The Definition of Points
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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.
From: MoeBlee on 13 Apr 2007 17:32 On Apr 13, 12:51 pm, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 13, 11:25 am, Tony Orlow <t...(a)lightlink.com> wrote: > > >> I've discussed that with you and others. It doesn't cover the cases I am > >> talking about. The naturals have a "measure" of 0, no? So, measure > >> theory doesn't address the relationship between, say, the naturals and > >> the evens or primes. It's not as general as it should be. So, what do > >> you want me to say? > > > Nothing, really, until you learn the mathematics you're pretending to > > know about. > > I didn't bring up "measure theory". Where do I begin: transitivity, ordering, recursion, axiom of infinity, non-standard analysis...on and on and on... > > Nothing to which you responded "pretends" that cardinality "can tell > > things it can't". What SPECIFIC theorem of set theory do you feel is a > > pretense of "telling things that it can't"? > > AC If you mean non-constructivity, then no one disputes that the axiom of choice is non-constructive. No one says that the axiom of choice proves the existence of a definable well ordering. But if you require constructivity then you can't without contradiction endorse Robinson's non-standard analysis. > >>> It has CARDINALITY aleph_0. If you take "size" to mean cardinality > >>> then aleph_0 is the "size" of the set of naturals. But it simply isn't > >>> true that "a set of naturals with 'size' y has maximum element y" if > >>> "size" means cardinality. > >> I don't believe cardinality equates to "size" in the infinite case. > > > Wow, that is about as BLATANTLY missing the point of what you are in > > immediate response to as I can imagine even you pulling off. > > > MoeBlee > > What point did I miss? The MAJOR point - the hypothetical nature of mathematical reasoning (think about the word 'if' twice in the poster's paragraph) and the inessentiality of what words we use to name mathematical objects and their properties. I've been trying to get you to understand that for about two years now. > I don't take transfinite cardinality to mean > "size". You say I missed the point. You didn't intersect the line. You just did it AGAIN. We and the poster to whom you responded KNOW that you don't take cardinality as capturing your notion of size. The point is then just for your to recognize that IF by 'size' we mean cardinality, then certain sentences follow and certain sentences don't follow and that what is important is not whether we use 'size' or 'cardinality' or whatever word but rather the mathematical relations that are studied even if we were to use the words 'schmize' or 'shmardinal' or whatever. MoeBlee
From: Virgil on 13 Apr 2007 17:45 In article <461fde40(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > cbrown(a)cbrownsystems.com wrote: > > On Apr 13, 10:13 am, Tony Orlow <t...(a)lightlink.com> wrote: > >> Virgil wrote: > >>> Sets in general have all sorts of orderings, but none which is as > >>> inherent in their being sets as the ordinal order is in sets being > >>> ordinals. > >> 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. > > > > Why is a recursive definition required? Given any set S, the set of > > all subsets of S can be (partially) ordered as follows: for subsets A, > > B of S, define A <= B iff every member of A is a member of B. What is > > recursive about that definition? > > That structure doesn't look like a tree to you, which each node having > as children the elements of its power set? That's the picture that > appears to me. Then TO cannot have looked very carefully. In a tree, no node has more than one parent but P({a,b}) = {{}, {a}, {b} and {a,b}} requires that, (1) if {} is the root, then {a} and {b} are both parents of {a,b}, or (2) if {a,b} is the root then {a} and {b} are both parents of {}. The appropriate mathematical name for such a structure is a /lattice/. > > > > > Alternatively, S be the set of all subsets of the naturals (note that > > S is not countable). > If A, B are in S, define A <= B if there is a > > natural number m in B such that m is not in A, and for all n < m, n in > > A and n in B. What is recursive about that definition? > > > > The Archimedean principle, as far as I can tell. Where does the Archimedean principle apply in that definition, and how is it recursive in that definition? > > >> In that sense, there is > >> no pure infinite set without some defining structure Previously claimed without proof and again claimed without proof. , .... > >> Um, that one is blatantly self-contradictory. x>y -> not y>x, always. > > > > I don't see how this follows only from your assertion "x < y and y < z > > -> x < z". You stated: > > > >>>>>> Order is defined by x<y ^ y<z -> x<z. > > > > Or do you mean that there is /more/ to the definition of an order "<" > > than "x < y and y < z -> x < z"? If so, that was exactly Virgil's > > point. > > > > Actually I corrected this response to Virgil. I misspoke a little, but > he's still wrong. :) Then according to TO mere transistivity make equality an order relation, as it satisfies TO's sole criterion. > > >> suppose you meant: > >> ((x>=y) and (y>=x)) -> x = y > >> or: > >> (~(x>y) and ~(y>x)) -> x = y > >> > > > > These two statements are not equivalent. In some situations, the first > > can hold, while the second does not. > > > > Please do elaborate. if "<" and "=" are the same relation of equality, for instance. > > >> Yes, if neither x<y or y<x is true, that is, if no order can be > >> determined, then x=y for the purposes of that order. > > > > Let's order the subsets of {a,b,c} by inclusion. Then: {a,b} < > > {a,b,c}. {a,c} < {a,b,c}. {a} < {a,b}. {a} < {a,c}. But not ({a,b} < > > {a,c}); and not ({a,c} < {a,b}). Does that mean that {a,b} = {a,c} > > "for the purposes of that order"? > > > > Where b<c, {a,b}<{a,c}. Not necessarily. An order relation on the members of a set does not necessitate any order relation of the subsets of that set. Nor vice versa. > > Could you state what the definition of "<= totally orders the set S" > > is again? There are three simply stated properties, IIRC. > > > > Lookitup. Transitivity does not define "total order", but is that start > of order. > >>> Also there are lots of transitive relations which are not orderings, at > >>> least as usually understood. E.g., universal relations, which hold true > >>> for all x and y in the relevant set. > >>> So that TO's notion of an ordering does not necessarily order anything. > >> You're missing the point. All I said was that one starts with inequality > >> defining the line itself. > > > > Is every set a line? Is the set of all triangles in the Euclidean > > plane a line? > > > > That's a bunch of lines. Then it is not one line. > > >> Then one defines equality. Defining equality > >> where there is no relative order doesn't make sense. > > > > So, it makes no sense to say that the set of all finite subsets of the > > naturals having a prime number of elements is equal to itself? > > > > Cheers - Chas > > > > Again, there is order to the elements themselves. Which of a number of possible orders is "the" order on "the set of all finite subsets of the naturals having a prime number of elements"
From: MoeBlee on 13 Apr 2007 17:46
On Apr 13, 12:11 pm, Tony Orlow <t...(a)lightlink.com> wrote: > I had said that, hoping you might give some explanation, but you didn't > really. Since you posted that, I wrote a long post about the axiom of choice. Now it's not showing up in the list of posts. Darn! I went into a lot of detail and answered your questions; I don't want to write it all again; maybe it will show up delayed. MoeBlee |