Interior points of the closure of A

[golriz]

Is it true?
" Set of interior points of the closure of A equals the set of interior points of A. "

 

[micromass]

Is it true?
" Set of interior points of the closure of A equals the set of interior points of A. "

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.

 

[golriz]

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.

 

[golriz]

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.

 

[micromass]

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.

This can be made easier. Just use that A\subseteq \overline{A}.

For the other direction, use the definition of closure. And use that A is an open set that doesn't intersect U.

 

[golriz]

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?

 

[golriz]

sorry!
what is?
" Just use that A\subseteq \overline{A}. "

 

[golriz]

oh! at first it was typed with vague characters! but now it is in correct form

 

[golriz]

Thank you very much!for your help