Another AC anomaly?
in [Logic]
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From: Bill Taylor on 25 Nov 2009 23:33 "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Well, I'm not at all sure that there's no problem with forward > supertasks. Surely, it is not difficult to come up with a > problematic case. Yes, the worth of supertasks as indicators of philosophical concerns is very much up in the air. Some seem relevant, others just stupid. Perhaps (temporally) well-ordered supertasks are more sensible than most. But I doubt that's all there is to it. One of my favourites is this; for naturals n:- compare... a) At each time 1 - 1/n, add balls numbered 2^(n-1) to 2^n - 1 to the pot, and remove ball number n. b) At each time 1 - 1/n, add 2^(n-1) - 1 balls to the pot, and replace the numbering stickers in agreement with case (a). After time 1: the final situation in case (a) is that the pot is empty. in case (b), the pot has infinitely many balls with no stickers! And yet at any intermediate time the two cases are indistinguishable! This sort of example shows that even omega-supertasks can be remarkably silly! > Now alter the situation slightly. At each step, again place 10 balls > into the vase and then remove one ball, but remove the ball > *randomly*. At the end, the vase may contain any number of balls Actually NOT. The pot will be empty(!) [with probability 1] For any ball, the probability of it being removed is like a harmonic series, which sums to oo, which by Borel-Cantelli means it will happen for sure. [meaning probability one, as always here] HOWEVER - if you add (say) 1, 4, 9, 16... balls per turn, and again remove one at random, each turn, then (Borel-Cantelli) each ball has a positive probability of being left behind. It is an interesting problem in probability generating functions to work out the individual proabilities for each ball, and the expected number, left in the pot at the end! -- Borellic Bill
From: Bill Taylor on 26 Nov 2009 00:26 > Apparently there's something wrong with backward supertasks (and not > with ordinary, 'forward' supertasks). But why should that be? So not only can Achilles not catch up with the tortoise, but he can't even get off the starting line!! :) -- Bionic Bill
From: Daryl McCullough on 26 Nov 2009 10:04 Bill Taylor says... >One of my favourites is this; for naturals n:- compare... > >a) At each time 1 - 1/n, add balls numbered > 2^(n-1) to 2^n - 1 to the pot, and remove ball number n. > >b) At each time 1 - 1/n, add 2^(n-1) - 1 balls to the pot, and > replace the numbering stickers in agreement with case (a). > >After time 1: > the final situation in case (a) is that the pot is empty. > in case (b), the pot has infinitely many balls with no stickers! > >And yet at any intermediate time the two cases are indistinguishable! > >This sort of example shows that even omega-supertasks >can be remarkably silly! Actually, I think that they are fun to think about. What this example (and similar ones) show is that for tasks involving a transfinite number of steps, you have to be more precise about what you are doing. In order for mathematics to tell us what the result is of performing some supertask, you have to describe the supertask using standard mathematical objects (typically sets). There are modeling choices that have to be made, and the answer can depend on the modeling choice. If it does depend on the modeling choice, that means that the original supertask was insufficiently specified. The big ambiguity is how to compute limit states. Associated with each ordinal alpha, there is a corresponding state of the system, S_alpha. The statement of the supertask explains how to go from S_alpha to S_{alpha+1}. But that tells us nothing about the state S_alpha when alpha is a limit ordinal. There are certain assumptions about the limit states that are so "obvious" that they seem to go without saying. For instance: "If a ball is added at stage alpha_1, and is never removed at any stage beta such that alpha_1 < beta < alpha_2, and alpha_2 is a limit ordinal, then the ball is present at stage alpha_2. But that's an assumption about the limit state. To really reason about supertasks, you have to state all the assumptions about limit states. > >> Now alter the situation slightly. At each step, again place 10 balls >> into the vase and then remove one ball, but remove the ball >> *randomly*. At the end, the vase may contain any number of balls > >Actually NOT. The pot will be empty(!) [with probability 1] > >For any ball, the probability of it being removed is like a harmonic >series, which sums to oo, which by Borel-Cantelli means it will >happen for sure. [meaning probability one, as always here] > >HOWEVER - if you add (say) 1, 4, 9, 16... balls per turn, >and again remove one at random, each turn, then (Borel-Cantelli) >each ball has a positive probability of being left behind. Very interesting. -- Daryl McCullough Ithaca, NY
From: WM on 26 Nov 2009 11:24 On 26 Nov., 05:33, Bill Taylor <w.tay...(a)math.canterbury.ac.nz> wrote: > It is an interesting problem in probability generating functions > to work out the individual proabilities for each ball, > and the expected number, left in the pot at the end! 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. 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. Instead of 100 balls you can use only 80 balls as well. The Romans also used to have 80 men in a Centuria- --- That pointed already to the final period of the empire. 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. Regards, WM
From: William Hughes on 26 Nov 2009 13:22
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] > 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. - William Hughes |