Separation Properties in Topology

Question

This is Remark 54.5 from Kasriel's Topology book pg. 110

"Suppose X is a metric space Then subsets A and B are mutually separated subsets of X IFF A and B are both closed (or equivalently both open) if A $\cup$ B are disjoint."

I am trying to understand this remark. First of all I thought there can be open subsets that are mutually separated but they're neither open or closed. Second, why in this case A and B being both open and closed are equivalent?

Munkres's book also states that if subsets A and B form a separation in Y then A is both open and closed. So there must be something I am misunderstanding.

Answer 1

This remark seems weird to me, at least it does not make complete sense. What I know is true: if $A$ and $B$ are disjoint and both closed or both open, then they are completely separated. But sets can be completely separated without being open or closed: $A = (0,1) \cap \mathbb{Q}$ and $B=(1,2) \cap \mathbb{Q}$ are neither open nor closed but still completely separated, as $\overline{A} \cap B = \emptyset = A \cap \overline{B}$. Maybe the book means something different by "mutually separated"?

ADDED From another question I answered later, I can see what caused the confusion I think: there the OP defined:

$A$ and $B$ are mutually separated iff $A$ and $B$ are disjoint and both open in $A \cup B$.

which is sort of similar what Firat was posting. I then showed that this is the same notion of being completely separated ($A$ and $B$ are completely separated iff $A \cap \overline{B} = \emptyset =\overline{A} \cap B$).

I'll repeat the proof here: if $A$ and $B$ are mutually separated then there is some open set $O$ of $X$ such that $O \cap (A \cup B) = A$. This $O$ then witnesses that $A \cap \overline{B} = \emptyset$ (we use $O$ as a neighbourhood disjoint from $B$ for all $x \in A$) and similarly for $B$.

If $A$ and $B$ are totally separated, then the closure of $A$ in $A \cup B$ is $\overline{A} \cap (A \cup B) = (\overline{A} \cap A) \cup (\overline{A} \cap B) = A$ so that $A$ is closed in $A \cup B$ and as $A,B$ are disjoint, $B$ (its complement in the union) is open in $A \cup B$, and we similarly show that $A$ is too.