On Thu, 6 Nov 1997 Hal Finney <hal@rain.org> Wrote:
>This is not quite right; it is easy to show that 2^B equals C, where
>B is aleph-null.
It could be true, but it won't be easy to prove it.
>Simply identify the points on the unit segment with their
>representation in base two.
But can you identify ALL the points or just some of them ?
>The number of possible binary representations of length n is 2^n,
OK.
>so the total number of binary representations of length aleph-null
>is 2^aleph-null,
OK.
>and that is the number of points on the line.
That's the problem. To prove it you'd have to find a one to one correspondence
between a set with 2^alepha-null members and the points on a line. You'd have
little difficulty with the rational numbers, those that can be expressed as
ratios of whole numbers, like 5/7, but then you'd also have to deal with
numbers that are not fractions but are the solution to some polynomial
equation, like the square root of 2, after that you'd also have to deal with
numbers that are not fractions or the solution to any polynomial equation
but can be expressed as the sum of an infinite sequence, such as PI or e.
After you've done all that your problems would have just begun because then
you'd have to deal with Chaitin's numbers such as... well, actually only one
of them has a name but there are an awful lot of them.
About 20 years ago Gregory Chaitin extended Turing's work on non computable
numbers and proved that for ever number that can be found by a deterministic
method, such as the ones I mentioned above, there are an infinite (class
unknown) number that can not. Almost all the points in a line are Chaitin
numbers, but none of us has ever seen one, or ever well. The only way to find
a Chaitin number is to let heads be one and tails be zero and then flip a
coin an infinite number of times. Because they can not be generated by any
deterministic process it's very hard to count them, that is, to put them in
a one to one correspondence with a set on a known size in a mathematical
proof, but a reliable method must be found to prove the Continuum Hypothesis.
Current number theory is not up to the challenge.
John K Clark johnkc@well.com
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