Re: Cantor Confusion


Newberry schrieb:

mueckenh@xxxxxxxxxxxxxxxxx wrote:
Newberry schrieb:

mueckenh@xxxxxxxxxxxxxxxxx wrote:
Newberry schrieb:


The edges have cardinal number aleph_0.

Certainly.

A
subset of the paths can be shown to be in bijection with the real
numbers of the interval [0,1].

Which subset? Path as a set of intervals or path as a set of eges? How
does the above follow from this?

The paths are binary representations of all real numbers of the
interval [0,1]. Some rational reals have double representations, namely
such ending by 000... and such ending by 111... like 1.000... and
0.111... . So there are not less real numbers in the interval [0, 1]
than paths in the binary tree. We have

aleph_0 = |{edges}| >= |{paths}| >= |{reals of [0,1]}| = 2^aleph_0

while according to set theory

aleph_0 < 2^aleph_0

Regards, WM

Well, a Cantorist would say that this part does not hold

|{edges}| >= |{paths}|

because you considered only the finite paths but not the infinite paths.

In the infinite tree there are no finite paths.

Yes, but you arrived at |{edges}| >= |{paths}| by calculating

lim{n-->oo} (2*2^n - 2)/2^n = 2

In this limit all the n are finite. So a Cantorist would claim that you
examined only the finite sub-paths,


Every real number (like, e.g., pi) has digits r_1, r_2, r_3, ..., only
at finite digit positions: 1,2, 3, ...(because an infinite digit
position could not be identified and the corresponding digit could not
be determined)..

and not the infinite paths. I do
not know if this Cantorist claim makes sense,

It helps them to veil the contradictions appearing in set theory,
because they can easily jump from "every natural number is finite" to
"there are infinitely many natural numbers".

but the question is
whether you can generate a direct contradiction of the type P & ~P in
ZFC.

I do not think that this is of any interest after it can easily be
shown that the results of ZFC are inacceptable. Compare for instance
the resuls I have recently prepared in a discussion with Virgil:

You accept a diagonal of a matrix longer than any line.
You accept more separated paths than splitting positions (= origins of
separated paths) [in the binary tree].
You accept that 1 + 1/2 + 1/4 + ... > 10 is possible.
And at last you accept the existence of an actually infinite finite
number.

Enough?

Regards, WM

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