Re: Cantor Confusion

On 23 Feb., 20:21, Virgil <vir...@xxxxxxxxxxx> wrote:
In article <1172253288.305584.116...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

mueck...@xxxxxxxxxxxxxxxxx wrote:
On 23 Feb., 14:35, "*** T. Winter" <***.Win...@xxxxxx> wrote:
Every tree T(n) contains only finite paths,
namely such with n nodes. There is no tree containing an infinite
path.

Then there can be no infinite trees.

Correct!

Every set of finite subsets of N is countable. The set of all finite
subsets of N is countable.

> There are countably many unions like U(p(n)).

Prove it.

If A is countable, then the set of all finite sequences of elements of
A is countable.

But neither extends to the infinite subsets of A or infinite sequences
of elements of A.

The theorem extends to the finite subsets or finite sequences of
elements of of U(T(n)) . As in case of the tree all elements of P(oo)
like p(oo) are unions of finite pathes, p(oo) = Up(n), q(oo) = Uq(m),
r(oo) = Ur(i), and the sets of finite paths like p(n), are countable,
the unions of sets of finite paths like U(p(n)) = p(oo) belong to a
countable set.

In short: Everything in the countable union of finite nodes and finite
paths of U(T(n)) is countable. Therefore the set of all possible
unions of finite paths is countable too. P(oo) is a subset of this set
of all possible unions of finite paths.

Regards, WM

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