Another AC anomaly?
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From: Alan Smaill on 27 Nov 2009 09:31 WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 27 Nov., 11:43, Alan Smaill <sma...(a)SPAMinf.ed.ac.uk> wrote: > > WM <mueck...(a)rz.fh-augsburg.de> writes: > > > 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. > > > > But you lose control at infinity! > So does Cantor. Good to know you agree that *you* lose control. Let's leave Cantor out of it. > > So your "hence" doesn't work. > > It works if there is anyone who does not lose control at infinity. > That's enough. If you are not in control, it doesn't work for you; you can't say anything about what does or doesn't happen at infinity, except by pretending that it must be the same as the finite case. That pretence is worthless when you admit you have lost control. > Regards, WM -- Alan Smaill
From: A on 27 Nov 2009 10:21 On Nov 27, 1:43 am, WM <mueck...(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. > > Regards, WM > > Regards, WM The set X described above is certainly the empty set, as one can easily prove, despite im x_n being nonempty for each n. This seems to be a rigorous statement of what you are describing by talking about balls, vases, etc. I have never seen any mention of "potential infinity" or "completed infinity" which is precise enough to even be considered mathematics, and talking about these ideas generally leads to nothing better than confusion--it might be better to speak precisely about the limits or sets that you mean, in any given situation, rather than worrying whether they represent "potential infinity" or "completed infinity."
From: Dik T. Winter on 27 Nov 2009 11:02 In article <ac583807-eb9d-4ae4-86f2-37f3296923f1(a)k17g2000yqh.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: .... > Therefore the vase is never empty - Similar in the sequence 1/n the elements value is never zero. > and the only error is your > assertion it would be empty after the last step and there was a state > of the vase "at infinity". Never heard about limits? We may consider the following for limits of sets (X_n etc are sets): lim sup X_n: x in lim sup X_n if and only if x in infinitely many X_n lim inf X_n: x in lim inf X_n if there is some n0 such that x in X_n for n > n0 lim X_n: exists when lim sup X_n = lim inf X_n and in that case is equal to them. You may verify that in the case you proposed lim X_n does exist and is equal to the empty set. Note: 1 = lim |X_n| != |lim X_n| = 0 but that should not come as a surprise. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: William Hughes on 27 Nov 2009 12:17 On Nov 27, 10:29 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 27 Nov., 14:48, William Hughes <wpihug...(a)hotmail.com> wrote: > > ********************* > 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 > ***************** > > > > > The argument why the change is the number > > of balls is not problem is simple. Outside Wolkenmuekenheim > > > There is a contradiction at the last step > > There is no last step > > There is no contradiction. > > Therefore the vase is never empty Nope, only in Wolkenmuekenheim, where we cannot have all steps without a last step, do we have The vase can only be empty if there is a last step. Ouside of Wolkenmeukenheim we can have all steps with no last step so The vase is empty after all steps. - William Hughes
From: WM on 27 Nov 2009 12:32
On 27 Nov., 18:17, William Hughes <wpihug...(a)hotmail.com> wrote: > On Nov 27, 10:29 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 27 Nov., 14:48, William Hughes <wpihug...(a)hotmail.com> wrote: > > > ********************* > > 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 > > ***************** > > > > The argument why the change is the number > > > of balls is not problem is simple. Outside Wolkenmuekenheim > > > > There is a contradiction at the last step > > > There is no last step > > > There is no contradiction. > > > Therefore the vase is never empty > > Nope, only in Wolkenmuekenheim, where we cannot > have all steps without a last step, do we have > > The vase can only be empty if there is > a last step. > > Ouside of Wolkenmeukenheim we can have all steps > with no last step so > > The vase is empty after all steps. So all steps have been done whereas the last is pending. And if we leave the vase with a ball inside and without touching it? After all none-steps have been done: the vase will be empty. Matheology at its best. Regards, WM |