If $A$ and $B$ are disjoint open sets in some metric space $X$, prove that they are separated.
My (Attempted) Proof:
$A$ and $B$ disjoint $\implies$ $A \cap B = \emptyset$
Since $A$ is open, every point $p \in A$ is an interior point of $A$. Thus for every $p$ in $A$, we can find a neighborhood $N(p)$, such that $N(p) \subset A$
Let $A'$ be the set of all limit points of $A$. We want to show $a' \in A' \implies a' \not\in B$
If $a' \in A'$ and $a' \in A$, then $a' \not\in B$ (as $A$ and $B$ are disjoint). Now if $a' \in A'$ and $a' \not\in A'$, then $a'$ must be a boundary point of $A'$
Assume $a' \in B$. Since $B$ is open, $a'$ must be an interoir point of $B$. Therefore there exists a neighborhood, $N(a') \subset B$, with radius $\delta$, i.e. $d(a', q)) < \delta$. But since $a'$ is a boundary point of $A$, that would imply there are some $q \in N(a')$, that are also in $A$. But those $q$ must also be in $B$, implying $A \cap B \neq \emptyset$. Thus $a' \not \in B$.
Therefore no $a' \in A'$ is an element of $B$ and $A' \cap B = \emptyset$. Let $\overline{A}$, denote the closure of $A$. It is clear that $(\overline{A} = A \cup A') \cap B = \emptyset$.
We repeat this proof for $B$ to show $(\overline{B} = B \cup B') \cap A = \emptyset$, and thus we have $A$ and $B$ as separated sets as desired. $\ \square$
Is my proof correct, and logically sound and rigorous? Also if you have any feedback on my proof writing skills, and suggestions as to where I can improve I would love to hear them!
Best Answer
Your proof is correct, but possibly unnecessarily convoluted - for example, you clearly can simply let $A$ itself be your neighbourhood $N(p)$ for $p\in A$.
You want to show $\overline A\cap B=\emptyset$. But simply as $A$ is open, we know the complement $B^c$ is closed and from $A\cap B=\emptyset$, we have $A\subseteq B^c$. Consequently, $\overline A\subseteq B^c$ and as $B^c\cap B=\emptyset$, also $\overline A\cap B=\emptyset$.