The Definition of Points
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From: Tony Orlow on 17 Apr 2007 14:07 Lester Zick wrote: > 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~~ Incorrect. One leg is both the same. So, start with uncertainty. Then, build truth from such statements. Start with the line, and determine the point of intersection. Just remember, when the lines are moving, so is the point. 01oo
From: Tony Orlow on 17 Apr 2007 14:08 Virgil wrote: > 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. Define "correct", ala Brouwer. :)
From: Tony Orlow on 17 Apr 2007 14:15 Alan Smaill wrote: > 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~~ > In all fairness to Lester, I am the one who said 0 for 0 is 100%. T'was a joke, ala L'Hospital's theft from the Bernoullis, and the division by 0 proscription. :) Tony
From: Tony Orlow on 17 Apr 2007 14:17 K_h wrote: > "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". > "y=x" Means x and y are not "distinct" ala Leibniz. > 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 > > T_o
From: Tony Orlow on 17 Apr 2007 14:18
Michael Press wrote: > 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. > You don't but you enjoy it. |