The Definition of Points
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From: Virgil on 14 Apr 2007 19:14 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.
From: Virgil on 14 Apr 2007 19:18 In article <vcGdnVSEYMdHxrzbnZ2dnUVZ_uqvnZ2d(a)comcast.com>, "K_h" <KHolmes(a)SX729.com> wrote: > So my only remaining question is the one about equality on > sets with more than one member. It is weak a partial order but not a total order, weak or strong, on such sets.
From: Lester Zick on 14 Apr 2007 19:34 On Sat, 14 Apr 2007 19:56:23 +0100, Alan Smaill <smaill(a)SPAMinf.ed.ac.uk> wrote: >Lester Zick <dontbother(a)nowhere.net> writes: > >> On Sat, 14 Apr 2007 13:56:37 +0100, Alan Smaill >> <smaill(a)SPAMinf.ed.ac.uk> wrote: >> >>>Lester Zick <dontbother(a)nowhere.net> writes: >>> >>>> On Fri, 13 Apr 2007 16:10:39 +0100, Alan Smaill >>>> <smaill(a)SPAMinf.ed.ac.uk> wrote: >>>> >>>>>Lester Zick <dontbother(a)nowhere.net> writes: >>>>> >>>>>> On Thu, 12 Apr 2007 14:23:04 -0400, Tony Orlow <tony(a)lightlink.com> >>>>>> wrote: >>>>>>> >>>>>>>That's okay. 0 for 0 is 100%!!! :) >>>>>> >>>>>> Not exactly, Tony. 0/0 would have to be evaluated under L'Hospital's >>>>>> rule. >>>>> >>>>>Dear me ... L'Hospital's rule is invalid. >>>> >>>> What ho? Surely you jest! >>> >>>Who, me? >>> >>>> Was it invalid when I used it in college? >>> >>>If you used it to work out a value for 0/0, then yes. >> >> Well the problem is that you didn't claim my application of >> L'Hospital's rule was invalid. You claimed the rule itself was >> invalid. So perhaps you'd like to show how the rule itself is invalid >> or why my application of the rule is? > >Or both: > >The rule is invalid because that's what you find in Hospitals. Haha. Next time remind me when to laugh. >Your use is invalid because the rule says nothing about the >value of 0/0. It doesn't? My mistake. Perhaps that's why we used it instead of 0/0. ~v~~
From: Alan Smaill on 14 Apr 2007 19:55 Lester Zick <dontbother(a)nowhere.net> writes: > On Sat, 14 Apr 2007 19:56:23 +0100, Alan Smaill > <smaill(a)SPAMinf.ed.ac.uk> wrote: > >>Lester Zick <dontbother(a)nowhere.net> writes: >> >>> On Sat, 14 Apr 2007 13:56:37 +0100, Alan Smaill >>> <smaill(a)SPAMinf.ed.ac.uk> wrote: >>> >>>>Lester Zick <dontbother(a)nowhere.net> writes: >>>> >>>>> On Fri, 13 Apr 2007 16:10:39 +0100, Alan Smaill >>>>> <smaill(a)SPAMinf.ed.ac.uk> wrote: >>>>> >>>>>>Lester Zick <dontbother(a)nowhere.net> writes: >>>>>> >>>>>>> On Thu, 12 Apr 2007 14:23:04 -0400, Tony Orlow <tony(a)lightlink.com> >>>>>>> wrote: >>>>>>>> >>>>>>>>That's okay. 0 for 0 is 100%!!! :) >>>>>>> >>>>>>> Not exactly, Tony. 0/0 would have to be evaluated under L'Hospital's >>>>>>> rule. >>>>>> >>>>>>Dear me ... L'Hospital's rule is invalid. >>>>> >>>>> What ho? Surely you jest! >>>> >>>>Who, me? >>>> >>>>> Was it invalid when I used it in college? >>>> >>>>If you used it to work out a value for 0/0, then yes. >>> >>> Well the problem is that you didn't claim my application of >>> L'Hospital's rule was invalid. You claimed the rule itself was >>> invalid. So perhaps you'd like to show how the rule itself is invalid >>> or why my application of the rule is? >> >>Or both: >> >>The rule is invalid because that's what you find in Hospitals. > > Haha. Next time remind me when to laugh. haha. >>Your use is invalid because the rule says nothing about the >>value of 0/0. > > It doesn't? My mistake. It doesn't. > Perhaps that's why we used it instead of 0/0. That's a mistake too; keep them coming. > ~v~~ -- Alan Smaill
From: K_h on 14 Apr 2007 20:58
"Virgil" <virgil(a)comcast.net> wrote in message news:virgil-9A74B0.17144214042007(a)comcast.dca.giganews.com... > In article <JvmdneShNPwdxLzbnZ2dnUVZ_tCtnZ2d(a)comcast.com>, > "K_h" <KHolmes(a)SX729.com> wrote: > > > 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. It depends on the set. Consider the set {A, B, C, D} with an order relation A=B=C=D. Clearly there is more than one member in the set and the order relation is clearly transitive: A=B and B=C --> A=C. For this particular set, the order relation always selects x=y to be true and both cases x<y and y<x are false. See http://en.wikipedia.org/wiki/Total_order for total order. Of course, one can construct sets where equality only holds for individual members: {E, F, G, H} with ordering E<F<G<H only has equality for E=E, F=F, G=G, and H=H. > 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. We agree, it is true but not useful. K_h |