Counterintuitions and the well-ordering theorem
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From: Daryl McCullough on 30 Oct 2009 09:21 Bill Taylor says... > >OHerman Jurjus <hjm...(a)hetnet.nl> wrote: > >> But: either player 1 has a winning strategy or he hasn't. >> Now what does it mean for player 1 to not have a winning strategy? >> >> I'd say that amounts to 'player 2 has some way to prevent player 1 from >> winning'. > >That sound fairly unimpeachable. It certainly doesn't to me. To say that player 1 has no winning strategy means that, no matter how brilliant player 1 is, there is a player 2 who can beat him. That's just different from saying that there is a player 2 who can beat any player 1. The fact that Bob as first player can always beat Carol, and Ted as first player can always beat Alice, does *not* imply that Bob can always beat Alice. So the equivalence "Player 1 has no winning strategy" does not *mean* the same thing as "Player 2 has a winning strategy". It just doesn't. Now, they might be equivalent for some games, but they don't *mean* the same thing, which is what Herman was implying. -- Daryl McCullough Ithaca, NY
From: James Burns on 30 Oct 2009 10:04 Daryl McCullough wrote: > 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. Cool! It might be interesting to point this out to the next person to prove that the reals are countable -- that /if/ the reals are countable, then Banach-Tarski is unavoidable. Jim Burns
From: Herman Jurjus on 30 Oct 2009 12:25 Daryl McCullough wrote: > Bill Taylor says... >> OHerman Jurjus <hjm...(a)hetnet.nl> wrote: >> >>> But: either player 1 has a winning strategy or he hasn't. >>> Now what does it mean for player 1 to not have a winning strategy? >>> >>> I'd say that amounts to 'player 2 has some way to prevent player 1 from >>> winning'. >> That sound fairly unimpeachable. > > It certainly doesn't to me. To say that player 1 has no winning > strategy means that, no matter how brilliant player 1 is, there is > a player 2 who can beat him. That's just different from saying > that there is a player 2 who can beat any player 1. The fact > that Bob as first player can always beat Carol, and Ted as > first player can always beat Alice, does *not* imply that > Bob can always beat Alice. > > So the equivalence "Player 1 has no winning strategy" > does not *mean* the same thing as "Player 2 has a winning > strategy". It just doesn't. Now, they might be equivalent > for some games, but they don't *mean* the same thing, which > is what Herman was implying. I didn't (intend to) imply that they conceptually /mean/ the same thing, but only that one 'should' imply the other, in a sense similar to the sense in which the Jordan curve theorem 'should' be true. (Something like "no doubt you can make a system in which the two are not equivalent, but then that system does not represent the 'real' universe of sets, the 'real' notion of sequence of numbers, etc.") As regards AD+AC being 'true' simultaneously: When pushed to the extreme, i can't take on a 'realist' point of view regarding sets, without either ignoring parts of my intuitions (and making an arbitrary choice about which parts), or accepting both AD and AC (which is not really an option, of course, unless we get extreme and drop FOL or so - but this would in itself constitute a rejection of realism). Hence, i can't be a 'realist' when it comes to 'the real universe of sets' without being intellectually dishonest to myself to some degree. I must take *some* part of it with a grain of salt. -- Cheers, Herman Jurjus
From: Bill Taylor on 31 Oct 2009 01:06 stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >>You gave the answer yourself,in terms of infinte-depth quantifiers > > But I really don't have an intuition about what infinitely many > alternations of quantifiers *mean*, Of course you do! (Else I have a quite different understanding of the words you use.) The mere fact that you *came up* with the idea, independently of me, shows you have a fair idea of what it means. Naturally we don't understand *every* nuance, implication, relation and what-not about them, but that could be said of most math things. > unless I can do the Skolem > trick of replacing them by a single quantifier over strategies. Sure you *can* do that if it helps. It's a parallel way of looking at the same thing, but not identical. As you say, Skolem's automatically raise questions of simultaneity and choice. > >> It's not *obvious* that chess or checkers has a > >> winning strategy; > >>It IS.(That is, a winning strategy or a drawing strategy for both) > > >It was obvious to me even before I started high school. > > I'm sure you never played infinite games before high school. I's talking about CHESS, you goose, when I mentioned high school! :) > I don't agree. Once you introduce infinite games, the intuition > disappears completely for me. Well, I could say exactly the same thing about choice, which you seem to find no problems with. How are we to resolve this quagmire!? > I know that some things that are > true for finite objects are not true for infinite objects, Apparently these do not include choice... > >Again, consider your own i-d quantifiers! > > That's a *notation*. Nothing follows from a notation. OCN. It is the concept that the notations (attempt to) represent that things follow from. > Using infinite quantifiers, we can't (as far as I know) make > the distinction between "Player 1 has no winning strategy" and > "Player 2 has a winning strategy". OCN! Because there isn't one, (or so I claim), for tie-free games. > It's not expressive enough to make that distinction. Nothing is expressive enough to make distinctions that don't exist! -- Wriggling William ** Set theory is a shotgun marriage - between well-ordering and power-set. ** The two parties get along OK; but they hardly seem made for each other.
From: Bill Taylor on 31 Oct 2009 01:15
stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > People assume the axiom of choice because reasoning without > is an incredible pain. That is a very wise observation! And you examples are all well-chosen. One might add that it's very slickly comforting to know that all vector spaces have bases, and many other similar uniformity results. But I note you have slipped into "ease and elegance" mode of reasoning, or rather of apologetics, as opposed to "truth and philosophical clarity" mode, which we started with. > For example, without choice, we can't reason > about cardinality based on bigger/smaller. There are many different > notions of cardinality that are *inequivalent*: X < Y can be defined > by the existence of an injection from X to Y, or the existence of a > surjection from Y to X, and the two definitions are inequivalent. As you say, these things are all a terrible pain for the mainstream mathie doing mainstream math. OTOH, they are a fascinating bywater tributary to the main stream, for those inclined to dawdle there! > and seemingly not in an interesting way. Not interesting to the mainstream sculler. But fascinating to the dawdler. -- Backwater Bill |