Counterintuitions and the well-ordering theorem
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From: Daryl McCullough on 2 Nov 2009 06:42 Bill Taylor says... > >stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> The reason I have little sympathy for this point of view is >> that I don't see that traditional mathematics ever makes the >> assumption that everything has an explicit name or description. > >OC it doesn't. It never needed to. Before Cantor everything DID. That isn't true. Real numbers did not have names before Cantor. Points and lines of Euclidean geometry didn't have names. >> It's an amazing accomplishment of ZF and the concept of the >> cumulative hierarchy that we can come so close to an explicit >> taxonomy of all the objects that will ever come into play in >> any mathematical argument, but I don't see that the existence >> of such a standard model has any relevance to what mathematicians >> *do* with sets. > >I don't follow this sentence at all, sorry. I'm saying that mathematicians do *not* need to assume that there is a way to construct, or define, every mathematical object. It is enough to be able to characterize them through axioms. To do topology, you start off with "There is a set with such and such properties", and you work out the consequences. You don't need to ask: "What is the name of that set?" "What are the names of its elements?" >> Instead, they start with certain basic sets >> that are *not* defined, > >In pure set theory they all are. They are the empty set. That's what I just said--in set theory, they *do* assume that all sets can be built up from the empty set (but not through applying definitions, though), but most of the *use* of sets does not rely on their being built up this way. >> To me, the astounding claim about set theory is that, whatever >> your basic objects are --- paths of particles, wavefunctions, > >That is all applied mat, applied set theory. >We are (I thought) speaking of pure set theory, as applied to >(at most) other pure math objects like N & R. Yes. Pure mathematics should (in my opinion) be a *superset* of applied mathematics. Whatever our foundation of mathematics should be, it should at least include the ability to model the sorts of problems that show up in physics and applied math. In applied math, there is no assumption that every object is definable. There is no reason to. >> >{ x | phi(x) } >> >> >should be taken as a model or paradigm for what a set actually IS. >> >> I sort of agree with this, > >Well! There's something. > >> but I don't feel that it is >> appropriate to assume that the only possible conditions >> phi are ones that are definable using pure set theory. > >Set theory as applied to physical objects is completely trivial. No, it's not. >> If we are talking about sets of naturals, then "the set >> of phone numbers of mathematicians" is a perfectly good > >No, it's perfectly bad! > >> In a sense, the insistence that all mathematical objects >> have an explicit description is more radical than constructivism. > >It's NOT an "insistence", as you call it, but merely an observation >of standard (pre-Cantor) math. It was *not* at all true, pre-Cantor. Where are you getting that from? Before Cantor, there was no asumption that all Euclidean points and lines had names. There was no assumption that every real number had a finite description. The *possibility* of giving every mathematical object a unique, finite description until after set theory was being developed. >And I feel that set theory should >be conducted along the lines of those traditions. There was no such tradition. I think you have things almost completely backwards. Before attempts to formalize set theory, there was no attempt to say that every mathematical object can be be given a unique description. At least not since the discovery of transcendental reals. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 2 Nov 2009 06:46 Bill Taylor says... > >stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> >AC produces sets, by fiat of existence, ... >> > ... - all these things simply don't exist, >> >in that no explicit set can be named, with those properties. >> >> There is something a little strange about this philosophical >> position: If you assume that all sets have an explicit >> "membership criterion" Phi(x), then choice is *automatically* >> true. You can well-order sets by their defining formulas. > >AHA! A very astute observation. This is the kind of thing >that had me bothered for a long time. And it is all bound up >with the UNDEFINABILITY of "DEFINABILITY", >as I have noted before. > >All sets will be definable, but this notion cannot be delimited >in advance. So e.g. once you have a bounded set of reals, >it must have a least upper bound, BUT this lub will not be >definable at the same level - if the reals in the set are >definable at level "a" and below, then the lub will be most >likely be at level a+1. There is no end to the levels, they >can be notated exactly as any initial segment of the recursive >ordinals. But not of the whole set itself - that would be >declaring existences BY FIAT again. omega_1^CK itself is >purely a creature of ZF, with its all-encompassing power set >operation, (without which it cannot be defined.) I think it's still true that the axiom of choice would hold for such definable sets. If a set of nonempty sets is definable at some level alpha, then there will be a definable choice function at that level (or a few levels higher, maybe). 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. -- Daryl McCullough Ithaca, NY
From: Herman Jurjus on 2 Nov 2009 06:55 Bill Taylor wrote: > Herman Jurjus <hjm...(a)hetnet.nl> wrote: > >> What are your /reasons/ for rejecting AC, > > Well! That would be a whole thread in itself. > And I don't wish to bore the old hands YET AGAIN. > So I'll try to be brief. > > 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? -- Cheers, Herman Jurjus
From: Bill Taylor on 2 Nov 2009 22:59 utch Malahide <fred.gal...(a)gmail.com> wrote: > What objects of Euclid's geometry have explicit names or descriptions? All of them.
From: Bill Taylor on 2 Nov 2009 23:01
On Nov 3, 12:46 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Bill Taylor says... > > > > > > >stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >> >AC produces sets, by fiat of existence, ... > >> > ... - all these things simply don't exist, > >> >in that no explicit set can be named, with those properties. > > >> There is something a little strange about this philosophical > >> position: If you assume that all sets have an explicit > >> "membership criterion" Phi(x), then choice is *automatically* > >> true. You can well-order sets by their defining formulas. > > >AHA! A very astute observation. This is the kind of thing > >that had me bothered for a long time. And it is all bound up > >with the UNDEFINABILITY of "DEFINABILITY", > >as I have noted before. > > >All sets will be definable, but this notion cannot be delimited > >in advance. So e.g. once you have a bounded set of reals, > >it must have a least upper bound, BUT this lub will not be > >definable at the same level - if the reals in the set are > >definable at level "a" and below, then the lub will be most > >likely be at level a+1. There is no end to the levels, they > >can be notated exactly as any initial segment of the recursive > >ordinals. But not of the whole set itself - that would be > >declaring existences BY FIAT again. omega_1^CK itself is > >purely a creature of ZF, with its all-encompassing power set > >operation, (without which it cannot be defined.) > > I think it's still true that the axiom of choice would hold > for such definable sets. If a set of nonempty sets is definable > at some level alpha, then there will be a definable choice > function at that level (or a few levels higher, maybe). 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. > > -- > Daryl McCullough > Ithaca, NY |