Another AC anomaly?
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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
From: WM on 27 Nov 2009 01:45 On 27 Nov., 02:50, William Hughes <wpihug...(a)hotmail.com> wrote: > 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 Why then do you expect the digits of Cantor's diagonal number to be "continuous" at infinity (contrary to being *not* at infinity)? Regards, WM
From: Virgil on 27 Nov 2009 02:41 In article <aa9e46c0-56da-4510-8345-8dee84745816(a)b2g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 27 Nov., 02:50, William Hughes <wpihug...(a)hotmail.com> wrote: > > 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 > > Why then do you expect the digits of Cantor's diagonal number to be > "continuous" at infinity (contrary to being *not* at infinity)? Why would anyone ever expect a numerical digit to be continuous? All the ones I am aware of are members of a finite set of discrete objects. And why would you expect to find a digit of any sort "at infinity", when there is no such a position as "at infinity".
From: Virgil on 27 Nov 2009 02:44
In article <9d0132ac-2c6b-447f-8515-22d5c69f1832(a)s20g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. Maybe in your muecked up worlds, but, fortunately, mathematics does not occur in such worlds. |