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Math Help - A Property of Boundary of a Set in a Topological Space

  1. #1
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    A Property of Boundary of a Set in a Topological Space

    Dear Colleagues,

    I want a help for proving the following property:

    Bd[Bd{Bd(A)}]=Bd[Bd(A)], where Bd denotes the boundary and A is subset of a toplogical space X.

    Actually I proved that Bd[Bd{Bd(A)}] is a subset of Bd[Bd(A)] but I could not prove the converse.


    Regards,
    Raed.
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  2. #2
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    Re: A Property of Boundary of a Set in a Topological Space

    Quote Originally Posted by raed View Post
    Bd[Bd{Bd(A)}]=Bd[Bd(A)], where Bd denotes the boundary and A is subset of a toplogical space X.
    Have a look at the web reference
    Note that \partial (A) is the boundary of A which is a closed set so that \partial \partial (A) = \partial \partial \partial (A)
    Thanks from raed
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    Re: A Property of Boundary of a Set in a Topological Space

    Quote Originally Posted by Plato View Post
    Have a look at the web reference
    Note that \partial (A) is the boundary of A which is a closed set so that \partial \partial (A) = \partial \partial \partial (A)
    Thank you for your reply actually using that the boundary is closed I proved one inclusion it is not clear for me how to get the converse.


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    Re: A Property of Boundary of a Set in a Topological Space

    Quote Originally Posted by raed View Post
    Thank you for your reply actually using that the boundary is closed I proved one inclusion it is not clear for me how to get the converse.
    Actually I proved that Bd[Bd{Bd(A)}] is a subset of Bd[Bd(A)] but I could not prove the converse.
    Well I have not have any idea what to tell you.
    It is clear on that webpage I gave you.
    It says "For any set S, ∂S ⊇ ∂∂S, with equality holding if and only if the boundary of S has no interior points, which will be the case for example if S is either closed or open. Since the boundary of a set is closed, ∂∂S = ∂∂∂S for any set S.

    If M is a set then the closure \overline{M}=\text{int}(M)\cup\partial(M).
    Thus \partial(M)\subseteq M if M is closed.

    Is it possible that \partial\partial(A) has any interior points?
    Thanks from raed
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