The Definition of Points
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From: Alan Smaill on 15 Apr 2007 16:34 Lester Zick <dontbother(a)nowhere.net> writes: > On Sun, 15 Apr 2007 00:55:33 +0100, Alan Smaill > <smaill(a)SPAMinf.ed.ac.uk> wrote: > >>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. > > So returning to the original point, would you care to explain your > claim that L'Hospital's rule is invalid? haha! > ~v~~ -- Alan Smaill
From: Virgil on 15 Apr 2007 17:43 In article <VZGdnTTzjvOPG7_bnZ2dnUVZ_hGdnZ2d(a)comcast.com>, "K_h" <KHolmes(a)SX729.com> wrote: > "Virgil" <virgil(a)comcast.net> wrote in message > news:virgil-B82679.22384714042007(a)comcast.dca.giganews.com... > > In article <qbOdnddPifcj97zbnZ2dnUVZ_rylnZ2d(a)comcast.com>, > > "K_h" <KHolmes(a)SX729.com> wrote: > > > >> "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. > > > > If "=" is the equality relation on {A,B,C,D} with A=B=C=D,then > > {A,B,C,D} = {A} = {B} = {C} = {D}, but is, in any case, a set with only > > one element. > > > > > > > >> Clearly there is more than one member in the set > > > > If all members are equal then there s only one member. > > I did not mean equality as in the same set. From Wikipedia, a set is > ordered if every pair of its members is mutually comparable. If, for any > two distinct members x y, we cannot tell if any of these are true (x<y, x=y, > y<x) then the set is not ordered. Consider a set of four people and the set > is ordered by height. Sam if 5 feet tall, Jane is 6 feet tall, John is 6 > feet tall, and Mike is 7 feet tall. The set {Sam, Jane, John, Mike} is > ordered by: > > Sam < Jane = John < Mike > > Clearly every pair of members of this set is comparable and Jane = John > under this ordering. Equality here means that Jane and John are comparable > and equal under the ordering; it does not mean that Jane is John. This is > an ordered set by Wikipedia's definition because all of its members are > mutually comparable under the relation. Are you claiming that an ordered > set requires x<y or y<x for any two distinct members x and y? A totally ordered set does. > If so, please > cite your sources for this since this requirement is not specified in the > definition given. If that were true then Wikipedia's definition, along with > many others, is ambiguous. For totally and strictly ordered sets, with order "<", one always has one and only one of x<y or y<x or x = y, which is called the law of trichotomy. For totally but weakly ordered sets, with order ">=", one has x >= y or y >= x. For partially ordered sets one need not have trichotomy. http://mathworld.wolfram.com/StrictOrder.html http://mathworld.wolfram.com/TotallyOrderedSet.html > > K_h
From: Lester Zick on 15 Apr 2007 18:47 On Sun, 15 Apr 2007 21:34:09 +0100, Alan Smaill <smaill(a)SPAMinf.ed.ac.uk> wrote: >>>>>>>>>Dear me ... L'Hospital's rule is invalid. >> >> So returning to the original point, would you care to explain your >> claim that L'Hospital's rule is invalid? > >haha! Why am I not surprized? Remarkable how many mathematiker opinions on the subject of mathematics don't quite hold up to critical scrutiny. ~v~~
From: K_h on 15 Apr 2007 20:06 "Virgil" <virgil(a)comcast.net> wrote in message news:virgil-19C008.15430215042007(a)comcast.dca.giganews.com... > In article <VZGdnTTzjvOPG7_bnZ2dnUVZ_hGdnZ2d(a)comcast.com>, > "K_h" <KHolmes(a)SX729.com> wrote: > >> "Virgil" <virgil(a)comcast.net> wrote in message >> news:virgil-B82679.22384714042007(a)comcast.dca.giganews.com... >> > In article <qbOdnddPifcj97zbnZ2dnUVZ_rylnZ2d(a)comcast.com>, >> > "K_h" <KHolmes(a)SX729.com> wrote: >> > >> >> "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: >> >> > >> >> > <removed for size> >> I did not mean equality as in the same set. From Wikipedia, a set is >> ordered if every pair of its members is mutually comparable. If, for any >> two distinct members x y, we cannot tell if any of these are true (x<y, >> x=y, >> y<x) then the set is not ordered. Consider a set of four people and the >> set >> is ordered by height. Sam if 5 feet tall, Jane is 6 feet tall, John is 6 >> feet tall, and Mike is 7 feet tall. The set {Sam, Jane, John, Mike} is >> ordered by: >> >> Sam < Jane = John < Mike >> >> Clearly every pair of members of this set is comparable and Jane = John >> under this ordering. Equality here means that Jane and John are >> comparable >> and equal under the ordering; it does not mean that Jane is John. This >> is >> an ordered set by Wikipedia's definition because all of its members are >> mutually comparable under the relation. Are you claiming that an ordered >> set requires x<y or y<x for any two distinct members x and y? > > A totally ordered set does. In mathworld's definition for a totally ordered set, I still don't see where there is a requirement that x<y or y<x for any two distinct members. The example I gave satisfied mathworld's definition. If x=Sam and y=Jane then x<y is satisfied and both x=y and y<x is false. If x=Jane and y=John then x=y is true and both x<y and y<x is false. In all these cases one and only one of (x<y, x=y, y<x) is true. Certainly for a set like the real numbers, under their usual ordering, two different real numbers can never be equal. Can you provide a source where the definition of a totally ordered set *requires* that x<y or y<x for any two distinct members x and y? If so then mathworld's definition for a totally ordered set is ambiguous. K_h
From: Virgil on 15 Apr 2007 19:23
In article <87WdnbF6VIxxMr_bnZ2dnUVZ_gqdnZ2d(a)comcast.com>, "K_h" <KHolmes(a)SX729.com> wrote: > "Virgil" <virgil(a)comcast.net> wrote in message > news:virgil-19C008.15430215042007(a)comcast.dca.giganews.com... > > In article <VZGdnTTzjvOPG7_bnZ2dnUVZ_hGdnZ2d(a)comcast.com>, > > "K_h" <KHolmes(a)SX729.com> wrote: > > > >> "Virgil" <virgil(a)comcast.net> wrote in message > >> news:virgil-B82679.22384714042007(a)comcast.dca.giganews.com... > >> > In article <qbOdnddPifcj97zbnZ2dnUVZ_rylnZ2d(a)comcast.com>, > >> > "K_h" <KHolmes(a)SX729.com> wrote: > >> > > >> >> "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: > >> >> > > >> >> > > > <removed for size> > > >> I did not mean equality as in the same set. From Wikipedia, a set is > >> ordered if every pair of its members is mutually comparable. If, for any > >> two distinct members x y, we cannot tell if any of these are true (x<y, > >> x=y, > >> y<x) then the set is not ordered. Consider a set of four people and the > >> set > >> is ordered by height. Sam if 5 feet tall, Jane is 6 feet tall, John is 6 > >> feet tall, and Mike is 7 feet tall. The set {Sam, Jane, John, Mike} is > >> ordered by: > >> > >> Sam < Jane = John < Mike > >> > >> Clearly every pair of members of this set is comparable and Jane = John > >> under this ordering. Equality here means that Jane and John are > >> comparable > >> and equal under the ordering; it does not mean that Jane is John. Then the set is not totally ordered, but only partially ordered. In a totally and strictly ordered set of any two different members, one MUST be larger and the other smaller. This > >> is > >> an ordered set by Wikipedia's definition because all of its members are > >> mutually comparable under the relation. Are you claiming that an ordered > >> set requires x<y or y<x for any two distinct members x and y? > > > > A totally ordered set does. > > In mathworld's definition for a totally ordered set, I still don't see where > there is a requirement that x<y or y<x for any two distinct members. Trichotomy! <quote> 4. Comparability (trichotomy law): For any a and x , either x >= y or y >= x. <unquote> This means that whenever x = y does NOT hold then either x>y or y>x. > Can you provide a source where the definition of a totally ordered set > *requires* that x<y or y<x for any two distinct members x and y? If so then > mathworld's definition for a totally ordered set is ambiguous. See above! Unless x = y, one must have either x<y or y<x !!!! |