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Interior points of the closure of A

  1. Oct 18, 2011 #1
    Is it true?
    " Set of interior points of the closure of A equals the set of interior points of A. "
     
  2. jcsd
  3. Oct 18, 2011 #2

    micromass

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    What did you try already???

    Maybe you can start by taking A open. Then your statement says that A equal the interior of the closure of A. Try it with some easy/less easy examples of open sets first.
     
  4. Oct 18, 2011 #3
    I want to prove this theorem:
    " Let U be an open subset of a metric space X,and A be an arbitrary subset of X. Prove that the intersection of U and closure of A is empty if and only if the intersection of U and A be empty. "
    I've proved one direction of this theorem:

    If the intersection of U and closure of A is empty then the intersection of U and A is empty too.
    The closure of A is equal to the union of A and the set of all limit points(accumulation points) of A. Then we can use this definition of the closure of A. then after substitution, we have:
    [ intersection of A and U ] U [intersection of U and the set of limit points of A] = empty set
    so it says that both the sets [ intersection of A and U ] and [intersection of U and the set of limit points of A] should be empty.
     
  5. Oct 18, 2011 #4
    But I don't know how to prove the opposite direction of this question.
    I have tried:

    using contradiction, it means the intersection of U and cl(A) is not empty.
     
  6. Oct 18, 2011 #5

    micromass

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    This can be made easier. Just use that [itex]A\subseteq \overline{A}[/itex].

    For the other direction, use the definition of closure. And use that A is an open set that doesn't intersect U.
     
  7. Oct 18, 2011 #6
    Then the intersection of set of interior points of U (that equals to U, since U is open) and set of interior points of A equals to the set of interior points of the intersection of (U and A). And the set of interior points of (U and cl(A) ) equals to intersection of U and the set of interior points of cl(A) that is not empty. can I use:
    " Set of interior points of the closure of A equals the set of interior points of A. "
    to show the contradiction?
     
  8. Oct 18, 2011 #7
    sorry!
    what is?
    " Just use that [itex]A\subseteq \overline{A}[/itex]. "
     
  9. Oct 18, 2011 #8
    oh! at first it was typed with vague characters! but now it is in correct form
     
  10. Oct 18, 2011 #9
    Thank you very much!!for your help
     
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