# Closed sets

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.

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gb7nash
Homework Helper
What have you tried?

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. "

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.

gb7nash
Homework Helper
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?

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...

is there any idea for continuing the above solution??

gb7nash
Homework Helper
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?