Re: Cantor Confusion

On 23 Mrz., 22:06, Virgil <vir...@xxxxxxxxxxx> wrote:
In article <1174670803.869854.53...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,


Cantor, unless working every digit simultaneously, will never finish.

Cantor's rule is not time dependent.

The following rule is not *time* dependent either:

By transpositions (as we know them from Cantor's work) bring
the n-th term (with n = 2, 3, 4, ...) into the first position and
after that bring the n+1-th term into first position. So for *every*
term number n you can determine, the number of transpositions required
to have it at the first position.

If Cantor can finish his replacement in no time, then the above rule
also finishes its task in no time.

In fact we need not know how to proceed, but it is sufficient to know
hat the sum remains 2 what ever we do.

But to get there, WM insists on patitioning that which has no parts.
Nodes are like points in that respect, they have no parts.

In other respects they have.

However, you may proceed as follows (but only after having read my
book)

That lets us off that hook. NO one need read WM's propagandizing.

> Because there is no "all paths".

Pray, provide a proof (within set theory, where we are arguing).

I am arguing within the tree.

Which tree?

That one which has all nodes but only finite paths.

In any CIBT or T(oo) there are sets of all paths, and those
sets are uncountable.

There is not at all an infinite path detectable in any CIBT.

That WM has imagined some other tree is irrelevant.

There is never an uncountable number of
separated paths.

As each path separates from every other at some node, there are
uncountably many pairwise separable paths, which is all that counts.

You say all paths were uncountable and separable.
Conclusion. there is no "all paths" within the tree.

By "separable" all we mean is that any two can be separated,

Everything that can be separated is separated in the infinite tree.

Regards, WM

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