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Boundedness in extended complex plane

  1. Oct 31, 2006 #1
    what does it mean to be bounded in extended complex plane (say C')?
    can we say all subsets of C' is bdd because C' itself bdd?
     
  2. jcsd
  3. Nov 1, 2006 #2

    matt grime

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    The extended complex plane, in my opinion, is not bounded. Bounded ought to mean, for a metric space, contained in some open ball of *finite* radius.
     
  4. Nov 1, 2006 #3
    to be more explicitly
    1)if i use spherical metric( i.e. i think on the sphere ) then C equivalently sphere without north pole is bdd .
    why adding one point namely inf. changes boundedness
    2)another point of view : i can prove C' is compact so it is closed and bdd

    maybe iam missing somethings but i cant find them
     
  5. Nov 1, 2006 #4

    matt grime

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    You are missing something. Just because the extended complex plane is homeomorphic to a sphere does not mean it is bounded. The real line is homeomorphic to (-1,1), and I doubt you think that R is bounded.

    Compact is not the same as closed and bounded. Cu{oo} is not a subset of R^3. It is homeomorphic to a subset of R^3, but 'size' is not invariant under homeomorphism.
     
  6. Nov 1, 2006 #5
    i didnt say it is bdd because S^2 is in R^3. for example an open ball with center at inf. and radius some small quantiity (say eps.) contains
    { z€C s.t. |z|>eps } U {inf} by the way it is defined .
    here what i stuck : why adding only one point(o.k. now it is a ball)
    changes boundedness.
    if you say i cant take inf. as a center i wont insist on that
    but i am pretty sure i saw somewhere compactness implies bddness.
     
  7. Nov 1, 2006 #6

    matt grime

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    It is not bounded. Boundedness applies to subsets of metric spaces. What is the metric you're using? How far from 2 is infinity? Compact is not the same as closed and bounded. The point is you cannot have balls of infinite radii. I said nothing about infinity being or not being the centre of an open set, but as it happens, you cannot have infinity as the centre of a ball defined by the metric because there is no metric given.
     
    Last edited: Nov 1, 2006
  8. Nov 1, 2006 #7
    now there is a new thing that comes to my mind.
    the reason that i cannot say such a thing because a set is bdd whenever that this set is a subset (proper)?.otherwise we would say C is bdd in C
    OR R is bdd in R e.t.c.
    still wrong?
     
  9. Nov 1, 2006 #8

    matt grime

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    You can't say it is bounded because you need to use a metric to define the size of your sets. You haven't put a metrc on the extended complex plane.
     
  10. Nov 1, 2006 #9
    :) i got it with one second difference . so i stop bothering.
    thanks
     
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