# Prove existence of disjoint open sets containing disjoint closed sets in a topology induced by a metric.

Question: Let $(X, d)$ be a metric space. Let $A$ and $B$ be disjoint subsets of $X$ that are closed in the topology induced by $d$. Prove that there exist disjoint open sets $U$ and $V$ such that $A\subset U$ and $B\subset V$.

Is this proof correct? : By assumption, $A\cap B=\emptyset$, so there is no pair like $(x_A,x_B)$ (i.e., $x_A\in A$ and $x_B\in B$), such that $x_A=x_B$. Because every metric space is Hausdorff, so for every pair $(x_A,x_B)$ there are two disjoint open sets $U_A\ni x_A$ and $V_B\ni x_B$; and thus union of all $U_A$s (which is an open set containing A) are disjoint from union of all $V_B$s (which is an open set containing A).

Thanks a lot for any help.

PS In special case of standard metric topology on $\mathbb{R}$, if $[a,b]$ and $[c,d]$ are disjoint (say $c>b$), intuitively it is apparent that there is a 'space' between $c$ and $b$ so we can have two disjoint open sets, each of them containing $[a,b]$ or $[c,d]$.

• I believe a good way to start this is to show that if $A$ consists of just one point $x$, and if $B$ is some closed set not containing $x$, then show the existence of open and disjoint $U$ and $V$, such that $x$ is an element of $U$, and $B$ is contained in $V$. Feb 4 '15 at 9:09
• Then for general $A$ and $B$, you know that each point $x\in A$ together with $B$ has two open and disjoint sets $U_x$ and $V_x$, such that $x\in U_x$ and $B\subseteq V_x$. Feb 4 '15 at 9:13

In fact, you don't quite have to assume that $$A$$ and $$B$$ are disjoint closed sets, or even that they have disjoint closures; it's enough to assume that they are separated sets, i.e., each is disjoint from the closure of the other, a condition which is plainly necessary as well as sufficient.
Theorem. Let $$A$$ and $$B$$ be subsets of a metric space $$(X,d)$$. If $$A\cap\overline B=\emptyset$$ and $$B\cap\overline A=\emptyset$$, then there are disjoint open sets $$U$$ and $$V$$ such that $$A\subseteq U$$ and $$B\subseteq V$$.
Proof. If one of the sets is empty, it's easy; e.g., if $$A=\emptyset$$, we can take $$U=\emptyset$$ and $$V=X$$. So we may assume that $$A$$ and $$B$$ are nonempty sets. In that case the functions $$x\mapsto d(x,A)=\inf\{d(x,a):a\in A\}$$ and $$x\mapsto d(x,B)=\inf\{d(x,b):b\in B\}$$ are well-defined and continuous. It follows that the sets $$U=\{x:d(x,A)\lt d(x,B)\}$$ and $$V=\{x:d(x,B)\lt d(x,A)\}$$ are open sets, and of course they are disjoint. Finally, we have $$A\subseteq U$$ because $$d(x,A)=0\lt d(x,B)$$ for all $$x\in A$$, and $$B\subseteq V$$ because $$d(x,B)=0\lt d(x,A)$$ for all $$x\in B$$.
• @LND Your proof is incorrect. Your sets called (in confusing notation) $U_A,V_B$ apparently depend only on the two points $x_A,x_B$. For instance, if $X=\mathbb R$, and if $x_A=5$ and $x_B=8$, you might choose $U_A=(4.9,5.1)$ and $V_B=(7.9,8.1)$. But what if I told you that $A=[4,5.02]$ and $B=[5.05,9]$? Now you've got $U_A\cap B\ne\emptyset$; that's not going to work, is it?