In a metric space,two disjoint closed sets can be separated by two disjoint open sets.

Question

Consider $2$ closed sets $A,B$ with $A\cap B=\emptyset$.Note that for $x\in A$,$d(x,B)>0$, because if for some $x\in A,d(x,B)=0$ then $x\in \overline B=B$, which contradicts the disjunction between $A$ and $B$. So, for each $x\in A$, we can choose $\eta_x$ such that $B(x,\eta_x)\cap B=\emptyset$. Consider $E=\cup_x B(x,\eta_x)$. Similarly construct $F=\cup_yB(y,\eta_y)$. Now, it is clear that $E\cap F=\emptyset$ and $A\subset E$ and $B\subset F$ and $E,F$ are both open. Is this correct?I just want to make sure whether my approach is correct or how can I proceed through this method.I am not looking for a solution to this problem.

Answer 1

Let $U = \cup_{a \in A} B(a,{1 \over 3} d_B(a))$, $V = \cup_{b \in B} B(b, {1 \over 3} d_A(b))$, note $U,V$ are open and $A \subset U, B \subset V$.

Also, for $a \in A, b \in B$ we have $d(a,b) \ge \max(d_A(b), d_B(a))$.

Suppose for contradiction that $x \in U \cap V$ then $x \in B(a,{1 \over 3} d_B(a)) \cap B(b, {1 \over 3} d_A(b))$ for some $a,b$ and so $\max(d_A(b), d_B(a)) \le d(a,b) \le d(a,x) + d(x,b) \le {1 \over 3} (d_A(b)+d_B(a)) \le {2 \over 3} \max(d_A(b), d_B(a))$, a contradiction.

Answer 2

For any non empty set $Y \subseteq X$, the distance function $d( \cdot, Y)$ is continuous. So define $f(x) = \frac{d(x, A)}{d(x, A) + d(x, B)}$. Then $f$ is continuous, because it is a quotient of continuous functions, where the denominator never vanishes. Take $E = f^{-1}(-1/2, 1/2)$ and $F = f^{-1}(1/2, 3/2)$. They are disjoint open sets that contain $A$ and B, respectively.