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Two subsets $A$ and $B$ in a topological space $X$ are said to be separated if $\overline{A} \cap B = \overline{B} \cap A = \emptyset$. Show that, if $F$ and $G$ are both open or both closed, then $A = F-G$ and $B = G-F$ are separated.

I tried to show that $p \in A$ plus $p \in B$ leads to an absurd, but I have had no luck.

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  • $\begingroup$ Hint: Try to express $A$ and B as intersections of sets which are both open and closed. In particular note that $\overline{F}$ and $\overline{G}$ are also both open an closed. $\endgroup$ – Nex Oct 17 '15 at 13:31
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We have always $$\overline {A\cap B}\subseteq \overline A \cap \overline B$$ So $$(\overline{F-G})\cap (G-F)=(\overline{F\cap G^c})\cap (G\cap F^c) \subseteq \overline F \cap \overline {G^c} \cap G\cap F^c$$ So for example if F is closed right hand side is empty. You should now complete the proof with this hint.

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Let $F$ and $G$ be both open or both closed in a topological space $X$. Note: For all subsets $H$ and $K$ of $X$: cl$\{H\cap K\} \subset$ cl$\{H\}\cap$ cl$\{K\}$.\

Let $A = F\setminus G$ and $B = G\setminus F$. Suppose there is an $x$ in $X$, say $x \in cl\{A\}\cap B= (cl\{F\setminus G\})\cap(G\setminus F)$, then clearly $x$ is in $G$, but not in $F$.

Now $x \in cl\{F\setminus G\}= cl(F\cap (X\setminus G)) \subset cl\{F\}\cap cl(X\setminus G) = cl\{F\}\cap (X \setminus int(G))$. This is $x \in cl\{F\}$ and $x \notin int(G)$.

Together $x \in clF\setminus F$ and $x \in G\setminus int(G)$. If $F$ and $G$ are both closed this implies $x \in clF\setminus F = \emptyset$ and if $F$ and $G$ are both open, this implies that $x \in G \setminus int(G) = \emptyset$, either way we obtain a contradiction.

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