The Definition of Points
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From: Tony Orlow on 18 Apr 2007 13:55 Virgil wrote: > In article <4625a9df(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Mike Kelly wrote: >> < snippery > >>> You've lost me again. A bad analogy is like a diagonal frog. >>> >>> -- >>> mike. >>> >> Transfinite cardinality makes very nice equivalence classes based almost >> solely on 'e', but in my opinion doesn't produce believable results. > > What's not to believe? > > Cardinality defines an equivalence relation based on whether two sets > can be bijected, and a partial order based on injection of one set into > another. > > Both the equivalence relation and the partial order behave as > equivalence relatins and partial orders are expected to behave in > mathematics, so what's not to believe? > > > What I have trouble with is applying the results to infinite sets and considering it a workable definition of "size". Mike's right. If you don't insist it's the "size" of the set, you are free to do with transfinite cardinalities whatever your heart desires. What I object to are statements like, "there are AS MANY reals in [0,1] as in [0,2]", and, "the naturals are EQUINUMEROUS with the even naturals." If you say they are both members of an equivalence class defined by bijection, then I have absolutely no objection. If you say in the same breath, "there are infinitely many rationals for each natural and there are as many naturals overall as there are rationals", without feeling a twinge of inconsistency there, then that can only be the result of education which has overridden natural intuition. That's my feeling. I'd rather acknowledge that omega is a phantom quantity, and preserve basic notions like x>0 <-> x+y>y, and extend measure to the infinite scale. >> So, >> I'm working on a better theory, bit by bit. I think trying to base >> everything on 'e' is a mistake, since no infinite set can be defined >> without some form of '<'. I think the two need to be introduced together. > > Since any set theory definition of '<' is ultimately defined in terms of > 'e', why multiply root causes? It's based on the subset relation, which is a form of '<'.
From: Tony Orlow on 18 Apr 2007 14:03 stephen(a)nomail.com wrote: > In sci.math Tony Orlow <tony(a)lightlink.com> wrote: >> Mike Kelly wrote: >>> The point that it DOESN'T MATTER whther you take cardinality to mean >>> "size". It's ludicrous to respond to that point with "but I don't take >>> cardinality to mean 'size'"! >>> >>> -- >>> mike. >>> > >> You may laugh as you like, but numbers represent measure, and measure is >> built on "size" or "count". > > What "measure", "size" or "count" does the imaginary number i represent? Is i a number? > The word "number" is used to describe things that do not represent any sort of "size". > > Stephen Start with zero: E 0 Define the naturals: Ex -> Ex+1 Define the integers: Ex -> Ex-1 Define imaginary integers: Ex -> sqrt(x) i=sqrt(0-(0+1)), so it's built from 0 and 1, using three operators. It's compounded from the naturals. A nice picture of i is the length of the leg of a triangle with a hypotenuse of 1 and a leg of sqrt(2), if that makes any sense. It's kind of like the difference between a duck. :) Tony
From: MoeBlee on 18 Apr 2007 14:16 On Apr 17, 10:02 am, Tony Orlow <t...(a)lightlink.com> wrote: > I am saying it's obvious that any countable set has a well ordering. It > is not obvious for uncountable sets. It's not only obvious, but it's trivial to prove that every countable set has a well ordering. So, just to be sure you understand, what is at stake with an axiom of countable choice is not at all that every countable set has a well ordering (since we don't need any choice principle to prove that) but rather that for any countable set there is a function that chooses exactly one member from every nonempty subset of that countable set (and actually, we need concern ourselves only with denumerable sets, since we don't need a choice principle to prove that for every finite set there is a function that chooses exactly one member from every nonempty subset of the finite set). MoeBlee
From: Tony Orlow on 18 Apr 2007 14:16 Virgil wrote: > 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? What drug are you on? Do you know any physics? Distance is in terms of d, say. Then area is in terms of d^2 and volume in terms of d^3, etc. So, it stands to reason that, since the derivative of a polynomial is one of the next lower power with respect to whichever variable one is considering, the x-dimensional boundary of a x+1-dimensional object be the derivative of the contained space. With the circles, it's clear. The volume of a sphere is 4/3 pi r^3, and the surface area its derivative, 4 pi r^2. The circumference of a circle is 2 pi r, whose integral, pi r^2, is the area contained. In this case, r appears to be the proper unit of measure to make these equations work. In a regular polygon or polyhedron, one might wonder which r to use. As it turns out, r is from the center of the object to the center of each face or side, and not to the vertices. Using that distance as one's unit of measure always makes the boundary equal to the derivative of the contained space. One of these days it might be nice to generalize this to irregular polygons and polyhedra. I have a feeling it might be a simple matter to determine the proper unit given the set of points, but then, it might get hairy.
From: Tony Orlow on 18 Apr 2007 14:18
Virgil wrote: > 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. Well, actually, I mean N*xN*, including infinite naturals in N*. Sorry, I should have been more specific. |