popper-logic-scientific-discovery

354
new appendices
It should be mentioned, in this connection, that Kolmogorov fails,
in his book, to offer an independence proof for his ‘axiom of continuity’
although he claims independence for it. But it is possible to reframe
our proof of independence so that it becomes applicable to
Kolmogorov’s axiom and his set-theoretic approach. This may be
done by choosing, instead of our S1 a system of intervals S3, exactly
like S1 but based upon a sequence C = c1, c2, ..., defined by cn = (0,
2 − n ] rather than upon the sequence A = a1, a2, ..., with an = (0,
1
2 + 2 − n ]. We can now show the independence of Kolmogorov’s axiom
by defining the probabilities of the elements of the sequence A as
follows:
p(c n) = l(c n) + 1 2 = p(a n)
Here l(cn) is the length of the interval cn. This definition is highly
counter-intuitive, since, for example, it assigns to both the intervals (0,
1
2] and (0, 1] the probability one, and therefore to the interval ( 1 2, 1] the
probability zero; and the fact that it violates Kolmogorov’s axiom
(thereby establishing its independence) is closely connected with its
counter-intuitive character. For it violates the axiom because lim
p(cn) = 1 2, even though p(c) = 0. Because of its counter-intuitive character,
the consistency of this example is far from self-evident; and so the
need arises to prove its consistency in order to establish the validity of
this independence proof of Kolmogorov’s axiom.
But this consistency proof is easy in view of our previous independence
proof—the proof of the independence of our own first definition
with the help of the example S1. For the probabilities p(an) and p(cn) of
the two examples S1 and S3 coincide. And since by correlating the two
sequences, A and C, we may establish a one-one correspondence
between the elements of S1 and S3, the consistency of S1 proves that of S3. It is clear that any example proving the independence of Kolmogorov’s
axiom must be equally counter-intuitive, so that its consistency
will be in need of proof by some method similar to ours. In other
words, the proof of the independence of Kolmogorov’s axiom will
have to utilize an example which is, essentially, based upon a
(Boolean) definition of product such as ours, rather than upon the
set-theoretic definition.