The Definition of Points
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From: Tony Orlow on 13 Apr 2007 16:07 cbrown(a)cbrownsystems.com wrote: > 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 > > Thanks for the reply, Chas. You should see mine to MoeBlee. I don't have a lot of time now, because I have to tool on outta here, but I took a quick look at your link, and found it rests on transfinite induction. I don't recall exactly how that works right now, but I definitely recall feeling it was very kludgy. So, I'm not sure I have to accept that proof. I'll try to take a look at it in more dtail when I get a chance. Just to recap what I said to MoeBlee, I can't help feeling that, since a countable union of countable sets is countable, either there will be an uncountable number of partitions of an uncountable set, in which case there will exist infinite descending chains before the first "limit ordinal" in the well order (besides the first element of the first partition), or there will exist at least one uncountable partition which will produce infinite descending chains within itself, after all countable partitions have been exhausted. DC or ACC seem acceptable, but not AC, to me. Tony
From: Tony Orlow on 13 Apr 2007 16:11 Lester Zick wrote: > 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~~ Varying is the opposite of being constant. Checkiddout! http://dictionary.reference.com/browse/constant 01oo
From: Tony Orlow on 13 Apr 2007 16:19 Lester Zick wrote: > On 13 Apr 2007 11:24:48 -0700, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: > >> On Apr 13, 10:56 am, Tony Orlow <t...(a)lightlink.com> wrote: >>> Well, of course, Moe's technically right, though I originally asked >>> Lester to define his meaning in relation to his grammar. Technically, >>> grammar just defines which statements are valid, to which specific >>> meanings are like parameters plugged in for the interpretation. >> That is completely wrong. You have it completely backwards. What you >> just mentioned is part of semantics not grammar. Grammar is syntax - >> the rules for formation of certain kinds of strings of symbols, >> formulas, sentences, and other matters related purely to the >> "manipulation" of sequences of symbols and sequences of formulas, and >> of such objects. On the other hand, semantics is about the >> interpretations, the denotations, the meanings of the symbols, strings >> of symbols, formulas, sentences, and sets of sentences. Mathematical >> logic includes the study of these two things - syntax and semantics - >> both separately and in relation to each other. >> >>> I asked >>> the question originally using truth tables to avoid all that, so that we >>> can directly equate Lester's grammar with the common grammar, on that >>> level, and derive whether "not a not b" and "not a or not b" were the >>> same thing. They seem to be. >> Truth tables are basically a semantical matter. Inspection of a truth >> table reveals the truth or falsehood of a sentential formula per each >> of the assigments of denotations of 'true' or 'false' to the sentence >> letters in the formula. > > If any and all these things are not demonstrably true and merely > represent so many assumptions of truth why would anyone care what you > think about what they are or aren't? I mean it really isn't as if > truth is on your side to the exclusion of what others claim, Moe(x). > > ~v~~ Define "assumption". Do you "believe" that truth exists? Is there a set of statements S such that forall seS s=true? Is there such a thing as truth, or falsity? Does logic "exist". There exists a set of assumptions, A, which are true. True? 01oo
From: Virgil on 13 Apr 2007 17:20 In article <461fd938(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > 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. But a "cyclical-type system" is not an "ordered system" in any standard mathematical sense. For any in which "<" is to represent the mathematical notion of an order relation one will always have ((x<y) and (y<x)) implies (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. Which "premise" of mine are you presuming is false? > > >> 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. But one can have transitivity in an order relation without its being an order relation. For example, the equality relation is clearly transitive, but is clearly NOT an order relation on any set of more than one member. > > >>> 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. It may be procrustean of mathematics to require such pickiness, but it is not anything I impose on mathematics or on those attempting to learn mathematics, it just is the way mathematics is. > > >>>> 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? Nowadays I only use spoons. > > >> 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. it is the inevitable consequence of defining "equinumerous for sets as meaning capable of being bijected with each other. Just another of those inevitable mathematical nits. > The rationals are defined by NxN, minus the redundancies in > quantity within the matrix. That "matrix" is a geometric interpretation, which is quite irrelevant. A better definition for the rationals, based on I as the set of integers and P as the set of strictly positive integers is the set IxP modulo the "==" relation defined by (a,b) == (c,d) iff a*d = b*c. > Equinumerous to those redundancies, which > are the vast majority of cells, are the irrationals. That's how it > actually works. Is TO actually claiming that the irrationals form a subset of the countable set NxN. That is NOT how it works in any standard mathematics.
From: Virgil on 13 Apr 2007 17:21
In article <461fd99d(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > 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 That convenience was noted long before TO was born, and will remain long after he is gone. |