The Definition of Points
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From: Tony Orlow on 20 Apr 2007 12:13 MoeBlee wrote: > On Apr 19, 7:06 am, Tony Orlow <t...(a)lightlink.com> wrote: >> MoeBlee wrote: > >>> You just need to learn it theorem by theorem from the axioms. All it >>> takes is an introductory set theory textbook, and all of them will >>> prove the well ordering theorem for you. > >> But, a well order cannot have infinite descending chains of any sort, so >> if there are an uncountable number of limit ordinals within the order, >> then there will exist an infinite descending chain within that sequence >> of limit ordinals itself. How does one get around that? > > Again, you're reasoning from vague ideas you have of what the > terminology means. If you were to pin down these terms you use such as > 'infinite descending chain' with the precise definitions, then tried > to derive a formal contradiction, you wouldn't succeed to prove any > such contradiction nor evidence of anything that we must "get around" > here. Ironically, your response is right after I reiterated that you > need to study this in a systematic and precise way, definition by > definition, theorem by theorem. It is just more effort down the drain > for me to untangle your confusions over terms of which you've not > bothered to learn their precise definitions from primitives and > axioms. > > A well ordering doesn't preclude that a member of the field of the > ordering may have infinitely many predecessors. What a well ordering > does demand is that in any nonempty set of predeccessors, there is a > least one, which doesn't preclude that there may be infinitely many > "on top" (as we're "working downwards") of that least one. You don't > even need to go to the uncountable; all you need is just one limit > ordinal in the field of the well ordering to see the point: Just take w > +1. The well ordering on w+1 by the membership relation has w itself > as member of the field of the ordering. But to "travel from" w "to" > the least member of w requires "travelling through" an infinte number > of members of the domain of the field of the ordering. As it was explained to me, because every natural is finitely far from the lest in w, there is no infinite descending chain of predecessors one can define. In other words, between any two limit ordinals can only a countable number of elements. Now, can a well order have an infinite descending chain of limit elements? If so, then an uncountable set could be partitioned into an uncountable set of countable partitions, and be well ordered, without a doubt, and it would already have been done. > > I'm using picture language such as "travelling through" to get this > across to you, even though I think it is reliance on picture language > that is a large part of your problem in failing to come to grips with > the precise mathematics. The alternative is for me to explicate such > terms as you misuse, but, as I said, that is effort down the drain if > you continue to refuse to do the minimal work yourself of learing the > definitions from the primitives. > > One thing I should say is that I was too glib by saying that one might > easily see the intuitive basis of well ordering from choice. While it > is true that choice from well ordering is a piece of cake, but getting > well ordering from choice is usually done with Zorn's lemma and/or > transfinite recursion, and those are not simple matters. But to at > least appease your curiosity about this (which you continue to refuse > to satisfy properly by actually studying the material), when you think > about limit ordinals, you're thinking about unions, and that's the > main connection with Zorn's lemma and transfinite recursion too. The > proofs in Stoll run about three pages and even then you have to fill > in some of the sub-steps mentally or on paper for yourself. The proof > in Enderton is about half a page, but it is based on a couple of > chapters worth of development of transfinite recursion and theorems > about ordinals. I can't properly compress all of that into a post. > Especially when such proofs are so complex, there comes a point at > which you just have to study a textbook and cannot expect to have it > explained to you in casual postings. (Or perhaps take a look at the > relevant parts of that paper by Kanamori I linked you too.) I went and looked those things up a bit, except for Zorn's lemma, which I just googled. I'm doing some homework here and there, though kind of casually. > > P.S. In another post I mentioned 'weak' versus 'strong' versions of > well ordering. Actually, that should be 'weak' versus 'strict'. > > MoeBlee > > > > I meant to google that and read up, but ran out of time and steam. TOEknee
From: MoeBlee on 20 Apr 2007 12:25 On Apr 20, 8:41 am, Tony Orlow <t...(a)lightlink.com> wrote: > Mike Kelly wrote: > > No interval in the ordering on the real line. But there are orderings > > where there are infinitely many naturals between each pair of > > rationals. You seem hung up on the idea of orderings on reals/ > > rationals/naturals that aren't the standard one. Maybe because you > > think about everything "geometrically"? > > I think about quantity in terms of measure, yes. Where we have one atom > of space we have a point, without form. Where we have two, we have a > measurable difference, a line. Half of mathematics arose from > geometrical concerns, largely for dividing land and building buildings. > The other half arose from financial concerns, keeping books regarding > accounts. The one starts from the notion of points and lines, "real" > objects, and the other from the notion of a tally or count, in units. > That modern mathematics considers the second to supersede the first is, > I think, a mistake. Any theory that integrates mathematics should > satisfy both notions, no? Whatever notions should be satisfied, you can have all the geometrically motivated mathematics you want, but it won't change that what Mike has told is correct: There existorderings on the naturals and rationals such that between two rationals there are denumerably many naturals. The is a 1-1 correspondence between the naturuals and the rationals. You can either come to understand why that is so and we can move on from this point, or you can continue to remain ignorant about it and instead elaborate upon your historical theories. > I never imagined it would take anyone two years to still not understand, > as I've repeatedly stated, that bijections alone are not sufficient for > proper relative measure of infinite sets. When I say that, doesn't that > sort of imply that I understand that cardinality is based on bijection > alone? The mapping functions describe the relative size, as long as both > sets follow quantitative order. Otherwise it's slightly more complicated. We understand that you think there is some as yet mathematically unarticulated notion you have about this subject. It's just that you haven't given any mathematical theory for it nor shown what its specific advantages would be (since as yet it remains a personal notion of yours while you still have not learned to articulate any of it mathematically), let alone how it would affect applied mathematics in any way whatsoever. > > I do say so. Does this mean you're going to stop claiming that > > relations in set theory aren't based solely on 'e'? > No, not when sequences are defined using a recursively defined successor > function, which is a relation between two elements, as opposed 'e', a > relation between an element and a set. In set theory, the definition of 'successor' and the proofs of the various definition by recursion theorems revert to the primitive 'e'. That is what I mean by you being a big mouth ignoramus on this subject. You just don't know anything about it, yet you continue and continue and continue to declare dogmatically about it. Morevover, what recursive definition of 'successor' are you referring to? Neither first order PA (where successor is primitive and not defined at all) nor in set theory (where successor is defined and such that it reverts to 'e') is there a recursive definition of 'successor'. > The combination of the two is > what produces an infinite set, no? No. MoeBlee
From: Tony Orlow on 20 Apr 2007 12:24 MoeBlee wrote: > On Apr 19, 7:01 am, Tony Orlow <t...(a)lightlink.com> wrote: >> Mike Kelly wrote: >>>> If you say in the same breath, "there >>>> are infinitely many rationals for each natural and there are as many >>>> naturals overall as there are rationals", >>> And infinitely many naturals for each rational. >> How do you figure? In each 1-unit real interval, there is exactly one >> natural, and an infinite number of rationals. Which interval has one >> rational and an infinite number of naturals? > > That there are denumerably many rationals but only one natural in a > unit interval defined by the standard ordering on rationals doesn't > refute that there exists the kind of correspondence Mike Kelly > mentioned. Several months ago I defined for you a dense linear > ordering on the set of natural numbers. You ignored it. > I did? I was out of touch with this stuff for a while. Maybe that was then. Sorry >>> Or, maybe, other people don't share the same intuitions as you!? Do >>> you really find this so hard to believe? >> Most people find transfinite cardinality "counterintuitive". Surely, you >> don't dispute that. > > Probably most people who are not familiar with the theorem by theorem > proofs of set theory would find uncountability unintuitive. But I know > of no evidence that most people in general can't grasp the idea of an > infinite set such as the set of natural numbers if the matter is > presented to them in a clear way. Then, if one accepts that there are > infinite sets, and one grasps the notion of a power set and some other > basic concepts about sets, uncountability follows. > Some people have problems with uncountability, but that's not what I'm talking about. Most people find it strange that half the naturals are even, but that half is considered the same "size". I really don't have any problem with cardinality, but I wouldn't call any transfinite cardinalities "quantities" by any means. I think referring to them as "numbers" is what makes the theory suspicious to people. >>> Well, maybe you'd like to do that. But you have made no progress >>> whatsoever in two years. Mainly, I think, because you have devoted >>> rather too much time to very silly critiques of current stuff and >>> rather too little to humbling yourself and actually learning >>> something. >> That may be your assessment, but you really don't pay attention to my >> points anyway, except to defend the status quo, so I don't take that too >> seriously. > > You're ridiculous. Certain people have even OVER-indulged you by > showing in detail, ad nauseam, exactly what's whack in your various > proposals. And it doesn't even matter WHO is telling you - that you > need to study this stuff from a good textbook is just plain, basic, > good adive that you do need to take seriously. Or continue to be a > nonsense spouting crank. Your choice. > There have been lots of objections, and a few valid points, but no major flaws detected in what I propose. It's just not compatible with ZFC. >>> Get this through your head : every relation between objects in set >>> theory is based on 'e'. It's really pathetic to keep mindlessly >>> denying this. Set theory doesn't just "try to base everything on 'e'". >>> It succeeds. > >> If you say so. > > No, not just "if he says so". He says so and he's RIGHT. And you can > VERIFY for yourself just by reading a textbook already. > > MoeBlee > So, defining N doesn't involve a successor relation between two elements, as well as a member relation between an element and a set? When you define N, doesn't the rule E x e N -> E succ(x) e N define a relation between two elements, as well as between those elements and N? TOEknee
From: MoeBlee on 20 Apr 2007 12:59 On Apr 20, 9:13 am, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > As it was explained to me, because every natural is finitely far from > the lest in w, there is no infinite descending chain of predecessors one > can define. I wouldn't put it that way myself, but okay. Yes, the converse of the membership relation on any natural number is finite. > In other words, between any two limit ordinals can only a > countable number of elements. "In other words." I love it! In 1964 Lyndon Johnson defeated Barry Goldwater in the U.S. presidential election. In other words, Mickey Rooney is a spy for an ancient species of shapeshifters from another galaxy. > Now, can a well order have an infinite > descending chain of limit elements? I asked you already to define your terminology: Define 'infinite descending chain'. No, nevermind. Better you should go back even further to understand that 'successor' is defined by 'e'. > If so, then an uncountable set could > be partitioned into an uncountable set of countable partitions, "partitioned into an uncountable set of countable partitions"? I'm not sure you're saying what you mean to say. > and be > well ordered, without a doubt, and it would already have been done. > I went and looked those things up a bit, except for Zorn's lemma, which > I just googled. I'm doing some homework here and there, though kind of > casually. I'ts better than nothing, but barely productive - hit and miss, smatterings of this and that. A proper understanding requires systematically working through the definitions and theorems. I don't know why you would settle for a haphazard, quite sloppy, thrown together hodgepodge of bits of information. I'd think your intellectual curiosity would demand that you have the real thing - a real solid and coherent grasp of the subject, even if your purpose in obtaining that understanding is to act as a critic. And even if not motivated by true intellectual curiosity, I'd think at least your vanity would require that you be able to talk about this subject with at least the authority of having grasped it at a first year level. I mean, you pretend to go toe to toe with your interlocuters here and it's nothing but a joke as you come so completely intellectually disarmed and you stay that way for years! MoeBlee
From: David R Tribble on 20 Apr 2007 13:12
Stephen wrote: >> You cannot define the sqrt(-1) geometrically. You are never going >> to draw a line with a length of i. > David R Tribble wrote: >> That's not entirely true. It's all a matter of definitions. For >> example, suppose I define "1" to be the length of a line one cm long >> drawn in a north/south direction. I can further define, arbitrarily >> mind you, that "i" is the length of a line one cm drawn in an >> east/west direction. But we both have to agree on these definitions, >> of course, if they are going to make any sense. > Lester Zick wrote: > Well I daresay we can find crackpots most anywhere who'll agree with > your definitions, David. By the way why exactly do you define "1" to > be a line of length "1"? I mean surely there must be other numbers you > could use? Don't be shy. Why not define "1" as the square root of 00? If you read carefully, you'll see that I defined the unit "1" to be a line of length 0.3937. |