The Definition of Points
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From: Tony Orlow on 17 Apr 2007 13:39 Lester Zick wrote: > On Fri, 13 Apr 2007 14:36:12 -0400, Tony Orlow <tony(a)lightlink.com> > wrote: > >> Lester Zick wrote: >>> On Thu, 12 Apr 2007 15:30:57 -0400, Tony Orlow <tony(a)lightlink.com> >>> wrote: >>> >>>> Lester Zick wrote: >>>>> On Mon, 2 Apr 2007 16:12:46 +0000 (UTC), stephen(a)nomail.com wrote: >>>>> >>>>>>> It is not true that the set of consecutive naturals starting at 1 with >>>>>>> cardinality x has largest element x. A set of consecutive naturals >>>>>>> starting at 1 need not have a largest element at all. >>>>>> To be fair to Tony, he said "size", not "cardinality". If Tony wishes to define >>>>>> "size" such that set of consecutive naturals starting at 1 with size x has a >>>>>> largest element x, he can, but an immediate consequence of that definition >>>>>> is that N does not have a size. > >>>>> Is that true? >>>>> >>>>> ~v~~ >>>> Yes, Lester, Stephen is exactly right. I am very happy to see this >>>> response. It follows from the assumptions. Axioms have merit, but >>>> deserve periodic review. >>> What follows from the assumptions, Tony? Truth? > >> "that N does not have a size." > > I wasn't commenting on whether your assumptions are consistent with > your axioms, Tony. I was asking whether your assumptions were true. > So, then. it's not true that every statement is either true or false. What about the statement that every statement is true or false? That's false? Perhaps it's not possible to determine the root of truth in any deductive manner, but that determining truth of statements is an infinite regress called "science". Have you considered that notion? >> If the assumptions >>> were true and could be demonstrated they wouldn't have to be assumed >>> to begin with. >> Can we assume that a statement is either true, or it's false? > > Sure. Happens all the time. However if you're asking whether a > statement must be one or the other the answer is no. There are > problematic exceptions to the so called excluded middle. > Please eloborate. >> Is that >> too much of an assumption to make, when exploring the meaning of truth? >> In ways yes, but for a start, no. > > Well your phrase "exploring the meaning of truth" is ambiguous, Tony, > because what you're really doing is exploring consequences of truth or > falsity given assumptions of truth or falsity to begin with, which is > an almost completely trivial exercise in comparison with the actual > determination of truth in mechanically exhaustive terms initially. > I am exploring the mechanics of truth, and its pursuit, which you are not, really, as far as I can tell. >> Mathematikers and empirics expect their students to use >>> the most rigorous, exhaustive mechanics in extrapolating theorems and >>> experimental methods from foundational assumptions. But the minute the >>> same requirements of rigorous mechanics are laid on them and their own >>> axioms and foundational assumptions they cry foul and claim no one can >>> prove their assumptions and that even their definitions are completely >>> arbitrary and can be considered neither true nor false. >>> >>> ~v~~ >> The question about axioms is whether each one is justifiable and >> sufficiently general enough to be accepted as "true" in some universal >> sense. > > No the actual question is whether each and every axiom is actually > true and demonstrably so in mechanically exhaustive terms. Otherwise > there's not much point to the exhaustively rigorous demonstration of > theorems in terms of axioms demanded of students if axioms themselves > are only assumed true. > > ~v~~ I am saying that one can assume axioms for the sake of deduction, but that the conclusions derived are only as reliable as the starting axioms, and so there is an inductive process in deciding which axioms to accept for the sake of one's "theory", expecially when looking for universal truths that serve as axions in a TOE, depending on whether the conclusions drawn fit the empirical evidence. 01oo
From: Tony Orlow on 17 Apr 2007 13:45 Virgil wrote: > In article <462117d1(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Mike Kelly wrote: > >>> Blind axiomatics? So you think ZFC was developed by blindly? People >>> picked the axioms randomly without any real consideration for what the >>> consequences would be? Please. ZF(C) provides a foundation for >>> virtually all modern mathematics. This didn't happen by accident. >>> >>> What's "blind" about ZF(C)? What great insight do you think is missed >>> that you are going to provide, oh mighty revolutionary? What >>> mathematics can be done with your non-existant foundation that can't >>> be done in ZF(C)? >>> >> Axiomatically, I think the bulk of the burden lies on Choice in its full >> form. Dependent or Countable Choice seem reasonable, but a blanket >> statement for all sets seems unjustified. > > Since it has been shown that if ZF is consistent then ZFC must be > consistent as well, what part of ZF does TO object to? > Well, I also put the onus on the Extensionality, as far as equating sets with the same general membership, but different rates of growth per iteration, but I haven't quite figured out how to formalize that statement, or at least, am not in a position to do so now. > >>>> Then how do you presume to declare that my statement is "not true"? >>>> >> No answer? Do you retract the claim? >> Yes? >>>>> It's very easily provable that if "size" means "cardinality" that N >>>>> has "size" aleph_0 but no largest element. You aren't actually >>>>> questioning this, are you? >>>> No, have your system of cardinality, but don't pretend it can tell >>>> things it can't. Cardinality is size for finite sets. For infinite sets >>>> it's only some broad classification. > > It is one form of size for all sets. One might use the physical analogy > that volume, surface area, and maximum linear dimension are all measures > of the size of a solid. So implying that one "size" fits all is false. > > Each of those is derivative of the last, given the proper unit of measure. Ask me more about that, if you're interested. :) >>> OK so all of the above comes down to you demanding that we don't call >>> cardinality "size". If we don't call cardinality "size" then all your >>> objections to cardinality disappear. >> It would also be nice to have an alternative to cardinality > > Why? > Because intuitions and details are not satisfied by cardinality. >> So, what's your opinion of infinite-case induction, IFR and N=S^L, and >> multilevel logics, again? I forget. > > AS presented by TO, garbage, garage and garbage. There is a transfinite > induction but it doesn't work the way TO would have it work, and there > are a wide variety of logics none of which seems to work at all like TO > would have them work. Hmmm.... Garbage? I don't think you've actually considered them, but you may be too old to do so at this point. That's okay. I appreciate you, Virgil. Have a nice day. (no signature, but at least my name is real)
From: Tony Orlow on 17 Apr 2007 13:58 Virgil wrote: > In article <46211c0a(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> 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. >> Times of day have no order? Pulllease!!! > > Does 12 midnight on a clock come before or after 12 noon on that clock? > > There is an assumed local order in the sense that if two clock times are > close enough together one usually assumes that one of them is "before" > and the other "after", but is one minute before midnight clocktime > before or after one minute after midnight clocktime? it could be either. Yes, it can be that x<y and y<x and y<>x. >>> 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) >>> >> Okay, I'm worried about you. You repeated the same erroneous statement. >> You didn't cut and paste without reading, did you? Don't you mean "<=" >> rather than "<". The statement "x<y and y<y" can only be true in two >> unrelated meanings of "<", or else "=" doesn't have usable meaning. > > TO betrays his lack of understanding of material implication in logic. > For "<" being any strict order relation, "(x<y) and (y<x)" must always > be false so that any implication with "(x<y) and (y<x)" as antecedent > for such a relation, regardless of conseqeunt, is always true. Oh, yes, well. Any false statement implies any statement, true or false, as long as you're not an intuitionist. If (x<y) -> ~ (y<x), then x<y ^ y<x is of the form P ^ ~P, or ~(P v ~P) which is false in classical logic, but not intuitionistically. There is debate on this topic. > > So that I have better cause to be worried about TO than he has to worry > about me. > I won't take it personally. >>>> 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. > > Then why does TO claim it? I am not. I am saying it is a set equal in magnitude to the redundancies in NxN.
From: Tony Orlow on 17 Apr 2007 14:07 Lester Zick wrote: > On Fri, 13 Apr 2007 13:49:29 -0400, Tony Orlow <tony(a)lightlink.com> > wrote: > >> Lester Zick wrote: >>> On Thu, 12 Apr 2007 14:30:32 -0400, Tony Orlow <tony(a)lightlink.com> >>> wrote: >>> >>>> Lester Zick wrote: >>>>> On Sat, 31 Mar 2007 21:14:27 -0500, Tony Orlow <tony(a)lightlink.com> >>>>> wrote: >>>>> >>>>>>>> You need to define what relation your grammar denotes, or there is no >>>>>>>> understanding when you write things like "not a not b". >>>>> What grammar did you have in mind exactly, Tony? >>>> Some commonly understood mapping between strings and meaning, basically. >>>> Care to define what your strings mean? :)1oo >>> What strings? Care to define what your "mappings" "between" "strings" >>> and "meaning" mean, Tony? Then we can get to the basis of grammar. >>> >>>>>>> Of course not. I didn't intend for my grammar to denote anything in >>>>>>> particular much as Brian and mathematikers don't intend to do much >>>>>>> more than speak in tongues while they're awaiting the second coming. >>>>>>> >>>>>> Then, what, you're not actually saying anything? >>>>> Of course I am. >>> ~v~~ >> You do know what "strings" are, don't you? And grammar? And language? >> And, um, meaning? > > I don't know what anything is, Tony. I'm still trying to come to terms > with "truth". You seem to think you've already come to terms with > "truth" "strings" "grammar" "language" and um "meaning". You're quite > fortunate in this respect. I should be so lucky. It might help if I > could just assume the truth of whatever I was babbling about without > having to demonstrate its truth in mechanically exhaustive terms like > you and Moe(x) but then I guess I'm just more particular. > >> What's the difference between a duck? > > 46. > > ~v~~ Incorrect. One leg is both the same. So, start with uncertainty. Then, build truth from such statements. Start with the line, and determine the point of intersection. Just remember, when the lines are moving, so is the point. 01oo
From: Tony Orlow on 17 Apr 2007 14:08
Virgil wrote: > In article <JvmdneShNPwdxLzbnZ2dnUVZ_tCtnZ2d(a)comcast.com>, > "K_h" <KHolmes(a)SX729.com> wrote: > >> "Virgil" <virgil(a)comcast.net> wrote in message >> news:virgil-89D080.15510414042007(a)comcast.dca.giganews.com... >>> In article <VfmdnegunffO17zbnZ2dnUVZ_sapnZ2d(a)comcast.com>, >>> "K_h" <KHolmes(a)SX729.com> wrote: >>> >>>> "Virgil" <virgil(a)comcast.net> wrote in message >>>> news:virgil-DCD754.15201113042007(a)comcast.dca.giganews.com... >>>>> In article <461fd938(a)news2.lightlink.com>, >>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>> >>>>>> 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) >>>> >>>> For a partially ordered set this is always true but why are you claiming >>>> it >>>> is always true for a strictly ordered set? An order relation on a set S, >>>> denoted by <, is defined by the following two properties: >>>> >>>> (1) If x and y are both members of S then only one of the following >>>> statements is true: x<y, x=y, y<x. >>>> (2) If x, y, and z are members of S then if x<y and y<z then x<z. >>>> >>>> By (1), (x = y) is only true if (x<y) and (y<x) are both false. If (x<y) >>>> is >>>> true and (y<x) is false then ((x<y) and (y<x)) is false and, by (1), >>>> (x=y) >>>> is false. >>>> >>>> >>>>>> 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. >>>> >>>> Why not? Suppose I have the set {A, B, C, D} with an order relation >>>> A=B=C=D >>>> which is consistent with both (1) and (2) in the above definition of an >>>> order relation. Clearly there is more than one member in the set and the >>>> order relation is clearly transitive. >>>> >>>> >>>> K_h >>> In formal logic, a statement of the form "if A then B" is true whenever >>> A is false, or B is true, or both, and is only false when A is true and >>> B is false. >>> >>> For a strict order relation "<", the 'A' statement, (x< y and y < x ) is >>> always false. >>> >>> Thus the "if A then B" compound statement is true for strict >>> inequalities regardless of what B says. >> >> The statement "if False then False" is a true statement, yes. But are you >> claiming that (x<y) and (y<x) can both be true for a strictly ordered field > > > NO! I am saying they CANNOT both be true for a strictly ordered set. > >> and are you standing by your claim that the equality relation cannot be an >> order relation on a set with more than one member? > > A total order relation "<" on a set has the property that for every x > and y in the set, at least one of x<y, y< x or x=y is true > For a set with two or more elements, equality does not satisfy this > requirement, since there are members which are NOT equal. >> Your statement: "For any in which < is to represent the mathematical notion >> of an order, ((x<y) and (y<x)) implies (x = y)" then becomes: "For any in >> which < is to represent the mathematical notion of an order, FALSE implies >> (x = y)" which is not very helpful especially since (x=y) can be false. > > The issue is not whether it is useful but whether it is true that for an > order relation "<" "((x<y) and (y<x)) implies (x = y)", and on that > issue I am correct. Define "correct", ala Brouwer. :) |