Re: Cantor Confusion

mueckenh@xxxxxxxxxxxxxxxxx wrote:
Tony Orlow schrieb:


Repeating a falsehood does not make it any less false.
Why then did you repeat always that here are no balls in the vase at
time 0? Why do you repeat without reasonable arguments that there are
more paths than edges in the binary tree, although no path springs off
without its own egde?
=========================

Again I agree. There are 9 n balls after n iterations, and there are
half as many paths as edges or nodes in the tree, despite supposed
bijection hat tricks. Banach-Tarski is ridiculous, and omega is a phantom.

Correct. The function f(n) = 9n will never yield the value 0.
Why he maintains this this somehow bijects the individual branches with
paths, is not clear.
Because every branch consists of two edges, one for the each of the
resulting paths.

=========================

To put it another way, each right branch may be considered a
continuation of the same original infinite path, like adding a 0 to the
right of the decimal point - it doesn't change the value. The left
branch from every node produces a new path, creating a new value with
the concatenation of a 1 to the string. Therefore, for every two
branches, there is one additional path.

Of course that is a possible way to construct a bijection. But set
theorist will argue that by this constructuibn you get only those paths
representing rational numbers. (That is even true, because there are no
other numbers.) But they do not see that there are no paths without
edge. They just believe that "in the infinite" their God will provide
the missing paths.

If every value in the tree is generated by an edge, then if 1/3 exists there is an edge where the path achieves that value. But, it cannot achieve that in any finite number of bits. If the depth of the tree is countably infinite, aleph_0 levels, then there are 2^aleph_0 nodes and edges. If not, then 1/3 does not exist in the tree, or anything else that isn't a binary rational (sum of a finite number of powers of 1/2).


Is the diagonal longer than any line? Nope.

By definition, it cannot exist, where no line is.


Agree.

Perhaps, with a countable number of bits. But, there are more than any
finite number of reals in the unit interval, and this is an infinite
number.

Where *are* they? They canot be addressed. They cannot be imagined.
They cannot be known. And, moreover, the proof proving their existence,
is false.

Quite a poor existence.

Regards, WM


Well, the proof is simple. Any finite number of subdivisions of any finite interval will only identify a finite number of real midpoints in that interval, between any two of which will remain more real midpoints. Therefore, there are more than any finite number of real points in the interval. Of course, most of them require infinite strings of bits to specify, but that doesn't mean the point doesn't exist.
.

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