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Math Help - Interior/Closure

  1. #1
    Super Member Deadstar's Avatar
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    Interior/Closure

    Is this correct?

    Prove that \overline{A} = X\setminus int(X\setminus A)

    Is it equivalent to prove that X\setminus\overline{A} = int(X\setminus A)?

    If so then heres my proof...

    Let x \in X\ \overline{A}, then B(x,r) \cap X\ A = \emptyset \forall r>0
    So then \exists r > 0 such that B(x,r) \subseteq X\A. Hence x \in int(X\A).

    Use [tex]X\setminus A[/tex] to get X\setminus A.
    Last edited by Plato; May 4th 2009 at 04:28 PM.
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    Is this correct?

    Prove that \overline{A} = X\setminus int(X\setminus A)

    Is it equivalent to prove that X\setminus\overline{A} = int(X\setminus A)?

    If so then heres my proof...

    Let x \in X\ \overline{A}, then B(x,r) \cap X\ A = \emptyset \forall r>0
    So then \exists r > 0 such that B(x,r) \subseteq X\A. Hence x \in int(X\A).

    Use [tex]X\setminus A[/tex] to get X\setminus A.
    Seems correct to me, except that it should be there exists instead of for all: there exists some s>0 such that the intersection is empty, that's the negation of for all r>0, the intersection is non-empty.
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