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Closed sets

  1. Oct 17, 2011 #1
    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.
     
  2. jcsd
  3. Oct 17, 2011 #2

    gb7nash

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    What have you tried?
     
  4. Oct 17, 2011 #3
    Please help me how can I Prove that " the intersection of U and closure of A is empty if and only if the intersection of U and A be empty. "
     
  5. Oct 17, 2011 #4
    I have proved just one direction of this question:
    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.
    But I don't know how to prove the opposite direction of this question.
    Please help me.
     
  6. Oct 17, 2011 #5
    could you please help me with this question
     
  7. Oct 17, 2011 #6

    gb7nash

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    I would approach this doing a proof by contradiction.

    Assume the intersection of U and A is empty and assume to the contrary that the intersection of U and the closure of A is not empty.

    What does this mean?
     
  8. Oct 17, 2011 #7
    it means that the intersection of U and the closure of A is a set S which contains a point z that is in U and closure of A both. Actually I don't know how to continue the rest of the prove...
     
  9. Oct 17, 2011 #8
    is there any idea for continuing the above solution??
     
  10. Oct 18, 2011 #9

    gb7nash

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    Yes, expand on this. If z is in U and the closure of A, what does this mean?
     
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