Let $X$ be a topological space, and $A, B \subset X$ be open in $X$.
Is $\overline{A \cap B}$ open in $\overline{A} \cap \overline{B}$?
Background:
- It can be shown that $\overline{A \cap B} \subset \overline{A} \cap \overline{B}$.
- In general, these sets are different, e.g. if $A = (0, 1)$ and $B = (1, 2)$ are open intervals in $\mathbb{R}$, then $\overline{A \cap B} = \emptyset$, while $\overline{A} \cap \overline{B} = \{1\}$. For our question, $\emptyset$ is open in $\{1\}$.
- Without the openness requirement for $A$ and $B$, one can give the following counter-example. Let $A = \mathbb{Q} \cup [0, 1]$, and $B = \mathbb{R} \setminus \mathbb{Q}$. Then $\overline{A \cap B} = [0, 1]$, and $\overline{A} \cap \overline{B} = \mathbb{R}$.
- Suppose $\overline{A} \cap \overline{B}$ is connected, and $\overline{A \cap B} \neq \emptyset$. Then $\overline{A \cap B}$ is open in $\overline{A} \cap \overline{B}$ if and only if $\overline{A \cap B} = \overline{A} \cap \overline{B}$.
Best Answer
No.
Let $$A=\{(x,y)\in\mathbb{R}^2 : y>0\}\cup B_r(0,0)$$ and $$B=\{(x,y)\in \mathbb{R}^2 : y<0\}\cup B_r(0,0)$$
where $B_r(0,0)$ is an open ball of radius $r$ around $0$ (choose $r=1$).
Then, $A\cap B = B_r(0,0)$, however $\overline{A}\cap \overline{B} = \{(x,0) : x\in\mathbb{R}\} \cup \overline{B_r(0,0)}$. The closed ball of radius $r$ around zero is not open in that set.