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Closed sets

  • Thread starter golriz
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  • #1
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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.
 

Answers and Replies

  • #2
gb7nash
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What have you tried?
 
  • #3
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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. "
 
  • #4
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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.
 
  • #5
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could you please help me with this question
 
  • #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?
 
  • #7
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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...
 
  • #8
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is there any idea for continuing the above solution??
 
  • #9
gb7nash
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a point z that is in U and closure of A
Yes, expand on this. If z is in U and the closure of A, what does this mean?
 

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