Another AC anomaly?
in [Logic]
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From: Jesse F. Hughes on 23 Nov 2009 11:02 Herman Jurjus <hjmotz(a)hetnet.nl> writes: > More conclusive (at least for me): you switch a light bulb on and > off; after infinitely many steps, is the light on or off? Yes, of course, that's much simpler and clearer. -- Jesse F. Hughes "You know, I -- when I speak, like right now, for example -- I'm speaking to the American people, of course, and I want them to know that I know how tough it is[...]" -- Reassuring words from Pres. Bush
From: Herman Jurjus on 23 Nov 2009 11:45 William Hughes wrote: > On Nov 23, 10:09 am, Herman Jurjus <hjm...(a)hetnet.nl> wrote: >> William Hughes wrote: >>> On Nov 23, 7:22 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. >>> Note you can avoid AC, by defining >>> a_n(m)= s(m) m>n, a_n(k) heads otherwise. >> What's s(m) ? >> >> As a matter of fact, what's a_n(m) ? >> The game involves tosses and guesses, i.e. two sequences with one index: >> toss(n), guess(n). >> >> -- >> Cheers, >> Herman Jurjus > > > s is the sequence of tosses, so s(n) = toss(n) > > As above, a_n(m) is equal to the toss if m>n, otherwise > is a head. Note that a_n is equal to s except perhaps on > a finite set. > > For any n, a_n(n) is heads, so guess(n)=a_n(n) is heads. That would mean you always guess heads. But if you always guess heads, and the tosses are random, you will make infinitely many mistakes with probability 1. > I don't think the fact that time is reversed has any > real bearing on this. If the tosses are independent, perfect > knowledge of the future is of no help for the present. > > From the quoted page > > Now, since the representative sequence and the actual > sequence of heads and tails that occurs are in the same > equivalence class, they must only differ at finitely many > places. > > Correct. Note this is true of sequence a_n. Huh? > So, if you guessed according to the representative > sequence, you have only made finitely many mistakes. > > Correct. However, you did not guess acconding to the > representative sequence. You only used a_n for guess(n). > For other guesses you used other sequences. What other guesses? At t=1/n, you guess 'heads'/'tails' once, after which one toss is done. And obviously the game lets you make each guess before the toss is done, otherwise it would be rather easy to make no mistakes at all. > > > - William Hughes -- Cheers, Herman Jurjus
From: William Hughes on 23 Nov 2009 12:25 On Nov 23, 12:45 pm, Herman Jurjus <hjm...(a)hetnet.nl> wrote: > William Hughes wrote: > > On Nov 23, 10:09 am, Herman Jurjus <hjm...(a)hetnet.nl> wrote: > >> William Hughes wrote: > >>> On Nov 23, 7:22 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. > >>> Note you can avoid AC, by defining > >>> a_n(m)= s(m) m>n, a_n(k) heads otherwise. > >> What's s(m) ? > > >> As a matter of fact, what's a_n(m) ? > >> The game involves tosses and guesses, i.e. two sequences with one index: > >> toss(n), guess(n). > > >> -- > >> Cheers, > >> Herman Jurjus > > > s is the sequence of tosses, so s(n) = toss(n) > > > As above, a_n(m) is equal to the toss if m>n, otherwise > > is a head. Note that a_n is equal to s except perhaps on > > a finite set. > > > For any n, a_n(n) is heads, so guess(n)=a_n(n) is heads. > > That would mean you always guess heads. > But if you always guess heads, and the tosses are random, you will make > infinitely many mistakes with probability 1. Correct. > > > I don't think the fact that time is reversed has any > > real bearing on this. If the tosses are independent, perfect > > knowledge of the future is of no help for the present. > > > From the quoted page > > > Now, since the representative sequence and the actual > > sequence of heads and tails that occurs are in the same > > equivalence class, they must only differ at finitely many > > places. > > > Correct. Note this is true of sequence a_n. > > Huh? a_n differs from the sequence of tosses at finitely many places. > > > So, if you guessed according to the representative > > sequence, you have only made finitely many mistakes. > > > Correct. However, you did not guess acconding to the > > representative sequence. You only used a_n for guess(n). > > For other guesses you used other sequences. > > What other guesses? > At t=1/n, you guess 'heads'/'tails' once, after which one toss is done. Take two different times, o and p. The guess for time o is based on sequence a_o. The guess for time p is based on sequence a_p [I think you may be confusing the sequence a_n, with the nth element of sequence a_n, a_n(n)]. For each guess you use a different sequence. - William Hughes
From: Mike Terry on 23 Nov 2009 13:41 "William Hughes" <wpihughes(a)hotmail.com> wrote in message news:0210e1fa-23fa-4af9-a59b-17e2f0b2d0fc(a)m33g2000vbi.googlegroups.com... > On Nov 23, 10:09 am, Herman Jurjus <hjm...(a)hetnet.nl> wrote: > > William Hughes wrote: > > > On Nov 23, 7:22 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. > > > > > Note you can avoid AC, by defining > > > a_n(m)= s(m) m>n, a_n(k) heads otherwise. > > > > What's s(m) ? > > > > As a matter of fact, what's a_n(m) ? > > The game involves tosses and guesses, i.e. two sequences with one index: > > toss(n), guess(n). > > > > -- > > Cheers, > > Herman Jurjus > > > s is the sequence of tosses, so s(n) = toss(n) > > As above, a_n(m) is equal to the toss if m>n, otherwise > is a head. Note that a_n is equal to s except perhaps on > a finite set. > > For any n, a_n(n) is heads, so guess(n)=a_n(n) is heads. > > I don't think the fact that time is reversed has any > real bearing on this. If the tosses are independent, perfect > knowledge of the future is of no help for the present. > > From the quoted page > > Now, since the representative sequence and the actual > sequence of heads and tails that occurs are in the same > equivalence class, they must only differ at finitely many > places. > > Correct. Note this is true of sequence a_n. Yes, but there is not one sequence a_n, but rather one for every n... In what follows, I'm using: g := our guess sequence r := the "representative" sequence. Perhaps you're wanting me to have r_n instead of r? But then we would just have r_n = r_m for all n,m, (since they differ only in finitely many places) So there is just one r sequence... (I think you've missed this point?) > > So, if you guessed according to the representative > sequence, you have only made finitely many mistakes. > > Correct. Good... you're agreeing that we're only going to guess wrong a finite number of times... (given the unachievable physical nature of the experiment in the first place!) > However, you did not guess acconding to the > representative sequence. Yes we did! That's our strategy: guess according to the representative sequence. Note that there is only *one* representative sequence r that we are using for ALL our guesses. (I.e. our g(n) = r(n) for all n.) > You only used a_n for guess(n). > For other guesses you used other sequences. Well, that's why your approach doesn't work, while the one suggested in the article does! (given the unachievable physical nature of the experiment in the first place etc.) In the article, r(n) is just one sequence, and our g(n) = r(n) for all n. So g(n) = s(n) for all but finitely many n (by definition of r). (So you can't claim to be eliminating the use of AC, since your approach doesn't work.) Regards, Mike.
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.) |