Re: Cantor Confusion

In article <1167493176.710056.286330@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
*** T. Winter schrieb:
....
At what point is 1/3 represented?

In the binary tree it is not represented, but it could be represented
by using a ternary tree.

Now you are arguing differently from before. When I stated that 1/3
was not in your tree, you argued that it was in your tree. Do you now
think differently?

There is no representation
of 1/3, unless it were in the infinite. But if you insist on an
existing decimal representation of 1/3 in the infinite, then also the
union of all finite binary trees contains such an existing binary
representation of 1/3.

I still do not understand what subtleties you are making. In the *complete*
binary tree 1/3 does exist. But that is *not* the union of your rational
trees.

> Take the union of all these rational trees. It contains all paths
> representing rational numbers.

No. It contains all paths representing rational numbers for which the
denominator is a power of two.

Every decimal number gives a power of 10.

A distraction again. Pray keep to the discussion.

> What distinguishes this union of rational trees from the complete tree?

That some rational numbers are missing. If the union contains all
rational numbers, there must be one of the rational trees that contains
(for instance) 1/3. There is none.

My question is only the following: Do you think that the union of all
rational binary trees is the complete infinite binary tree or not?

It is not. But I already did answer that when I answered your question
what the distinction was between the union of rational trees and the
complete tree. So why do you ask again? I state they are distinct,
and indicate in what way they are distinct, and you ask whether I
think they are distinct?

Where is 1/3? At what point does it get into the union of rational
trees?

My question is only the following: Do you think that the union of all
rational binary trees is the complete infinite binary tree or not?

How many times must I answer that question?

> That is the reason why the cardinal number of the paths in both trees
> cannot be different.

That is the reason why the cardinal number of the paths in both trees
*is* different.

With a different number and configuration of nodes (or edges) or with
exactly the same?

You lost me here. We were talking about the union of finite trees and
the complete tree.

> But if we subtract the rational paths from the complete tree, then
> nothing remains but the empty set which obviously is the set of all
> paths representig irrational numbers.
>
> This shows that irrational numbers are nothing (but a chimera).

And apparently a lot of rational numbers are nothing?

A lot of binary representations, yes. But every rational number has
some representation - in contrast to irrational numbers.

You are wrong. sqrt(2) has a pretty good representation: "sqrt(2)".
--
*** 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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