Let $A$ and $B$ be disjoint closed subsets of a metric space $(X,d)$. Give a direct proof for the existence of disjoint open subsets $U_a$ and $U_b$ of $X$ such that $A \subset U_a$ and $B \subset U_b$.
My approach: I found this problem a bit trivial, but maybe I was wrong. Since $A$ and $B$ are disjoint closed subsets, there must exist a set of points $\left\{x_1,…,x_n\right\}$ that do not belong to both $A$ and $B$. Take $U_a = A\cup \left\{x_1, x_2, x_3\right\}$ and $U_b = B\cup \left\{x_4,x_5,…x_n\right\}$ such that for each $x_i$ in $U_a$ or $U_b$, exist some $\epsilon>0$ that contains the open ball $B_{\epsilon}(x_i)$. So $U_a$ and $U_b$ are open and disjoint (Q.E.D)
Best Answer
This answer is from Munkres Theorem 32.2 (metrizable spaces are normal)
For any $a\in A$ choose $\epsilon_a>0$ s.t. $B_{\epsilon_a}(a)\cap B=\emptyset$
For any $b\in B$ choose $\epsilon_b>0$ s.t. $B_{\epsilon_b}(b)\cap A=\emptyset$
Define $U=\bigcup_{a\in A}B_{\frac12\epsilon_a}(a)$ and $V=\bigcup_{b\in B}B_{\frac12\epsilon_b}(b)$.
They are open with $A\subseteq U$ and $B\subseteq V$.
It remains to be shown that the sets are disjoint.
Assume that $x\in U\cap V$.
Then $a\in A$ and $b\in B$ exist with $d(a,x)<\frac12\epsilon_a$ and $d(b,x)<\frac12\epsilon_b$ leading to $d(a,b)<\frac12(\epsilon_a+\epsilon_b)$.
If $\epsilon_b\leq\epsilon_a$ then $d(a,b)<\epsilon_a$ contradicting that $B_{\epsilon_a}(a)\cap B=\emptyset$.
If $\epsilon_a\leq\epsilon_b$ then $d(a,b)<\epsilon_b$ contradicting that $B_{\epsilon_b}(b)\cap A=\emptyset$.
Hence no such x exists $\implies U\cap V=\phi$