that happens
even *after* any finite distance from the root node, nevertheless
happens in the tree - unless it does not happen at all.

Nothing happens after any finite distance. There is *no* level where
all paths are separated from each other. Why you think that there should
be such a level escapes me.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.