The Definition of Points
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From: Virgil on 17 Apr 2007 17:33 In article <462507cf(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <462117d1(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > >> Axiomatically, I think the bulk of the burden lies on Choice in its full > >> form. Dependent or Countable Choice seem reasonable, but a blanket > >> statement for all sets seems unjustified. > > > > Since it has been shown that if ZF is consistent then ZFC must be > > consistent as well, what part of ZF does TO object to? > > > > Well, I also put the onus on the Extensionality, as far as equating sets > with the same general membership, but different rates of growth per > iteration, but I haven't quite figured out how to formalize that > statement, or at least, am not in a position to do so now. I.e., vaporware! > > > > >>>> Then how do you presume to declare that my statement is "not true"? > >>>> > >> No answer? Do you retract the claim? > >> > > Yes? > > >>>>> It's very easily provable that if "size" means "cardinality" that N > >>>>> has "size" aleph_0 but no largest element. You aren't actually > >>>>> questioning this, are you? > >>>> No, have your system of cardinality, but don't pretend it can tell > >>>> things it can't. Cardinality is size for finite sets. For infinite sets > >>>> it's only some broad classification. > > > > It is one form of size for all sets. One might use the physical analogy > > that volume, surface area, and maximum linear dimension are all measures > > of the size of a solid. So implying that one "size" fits all is false. > > > > > > Each of those is derivative of the last, given the proper unit of > measure. So does one use distance, area, volume? What is the maximum length of a solid in cubic meters, or in square meters? What is its surface area in cubic meters or in linear meters?
From: Virgil on 17 Apr 2007 17:41 In article <46250ad6(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > Yes, it can be that x<y and y<x and y<>x. Then that is not an order relation. In any form of order relation "<" allowed in mathematics (x<y and y<x) requires x=y. > > >>> 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. > > Oh, yes, well. Any false statement implies any statement, true or false, > as long as you're not an intuitionist. If (x<y) -> ~ (y<x), then x<y ^ > y<x is of the form P ^ ~P, or ~(P v ~P) which is false in classical > logic, but not intuitionistically. There is debate on this topic. I very much doubt that any intuitionist would say that for an order relation,"<", on any set one could have x<y and y<x without having x=y. > > >>> 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? > > I am not. I am saying it is a set equal in magnitude to the redundancies > in NxN. The set of "redundancies in NxN" must be a subset of NxN itself, so must be countable, so that TO is saying that the set of irrationals is "equal in magnitude" to a countable set.
From: Virgil on 17 Apr 2007 17:43 In article <46250d29(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > 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. :) Let Brouwer do it if he disagrees with standard logic. For standard logic it is correct.
From: Virgil on 17 Apr 2007 17:44 In article <46250ea8(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > In all fairness to Lester Why bother to be fair to one who is so compulsively unfair?
From: Virgil on 17 Apr 2007 17:48
In article <46251053(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > Why would I respond to that kind of abuse? If I am a smartass, well, > I've got more smarts than you have in your little finger. (Did that come > out right?) ;) As Moblee is almost certain not to keep much of his smarts in his little finger, that leaves him easily smarter than TO. Even if there were not otherwise overwhelming evidence supporting it. |