The Definition of Points
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From: Tony Orlow on 14 Apr 2007 22:08 MoeBlee wrote: > On Apr 13, 12:51 pm, Tony Orlow <t...(a)lightlink.com> wrote: >> MoeBlee wrote: >>> On Apr 13, 11:25 am, Tony Orlow <t...(a)lightlink.com> wrote: >>>> I've discussed that with you and others. It doesn't cover the cases I am >>>> talking about. The naturals have a "measure" of 0, no? So, measure >>>> theory doesn't address the relationship between, say, the naturals and >>>> the evens or primes. It's not as general as it should be. So, what do >>>> you want me to say? >>> Nothing, really, until you learn the mathematics you're pretending to >>> know about. >> I didn't bring up "measure theory". > > Where do I begin: transitivity, ordering, recursion, axiom of > infinity, non-standard analysis...on and on and on... > That sounds more like a question of when to end, but anyway... >>> Nothing to which you responded "pretends" that cardinality "can tell >>> things it can't". What SPECIFIC theorem of set theory do you feel is a >>> pretense of "telling things that it can't"? >> AC > > If you mean non-constructivity, then no one disputes that the axiom of > choice is non-constructive. No one says that the axiom of choice > proves the existence of a definable well ordering. > > But if you require constructivity then you can't without contradiction > endorse Robinson's non-standard analysis. > Maybe I don't understand Robinson as well as I should, but it seems to me the basis of his analysis was semantic, regarding statements that would be considered true of *N if true of N. But, do go on... >>>>> It has CARDINALITY aleph_0. If you take "size" to mean cardinality >>>>> then aleph_0 is the "size" of the set of naturals. But it simply isn't >>>>> true that "a set of naturals with 'size' y has maximum element y" if >>>>> "size" means cardinality. >>>> I don't believe cardinality equates to "size" in the infinite case. >>> Wow, that is about as BLATANTLY missing the point of what you are in >>> immediate response to as I can imagine even you pulling off. >>> MoeBlee >> What point did I miss? > > The MAJOR point - the hypothetical nature of mathematical reasoning > (think about the word 'if' twice in the poster's paragraph) and the > inessentiality of what words we use to name mathematical objects and > their properties. I guess, by "inessentiality", you mean any attribute that one could assign to any object... IF that's what you mean, or not.... > > I've been trying to get you to understand that for about two years > now. > Perhaps what you mean is exactly what I am trying to get across to Lester. If the truth table is the same, then it's the same logical function. It doesn't matter what the parameters are, it always works the same way. The pattern defines the relationship. On the subject of ifs, "if this then that" means logical implication, or causality, and sometimes it's hard to tell which is meant, or if they're being confused. Is that what you "mean"? TOEknee >> I don't take transfinite cardinality to mean >> "size". You say I missed the point. You didn't intersect the line. > > You just did it AGAIN. We and the poster to whom you responded KNOW > that you don't take cardinality as capturing your notion of size. The > point is then just for your to recognize that IF by 'size' we mean > cardinality, then certain sentences follow and certain sentences don't > follow and that what is important is not whether we use 'size' or > 'cardinality' or whatever word but rather the mathematical relations > that are studied even if we were to use the words 'schmize' or > 'shmardinal' or whatever. > > MoeBlee > >
From: Virgil on 15 Apr 2007 00:38 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.
From: Tony Orlow on 15 Apr 2007 11:13 Virgil wrote: > In article <461fde40(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> cbrown(a)cbrownsystems.com wrote: >>> On Apr 13, 10:13 am, Tony Orlow <t...(a)lightlink.com> wrote: >>>> Virgil wrote: > >>>>> Sets in general have all sorts of orderings, but none which is as >>>>> inherent in their being sets as the ordinal order is in sets being >>>>> ordinals. >>>> Once they are ordered in whatever manner, they become sequences, trees, >>>> or other structures, and it is only with such a recursive definition >>>> that such an infinite structure can be created. >>> Why is a recursive definition required? Given any set S, the set of >>> all subsets of S can be (partially) ordered as follows: for subsets A, >>> B of S, define A <= B iff every member of A is a member of B. What is >>> recursive about that definition? >> That structure doesn't look like a tree to you, which each node having >> as children the elements of its power set? That's the picture that >> appears to me. > > Then TO cannot have looked very carefully. > > In a tree, no node has more than one parent but > P({a,b}) = {{}, {a}, {b} and {a,b}} requires that, > (1) if {} is the root, then {a} and {b} are both parents of {a,b}, or > (2) if {a,b} is the root then {a} and {b} are both parents of {}. > Well, sure, it's a redundant tree. > The appropriate mathematical name for such a structure is a /lattice/. >>> Alternatively, S be the set of all subsets of the naturals (note that >>> S is not countable). >> If A, B are in S, define A <= B if there is a >>> natural number m in B such that m is not in A, and for all n < m, n in >>> A and n in B. What is recursive about that definition? >>> >> The Archimedean principle, as far as I can tell. > > Where does the Archimedean principle apply in that definition, and how > is it recursive in that definition? There is always a natural greater than A. >>>> In that sense, there is >>>> no pure infinite set without some defining structure > > Previously claimed without proof and again claimed without proof. > > What axioms would you allow for such a proof? Certainly none less than twice your age. > ... > >>>> Um, that one is blatantly self-contradictory. x>y -> not y>x, always. >>> I don't see how this follows only from your assertion "x < y and y < z >>> -> x < z". You stated: >>> >>>>>>>> Order is defined by x<y ^ y<z -> x<z. >>> Or do you mean that there is /more/ to the definition of an order "<" >>> than "x < y and y < z -> x < z"? If so, that was exactly Virgil's >>> point. >>> >> Actually I corrected this response to Virgil. I misspoke a little, but >> he's still wrong. :) > > Then according to TO mere transistivity make equality an order relation, > as it satisfies TO's sole criterion. I finally did get your point about that, but you made it circuitously, it seemed to me. You're right. Transitivity is trivially true for equality, in any order of statements. Equality is reflexive. So, order doesn't matter. Equality certainly doesn't define order. And yet, it's transitive. But, it's not the primitive '<', and is not mentioned in "a<b ^ b<c ->a<c". It's the result of "not a<b and not b<a". Then, okay, they're as good as equal, ala Leibniz. Ain't no difference. >>>> suppose you meant: >>>> ((x>=y) and (y>=x)) -> x = y >>>> or: >>>> (~(x>y) and ~(y>x)) -> x = y >>>> >>> These two statements are not equivalent. In some situations, the first >>> can hold, while the second does not. >>> >> Please do elaborate. > > if "<" and "=" are the same relation of equality, for instance. Yes, if only transitivity is considered, you're right, equality could be the interpretation of '<', and satisfy "a<b ^ b<c -> a<c". Maybe this gets back to Lester's core topic. We make a distinction between the line defined by '<' and the point defined by '='. Which comes first? >>>> Yes, if neither x<y or y<x is true, that is, if no order can be >>>> determined, then x=y for the purposes of that order. >>> Let's order the subsets of {a,b,c} by inclusion. Then: {a,b} < >>> {a,b,c}. {a,c} < {a,b,c}. {a} < {a,b}. {a} < {a,c}. But not ({a,b} < >>> {a,c}); and not ({a,c} < {a,b}). Does that mean that {a,b} = {a,c} >>> "for the purposes of that order"? >>> >> Where b<c, {a,b}<{a,c}. > > Not necessarily. An order relation on the members of a set does not > necessitate any order relation of the subsets of that set. Nor vice > versa. > > > There is certainly a simple onstructive answer to the question, no? Lexicographic order? >>> Could you state what the definition of "<= totally orders the set S" >>> is again? There are three simply stated properties, IIRC. >>> >> Lookitup. Transitivity does not define "total order", but is that start >> of order. > >>>>> Also there are lots of transitive relations which are not orderings, at >>>>> least as usually understood. E.g., universal relations, which hold true >>>>> for all x and y in the relevant set. >>>>> So that TO's notion of an ordering does not necessarily order anything. >>>> You're missing the point. All I said was that one starts with inequality >>>> defining the line itself. >>> Is every set a line? Is the set of all triangles in the Euclidean >>> plane a line? >>> >> That's a bunch of lines. > > Then it is not one line. Right. It's built upon the concept of a line. >>>> Then one defines equality. Defining equality >>>> where there is no relative order doesn't make sense. >>> So, it makes no sense to say that the set of all finite subsets of the >>> naturals having a prime number of elements is equal to itself? >>> >>> Cheers - Chas >>> >> Again, there is order to the elements themselves. > > Which of a number of possible orders is "the" order on "the set of all > finite subsets of the naturals having a prime number of elements" It's easy to define an order on that set of subsets as a subset of the lexicographic order of the power set, no?
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 |