Suppose \(A\), \(B\) are disjoint non-empty compact subsets of \(\RR^n\). Prove that there exist \(a \in A\) and \(b \in B\) satisfying \(\norm{a - b} = \inf \{ \norm{x - y} : x \in A, y \in B \}\).
Solution
Define \[ D:A\times B\longrightarrow\RR, \qquad D(x,y)=\norm{x-y}. \]
1. The space \(A\times B\) is nonempty and compact, and \(D\) is continuous.
The sets \(A\) and \(B\) are nonempty, so their product is nonempty. Since both are compact subsets of \(\RR^n\), their finite product \(A\times B\) is compact.
The subtraction map \[ (x,y)\longmapsto x-y \] is continuous, and the Euclidean norm is continuous. Hence their composition \(D\) is continuous.
2. There exist \(a\in A\) and \(b\in B\) such that\[ \boxed{ \norm{a-b} = \inf\{\norm{x-y}:x\in A,\ y\in B\} }. \]
By step 1, \(D\) is a continuous real-valued function on the nonempty compact space \(A\times B\). The extreme value theorem therefore gives \[ (a,b)\in A\times B \] at which \(D\) attains its minimum. Thus \[ D(a,b) = \min_{(x,y)\in A\times B}D(x,y) = \inf_{(x,y)\in A\times B}D(x,y), \] which is the displayed equality.
3. Q.E.D.
Step 2 supplies the required points.