Showing that if closed subsets don't intersect then there exists open sets in which they exist that also don't intersect

Question

Let $A$ and $B$ be closed subsets of a normed linear space $V$ . Suppose that $A\cap B=\emptyset$. Show that there exist open sets $U$ and $V$ such that $U\cap V =\emptyset$, $A \subset U$ and $B \subset V$ .

It seems intuitively true but how is it intended to be proven? It's known that since $A$ and $B$ are closed and their intersection is empty then they each should exist in a larger open set but not why the intersection of those two sets is empty.

Answer 1

Let $d(x,y)=\|x-y\|$ (or work in the more general case of metric spaces to begin with). For each $a\in A$, we know that $$r(a):=\frac12\inf_{b\in B}d(a,b)>0$$ because otherwise we'd find a sequence $\{b_n\}_n$ with $b_n\to a$. Note that this means $d(a,b)\ge \frac12r(a)$ for all $a\in A$, $b\in B$. Likewise, for each $b\in B$, $$s(b):=\frac12\inf_{a\in A}d(a,b)>0$$ and thereby $d(a,b)\ge \frac12s(b)$ for all $a\in A$, $b\in B$. The sets $$U:=\bigcup_{a\in A}{B_{r(a)}(a)},\qquad V:=\bigcup_{b\in B}{B_{s(b)}(b)} $$ are open and contain $A$ and $B$, respectively. Suppose $c\in U\cap V$. Then $c\in B_{r(a)}(a)$ for some $a\in A$ and $c\in B_{s(b)}(b)$ for some $b\in B$. Then $$d(a,b)\le d(a,c)+d(c,b)< r(a)+r(b) \le \frac12d(a,b)+\frac12d(a,b)=d(a,b),$$ contradiction. We conclude that $U\cap V=\emptyset$.

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Note that we do not have $$ \inf_{a\in A,b\in B}d(a,b)>0.$$ Consider for example the closed sets $A=\{\,(x,y)\mid xy=0\,\}$, $A=\{\,(x,y)\mid xy=1\,\}$ in the normed space $\Bbb R^2$.

Answer 2

Hint: Let's call the normed space $X$. For $x\in X$, consider the function $$d(x,A)=\inf_{y\in A}\|x-y\|.$$ Then one can show that $d$ is a continuous function on the normed space. Then define $$U=\{x\in X:d(x,A)<d(x,B)\},$$ $$V=\{x\in X:d(x,A)>d(x,B)\}.$$ Then both $U,V$ are open, $A\subseteq U, B\subseteq V$ and $U\cap V=\emptyset$.