Counterintuitions and the well-ordering theorem
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From: Bill Taylor on 30 Oct 2009 00:18 OK, these are my last 2 responses for today, bringing me up to date for posts in this thread that appeared on or before yesterday. Apologies if I'm repeating stuff, or if I'm making points that have already been answered. Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: >> But: either player 1 has a winning strategy or he hasn't. [etc] > I think it's a good idea to explicitly introduce quantifiers, I agree! Wholeheartedly. It was the consideration of quantifiers that convinced me, almost immediately, that AD just had to be true. Of course, I like to consider sequential games *sequentially*, and that means being able to extend quantifiers to omega depth, as Daryl has already noted. Thanks Daryl (!) - you saved me a lot of preliminary work. So here goes:- Either Ea Ab Ec Ad Ee Af.... <a, b, c, d, e, f ... > in WIN-1 (*) is true, or it is false. This is the LEM that was adduced earlier. If * is true, it says precisely that player-1 has a winning strategy, now given sequentially (as seems most natural), rather than simultaneously (as required by FOL-ZF)). If it is false, then it is true that ~ Ea Ab Ec Ad Ee Af.... <a, b, c, d, e, f ... > in WIN-1; and this is DeMorgan transformable (quite legitimately by initial segements!) into... Aa Eb Ac Ed Ae Ef.... <a, b, c, d, e, f ... > in WIN-1^C; ....which is precisely the statement that P-2 has a winning strategy. Q.E.D. ......... Now OC, we do not (yet) have a formalism in which we can treat with such infinitely long statements, but attempts have been made from time to time with varying success and domains of application. I would claim, that these statements are of such AMAZINGLY simple and "uniform" types, that they would be part of any reasonable successful formalism, and that thus AD would be provable in that context. This may well be a pipe dream, as Aatu suggests, but it is a clear-cut intuition in favour of AD, it uses quantifiers clearly and concisely, (I have removed simplifying subscripts in favour of a,b,c,d... for typographic clarity), and is far more believable than considerations based on infinite sets of ridiculously complex and inexpressible types. -- Battling Bill.
From: Bill Taylor on 30 Oct 2009 00:47 Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > This is a nice example of divergent intuitions! It is clearly so. > My intuitions tell me nothing whatever about the existence of > winning strategies for chess or checkers. You have been saying so all along. Perhaps game players have better-informed intuitions. (Perhaps not.) The rest of this article is irrelevantto the thread, but may be interesting as a sidelight. - - - - - - > I have played chess about five > times in my life, the plays consisting of my moving the pieces > essentially at random, the opponent declaring at some arbitrary, > or so it seemed to me, point that I'd lost. Don't you just HATE opponents like that!? It's even worse at Go, where the learner is expected to believe all his "dead" stones can just be removed at the end, without any scoring penalty! The proper thing is for the better player to ask, "You agree you're hopelessly lost now?", and if there is any demurral or hesitation, just play on till checkmate is actually thrashed out on the board. It may be that the better player was just trying to save you from this apparent humiliation, but in the face of even the slightest demurral, it must be done. Get your nose rubbed in it! It might even make you learn quicker! > I'm also utterly tone deaf, and > have no aptitude for crossword puzzles; and mentally adding two > two-digit numbers takes about half a minute for me. Yikes! These are VERY damaging admissions, IMHO. > I'm very proud of and pleased with all these shortcomings, YIKES AGAIN! You transform them from an *admission* into a CLAIM! This reminds me of the perfect squelch, to those cocktail-party matrons who sweetly declare with faux-superior innocence, "Oh I was always TERRIBLE at math, at school!" ....as if it were a prideful thing. The proper reply is... "Oh don't worry, I never learned to read!" Still, Aatu, at least it seems you can read; unless you're getting a reader-writer to do all the heavy lifting for you, (like those pampered, allegedly dyslexic students we see more and more of these days, who must think they're going to have reader-writers throughout their working life!) - - - - - For my part:- I'm quite good at chess, even better at Go, and many chess variants and a great many other board games; have excellent ear for tunes and songs, recall a great many, can play a few (very slightly), and sing most; am excellent at crossword puzzles, especially in tandem with my wife; also sudoku; am great at cards and Scrabble; and can do mental arithmetic up to factorizing three-digit numbers very quickly indeed, (due to long practice). I have also a large, self-instructed familiarity with most forms of art and philosophy, to the point where I can easily hold my own with professionals in these topics, even though I regard the topics as mostly rubbish. There! How am I doing in the claims stakes? Oh - I also have a very long-lasting marriage to a devoted wife, with two very happy and societally effective children.... -- Well-rounded William.
From: Rupert on 30 Oct 2009 00:54 On Oct 30, 12:54 pm, Bill Taylor <w.tay...(a)math.canterbury.ac.nz> wrote: > stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > I don't understand why you think that "all games are determined" is > > intuitively true. > > You gave the answer yourself, in terms of infinite-depth quantifiers. > > > It's not *obvious* that chess or checkers has a > > winning strategy; it's *provable*. > > It IS. (That is, a winning strategy or a drawing strategy for both.) > > This just plain OBVIOUS to any game player. > It was obvious to me even before I started high school. > > >To prove it, you have to use the > > fact that they are finite-length games > > Yes, the proof requires finitude, but the intuition does NOT. > Again, consider your own i-d quantifiers! > > > What reason is there for believing that the principle > > applies to games for which is not provable? > > I think you are asking - whose intuitions about infinite-depth > games is more reliable. Is it the man in the street, > or the hardened game-player, or the hardened math-logician? > > Obviously mileage will vary! > > -- Board-gaming Bill Sorry, I know you've addressed this elsewhere in the thread, but I wonder if we could just go through this. I don't believe that AD is true. I have some experience playing games, but maybe not as much as you. Do you want to have a go at convincing me that it's true?
From: Daryl McCullough on 30 Oct 2009 08:29 Bill Taylor says... > >stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> As a matter of fact, I think it is provably *false* for some natural >> ways of formulating continuity. > >Whoa! That IS interesting! Would you do us all a kindness, Daryl, >and teach us about some of these natural alternatives. Thinking about it, I think the ambiguity is with how "connected" is defined. Okay, so here's a claim that I think is equivalent to the Jordan curve theorem, or at least it's similarly intuitively true: Consider the square with corners at (x,y)= (-1,-1), (-1,1), (1,1) and (1,-1). Let C1 and C2 be two connected curves such that C1 connects points (-1,1) and (1,-1), and C2 connects points (-1,-1) and (1,1). Then C1 and C2 must intersect. (assuming that they never leave the square). Now, the counterexample is this: Let C1 be constructed as follows: C1 = the line segment from (-1,1) to (0,1/2) together with the set of points (x,y) such that 1 >= x > 0 and y = 1/2 + 1/2 sin(pi/(2x)). Let C2 be constructed as follows: C2 = the line segment from (-1,-1) to (0,0) together with the set of points (x,y) such that 1 >= x > 0 and y = -1/2 + 1/2 sin(pi/(2x)), together with the line segment from (1,0) to (1,-1). C1 and C2 are connected curves in the sense that they are one-dimensional, connected subsets of the plane, and they don't intersect. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 30 Oct 2009 08:58
Bill Taylor says... >And I meant to add - this was what was intended to happen with >the Banach-Tarski decomposition, or so I have read. >Allegedly, one of them hoped it would be so ludicrous that >mathies as a whole would thus drop AC, but that didn't happen. Actually, the hard work of the Banach-Tarski doesn't involve choice at all. If we parameterize the points on the sphere by longitude and lattitude, consider the set of points such that the longitude and lattitude are rational multiples of pi radians. This object, a "holy" sphere, can be decomposed into a finite number of pieces that can be reassembled by translations and rotations to form two identical copies of the original holy sphere. That by itself is pretty amazing to me. And it doesn't involve the axiom of choice (basically because choice already holds for a countable set). Choice just allows basically the same proof to go through for all points on the sphere, not just a countable dense subset. -- Daryl McCullough Ithaca, NY |