Counterintuitions and the well-ordering theorem
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From: Bill Taylor on 2 Nov 2009 23:07 O stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > I think it's still true that the axiom of choice would hold > for such definable sets. It would, if you could gather enough together to talk about them all simultaneously, but you can't. > The only way to *falsify* the axiom of choice is to have a set > whose elements are *not* definable (at any level). At least, > that's the way it seems to me. Well, were not trying to falsify it formally, in any sense; that would probably be impossible. We're just noticing that it's not true, of the "sets" that people use it on. The real problem, OC, is not AC itself, but the attempt to talk about "all real numbers at once" - this is where ZF breaks down. However, we are stuck with ZF for various reasons, and that is OK; but if we have to swallow Powerset, we should not go the whole hog, and throw good money after bad, and swallow AC itself - it just ISN"T true, of sets of subsets of reals of the type it is used for. Swallowing the C after the ZF will just painfully stick in our craws. -- Bronchial Bill
From: Bill Taylor on 2 Nov 2009 23:13 Herman Jurjus <hjm...(a)hetnet.nl> wrote: > > AC produces sets, by fiat of existence, which simply DON'T EXIST! > > > Non-measurable sets in R,free ultrafilters on N, partitions of R^3 > > into non-parallel lines - all these things simply don't exist, > > in that no explicit set can be named, with those properties. > > And AD doesn't produce sets (strategies, certain large cardinals) that > also "simply don't exist" in this same sense? Well, it's odd, isn't it, but it doesn't seem to!? I would be delighted to see a counterexample. That is, can you produce a well-defined set, A, such that there ISN'T, on the face of it, any clear winning strategy for game A? As I say, I'd LOVE to see one. I haven't been able to find one. -- Baffled Bill
From: David C. Ullrich on 3 Nov 2009 06:44 On Mon, 2 Nov 2009 20:13:15 -0800 (PST), Bill Taylor <w.taylor(a)math.canterbury.ac.nz> wrote: >Herman Jurjus <hjm...(a)hetnet.nl> wrote: > >> > AC produces sets, by fiat of existence, which simply DON'T EXIST! >> >> > Non-measurable sets in R,free ultrafilters on N, partitions of R^3 >> > into non-parallel lines - all these things simply don't exist, >> > in that no explicit set can be named, with those properties. >> >> And AD doesn't produce sets (strategies, certain large cardinals) that >> also "simply don't exist" in this same sense? > >Well, it's odd, isn't it, but it doesn't seem to!? > I would be delighted to see a counterexample. This is your lucky day. AD implies the existence of a model of ZF. That model "simply doesn't exist", if we take that to mean "no explicit set can be named which is a model of ZF" as above. >That is, can you produce a well-defined set, A, such that >there ISN'T, on the face of it, any clear winning strategy for game A? > >As I say, I'd LOVE to see one. I haven't been able to find one. > >-- Baffled Bill David C. Ullrich "Understanding Godel isn't about following his formal proof. That would make a mockery of everything Godel was up to." (John Jones, "My talk about Godel to the post-grads." in sci.logic.)
From: Daryl McCullough on 3 Nov 2009 06:49 Bill Taylor says... > >O stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> I think it's still true that the axiom of choice would hold >> for such definable sets. > >It would, if you could gather enough together to talk about >them all simultaneously, but you can't. I'm not sure what that means. In the case of reals, presumably if every real is definable at some level, then the collection of all such levels has a supremum, alpha. In which case, there would be a definable choice function on reals at level alpha+1. The alternative, it seems to me, is to say that there is no *set* of all reals (because every level produces new reals, and the levels are never completed). >> The only way to *falsify* the axiom of choice is to have a set >> whose elements are *not* definable (at any level). At least, >> that's the way it seems to me. > >Well, were not trying to falsify it formally, in any sense; that >would probably be impossible. We're just noticing that it's >not true, of the "sets" that people use it on. I'm noting just the opposite: *if* people only use definable sets, *then* there is a definable well-ordering of those sets. We can well-order the reals as follows: Let level(r) = the first level at which r becomes definable. Then we can say r1 < r2 if level(r1) < level(r2) or if level(r1) = level(r2) and the formula defining r1 is less complicated than the formula defining r2. >The real problem, OC, is not AC itself, but the attempt to talk about >"all real numbers at once" - this is where ZF breaks down. Then you *are* saying that the reals don't form a set. Well, then AC is definitely not the problem, separation is. Whenever we talk about the set of all reals r such that Phi(r), that's an illegitimate operation if the reals don't form a set. It would be a proper class, perhaps. >However, we are stuck with ZF for various reasons, and that >is OK; but if we have to swallow Powerset, we should not go >the whole hog, and throw good money after bad, and swallow >AC itself - it just ISN"T true, of sets of subsets of reals of >the type it is used for. Swallowing the C after the ZF will >just painfully stick in our craws. That doesn't make a bit of sense to me. You are saying that AC s bad because it produces undefinable sets of reals. But it doesn't *unless* there are already undefinable reals. AC would not be the *source* of the undefinability. It's the innocent bystander. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 3 Nov 2009 06:51
Bill Taylor says... >utch Malahide <fred.gal...(a)gmail.com> wrote: > >> What objects of Euclid's geometry have explicit names or descriptions? > >All of them. No, they do not. Lines and points don't have names in Euclidean geometry. -- Daryl McCullough Ithaca, NY |