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Metrics on the plane and the sphere

  1. Aug 2, 2014 #1
    Are plane and surface of sphere different metric spaces?
    Can distance function of plane be applied as distance function of surface of sphere?
    Please correct my question if needed?
     
  2. jcsd
  3. Aug 2, 2014 #2

    HallsofIvy

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    They certainly are different metric spaces! One is unbounded while the other is bounded. I presume that by "distance function of plane" you mean the "standard metric" in three dimension" [itex]d((x_0, y_0, z_0), (x_1, y_1, z_1))= \sqrt{(x_1- x_0)^2+ (y_1- y_0)^2+ (z_1- z_0)^2}[/itex] where (x, y, z) satisfy the equation of the plane.

    Yes, with the provision that (x, y, z) satisfy the equation of the surface of the sphere instead of that of a plane, that same formula defines a metric on the surface of the sphere. But they are still different metric spaces. There is, after all, a difference between a metric formula and a metric space.
     
  4. Aug 2, 2014 #3

    WWGD

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    Like HIvy said, they are not homeomorphic, one, the sphere is compact, plane is not.
     
  5. Aug 2, 2014 #4
    1.Plane is a Euclidean metric space and sphere surface ,I think is non Euclidean metric space.
    2.How (x,y,z) of plane can satisfy surface of sphere,except in the condition when their intersection is the circle.?
    3. Please bear with me as I am a learner.
     
  6. Aug 2, 2014 #5

    HallsofIvy

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    This doesn't mean anything until you have specified which metric you are using.

    Yes, the intersection of a given plane with a given sphere is either
    1) empty
    2) a point
    3) a circle

    But that is not relevant to the original question:
    which is a question about homeomorphism of metric spaces, NOT a question about the intersection of two sets.

     
  7. Aug 2, 2014 #6

    WWGD

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    Yes, a plane is Euclidean, in the sense that it is a manifold with a single chart , e.g., the identity, but the circle does not admit a global Euclidean structure as ## \mathbb R^n ## for any n (because, e.g., this would imply that ## \mathbb R^n ## is compact, which it is not ). For 2, see Ivy's reply.
     
  8. Aug 2, 2014 #7
    Thankyou very much dear friends! Your answers motivate me to study more.Will be back after more study.
     
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