Re: Continuum hypothesis
- From: Han.deBruijn@xxxxxxxxxxxxxx
- Date: 5 Jan 2006 13:17:27 -0800
cbrown@xxxxxxxxxxxxxxxxx wrote:
>
> http://en.wikipedia.org/wiki/Probability_axiom
>
> gives (under "kolmogorov axioms") a set of axioms that a probability
> obeys.
OK. Here goes.
1. For any event E the probability P(E) >= 0 is a real number. True:
for any natural in the finite interval (1..n) let the probability
be 1/n .
Note that the Kolgomorov axioms do not specify anything else about
the probabilities of elementary events. So I can choose them as if I'm
throwing a perfect die with n sides all numbered differently. Right?
And again, if you cannot accept a die with n sides, then you should be
consequent and remove all playing with 6-dice from probability theory.
(As well as shuffling with cards, which presents analogous problems.)
2. The probability that some elementary event in the entire sample set
will occur is 1. More specifically, there are no elementary events
outside the sample set. True: the probability that _some_ number
in the interval (1..n) will be "thrown" is 1. Because it is not
allowed
to use other naturals, as defined by _this_ setup.
3. Any countable sequence of pairwise disjoint events E_1,E_2, ...
satisfies P(E_1 u E_2 u E_3 u ... ) = Sum_i P(E_i) . True. Just
let
P(E_i) = P(a natural in (1..n) equals i _in_ (1..n)) = 1/n and
proceed.
Therefore the Kolgomorov axioms of probability are valid for ANY finite
interval of naturals (1..n), provided that the rules of the game _are
as_
defined above. Now consider the case n is _large_. Then everything will
remain the same as before, except that some probabilities will become
defined more unambiguously in some cases. See below.
Consider as an example the probability that a natural is divisible by a
..
At the finite interval (1..n) that probability is calculated by
counting as:
(a-1) no, 1 yes, (a-1) no, 1 yes, ... giving (n/a) numbers with
remainder.
The remainder is at most (a-1). Multiplying with the elementary P(E_i)
gives 1/a <= P(i divisible by a) < 1/a + a/n . This is well defined for
the
limiting case n -> oo : P(i divisible by a) = 1/a . The limiting case
being
"the" naturals "as a whole". Again, I can _not_ understand what could
possibly be wrong here. My approach is constructive: start with a
finite
set and let it grow. More precise: give me an interval (1..n) and I'll
give
you a larger interval. Potential infinite instead of actual infinite.
Limits,
just like in the classical calculus books.
But it is very clear that the mainstream reasoning with "all" naturals,
as
a completed infinity, gives a _very_ different answer, namely _zero_,
i.e.
P(natural is divisible by a) = 0 . Which I find utterly absurd.
Han de Bruijn
.
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