
Only for mathematicians!
Posted:
Aug 16, 2014 6:41 AM


The following postings have raised a lot of noise. Therefore I repost them, asking only mathematicians (i.e., those who accept the mathematics of infinite sequences in real analysis) to comment and asking matheologians to stay away.
Cantor's enumeration of the positive rational numbers Q+ (mentioned in a letter to Lipschitzon 19. Nov.1883) is ordered by the ascending sum (a+b) of numerator a and denominator b of q = a/b, and in case of equal sum, by ascending numerator a. Since all fractions will appear infinitely often, all beyond the first one will be dropped. This yields the sequence 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1 1/6, 2/5, 3/4, 4/3, 5/2, 6/1 ... It is easy to see that at least half of all fractions of this sequence belong to the first unit interval (0, 1].
While every positive rational number q gets a natural index n in a finite step of this sequence there remains always a set s_n of positive rational numbers less than n which have not got an index less than n. s_(n+1) = (s_n U { q  n < q =< n+1 }) \ {q_(n+1)} with s_1 = { q  0 < q =< 1 } \ {q_1} (cp. § 529)
All s_n are infinite s_n = oo. But also their geometric measure is increasing beyond every bound. This is shown by the following
Theorem. For every k in N there is n_0 in N such that for n >= n_0: (nk, n] c s_n.
Proof: Let a/1 be the largest fraction indexed by n. Up to every such n at least half of the natural numbers are mapped on fractions of the first unit interval. a is continuously increasing, i.e., without gaps. Therefore n must be about twice as a, precisely: n1 >= 2(a1) or n >= 2a  1.
Examples: a = 1, n = 1 a = 2, n = 3 a = 3, n = 5 a = 4, n = 10 a = 5, n = 12 a = 6, n = 17 ...
Therefore for k we can take n_0 = 2k. Then the interval (n_0  k, n_0] c s_(n_0). This is satisfied for every n >= n_0 too.
This means, there are arbitrarily large sequences of undefiled unit intervals (containing no rational number with an index n or less) in the sets s_n.
Remark: It is easy to find a completely undefiled interval of length 10^1000^100000000000 or every desired multiple in some set s_n. Everybody may impartially examine himself whether he is willing to believe that nevertheless all rational numbers can be enumerated.
Remark: Cantor does neither assume nor prove that the whole set N is used for his enumeration (in fact it cannot be proved). Cantor's argument is this: Every natural number is used, so no natural is missing. He and most set theorists interpret this without further ado as using N.
Remark: Although more than half of all naturals are mapped on fractions of the first unit interval, never (for no n) more than 1 % of all fractions of this interval will become enumerated. In fact it can be proven for ever natural number n, that not the least positive interval (x, y] of rational numbers is ever completely enumerated.
Set theorists claim that all rational numbers can be indexed by all natural numbers. Above I have shown not only that every natural number n fails but even that with increasing n the number of unit intervals of rationals without any rational indexed by a natural less than n increases without bound, i.e., infinitely. Since nothing but finite natural numbers are available for indexing, and provably all fail, this task cannot be accomplished.
I don't know what goes on in the heads of matheologians. But I know that it is deliberately contradicting the magnificent, powerful, and, for all nonmatheological purposes, extremely useful mathematics of the infinite that has been devised by Euler, Gauss, Cauchy, and Weierstrass.
Rational arguments to straighten these matheological assertions are not available.
Regards, WM

