# Closure and nowhere dense

Let $X$ be a topological space and $A,B \subseteq X.$

1. If $\overline{X\cap A} \cap \overline{X\setminus A}$ is nowhere dense in $X$, is it true that for each closed set $F$ in $X$, $\overline{F\cap A}\cap \overline{F\setminus A}$ is nowhere dense in $F$?

2. If $\overline A \cap \overline{X\setminus A}\neq X$ and $\overline B \cap \overline{X\setminus B} \neq X,$ is it true that $\overline{A\cup B}\cap \overline{X\setminus(A\cup B)} \neq X$?

Any hint would be appreciated. Thanks in advance.

• 1.If $U$ is nonemptyset and open in $F$, then $U=O\cap F$, with $O$ open in $X$. Since $T=\overline{X\cap A}\cap \overline{(X\setminus A)}$ nowhere dense in $X,$ there is open set $V$ in $X$ such that $V\subseteq O$ and $V \cap T=\emptyset.$ Then $V\cap F\subseteq O\cap F=U$ and $$V\cap F\cap (\overline{(F\cap A)} \cap \overline{(F \setminus A)})=\emptyset.$$ I'm not sure here that $V\cap F$ is a nonempty set. – Maurten Erik Nov 24 '17 at 10:40
• 2. Assume that $(\overline{A\cup B}\cap \overline{X\setminus (A\cup B)}=X.$ Then we get $A\cup B$ dense in $X$ and $X\setminus (A\cup B)$ dense in $X$. If $X\setminus(A\cup B)$ dense in $X,$ then we get $X\setminus A$ and $X\setminus B$ dense in $X.$ Since $A\cup B$ dense in $X$, then $\overline{A}\cup \overline{B}=X.$ I get stuck here. – Maurten Erik Nov 24 '17 at 10:42
• 1. Note: $V$ is nonempty set – Maurten Erik Nov 24 '17 at 11:14