Re: Cantor Confusion

In article <1175177285.625501.231650@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:



It is a sequence according to mathematical definitions, if you read it
from the left hand side.

But since nodes are, like points, that which has no part, anything
requiring nodes to have parts is fallacious.

For example, which "part" of a node has the left child and which part
the right child? Or wold you, like Solomon, command the children to be
split.


Further you can determine a unique limit value by lim{n-->oo} (1/2^n
+ ... + 1/8 + 1/4 + 1/2 + 1}.


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

You know, this node cannot be determined.

Then the whole thing is a chimera.


Do you know of a way how to finish Cantor's diagonal?

When the rule is stated, the diagonal is finished.


Or do you think
you can do your exchanges all at once?

I do not think it, I know it. It is ridiculous to claim you could do
infinitely many exchanges simultaneously.

If the order in which operations are to be performed /does/ affect the
result, one cannot perform them simultaneoulsy, as in WM's rule for
permuting positions of terms in an infinite sequence.

If the order in which operations are to be performed does not affect
the result, one can perform them simultaneoulsy, as in Cantor's rule for
determining the digits of the "anti-diagonal".

That WM cannot see the difference is a measure of his mathematical
incompetence.


> > > 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.

1 + 1 = 2. Good so?

Which proves nothing, not even that 1 + 1 = 2.

> > > 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.

Therefore we know that only countably many are there.


Is that a royal "we"or an editorial "we"? It is certainly not plural.



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.

It is very simple. Every digit of a real number is indexed by a finite
natural number.

Yet the number of such real numbers exceeds the number of natural
numbers.

The set of paths in an infinite tree is like the set of binary infinite
sequences, and both are uncountable.

There are numerous proofs of both of these facts, and WM has not ever
been able to fault any of these proofs in ZFC or NBG, or in any system
which does not presume a priori that no such things as uncountable sets
can exist.

So that WM is not arguing about any mathematical system, but only some
private system of his own which he will not, or can not, specify.
.

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