Re: Cantor Confusion

In article <1168445349.094021.127750@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
*** T. Winter schrieb:
....
> A ghost path is a path which does not exist in a tree which contains
> the paths of all real numbers the bits of which can be indexed by
> natural numbers.

That is something I do not understand. Please elaborate.

See my recent posting. You need to assume that the complete union of
levels of the binary tree does not contain all the paths representing
real numbers. Therefoer something must be added, in order to get the
complete set of paths. But there is nothing to be added - but ghosts.

Oh. But depends on what is done. But even if you unite two finite levels,
the set of paths in the union is not the union of the sets of paths of
the separate levels.

> All path the bits of which can be indexed by natural numbers are in the
> union of all finite trees.

But *not* in the union of the sets of paths in the finite tree.

Yes. The union of all finite trees is not an (actally) infinite tree.
See my due explanations of the union of all lines of the EIT:

And again you *fail* to argue with the union of sets of paths in mind.
But that is something that you actually do use when you are talking
about the cardinalities of those sets and draw conclusions from that.

0.1
0.11
0.111
...
As long as there is not a line with infinitely many digits 1 there is
no diagonal with infinitely many digits 1. Conclusion: There is no
diagonal with infinitely many digits 1. The axiom of infinity leads to
a contradiction.

That is your (wrong) thinking.

> Try to complete the union of all finite trees to the complete infinite
> tree. Which level, node or edge would you add?

Pray, properly define the union of trees. In my opinion a tree consists
of three sets. First the set E of edges, next the set N of nodes and
finally the set P of paths. What is the union of two trees?

With the set of nodes and the definition of edges all other sets are
defined.

Right. And in that case the union of the sets of paths in individual trees
is *not* equal to the set of paths of the united trees.

My proof was about the set of paths in the union of finite trees not being
the union of the sets of paths in the finite trees. What was wrong about
that proof? Please, once come up with a proper definition of the union
of two trees. Before you come up with such a definition it is impossible
to even entertain a discussion.

Definition: The nodes of the tree are denoted by
(0,0)
(1,0) (1,1)
(2,0a) (2,1a) (2,0b) (2,1b)
...
(n,0a) (n,1a) ...


The union of all trees up to the n-levels tree is


{(0,0)} U {(0,0), (1,0), (1,1)} U .. U {(0,0), (1,0), (1,1), ...,
(n,0a) (n,1a) ...}

End of definition.

Example: The union of the one-level tree and the two-levels tree is

{(0,0), (1,0), (1,1)} U {(0,0), (1,0), (1,1), (2,0a), (2,1a), (2,0b),
(2,1b)}

Ok. Now show that the union of the sets of paths is the set of paths
of the union.

> Which edge of any infinite path is missing? Which bit could be added to
> one of the numbers represented?

What is the relevance? If you can state that there is some infinite path
in the union of the *sets* of paths in finite trees, you should also be
able to point to a finite tree that contains that infinite path.

There is no infinite initial segment of N of the from {1,2,3,...,n}.
Nevertheless the union is said to be {1,2,3,...}.

Yes. There is no infinite *set* of paths in the finite trees, nevertheless
is the *set* of paths in the infinite tree infinite.
On the other hand, there is no infinite natural in the infinite segments
and so there is also no infinite number in their union: N. As there is
no infinite path in any of the *sets* of paths of the finite trees, so
there is also no infinite path in their union.

There is
no edge missing, it is only that your infinite path is not a path in any
of the finite trees, so it is not in the union of the sets of paths in
finite trees.

So, how can an infinite path be described? How can the corresponding
real number be described, if not by all the levels n e N available in
the union?

There are infinite paths in the union of the trees. There is *no* infinite
path in the union of the sets of paths. Those two things are different.

> > Ok, so you do *not* use the union of the sets of paths but
> > something else.
>
> The material from which the paths are constructed.

Makes no sense.

But is fact.

How than can you draw conclusions about the cardinality of the sets of
paths in the union from the cardinalities of the sets of paths in the
finite trees?
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.

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