From: Hal Finney (hal@rain.org)
Date: Fri Nov 07 1997 - 14:58:42 MST
John K Clark, <johnkc@well.com>, writes:
> On Thu, 6 Nov 1997 Hal Finney <hal@rain.org> Wrote:
>
> >>Me
> >>But can you identify ALL the points or just some of them ?
>
> >This seems to be the main point. You can identify all of the points.
> >Every real number in [0,1] corresponds to an infinite binary fraction.
>
> Infinite? OK, but the big question is what order infinity, just how many
> decimal places do you need to represent all the points on that line? We know
> it can't be the smallest infinity, that of the integers (call it I) because
> we already know that I *I = I so the line would only have I points on it and
> Cantor proved that's not true.
The number of bits in an infinite binary fraction is a countable infinity.
There's a first bit, a second bit, a third bit, etc. Clearly if you
count them like this you won't miss any. So there is no question but
that an infinite binary (or decimal) fraction has a countably infinite
number of bits (digits).
The formula I * I (where I = aleph-null) does not come into play here.
The number of values that can be represented by an n bit fraction is 2^n.
If you are talking about infinite fractions, the number of values is 2^I.
That is the whole idea of the argument - that the number of infinite
binary fractions is 2^I, and that this is also the number of real numbers
in that range.
So, you need to disagree with one (or both?) of these two points:
- The real numbers correspond to infinite binary (or decimal) fractions
- The number of infinite binary fractions is 2^aleph-null.
I don't see how either of these can be wrong. For the first to be
wrong you'd have to have a real number which did not correspond to any
infinite decimal. In effect, it would have no value. This does not seem
coherent. (In fact aren't the real numbers sometimes *defined* as all
the values taken by infinite decimals?)
And the second is a simple counting argument. Of course to go from the
fact that there are 2^n n-bit fractions to 2^aleph-null infinite ones
is a bit facile. Probably a rigorous mathematical proof would have to
make sure that taking the limit like this made sense. But certainly we
do this kind of substitution all the time when dealing with infinities.
Hal
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