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.

  • $\begingroup$ What is the definition of being mutually separated? $\endgroup$ Oct 7 '18 at 17:22
  • 1
    $\begingroup$ How can a single set $A \cup B$ be disjoint? $\endgroup$ Oct 7 '18 at 17:23
  • 6
    $\begingroup$ 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 !!!! $\endgroup$
    – Andreas
    Oct 9 '18 at 20:00
  • 4
    $\begingroup$ @MichaelHoppe Yes it is question number 1,000,000. See my previous comment. $\endgroup$
    – Andreas
    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"?

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.