Another AC anomaly?
in [Logic]
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From: Tim Little on 26 Nov 2009 20:02 On 2009-11-27, LauLuna <laureanoluna(a)yahoo.es> wrote: > First note that what the relevant equivalence class is, is already > determinate at all 1/n points. > > Then assume that at any 1/n point I construct it -by guessing at > random the future results- iff I haven't constructed it at any 1/m > with m>n. Hence you contradict your original strategy which specified that you guess at random at every 1/n point. This new strategy is exactly the AC strategy. It cannot be proven in ZF alone that there exists a set of representatives covering all equivalence classes, so the strategy can fail if AC is not assumed. > As I see it, the essential paradox here is that the relevant > equivalence class is already fixed at each 1/n point, so that no > toss contributes to its determination. Have you seen the "infinite line of prisoners with hats" version illustrating the same use of AC? In this version, there is an queue of prisoners, having a back of the queue but extendingly infinitely forward. Each prisoner can see all the colours of hats worn by those ahead. Their task is to guess their own hat colour; if only finitely many guess incorrectly, they all go free. They get to confer beforehand, but cannot communicate with each other in any way after the hats are placed on their heads, and they only get one guess. In this formulation, there is no question of inheritance or tasks that cannot be begun. - Tim
From: William Hughes on 26 Nov 2009 20:50 On Nov 26, 4:51 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 26 Nov., 19:22, William Hughes <wpihug...(a)hotmail.com> wrote: > > Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim > > you will have an empty set. > > Besides your assertion, you have arguments too, don't you? > In particular you can explain, how the empty set will emerge while > throughout the whole time the minimum contents of the vase is 1 ball? > Since outside of Wolkenmuekenheim there is no reason to expect the number of balls to be continuous at infinity outside of Wolkenmuekenheim there is no problem. Inside of Wolkenmuekenheim we can use "logic" like All FISONS have a fixed last element. N does not have a fixed last element. N is a FISON and prove anything. - William Hughes
From: A on 26 Nov 2009 21:50 On Nov 26, 3:51 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 26 Nov., 19:22, William Hughes <wpihug...(a)hotmail.com> wrote: > > > On Nov 26, 12:24 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > Here is another interesting task: Use balls representing the positive > > > rationals. The first time fill in one ball. Then fill in always 100 > > > balls and remove 100 balls, leaving inside the ball representing the > > > smallest of the 101 rationals. > > > [at random with any measure that gives a positive probability > > to each rational] > > Simply take the first, seconde, third ... Centuria according to > Cantor's well-ordering of the positive rationals. Then there is no > need for considering any probabilities. > > > > > > If you get practical experience, you > > > will accomplish every Centuria in half time. So after a short while > > > you will have found the smallest positive rational. > > > Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim > > you will have an empty set. > > Besides your assertion, you have arguments too, don't you? > In particular you can explain, how the empty set will emerge while > throughout the whole time the minimum contents of the vase is 1 ball? > > Regards, WM Let S denote a set with exactly 101 elements. Let Q+ denote the positive rational numbers. Let inj(S,Q+) denote the set of injective functions from S to Q+. Let {x_n} denote a sequence of elements of inj (S,Q+) with the following properties: 1. Let im x_n denote the image of x_n. Then the union of im x_n for all n is all of Q+. 2. For any n, the intersection of im x_n with im x_(n+1) consists of exactly one element, which is the minimal element (in the standard ordering on Q+) in im x_n. Let X denote the subset of Q+ defined as follows: a positive rational number x is in X if and only if there exists some positive integer N such that, for all M > N, x is in the image of x_M. We are talking about X, right? This is the set of balls which I, and others, claim is empty, and WM claims is non-empty? Or does WM dispute one of the definitions above?
From: WM on 27 Nov 2009 01:39 On 27 Nov., 01:40, LauLuna <laureanol...(a)yahoo.es> wrote: > On Nov 26, 5:24 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > Of course the game is only possible, if there is an actual infinity > > waiting to be finished. --- And that points to the final period of > > mathematics. > > Yes, but that existence is not enough for the game to become possible. But it is enough for the game to be discussed as I did. The result is that the result is impossible (must be empty set but cannot be empty set). And that is the same as in case of any bijection including an infinite set like N or Q or both. Regards, WM
From: WM on 27 Nov 2009 01:43
On 27 Nov., 03:50, A <anonymous.rubbert...(a)yahoo.com> wrote: > On Nov 26, 3:51 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 26 Nov., 19:22, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On Nov 26, 12:24 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > Here is another interesting task: Use balls representing the positive > > > > rationals. The first time fill in one ball. Then fill in always 100 > > > > balls and remove 100 balls, leaving inside the ball representing the > > > > smallest of the 101 rationals. > > > > [at random with any measure that gives a positive probability > > > to each rational] > > > Simply take the first, seconde, third ... Centuria according to > > Cantor's well-ordering of the positive rationals. Then there is no > > need for considering any probabilities. > > > > > If you get practical experience, you > > > > will accomplish every Centuria in half time. So after a short while > > > > you will have found the smallest positive rational. > > > > Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim > > > you will have an empty set. > > > Besides your assertion, you have arguments too, don't you? > > In particular you can explain, how the empty set will emerge while > > throughout the whole time the minimum contents of the vase is 1 ball? > > > Regards, WM > > Let S denote a set with exactly 101 elements. Let Q+ denote the > positive rational numbers. Let inj(S,Q+) denote the set of injective > functions from S to Q+. Let {x_n} denote a sequence of elements of inj > (S,Q+) with the following properties: > > 1. Let im x_n denote the image of x_n. Then the union of im x_n for > all n is all of Q+. > > 2. For any n, the intersection of im x_n with im x_(n+1) consists of > exactly one element, which is the minimal element (in the standard > ordering on Q+) in im x_n. > > Let X denote the subset of Q+ defined as follows: a positive rational > number x is in X if and only if there exists some positive integer N > such that, for all M > N, x is in the image of x_M. > > We are talking about X, right? We are talking about a vase which is never emptied completely! Hence it cannot be empty unless "infinity" is identical to "never". But this describes potential infinity and excludes phantasies like Cantor's finished diagonal number. Regards, WM Regards, WM |