Counterintuitions and the well-ordering theorem
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From: Bill Taylor on 4 Nov 2009 23:34 Well, I think we've got to the point where we're starting to repeat ourselves, and not really listening to what the other has to say. So maybe this will be brief... stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > In the case of reals, presumably > if every real is definable at some level, Yes. > then the collection of all such levels has a supremum, alpha. No; as I said, w_1^CK doesn't really exist in the defineability sense. > The alternative, it seems to me, is to say that there is no *set* > of all reals Correct; taking things strictly. > (because every level produces new reals, and the > levels are never completed). Ditto. As I said, it's powerset that is the real villain, but it can be adopted for smoothness, provided one is subsequently careful, meaning avoiding AC (and similar). > Then you *are* saying that the reals don't form a set. > Well, then AC is definitely not the problem, separation is. Powerset. > 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. You're speaking of Unlimited Comprehension, not Separation. > You are saying that AC > is bad because it produces undefinable sets of reals. But it > doesn't *unless* there are already undefinable reals. This is clearly wrong, and I suspect not what you meant to say. It would be perfectly possible to work entirely with definable reals, but (shun to) have undefineable collections of them. Compare - all naturals are defineable, but (you said above) there may be undefineable collections of them (i.e. reals). > AC would > not be the *source* of the undefinability. It's the innocent bystander. I would rather put it, the executoror enforcer of an immoral law: powerset. -- Worried William
From: Marshall on 4 Nov 2009 23:48 On Nov 4, 8:34 pm, Bill Taylor <w.tay...(a)math.canterbury.ac.nz> wrote: > > I would rather put it, the executoror enforcer of an immoral law: > powerset. Pardon me; am I to understand that you are objecting to the powerset axiom? Marshall
From: Peter Webb on 5 Nov 2009 06:04 "Marshall" <marshall.spight(a)gmail.com> wrote in message news:40b03443-d7d3-4ce2-9cdc-f8e2f5ba4309(a)u16g2000pru.googlegroups.com... On Nov 4, 8:34 pm, Bill Taylor <w.tay...(a)math.canterbury.ac.nz> wrote: > > I would rather put it, the executoror enforcer of an immoral law: > powerset. Pardon me; am I to understand that you are objecting to the powerset axiom? Marshall __________________________ I was there when the crime occurred, and it was clearly the Axiom of Infinity who was the mastermind. Without him there, all the other axioms are completely well behaved. Even the Axiom of Choice is always true, even if you can't quite prove it ... I say get rid of the Axiom of Infinity, and our problems with disruly theorems will disappear completely.
From: Daryl McCullough on 5 Nov 2009 09:08 Bill Taylor says... >> In the case of reals, presumably >> if every real is definable at some level, >Yes. >> then the collection of all such levels has a supremum, alpha. >No; as I said, w_1^CK doesn't really exist in the defineability sense. To me, that's a *completely* meaningless claim. w_1^CK is perfectly definable: It's defined to be the supremum of all ordinals alpha such that there exists a recursive well-ordering of the naturals of order type alpha. You can certainly talk about cutting off the cumulative hierarchy at w_1^CK, so that it doesn't exist in your model, but to say it doesn't exist, period, is nonsensical to me. We can talk about, reason about it consistently. It exists in the same sense that any other abstract mathematical object exists: the square-root of 2, imaginary numbers, etc. >> You are saying that AC >> is bad because it produces undefinable sets of reals. But it >> doesn't *unless* there are already undefinable reals. > >This is clearly wrong, and I suspect not what you meant to say. No, it's clearly true, and it is exactly what I meant to say. If X is a set of nonempty sets of definable reals, then there will *always* be a choice function on X. >Compare - all naturals are defineable, but (you said above) >there may be undefineable collections of them (i.e. reals). Sure, but that has nothing to do with choice. AC does not produce any undefinable sets of reals unless there are *already* undefinable reals. >> AC would >> not be the *source* of the undefinability. It's the innocent bystander. > >I would rather put it, the executoror enforcer of an immoral law: >powerset. I have trouble making any sense of concern about the power set. If A is a set, then the collection of all subsets of A exists as a *concept*. It exists as a proper class (since proper classes are basically just formulas with free variables ranging over some set). To deny power set is to say that this collection doesn't exists *AS* *A* *SET*. But what does that even mean? It certainly makes sense to say *relative* to a model---certain collections appear in the model and other collections do not. But I don't see how it makes sense to talk about P(A) not existing as a set in any absolute sense. Unless you want to say that there is a *standard*, God-given (or Bill-given, since you're an atheist) model for set theory, and it doesn't happen to contain P(A). I still can't grasp what that could mean. If you have in mind a model M that has no P(A) for some particular A, then I can certainly imagine another model M' that is obtained by extending M to a new model that includes P(A). I have a hard time knowing what in the world you could mean by saying that P(A) does not exist, for some A. Do you have some Platonic universe of sets to appeal to, or what? -- Daryl McCullough Ithaca, NY
From: Bill Taylor on 5 Nov 2009 22:40
stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> That isn't true. Real numbers did not have names before Cantor. > >Oh heavens, of course they did! > No, they did not. *Some* reals had names. Oh. It would have been clearer if you'd prefaced the above with "some" or "all" . No matter. Go ahead with your evidence that (pre-Cantor) there were any reals that did not have names. By "name", OC, I mean a definition. > > I'll ignore the trivial fact that all Euclid's diagrams had letters > You don't understand the difference between a *variable* and a > unique name? When someone says "Let A be a point, and let R be OK OK, it was just a flippancy. I won't make any more! > >No-one doubts he [Descartes] was doing Euclidean geometry, and > >extending it, and all his points were real number pairs, see just above. > > Not every real has a name, then not every pair of reals has a name. So, we agree as regard geometry - that it comes down to naming reals. So we cannot proceed, until we have your answer to the above request. > That is completely ridiculous. Of course, if a real comes up > in an application, then you *give* it a name. That's where "e" > came from. That's where "pi" came from. OC. So where does any real come up that is NOT nameable, in this sense? This is still what I'm wondering about. -- Wondering William |