Counterintuitions and the well-ordering theorem
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From: George Greene on 23 Nov 2009 08:16 On Nov 22, 6:21 pm, Rupert <rupertmccal...(a)yahoo.com> wrote: > There is one sense in which the difference is clear. Let RA be a > theory in a two-sorted language, one sort of variables for real > numbers and one sort of variables for sets of real numbers, and take > the axioms for a complete ordered field and unrestricted comprehension This is a two-sorted FIRST-order language up to this point, right? > and take the deductive closure in second-order logic. But now it has just become second-order?? > Con(RA) is strictly stronger than Con(PA) in PA, Con(PA) is *in* PA, according to this. But is Con(RA) supposed to be in RA or in PA? > assuming that PA+Con(PA) is consistent. Again, are we in a 2nd-order language here or a 2-sorted 1st-order one? And is the answer to that question the same or different for PA vs. RA? > But you can easily formulate a second-order theory of > complex numbers and show that it is equiconsistent and indeed bi- > interpretable with RA. You can easily formulate a 2nd-order theory of PA (of natural numbers), too. It is a much different/bigger/stronger theory than any 2-sorted 1st- order amplification of PA. Does THAT increase NOT occur for 2-sorted 1st- order theories of real or complex numbers? > So in one sense the difference is clear; namely > that we have an increase in consistency strength. Clarity in general is easy to lose when talking about 2-sorted 1st- order theories and 2nd-order theories over the same signature.
From: George Greene on 23 Nov 2009 08:19 On Nov 22, 10:14 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > But with more knowledge about the > brain functions everything can be reduced to a physical object. In > particular because of this fact the assumption of unidentifyable > numbers having an independent existence is medieval superstition. Nobody is REALLY "assuming" that the letter 'a' has an "indepenent existence". That question doesn't even NEED to be addressed! The brute fact of the matter remains that people WILL CONTINUE, regardless of your personal philosophical objections, to speak AS THOUGH symbols existed. The sign on the door above the room SAYS sci.logic. In HERE, we absolutely HAVE reified all consistent abstractions.
From: Rupert on 23 Nov 2009 16:11 On Nov 24, 12:16 am, George Greene <gree...(a)email.unc.edu> wrote: > On Nov 22, 6:21 pm, Rupert <rupertmccal...(a)yahoo.com> wrote: > > > There is one sense in which the difference is clear. Let RA be a > > theory in a two-sorted language, one sort of variables for real > > numbers and one sort of variables for sets of real numbers, and take > > the axioms for a complete ordered field and unrestricted comprehension > > This is a two-sorted FIRST-order language up to this point, right? > Yes, correct. > > and take the deductive closure in second-order logic. > > But now it has just become second-order?? > This is quite a rare moment because I'm actually acknowledgeing a correction from you. I meant to say "first-order". Very sorry, and thanks for the clarification. > > Con(RA) is strictly stronger than Con(PA) in PA, > > Con(PA) is *in* PA, according to this. > But is Con(RA) supposed to be in RA or in PA? > Con(RA) is a sentence in the first-order language of arithmetic, right? My claim is that PA+Con(RA) proves Con(PA) but PA+Con(PA) does not prove Con(RA). > > assuming that PA+Con(PA) is consistent. > > Again, are we in a 2nd-order language here or a 2-sorted 1st-order > one? Let's say a two-sorted first-order theory. I would say "it doen't really matter" but I don't want you to explode. I know it makes a big difference as far as the *semantics* is concerned but I am just trying to specify a formal theory. I could have said "the deductive closure of the axioms for complete ordered field in second-order logic", and I would have said that before Keith Ramsay educated me about the fact that you need to make it clear *which* second-order logic. I don't mean *full* second-order logic, because I want a recursively enumerable theory. So I am describing it as a theory in a two-sorted first-order language and specifying a bunch of axioms. > And is the answer to that question the same or different for PA vs. > RA? > Well, of course the language for PA is the first-order language of arithmetic. > > But you can easily formulate a second-order theory of > > complex numbers and show that it is equiconsistent and indeed bi- > > interpretable with RA. > > You can easily formulate a 2nd-order theory of PA (of natural > numbers), too. Yes, but which second-order logic do you want to use? Are you talking about the deductive closure of the Peano axioms in *full* second-order logic? > It is a much different/bigger/stronger theory than any 2-sorted 1st- > order > amplification of PA. If you mean the deductive closure in *full* second-order logic, yes. I only wanted to talk about recursively enumerable theories. > Does THAT increase NOT occur for 2-sorted 1st- > order > theories of real or complex numbers? > Yes, it does, but what I am doing is specifying a *partial* axiomatisation of second-order logic and using that for my theories, so that I have recursively enumerable theories. > > So in one sense the difference is clear; namely > > that we have an increase in consistency strength. > > Clarity in general is easy to lose when talking about 2-sorted 1st- > order > theories and 2nd-order theories over the same signature. Well, you caught me out in a typo, but I think it should now be reasonably clear which formal theories I want to discuss. I say "two- sorted first-order language" *because* I want to make it clear that I don't wish to use full second-order logic.
From: Rupert on 23 Nov 2009 16:20 On Nov 6, 1:08 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > 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 Geoffrey Hellman interprets set existence claims as claims about the *possibility* of some structure. I can understand the claim perfectly well from that angle. It is a claim that a certain second-order sentence is not in fact satisfiable, even though ZFC says it is.
From: Rupert on 23 Nov 2009 22:10
On Nov 21, 1:14 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 20 Nov., 13:37, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > > Bill Taylor says... > > > >stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >You admit doubts about the O.S. of sets; I presume > > >(maybe wrongly?) you have no, or at least much lesser, > > >doubts about the O.S. of natural numbers. > > > I don't see any big difference between the two. The set > > of naturals and the set of reals are both abstractions. > > I don't understand in what sense either exists, other > > than exists as a coherent topic of study. > > None of them does exist other than as a name and a wrong, i.e., self- > contradictory idea. We can write sequences of symbols that allow us to > talk about numbers and to manipulate numbers. That's all that exists - > and it's enough to do mathematics. Everything else is a useless object > for useless Fools Of Matheology. > > Regards, WM That's not such an uplifting take on it. If you're really smart and study really hard, then you can learn to write down symbols in ways that a small handful of other people find interesting and worthy of respect, but which just about everyone has no hope of understanding and couldn't care less about, and fortunately the government is prepared to subsidise it. We like to think that the story is more uplifting than that. |