The Definition of Points
in [Math]
|
Prev: Guide to presenting Lemma, Theorems and Definitions
Next: Density of the set of all zeroes of a function with givenproperties
From: Lester Zick on 15 Apr 2007 13:57 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? ~v~~
From: K_h on 15 Apr 2007 17:07 "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? 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. http://en.wikipedia.org/wiki/Totally_ordered_set K_h
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~~ |