Proving that U∩A is Empty iff U∩Cl(A) is Empty

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In summary: It means that z is either in A or is a limit point of A. Now, since we assumed that the intersection of U and A is empty, this means that z cannot be in A. But this also means that z is a limit point of A, which implies that every neighborhood of z contains a point in A (by definition of a limit point). However, since z is also in U, this means that there exists a neighborhood of z that is contained entirely in U. But this contradicts the fact that every neighborhood of z contains a point in A, since U and A have no points in common. Therefore, our initial assumption that the intersection of U and the closure of A is not empty must be
  • #1
golriz
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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.
 
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What have you tried?
 
  • #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. "
 
  • #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.
 
  • #5
could you please help me with this question
 
  • #6
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
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
is there any idea for continuing the above solution??
 
  • #9
golriz said:
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?
 

FAQ: Proving that U∩A is Empty iff U∩Cl(A) is Empty

What does "U∩A" and "U∩Cl(A)" represent in this statement?

"U∩A" represents the intersection of sets U and A, while "U∩Cl(A)" represents the intersection of set U and the closure of set A.

Why is it important to prove that U∩A is empty if and only if U∩Cl(A) is empty?

This statement is important because it shows that the closure of a set is the smallest closed set that contains the original set. It also helps to understand the relationship between the closure of a set and the intersection of sets.

How can you prove that U∩A is empty if U∩Cl(A) is empty?

To prove this, we can use proof by contradiction. Assume that U∩A is not empty, which means there is an element that belongs to both U and A. This element must also belong to the closure of A, which contradicts the assumption that U∩Cl(A) is empty. Therefore, U∩A must be empty.

Can you provide an example to illustrate this statement?

Suppose U = {1, 2, 3} and A = {4, 5}. The closure of A, Cl(A), would then be {4, 5} as well. The intersection of U and A, U∩A, would be an empty set, since there are no elements that belong to both U and A. This also means that the intersection of U and Cl(A), U∩Cl(A), is also an empty set.

How does this statement relate to other mathematical concepts?

This statement is related to the concept of closed sets and closure in topology. It also connects to the concept of set intersection and the idea of proving statements by contradiction. Additionally, it can be applied to other fields such as graph theory and linear algebra.

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