Re: An uncountable countable set

In article <1157465168.349273.136370@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
*** T. Winter schrieb:
In article <1157367467.816725.158560@xxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
....
> Cantor's list has as many lines as my tree. The diagonal has as many
> digits as each path of my tree. The only difference is that the paths
> split while the diagonal does not.

And the difference is that given the list we can immediately state what
the n-th digit of the diagonal is for arbitrary n. You can not state what
the n-th path is for arbitrary n.

You intermingle numbers (paths) with digits. I can state the n-th digit
of any path you want.

I do not. The digits are countable because I can immediate state the
n-th digit when n is given. The paths are not countable because I can
not state the n-th path when n is given. The edges within a given path
are countable because you can state which is the n-th edge when n is
given.

> The same is true for my tree.

What is the tenth path in your tree?

You intermingle numbers (paths) with digits. I can state the n-th digit
of any path you want. (I do not ask what is the tenth entry of Cantor's
list.)

No, of course you do not ask that. You should ask that to the one who
provided the list. Which is *not* Cantor, because he does start with
an arbitrary list, not a specific list. And shows that given an
*aribitrary* list, the anti-diagonal produced is not in the list.
As this proof holds without any reference to the particulars of the
list, it holds for *all* lists. So, if someone gives me a list and
tells me that it contains all infinite sequences of two symbols, I
can show him such a sequence not in the list.

What is the tenth path of that tree?

What is the tenth entry of the list?

That is completely irrelevant. See above. But also, the list is a
*given* thing with some assumptions. And a list is *by definition*
a function from N to a (sub)set something. So in the case of Cantor,
the list is the function
f: N -> S
where S is the set the set of infinite sequences of two symbols. You
ask what the tenth element is, and I answer: f(10). I need not go
further because the list is a given entity.

On the other hand, you *claim* that your set of paths is countable,
so you should be able to state what the tenth path is. But you refuse
to do so.

> Give my a tree of infinite paths consisting of 0's and 1's, and I show
> that there are not less edges than paths.

Indeed. If all edges terminate, also all paths terminate, and both are
countable, and 1/3 is not in the tree. If edges do *not* terminale,

What are you talking about? Every edge terminates because it is the
connectio between subsquent nodes of a path.

And so all paths terminate, and 1/3 is not in your tree.

> You know that sets of order 2^omega are countable?

No. Can you prove it? It is precisely Cantor's diagonal proof that
shows that it is not countable.

You intermingle 2^aleph_0 and 2^omega.

No, I did not know the strange exponentiation used in ordinals. Apparently
it can be defined as the set of functions from a set with ordinal number
omega can be mapped to a set with ordinal number 2, with the proviso that
only finitely elements are mapped to the second element.

A well known example is the set
of the unit fractions 1/n used for the proof of the divergence of the
harmonic series:

1 + 1/2 + (1/3 + 1/4) + (1/5 +...+ 1/8) + ...

It has omega terms in parentheses and 2^omega fractions which biject to
the natural numbers.

How do you come at 2^omega fractions? (Yes, I sort of understand.)
Exponentiation can also defined as:
(1) a^0 = 1
(2) a^(b+1) = (a^b).a (note right handed multiplication)
(3) if d is a limit ordinal, a^d = lim{b -> d} a^b.
So this implies that 2^omega = lim{n -> omega} 2^n, which is omega.
--
*** 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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