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Set Theory Proof

  1. Sep 2, 2009 #1
    1. The problem statement, all variables and given/known data
    I have been asked to prove the following statement:
    An accumulation point of a set S is either an interior point or a boundary point of S.

    Here is my attempt at a solution:
    I started from the definition of accumulation points:
    A point is an accumulation point ([tex]x_0[/tex]) when [tex] \forall\epsilon > 0, N_{\epsilon}'(x_0)\capA\neq[/tex]empty set

    By using the definition of a deleted neighborhood:
    [tex](N_\epsilon(x_0)-\{x_0\})\capA\neq[/tex]empty set

    so [tex]\forall a \in S' a\inN_\epsilon(x_0), a \notin\{x_0\}, a \inA[/tex]


    and that's as far as I got. I'm not sure where to go from here.
     
  2. jcsd
  3. Sep 2, 2009 #2
    I'm not sure of all your notation but what would happen if [tex]x_{0}[/tex] were not contained in [tex]S[/tex] or its boundary? Specifically, what would the intersections of sufficiently small neighborhoods of [tex]x_{0}[/tex] with [tex]S[/tex] be?
     
  4. Sep 2, 2009 #3
    It appears you are speaking of relative deleted neighborhoods. I.e. [tex]N'_\epsilon(x_0) = \{x \in S-\{x_0\}: |x-x_0| < \epsilon \}[/tex].

    The last line in the original post unfortunately is not syntactically well formed. What was the intended meaning? Was it close to:

    [tex]\forall a \in S' a \in N_\epsilon(x_0), a \notin \{x_0\}, a \in A[/tex]? (Modified from OP)


    Recommendation: Attack this by contradiction. You are asked to show that an accumulation point of S must either lie in the interior of S or on its boundary. The only other possible place it could be is in its exterior (the complement of the closure of S). I'd start with:

    "Assume otherwise. Let x be an accumulation point of S that lies in the exterior of S..." and show this leads to contradiction. (Hint: think closed vs. open sets).

    --Elucidus
     
  5. Sep 3, 2009 #4

    HallsofIvy

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    Given a set S, every point, p, in the "universal set" is either an interior point of S (there is a neighborhood of p contained in S) or an exterior point of S (there is a neighborhood of p that contains no point of S) or a boundary point of S (neither of the above is true). To show that such a point is "either an interior point or a boundary point of S" means to show it is NOT an exterior point of S. You must show that every neighborhood of the point contains at least one point in S.
     
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