The Definition of Points
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From: K_h on 14 Apr 2007 18:40 "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
From: Virgil on 14 Apr 2007 17:51 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.
From: K_h on 14 Apr 2007 19:45 "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 and are you standing by your claim that the equality relation cannot be an order relation on a set with more than one member? 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. K_h
From: K_h on 14 Apr 2007 19:55 "K_h" <KHolmes(a)SX729.com> wrote in message news:JvmdneShNPwdxLzbnZ2dnUVZ_tCtnZ2d(a)comcast.com... > > "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 and are you standing by your claim that the equality relation cannot > be an order relation on a set with more than one member? Forget the first part of the question, I see you agreed that (x<y) and (y<x) is always false. So my only remaining question is the one about equality on sets with more than one member. K_h
From: Lester Zick on 14 Apr 2007 19:10
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~~ |