Another AC anomaly?
in [Logic]
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From: Virgil on 24 Nov 2009 14:57 In article <87pr77zpng.fsf(a)phiwumbda.org>, "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote: > George Greene <greeneg(a)email.unc.edu> writes: > > > On Nov 23, 9:48�am, Herman Jurjus <hjm...(a)hetnet.nl> wrote: > >> More conclusive (at least for me): you switch a light bulb on and off; > >> after infinitely many steps, is the light on or off? > > > > Well, there is more than one infinity. > > Ordinally, there are even successor ordinals after > > the first infinity. The reversed order (with no first element) can > > also have a last one. Basically, any infinity of steps that has > > a last element will have an answer to this question. > > Any infinite sequence that does not have a last element > > needs to get its "answer" from some NON-standard convention. > > No, that's not enough. > > Suppose that the light begins "on", at each step, I toggle the state. > It seems that you agree that after omega-many steps, we do not know > whether the light is on or off. But if we do not know at omega > whether the light is on or off, then surely we do not know whether it > is on or off at omega + 1. > > Right? I'm not sure that going from omega to omega + 1 is a "step" in the same sense as going from, say, 5 to 6 is a step. > > To put it differently, you claim "any infinity of steps that has > a last element will have an answer to this question." w + 1 is an > "infinity of steps" with a last element, but if we have an answer at > w + 1, then we also have an answer at w.
From: Daryl McCullough on 24 Nov 2009 22:52 Jesse F. Hughes says... >Let's use the usual trick: For n in w, step n occurs at t - 1/n. >Let's also let step w + n occur at t + 1 - 1/n (for n > 0), so step=20 >w + 1 occurs at time t. Thus, we have actions occurring at > >t - 1, t - 1/2, t - 1/3, ... t, t + 1/2, t + 2/3, .... > >Seems to me if we're willing to grant the original supertask, then >there's nothing less plausible about the new supertask. > >At t - 1, the bulb is on. Immediately after t - 1 (and up to and >including t - 1/2), the bulb is off. Immediately after t - 1/2, it's >on again, and so on. > >Now, according to George, we should have no trouble figuring out what >happens immediately after time t, if I read him right. But in the >original task (when nothing was done *at* t), we didn't know the state >of the bulb at t. I don't see how we know the state of the bulb >immediately after t either. It's not specified by the "theory" of light switches. What we are assuming about light switches is that if it is switched on at time t_1, and it is not switched off between t_1 and t_2, then it is still on at time t_2. If it is switched off at time t_1, and is not switched on between t_1 and t_2, then it is still off at time t_2. Now, we add to this the assumption that the light switch is turned on at times t - 1, t - 1/3, t - 1/5, ..., and that is switched off at times t - 1/2, t - 1/4, ... So we have a bunch of statements about what the state of the light is at various times, and those statements don't determine what the state of the light is at time t. -- Daryl McCullough Ithaca, NY
From: LauLuna on 25 Nov 2009 15:49 On Nov 23, 12:22 pm, 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 In fact, you don't need the axiom of choice. At any 1/n hour past 12pm it is already determinate what equivalence class the eventual sequence is in. Since you know what the previous results are and there are only finitely many outstanding results, you can complete the sequence at random and take it as your representative. Which means that from any 1/n hr past 12pm on you can guess at random. That is, you can make all your guesses at random and still be certain to guess wrong only finitely many times. I'd say the paradox arises from the assumption that an infinite sequence of past events is terminated. So it seems akin to some versions of Benardete's and Yablo's paradoxes. The following is a version of Benardete's. Consider an infinite past with a gong peal occurring at each day and a hearer being deafened by it iff no previous gong peal has deafened it. The hearer must be deaf from eternity, which, paradoxically, implies that no gong peal deafens it. The problem is that at any day it is already determinate whether the hearer is deaf or not and in such a way that no event at no day can contribute to the fact. Similarly, what the equivalence class of the eventual sequence is, is determinate at any 1/n, no matter how big n is; hence no toss at no 1/ n contributes to the fact. Of course, this is paradoxical. Regards.
From: Mike Terry on 25 Nov 2009 16:07 "LauLuna" <laureanoluna(a)yahoo.es> wrote in message news:36cb4819-04fd-4dbe-b204-7168ae550e23(a)p8g2000yqb.googlegroups.com... > On Nov 23, 12:22 pm, 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 > > In fact, you don't need the axiom of choice. > > At any 1/n hour past 12pm it is already determinate what equivalence > class the eventual sequence is in. Since you know what the previous > results are and there are only finitely many outstanding results, you > can complete the sequence at random and take it as your > representative. Which means that from any 1/n hr past 12pm on you can > guess at random. ...but then how would you prove that there have been only a finite number of incorrect guesses? If the guesses always follow the representative sequence (whose existence follows from AC) it's this that allows us to deduce at the end that we've only made finitely many mistakes. > > That is, you can make all your guesses at random and still be certain > to guess wrong only finitely many times. > Your proof for this doesn't work... Mike.
From: Tim Little on 25 Nov 2009 19:24
On 2009-11-25, LauLuna <laureanoluna(a)yahoo.es> wrote: > At any 1/n hour past 12pm it is already determinate what equivalence > class the eventual sequence is in. Since you know what the previous > results are and there are only finitely many outstanding results, > you can complete the sequence at random and take it as your > representative. Which means that from any 1/n hr past 12pm on you > can guess at random. That strategy fails for the same reason as the previous non-choice strategy. The choice sequence works only because you provably used the same sequence in the *past*, and hence at the time of your decision had already made only finitely many errors. > That is, you can make all your guesses at random and still be > certain to guess wrong only finitely many times. No, in your case you can't prove anything at all about how many of your past guesses were correct. > The following is a version of Benardete's. Consider an infinite past > with a gong peal occurring at each day and a hearer being deafened by > it iff no previous gong peal has deafened it. The hearer must be deaf > from eternity, which, paradoxically, implies that no gong peal deafens > it. Yes, mixing in other types of paradox is one reason why I prefer other formulations of this kind of AC problem. - Tim |