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Math Help - Closed set

  1. #1
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    Closed set

    If a subset of a top. space is closed, then it contains all its limit points.

    So I need to show that if A\subset X is closed then A=\bar{A}.

    Since A is closed, X-A is open. Also, \bar{A}=A\cup A'

    A' is the set of all limit points of A.

    I am not sure if this helps, but

    X-\bar{A}=X-(A\cup A')=(X-A)\cup (X-A')
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Closed set

    Quote Originally Posted by dwsmith View Post
    If a subset of a top. space is closed, then it contains all its limit points.

    So I need to show that if A\subset X is closed then A=\bar{A}.

    Since A is closed, X-A is open. Also, \bar{A}=A\cup A'

    A' is the set of all limit points of A.

    I am not sure if this helps, but

    X-\bar{A}=X-(A\cup A')=(X-A)\cup (X-A')
    I'm confused, you're trying to prove that if A is closed then \overline{A}=A where \overline{A} is DEFINED to be A\cup A'? Well, isn't A'\subseteq A so that trivially A\cup A'=A?
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    Re: Closed set

    Quote Originally Posted by Drexel28 View Post
    I'm confused, you're trying to prove that if A is closed then \overline{A}=A where \overline{A} is DEFINED to be A\cup A'? Well, isn't A'\subseteq A so that trivially A\cup A'=A?
    No. The limit points don't have to to be in A.
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    Re: Closed set

    Quote Originally Posted by dwsmith View Post
    The limit points don't have to to be in A.
    They must belong to A if A is closed.
    In the OP you wrote: if A is closed then A=\overline{A}.

    Note that one way of defining \overline{A} as the smallest closed set that contains A.
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    Re: Closed set

    I think there is an issue. The also part is clarifying notation not part of the statement to prove.
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    Re: Closed set

    Quote Originally Posted by dwsmith View Post
    I think there is an issue. The also part is clarifying notation not part of the statement to prove.
    This is your OP.
    Quote Originally Posted by dwsmith;682765
    So I need to show that if [TEX
    A\subset X[/TEX] is closed then A=\bar{A}. Since A is closed,
    There it is. You said it twice.
    Because X\setminus A is open then by definition of limit point, all A'\subseteq A.
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