The Definition of Points
in [Math]
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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 !!!!
From: Alan Smaill on 15 Apr 2007 20:03 Lester Zick <dontbother(a)nowhere.net> writes: > 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. knew you wouldn't get it, irony is not a strong point with Ziko. nor indeed do you bother defending your own view that you can use Hospital to work out the value for 0/0. well, there you go. > ~v~~ -- Alan Smaill
From: K_h on 15 Apr 2007 22:30 "Virgil" <virgil(a)comcast.net> wrote in message news:virgil-295022.17233815042007(a)comcast.dca.giganews.com... > 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. I am still looking for an authoritative source for this. The example set I gave was totally ordered, in the sense that every member of the set is mutually comparable, but, just to make it absolutely clear, I will enumerate the entire order: Sam < Jane Sam < John Sam < Mike Jane = John Jane < Mike John < Mike This total order satisfies the law of trichotomy: For any x and y in the ordered set, either (x >= y) or (y >= x). > 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. Sam < Jane Sam < John Sam < Mike Jane = John Jane < Mike John < Mike In the above ordered set, x=y does not hold for the below cases: Sam < Jane Sam < John Sam < Mike Jane < Mike John < Mike These cases satisfy 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 !!!! You seem to be saying that, for totally ordered sets, equality only exists in the case of identity and therefore equality among two distinct members cannot be defined in the order. If that is the case then what you are saying is correct. So what I would like is an authoritative source that says that. Both Wolfram and Wikipedia do not say that. If they did then the definition would say something like "...for any distinct members x y, x<y or y<x" and there wouldn't be statements like "...for any x and y, x<y or y<x or y=x". I am totally prepared to admit that your understanding is correct and both Wolfram and Wikipedia are badly worded. There is lots of misinformation on the internet. What troubles me is that the books on set theory I have also have their definitions worded like wolfram and Wikipedia. All I am asking from you is an authoritative source whose definition of a totally ordered set clearly states that equality among two distinct members cannot be defined in the order. K_h
From: Michael Press on 16 Apr 2007 00:10
In article <1176500762.035139.250710(a)n59g2000hsh.googlegroups.com>, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: > On Apr 13, 12:11 pm, Tony Orlow <t...(a)lightlink.com> wrote: > > I had said that, hoping you might give some explanation, but you didn't > > really. > > Since you posted that, I wrote a long post about the axiom of choice. > Now it's not showing up in the list of posts. Darn! I went into a lot > of detail and answered your questions; I don't want to write it all > again; maybe it will show up delayed. * Composer your messages in a local file. * Set a preference in your news reader to write to a local file a copy, complete with headers, of all your posted messages. * Set a preference in your news reader to mail to a specified address, a copy, complete with headers, of all your posted messages. I am surprised I have to explain this. -- Michael Press |