0
$\begingroup$

I have this exercise

Let $U$ a subset from a metric space $X$, the following two statements are equivalent :

1) $U$ is an open set,

2) For all $A\subset X,$ $\overline{A\cap U}=\overline{\overline{A}\cap U}$

All what I see is that $\overline{A\cap U}\subset\overline{\overline{A}\cap U}$

Edit1: I proved that 1) implies 2) , now i don't know how to prove that 2) implies 1)

Thank you

  • 0
    At least OP has done sth by observing a set inclusion. – 2017-02-24
  • 0
    Yeah but at the same time this is not the first homework question posted today by them. – 2017-02-24
  • 0
    @Noah i proved 1) implies 2) – 2017-02-24

1 Answers 1

2

1 to 2:

Always $A \subset \overline{A}$, so monotonicity of closure gives $\overline{A \cap U} \subset \overline{\overline{A} \cap U}$, without any assumptions. So assume $x \in \overline{\overline{A} \cap U}$. We want to show that $x \in \overline{A \cap U}$. So let $O$ be an open neighbourhood of $x$. Then $O$ intersects $\overline{A} \cap U$. So for any $y$ in that intersection $O \cap \overline{A} \cap U$ we know that $O \cap U$ is a neighbourhood of $y$ (here we use $U$ is open), and so $y \in \overline{A}$ implies that $A \cap (O \cap U) \neq \emptyset$. So $O \cap (U \cap A) \neq \emptyset$, and as $O$ was an arbitrary neighbourhood of $x$, $x \in \overline{A \cap U}$. So we have the reverse inclusion as well.

2 to 1: suppose the equality holds for all $A$. We need to see that $U$ is open. Idea: choose $A =X \setminus U$. Then the left hand side becomes $\overline{\emptyset} = \emptyset$, while the right hand side becomes $\overline{\overline{X \setminus U} \cap U}$. But this implies that no point can be in $\overline{X\setminus U} \cap U$. Or $\overline{X \setminus U} \subset X\setminus U$, which makes $X\setminus{U}$ closed and thus $U$ open.