Another AC anomaly?
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From: master1729 on 23 Nov 2009 06:59 > Jesse F. Hughes wrote: > > Herman Jurjus <hjmotz(a)hetnet.nl> writes: > > > >> Apparently there's something wrong with backward > supertasks (and not > >> with ordinary, 'forward' supertasks). But why > should that be? > > > > 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. > > > > For instance, take our favorite example: at each > time t - 1/n, place > > balls 10(n-1) to 10n - 1 in a vase and then remove > ball n. At the end > > the vase is empty. > > > > 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. > > This strikes me as suitably counterintuitive to say > that the forward > > supertask has something wrong with it. Or, > perhaps, with my > > intuitions. > > More conclusive (at least for me): you switch a light > bulb on and off; > after infinitely many steps, is the light on or off? > (Or: you put one > and the same ball in the vase, out of the vase, in > the vase, out of the > vase, etc. What's in the vase/where's the ball after > infinitely many steps?) > If you can't trust supertask-reasoning in this case, > why should you > trust it in other, seemingly less problematic cases? > > For the sake of clarity: I'm not endorsing or > advocating the > backward-supertask paradox as extremely important. > Just wanted to share > it here, because I had never heard of it, and thought > perhaps others > also hadn't. > > -- > Cheers, > Herman Jurjus thompsons lamp. i assumed people on sci.math knew about the term. apparantly i was to optimistic again. basicly thompsons lamp can be well expressed by ordinary number theory : lim n -> oo n mod 2 = ??? there is no convergeance , so the answer is not 0 or 1 but ' div ' and in terms of my 3-valued logic : '0' truth value. btw all these things have been said a billion times on sci.math , both my own comments and analogues of thompsons lamp. so i wonder why they are posted again , but i will look at the good side ; this gives me an opportunity to give a good reply and a good reminder of my ideas. (and btw , yes an AC anomaly) regards tommy1729
From: Butch Malahide on 23 Nov 2009 17:18 On Nov 23, 5:22 am, Herman Jurjus <hjm...(a)hetnet.nl> wrote: > Has anyone seen this before? > > http://possiblyphilosophy.wordpress.com/2008/09/22/guessing-the-resul... > > I'm not sure yet what to conclude from it; that AC is horribly wrong, or > that WM is horribly right, or something else altogether. > > In short the story goes like this: > > A game is played, in which infinitely many coins are tossed, and there's > one player, who makes infinitely many guesses. Both are done over a > finite period of time. The tosses and guesses are not made faster and > faster, however, but slower and slower: at t = 1/n. There's no 'first' > move. > > Claim: > There exists a strategy with which you're certain to guess all entries > correctly except for at most finitely many mistakes. Not 'certain' as in > 'probability is 100%', but absolutely certain. > > Reasoning: > On 2^w, consider the equivalence relation that makes x equivalent to y > when x(n) =/= y(n) for at most finitely many n. Next, using AC, create a > set S that contains precisely one element from every equivalence class. > Strategy: at every move, you already know the results of the previous > tosses, which is an infinite tail of some sequence in 2^w. Now take the > unique element from S associated to that tail, take the n'th element of > that sequence from S, and deliver that as your move. > After some thinking, you will see that with this strategy, you're indeed > certain to guess wrong at most finitely many times. > > Thanks, AC! Another nice mess you've gotten us into. In plain mathematical terms: for any set X, there is a function f:X^N- >X such that, for each sequence (x_1, x_2, ...} in X^N, the equality x_n = f(x_{n+1}, x_{n+2}, ...) holds for all but finitely many n. This is old hat: see Problem 5348, American Mathematical Monthly, volume 72 (1965), p. 1136. (This has been discussed on sci.math in the past.)
From: A on 23 Nov 2009 17:19 On Nov 23, 6:22 am, Herman Jurjus <hjm...(a)hetnet.nl> wrote: > Has anyone seen this before? > > http://possiblyphilosophy.wordpress.com/2008/09/22/guessing-the-resul... > > I'm not sure yet what to conclude from it; that AC is horribly wrong, or > that WM is horribly right, or something else altogether. > > In short the story goes like this: > > A game is played, in which infinitely many coins are tossed, and there's > one player, who makes infinitely many guesses. Both are done over a > finite period of time. The tosses and guesses are not made faster and > faster, however, but slower and slower: at t = 1/n. There's no 'first' > move. > > Claim: > There exists a strategy with which you're certain to guess all entries > correctly except for at most finitely many mistakes. Not 'certain' as in > 'probability is 100%', but absolutely certain. > > Reasoning: > On 2^w, consider the equivalence relation that makes x equivalent to y > when x(n) =/= y(n) for at most finitely many n. Next, using AC, create a > set S that contains precisely one element from every equivalence class. > Strategy: at every move, you already know the results of the previous > tosses, which is an infinite tail of some sequence in 2^w. Now take the > unique element from S associated to that tail, take the n'th element of > that sequence from S, and deliver that as your move. > After some thinking, you will see that with this strategy, you're indeed > certain to guess wrong at most finitely many times. > > Thanks, AC! Another nice mess you've gotten us into. > > -- > Cheers, > Herman Jurjus Am I misunderstanding this? Look at the following paragraph: "The stategy you should adopt runs as follows. At 1/n hrs past 12 you should be able to work out exactly which equivalence class the completed sequence of heads and tails that will eventually unfold is in. You have been told the result of all the previous tosses, and you know there are only finitely many tosses left to go, so you know the eventual completed sequence can only differ from what you know about it at finitely many places. Given you know which equivalence class youre in, you just guess as if the representative of that equivalence class was correct about the current guess. So at 1/n hrs past 12 you just look at how the representative sequence says the coin will land and guess accordingly." Apparently we do not choose a strategy before time begins to pass, but rather, we have only ever chosen a strategy for playing the game once some nonzero amount of time (1/n hours past 12) has passed. But after any nonzero amount of time has passed, we have made infinitely many coin tosses already, and there are only finitely many remaining coin tosses; so one of two things is true: either A) we have already guessed infinitely many coin tosses incorrectly, and hence there is no winning strategy for us, or B) we have already guessed only finitely many coin tosses incorrectly, in which case any strategy we choose will be a winning strategy, since even if we guess all our remaining tosses wrong, we will still have only guessed wrong a finite number of times. In other words, the author of the original article on Wordpress does not seem to be telling us how to choose a winning strategy from the beginning of this game, but rather, how to choose a winning strategy once some finite amount of time has already passed; and in that case, either there exists no winning strategy, or any strategy we choose is a winning strategy. Perhaps I am misunderstanding something here, but this doesn't seem like very novel stuff, and it doesn't seem to have anything to do with the axiom of choice.
From: Butch Malahide on 23 Nov 2009 17:56 On Nov 23, 4:18 pm, Butch Malahide <fred.gal...(a)gmail.com> wrote: > > In plain mathematical terms: for any set X, there is a function f:X^N->X such that, for each sequence (x_1, x_2, ...} in X^N, the equality > > x_n = f(x_{n+1}, x_{n+2}, ...) holds for all but finitely many n. > > This is old hat: see Problem 5348, American Mathematical Monthly, > volume 72 (1965), p. 1136. (This has been discussed on sci.math in the > past.) Alan D. Taylor and C. Hardin have some recent papers on this subject: http://www.math.union.edu/people/faculty/publications/taylora.html A peculiar connection between the axiom of choice and predicting the future (with C. Hardin), American Mathematical Monthly 115 (2008), 91-96. An introduction to infinite hat problems (with C. Hardin), Mathematical Intelligencer 30 (2008), 20-25. Limit-like predictability for discontinuous functions (with C. Hardin), Proceedings of the American Mathematical Society 137 (2009), 3123-3128.
From: Herman Jurjus on 23 Nov 2009 18:01
Butch Malahide wrote: > On Nov 23, 5:22 am, Herman Jurjus <hjm...(a)hetnet.nl> wrote: >> Has anyone seen this before? >> >> http://possiblyphilosophy.wordpress.com/2008/09/22/guessing-the-resul... >> > > In plain mathematical terms: for any set X, there is a function f:X^N- >> X such that, for each sequence (x_1, x_2, ...} in X^N, the equality > x_n = f(x_{n+1}, x_{n+2}, ...) holds for all but finitely many n. > > This is old hat: see Problem 5348, American Mathematical Monthly, > volume 72 (1965), p. 1136. (This has been discussed on sci.math in the > past.) Good to know that! When was that, approximately (I mean the sci.math discussion)? And under what name was/is it known? -- Cheers, Herman Jurjus |