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Proof: open sphere is an open set

  1. Apr 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that an open sphere in [itex]\mathbb{R}^m[/itex] is an open set.

    2. Relevant equations


    3. The attempt at a solution


    To show that an open sphere is an open set, any point inside the sphere has to be an interior point:
    Let us have a sphere [itex]B(P_0, r), r > 0[/itex], where [itex]P_0[/itex] is the centerpoint and r is the radius of the sphere and also an arbitrary point [itex]P[/itex] inside the sphere. Hence [itex]\exists\varepsilon>0\colon B(P,\varepsilon)\subset B(P_0, r) [/itex] and as we assumed the [itex]B(P_0, r)[/itex] to be open, then it does not contain its boundary.

    The idea is to use the triangle inequality by taking another random point in a sphere around the already random point P in the large sphere: [itex]S\in B(P, \varepsilon)[/itex]. The distance between two points [itex]d(A,B)[/itex] (Pythagorean)
    [itex]d(S, P_0)\leq d(S,P) + d(P,P_0) <\varepsilon + d(P,P_0)[/itex]. This is a shaky part - how can I justify this inequality is always correct? I can visualize it in my head, but is it accurate?

    Per that assumption I would fix [itex]\varepsilon\colon = r - d(P,P_0) > 0[/itex] since [itex]P[/itex] is an interior point and its distance cannot be greater than or equal to the radius of the large sphere, therefore [itex]\varepsilon[/itex] is always strictly positive and [itex]\forall S\in B(P, \varepsilon_1)[/itex] I can state that the distance [itex]d(S,P_0) < \varepsilon + d(P,P_0) = r[/itex] which would mean that the open sphere is an open set as all its points are interior points.
     
  2. jcsd
  3. Apr 1, 2015 #2

    PeroK

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    The argument is essentially correct, but things are a bit mixed up and it doesn't flow. I suggest trying to tidy it up. I would start with something like this:

    Let ##B(P_0, r)## be an open sphere in ##\mathbb{R}^m##

    Let ##P \in B(P_0, r)##

    ##d(P_0, P) < r## hence ##\exists \epsilon > 0## ...
     
  4. Apr 1, 2015 #3

    micromass

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    A silly comment, but I don't like the term "open sphere". The terminology "open ball" is standard, because a sphere is usually the name given to the boundary of the ball, not something in the interior.
     
  5. Apr 1, 2015 #4
    I knew there was something off in the translation. I'll use ball to describe the set and sphere to describe its boundary, then.
     
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