The Definition of Points
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From: cbrown on 13 Apr 2007 15:10 On Apr 13, 10:38 am, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 12, 2:36 pm, "MoeBlee" <jazzm...(a)hotmail.com> wrote: > > >> Zermelo's motivation was to prove that every set is well ordered. > > > Since that phrasing might be misunderstood, I should say that I mean: > > Zermelo's motivation was to prove that for every set, there exists a > > well ordering on it. > > > MoeBlee > > I am not sure how the Axiom of Choice demonstrates that. > AoC says (roughly) that we have a way to unambiguously choose an element from any set: i.e., for any set S, there exists a function f such that for any subset A of S, f(A) is a member of A (and of course therefore, a member of S). "S is well ordered by <=" states that "<=" is a total order on S, and for any subset A of S, there is a specific least element of A. So if "<=" well-orders S, then "the least element of A" is a function like the one that AoC tells us exists: it chooses a specific element from any subset A of S. So "every set can be well-ordered" implies "for every set S, there exists a choice function for S". The converse ("for every set S, there exists a choice function for S" implies "every set can be well-ordered") is a bit more complicated. A proof online is at http://planetmath.org/?op=getobj&from=objects&id=3359 but you'll have to accept certain facts about ordinals (e.g., that they exist, that any set of them is well-ordered by inclusion, that transfinite induction over a well-ordered set is possible, etc.) which you have previously balked at. The basic idea is to let f(S) be the smallest element of S, then f(S\ {f(S)}) be the next smallest element, and so on. Of course, transfinite induction is required for sets which are not countable. Cheers - Chas
From: Tony Orlow on 13 Apr 2007 15:11 MoeBlee wrote: > On Apr 13, 10:38 am, Tony Orlow <t...(a)lightlink.com> wrote: >> MoeBlee wrote: > >>> Zermelo's motivation was to prove that for every set, there exists a >>> well ordering on it. > >> I am not sure how the Axiom of Choice demonstrates that. > > You don't know how the axiom of choice is used to prove that for every > set there exists a well ordering of the set? Virtually any set theory > textbook will give a cycle of proofs showing equivalence of (not > necessarily in order) the axiom of choice (in its various > formulations), Zorn's lemma, the well ordering theorem, the numeration > theorem, etc. > > Among those textbooks I recommend Stoll's 'Set Theory And Logic' as it > accomplishes some of the proofs without using the axiom schema of > replacement while other textbooks do use the axiom schema of > replacement for certain of the proofs, though, I don't recommend > Stoll's book for an overall systematic treatment since it jumps around > topics too much and doesn't have the kind of "linear" format that > Suppes does so well. > > Anyway, even if you don't know the details of the proofs, don't you at > least have an intuition how a choice function would come in handy > toward proving the well ordering theorem? > > MoeBlee > Hi MoeBlee - I had said that, hoping you might give some explanation, but you didn't really. However, before I sent that response I reminded myself on Wikipedia exactly what AC says, and looked at the definitions of Dependent Choice and Countable Choice as well, and descriptions of the relationships between them. I didn't find anything objectionable in ACC or DC. I think it is the broad statement of AC that any set is well orderable that offends my sensibilities. Remember, we went through a lot of this in Well Ordering the Reals, where I attempted to offer an explicit well order of the reals using the H-riffic number system. My conclusion was that any such attempt would lead to an infinite regression. Here's my intuition. If you have a set, and can partition it into mutually exclusive subsets, within which and between which exists an order, then you can linearly arrange the elements of the set by choosing the first element of the first partition, and repeatedly choosing the first unchosen element from the next partition, returning to the first partition when the last is used, and skipping any partitions from which all elements have been chosen. Where we have a countably infinite set, we may choose a finite number of partitions, at least one of which has a countably infinite number of elements, or we may choose all finite partitions, but a countably infinite number of them. In such a case, indeed, it's possible to define a well order. However, where we have an uncountable set, I don't see that we can divide it into a countable number of partitions, each of which is countable. R>N*N, no? We may have some finite partitions, and some countably infinite, but we must have at least one uncountably infinite partition, in which case the problem regresses, because then we must well order that partition. Whether we have any countably infinite partitions may determine whether there exists some "limit ordinal" in the well order, within the uncountable partition, but thereafter, within that partition, the well ordering problem remains as it did before. So, my intuition tells me that explicit well ordering of an uncountable set is completely impossible, and the assertion that a well ordering is possible for every set is unfounded. When it comes down to AC, I think the problem lies in the statement that any element from each partition may be chosen, while ignoring the question as to whether all objects *within* an uncountable partition may be chosen, and the entire set included in the well order of the uncountable set. There is an infinite regression here that either allows for infinite strings or infinite alphabets, if that makes any sense to you. :) Whew! Well, I hope that was worth it. I feel a couple of neurons getting to know each other. Thanks. TOEKnee
From: Tony Orlow on 13 Apr 2007 15:25 Virgil wrote: > In article <461fbab9(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Once they are ordered in whatever manner, they become sequences, trees, >> or other structures, and it is only with such a recursive definition >> that such an infinite structure can be created. > > That is a broad claim. Can you prove it? > > Oh! I forgot! You never prove anything. > Counterexample? > >> Does Virgil forget what he cuts from the post? What do you think we were >> discussing? I thought it was N specifically. > > TO's discusssions skip so wildly from one thing to another, it is easy > to lose track. Nolo contendere. >>>>>> Order is defined by x<y ^ y<z -> x<z. >>>>> Transitivity is one of the properties of most of the orderings we're >>>>> talking about. But transitivity is not the only property that defines >>>>> such things as 'partial order', 'linear order', 'well order'. >>>>> >>>> It defines order, in general. >>> Only to TO. For everyone else, other properties are required. >>> >>> For example, in addition to transitivity, >>> ((x>y) and (y>x)) -> x = y >>> is a necessary property /every/ ordering. >> Um, that one is blatantly self-contradictory. x>y -> not y>x, always. I >> suppose you meant: >> ((x>=y) and (y>=x)) -> x = y >> or: >> (~(x>y) and ~(y>x)) -> x = y > > Nothing in TO's definition of "<" prohibits '(x>y) and (y>x)' from being > true, so if he wishes to require such a prohibition, he must > specifically add it to his transistivity requirement. > Yeah, actually, I misspoke, in a way. Your statement is still blatantly false, in any case. It's possible for x<y and y<x in a cyclical-type system, but those two facts together do not imply x=y. It may be the case, for every x and y, even when x=y, that x<y and y<x, but that doesn't mean x=y. The statements I gave you are correct, assuming your premise is false above. >> You're missing the point. > > > MY point is that requiring only transistivity of a relation is not > enough by itself to assure that one has an order relation. > > TO insists that transitivity is enough, which is wrong. > > > It is the start of order. It doesn't define whether the set is dense in that order, and it doesn't define whether a<b and b<a is necessarily false, but it does establish trichotomy, with the addition of a "not". >>> The mechanics of "less than" depends on what standard of measurement one >>> is using, so claiming that one measure measures all is a procrustean >>> fallacy. >> You have a very negative attitude. > > Mathematics involves a lot of very careful nit picking. Those who regard > such nit picking as "a very negative attitude" often have great problems > with mathematics. > > How procrustean of you. >>>> There can always be a 1-1 correspondence defined between a set >>>> with no end and its proper subset with no end, even if that >>>> correspondence is so complicated so as to defy all attempts to define >>>> it. >>> Trivially false. >>> >>> Neither the set of reals nor the set of rationals has an end, and the >>> rationals are a proper subset of the reals, but there is no bijection >>> between them. >> Golly! Wasn't it you among others that was telling me how R was derived >> from Q which was derived from N, but that they were all distinct sets, >> and N and Q WEREN'T subsets of R? > > See, TO can pick a nit when it pleases him. > > In any model of the reals there is a unique minimal subfield which is > field- isomorphic to the rationals. We might label that subfield as the > rational reals, in which case: > > Neither the set of reals nor the set of rational reals has an end, > and the rational reals are a proper subset of the reals, but there > is no bijection between them. > Is it nitpicking to point out the forks of your tongue? But, anyway... >> Isn't R a set defined using Dedekind >> cuts or Cauchy sequences, which neither naturals nor rationals are? But, >> I disputed that, anyway, so you're right. There remains the difference >> between countable and uncountable infinity, but that's just a >> distinction between potential and actual infinity. > > The set of reals and the set of rational reals are equally potential and > equally actual at being infinite. Yes, there is a strange discrepancy there. I've never been able to accept that, for every unit of quantity on R, there are aleph_0 rationals and 1 natural, and yet, N and Q are "equinumerous". It's hogwash. The rationals are defined by NxN, minus the redundancies in quantity within the matrix. Equinumerous to those redundancies, which are the vast majority of cells, are the irrationals. That's how it actually works. >>> And, given the axiom of choice, any well ordered uncountable set even >>> has well ordered countable subsets with which it does not biject. >> Sure, uncountable vs. countable. I stand corrected. > > You were probably sitting, not standing, as you wrote that. Sto correctato, or estoy correctado. Pero lo hago en el culo, claro que si, Senior. Antonio
From: Tony Orlow on 13 Apr 2007 15:27 Virgil wrote: > In article <461fc017(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> MoeBlee wrote: >>> On Apr 12, 2:36 pm, "MoeBlee" <jazzm...(a)hotmail.com> wrote: >>> >>>> Zermelo's motivation was to prove that every set is well ordered. >>> Since that phrasing might be misunderstood, I should say that I mean: >>> Zermelo's motivation was to prove that for every set, there exists a >>> well ordering on it. >>> >>> MoeBlee >>> >> I am not sure how the Axiom of Choice demonstrates that. >> >> Well Order the Reals! > > TO misses the point again. Existence proofs do not have to actually > instantiate what they are proving exists. > > And the AOC allows an existence proof of a well ordering of any set > without requiring that any such well orderings be actually created. How convenient!!! :D
From: Lester Zick on 13 Apr 2007 15:27
On Fri, 13 Apr 2007 13:40:24 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >Lester Zick wrote: >> On Thu, 12 Apr 2007 14:12:56 -0400, Tony Orlow <tony(a)lightlink.com> >> wrote: >> >>> Lester Zick wrote: >>>> On Sat, 31 Mar 2007 18:05:25 -0500, Tony Orlow <tony(a)lightlink.com> >>>> wrote: >>>> >>>>> Lester Zick wrote: >>>>>> On Fri, 30 Mar 2007 12:06:42 -0500, Tony Orlow <tony(a)lightlink.com> >>>>>> wrote: >>>>>> >>>>>>> Lester Zick wrote: >>>>>>>> On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <tony(a)lightlink.com> >>>>>>>> wrote: >>>>>>>> >>>>>>>>>>> You might be surprised at how it relates to science. Where does mass >>>>>>>>>>> come from, anyway? >>>>>>>>>> Not from number rings and real number lines that's for sure. >>>>>>>>>> >>>>>>>>> Are you sure? >>>>>>>> Yes. >>>>>>>> >>>>>>>>> What thoughts have you given to cyclical processes? >>>>>>>> Plenty. Everything in physical nature represents cyclical processes. >>>>>>>> So what? What difference does that make? We can describe cyclical >>>>>>>> processes quite adequately without assuming there is a real number >>>>>>>> line or number rings. In fact we can describe cyclical processes even >>>>>>>> if there is no real number line and number ring. They're irrelevant. >>>>>>>> >>>>>>>> ~v~~ >>>>>>> Oh. What causes them? >>>>>> Constant linear velocity in combination with transverse acceleration. >>>>>> >>>>>> ~v~~ >>>>> Constant transverse acceleration? >>>> What did I say, Tony? Constant linear velocity in combination with >>>> transverse acceleration? Or constant transverse acceleration? I mean >>>> my reply is right there above yours. >>>> >>>> ~v~~ >>> If the transverse acceleration varies, then you do not have a circle. >> >> Of course not. You do however have a curve. >> >> ~v~~ > >I thought you considered the transverse acceleration to vary >infinitesimally, but that was a while back... Still do, Tony. How does that affect whether you have a curve or not? Transverse a produces finite transverse v which produces infinitesimal dr which "curves" the constant linear v infinitesimally. ~v~~ |