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The Universal Geometric Set

  1. Jun 11, 2004 #1
    A simple[trivial?] postulate that gives a "Universal Set" and resolves the "set of all sets" paradox[in the geometric sense]:

    A circle of radius R, is isomorphic to a circle of radius 1/R.

    [1/R]<--->[R]

    For any arbitrarily large circle of radius R, there is an exact correspondence with a circle of radius 1/R, such, that the product R*[1/R] = 1
     
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  3. Jun 11, 2004 #2

    matt grime

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    and this resolves russell's paradox? so where is the set of all sets that do not contain themselves in this construction? and in what sense are you using isomorphism? in what category are your morphisms?
     
  4. Jun 11, 2004 #3
    I don't think it resolves "russell's" paradox without some more work.

    All circles are isomorphic to each other because they have the same shape. Likewise, all squares are isomorphic to each other. Now if sets can be transformed into geometric shapes, more specifically, circles, or "geometric shape-equivalents", the largest possible set with a geometric radius R, has a corresponding twin with radius 1/R.
     
  5. Jun 11, 2004 #4

    matt grime

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    well, when you've figured out what it is you're trying to prove let us know.
     
  6. Jun 11, 2004 #5
    Here is a definition of the "Euler characteristic":

    http://en.wikipedia.org/wiki/Euler_characteristic


    Graph Theory:

    http://en.wikipedia.org/wiki/Graph_theory


    If a polyhedron has V vertices, F faces, E edges, and is topologically equivalent to the sphere, the equation is:

    V + F - E = 2

    2 is the "Euler characteristic" of the polyhedron.

    Sets that are members of themselves correspond to a geometric form. Sets that are not members of themselves correspond to a different? geometric form.

    Interesting.
     
  7. Jun 11, 2004 #6

    Gokul43201

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    I thought circles were isomorphic with squares - they don't have to have the same shape.

    And the elements of the set transform into...?

    And I thought the paradox involved the cardinality of the Power Set being bigger than the cardinality of the Set (of all sets). I may be wrong...but what does this have to do with isomorphisms ?
     
  8. Jun 12, 2004 #7
    Circles are homeomorphic to squares, not isomorphic...?

    http://www.rdrop.com/~half/Creations/Puzzles/TriangleShapes/

    Elements of a set can be characterized as sets. All sets can be associated to geometric forms...?


    Any circle of arbitrarily large radius R, is isomorphic to a circle of radius 1/R.

    The magnitude of R corresponds to the cardinality of the powerset.

    Is the set of all geometric forms, a geometric form?

    Can Venn diagrams correspond to light cone cross sections?
     
    Last edited: Jun 12, 2004
  9. Jun 12, 2004 #8

    Gokul43201

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    Can I answer your questions with more questions ?

    PS : Yes that should have been homeomorphic. But I'm still not getting the point. What is the resolution of the paradox ?
     
    Last edited: Jun 12, 2004
  10. Jun 12, 2004 #9

    Set intersection is a type of multiplication of sets.

    The intersection of two circles of radius R and 1/R, respectively:

    R*[1/R] = 1

    R[<-[->[<-[1/R]->]<-]->]


    The "Universal Set"

    For the continual expansion of power set circle R, there corresponds circle[infinitesimal?] 1/R.
     
    Last edited: Jun 12, 2004
  11. Jun 12, 2004 #10

    Hurkyl

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    Not as far as I know.
     
  12. Jun 12, 2004 #11

    Gokul43201

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    No, it's not ! It's just a process of picking the common elements.

    You can have a set A containing millions of even numbers, and a set B containing thousands of odd numbers and you "multiply" them to get a null set ?
     
  13. Jun 12, 2004 #12
    Set intersection obeys the distributive law, which is a multiplicative law:

    http://www.jgsee.kmutt.ac.th/exell/Logic/Logic31.htm#13

    Two sets without common elements are disjoint.
     
  14. Jun 12, 2004 #13

    Venn diagrams are circles...

    Light cone cross sections are circles, ellipses, etc.
     
  15. Jun 12, 2004 #14

    Hurkyl

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    That doesn't mean set intersection has anything to do with arithmetic multiplication.


    And what does this have to do with associating all sets to geometric forms?
     
  16. Jun 13, 2004 #15

    Since the circle of radius R is isomorphic to the circle of radius 1/R, the cardinality of Circle with radius R is on the same line[radius] as the infinitesimal 1/R

    1/R 0--------0 R

    Since they are on the same line, they intersect. But perhaps a new type of set multiplicative identity needs to be derived?





    When two light cones intersect, they become "phase entangled". The intersection is much like a "set" intersection.


    In ordinary quantum mechanics, configuration space is space itself
    {i.e.,to describe the configuration of a particle, location in space
    is specified}. In general relativity, there is a more general kind of
    configuration space: taken to be the space of 3-metrics {"superspace",
    not to be confused with supersymmetric space} in the geometrodynamics formulation. The wavefunctions[Venn diagrams-light cones] will be
    functions over the abstract spaces, not space itself-- the
    wavefunction defines "space itself".


    The resultant metric spaces are thus defined as being diffeomorphism
    invariant. Intersecting cotangent bundles{manifolds} are the set of
    all possible configurations of a system, i.e. they describe the phase
    space of the system. When the "wave-functions/forms"
    intersect/entangle, and are "in phase", they are at "resonance",
    giving what is called the "wave-function collapse" of the Schrodinger
    equation. the action principle is a necessary consequence of the
    resonance principle.
     
  17. Jun 13, 2004 #16

    Hurkyl

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    Although you're using the words, you don't seem to be doing mathematics, so I'll move this thread over here.
     
  18. Jun 13, 2004 #17

    matt grime

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    no they aren't. what idiot told you that?
     
  19. Jun 13, 2004 #18
    You appear to be acting like an ignorant troll.

    http://mathworld.wolfram.com/VennDiagram.html


    Actually, spacetime does not really need to be "sliced up" in that it can proceed in discrete steps, yet, still be continuous.

    [density 1]--->[density 2]--->[density 3]---> ... --->[density n]


    [<-[->[<-[-><-]->]<-]->]
    Intersecting wavefronts = increasing density of spacelike slices

    As the wavefronts[circles/Venn diagrams] intersect, it becomes a mathematical computation:

    2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, ...2^n


    If the universe includes everything that is real and excludes that which is not real, then the universe is the "Universal" set.

    You cannot refute the above logic...
     
  20. Jun 14, 2004 #19

    matt grime

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    Since when is a simple closed curve necessarily a circle? As you aren't the ignorant one you must surely know that in order to demonstrate all the possible intersections of 4 sets in a venn diagram you cannot use circles. Moreover, surely you, still not being the ignorant one, must also recognise that a venn diagram is not an element of itself, and thus to take the definition you give, and then deduce that a venn diagram is a circle is most definitely not a logical conclusion?


    They, circles and closed curves in the plane, certainly aren't even isomorphic, using your particular definition of isomorphic which appears to mean related be some affine transformation when embedded in the plane.

    But Im the ignorant troll, so what do I know about affine transformations? Proved Fermat's last theorem yet?
     
  21. Jun 15, 2004 #20
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