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Thread: Interior/Closure

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

    Is this correct?

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

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

    If so then heres my proof...

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

    Use [tex]X\setminus A[/tex] to get $\displaystyle X\setminus A$.
    Last edited by Plato; May 4th 2009 at 03:28 PM.
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  2. #2
    Member
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    Quote Originally Posted by Deadstar View Post
    Is this correct?

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

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

    If so then heres my proof...

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

    Use [tex]X\setminus A[/tex] to get $\displaystyle X\setminus A$.
    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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