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Accumulation points must be interior or boundary points.

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove the following: an accumulation point of a set S is either an interior point of S or a boundary point of S.

    2. Relevant equations

    None

    3. The attempt at a solution

    Suppose x is not an interior point. Then you cannot find a neighborhood around x such that N is a subset of S. Suppose further that x is not a boundary point, then there exists a neighborhood N' around x such that N' intersecting S is empty OR N' intersecting ℝ\S is empty. But recall that if x is an accumulation point, all deleted neighborhoods N'* around x contain a point in S, which means N'* intersecting S is nonempty. So indeed, N' intersecting S is nonempty. So we must have that N' intersecting ℝ\S is empty. But this means that N' must be a subset of S, which contradicts our statement earlier, saying we cannot find a neighborhood N around x such that N is a subset of S. Thus, if x is not an interior point and x is an accumulation point, x is a boundary point.

    If x is not a boundary point, then there exists a neighborhood N around x such that N intersecting S is empty OR N intersecting ℝ\S is empty. But since x is an accumulation point, for all deleted neighborhoods N* around x, there exists a point in both S and N*. So N* intersecting S is nonempty, thus N intersecting S must be nonempty. So we must have that N intersecting ℝ\S is empty. Suppose further that x is not an interior point. Then there does not exist any neighborhood N' around x such that N is a subset of S.If N is never a subset of S, then N intersecting with ℝ\S is nonempty, a contradiction. Thus, if x is not a boundary point and x is an accumulation point, x is an interior point. ♦

    Not sure if my proof makes sense or not...it's so long that even i've gotten lost along the way. Any help would be greatly appreciated.

    Thank you in advance!
     
  2. jcsd
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