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Math Help - Proof: Accumulation points and closed sets.

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    Proof: Accumulation points and closed sets.

    I need help with this question cause I don't know where to start off:

    Prove that if a set contains each of its accumulation points, then it must be a closed set.
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    Quote Originally Posted by Porter1 View Post
    I need help with this question cause I don't know where to start off: Prove that if a set contains each of its accumulation points, then it must be a closed set.
    What sort of space are you working in?
    Is it a simple metric space, a Hausdorff space, or just a general topological space?
    It does make a difference is the answer.
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    Oh sorry its just a general topological space
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    Quote Originally Posted by Porter1 View Post
    Oh sorry its just a general topological space
    In that case it may not be true.
    So what topology do you have on this space?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Porter1 View Post
    I need help with this question cause I don't know where to start off:

    Prove that if a set contains each of its accumulation points, then it must be a closed set.
    Quote Originally Posted by Plato View Post
    In that case it may not be true.
    So what topology do you have on this space?
    Maybe I am overlooking something here, but I thought that in a topological space X A set E is closed iff E=\bar{E} and \bar{E}=E\cup D\left(E\right), where D\left(E\right) is the derived set (set of all limit points of E, and so D\left(E\right)\subseteq E\implies E=E\cup D\left(E\right)=\bar{E} and so E is closed?

    I am relatively new to general topology.
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    Quote Originally Posted by Drexel28 View Post
    Maybe I am overlooking something here, but I thought that in a topological space X A set E is closed iff E=\bar{E} and \bar{E}=E\cup D\left(E\right), where D\left(E\right) is the derived set (set of all limit points of E, and so D\left(E\right)\subseteq E\implies E=E\cup D\left(E\right)=\bar{E} and so E is closed?
    All of that is very true. But that is not the question.
    The question is: Prove that the derived set of a set is closed. And that is not true.
    Think of the space of real numbers, \mathbb{R} with a topology generated by the collection \left\{ {( - \infty ,a):a \in \mathbb{R}} \right\}.
    The derived set of \{0\} is the set (0,\infty) which is not closed because 0 is limit point of that set.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Plato View Post
    All of that is very true. But that is not the question.
    The question is: Prove that the derived set of a set is closed. And that is not true.
    Think of the space of real numbers, \mathbb{R} with a topology generated by the collection \left\{ {( - \infty ,a):a \in \mathbb{R}} \right\}.
    The derived set of \{0\} is the set (0,\infty) which is not closed because 0 is limit point of that set.
    Sorry! Question confusion.
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