The Definition of Points
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From: Lester Zick on 14 Apr 2007 15:46 On 14 Apr 2007 00:09:21 -0700, "Brian Chandler" <imaginatorium(a)despammed.com> wrote: >What a good job then, that mathematics is not about "your >sensibilities". If you accept (use, adopt, whatever) the Axiom of >Choice then there is a proof that any set has a well-ordering. >Sensibilities don't come into it. I wonder though, Brian, if there is an axiom of forgiveness in modern math and how I might go about using, adopting, or whatever that axiom in my SOAP opera proofs? For that matter I wonder if there might also be an axiom of truth in modern math or are we just forced to wing it? ~v~~
From: Virgil on 14 Apr 2007 15:53 In article <46211c0a(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <461fd938(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > > > >>> Nothing in TO's definition of "<" prohibits '(x>y) and (y>x)' from being > >>> true, so if he wishes to require such a prohibition, he must > >>> specifically add it to his transistivity requirement. > >>> > >> 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. > > Times of day have no order? Pulllease!!! Does 12 midnight on a clock come before or after 12 noon on that clock? There is an assumed local order in the sense that if two clock times are close enough together one usually assumes that one of them is "before" and the other "after", but is one minute before midnight clocktime before or after one minute after midnight clocktime? it could be either. > > > > > 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) > > > > Okay, I'm worried about you. You repeated the same erroneous statement. > You didn't cut and paste without reading, did you? Don't you mean "<=" > rather than "<". The statement "x<y and y<y" can only be true in two > unrelated meanings of "<", or else "=" doesn't have usable meaning. TO betrays his lack of understanding of material implication in logic. For "<" being any strict order relation, "(x<y) and (y<x)" must always be false so that any implication with "(x<y) and (y<x)" as antecedent for such a relation, regardless of conseqeunt, is always true. So that I have better cause to be worried about TO than he has to worry about me. .... > >> The rationals are defined by NxN, minus the redundancies in > >> quantity within the matrix. > > > > That "matrix" is a geometric interpretation, which is quite irrelevant. > > > > A better definition for the rationals, based on I as the set of integers > > and P as the set of strictly positive integers is the set IxP modulo the > > "==" relation defined by (a,b) == (c,d) iff a*d = b*c. > > > > > > > >> Equinumerous to those redundancies, which > >> are the vast majority of cells, are the irrationals. That's how it > >> actually works. > > > > Is TO actually claiming that the irrationals form a subset of the > > countable set NxN. > > > > That is NOT how it works in any standard mathematics. Then why does TO claim it?
From: K_h on 14 Apr 2007 18:40 "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
From: Virgil on 14 Apr 2007 17:51 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.
From: K_h on 14 Apr 2007 19:45
"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 and are you standing by your claim that the equality relation cannot be an order relation on a set with more than one member? 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. K_h |