Separation Properties in Topology

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.

• What is the definition of being mutually separated? Oct 7 '18 at 17:22
• How can a single set $A \cup B$ be disjoint? Oct 7 '18 at 17:23
• At the time when the 1-millionth question was actually posted, a screenshot was taken, see (scroll down) math.meta.stackexchange.com/questions/7021 and it is THIS question. So, yes, congratulations to @Firat Celebi and even more congratulations to MSE and all the contributors !!!! Oct 9 '18 at 20:00
• @MichaelHoppe Yes it is question number 1,000,000. See my previous comment. Oct 9 '18 at 20:13

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"?
$$A$$ 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.