Show that if $A$ and $B$ are compact subsets of $(\mathbb{R}^m,||.||_2)$ not empty and disjointed, then $\inf\{||a-b||_2:a\in A,b\in B\} > 0$

Question

Show that if $A$ and $B$ are compact subsets of $(\mathbb{R}^m,||.||_2)$ not empty and disjointed, then $$\inf\{||a-b||_2:a\in A,b\in B\} > 0$$

I know the definitions and I been trying for a while but I'm stuck with this proof. Any suggestions would be great!

Answer 1

Consider a function $A\times B \to \mathbb{R}$; $(a, b) \mapsto \|a-b\|_2$.

Since this function is continuous and the domain $A\times B $ is compact, there exist $a_0 \in A$, $b_0 \in B$ which satisfies $\|a_0 - b_0\|_2 = \inf \operatorname{im} f$. If this is zero, we have $\|a_0- b_0\|_2 = 0$, which implies $a_0 = b_0$, which implies $a_0 = b_0 \in A \cap B $ which is a contradiction.

Answer 2

Since the infimum of a set of real numbers is always the limit of a sequence of elements of that set, we have $$d(A, B) = \lim_{k \to \infty} |x_k - y_k|$$ for sequences $x_k \in A$ and $y_k \in B$. Since both $A$ and $B$ are compact, both these sequences are bounded and we can assume without loss of generality that they converge to $x_0$ and $y_0$ respectively. From the compactness of $A$ and $B$ we have $x_0 \in A$ and $y_0 \in B$. Since $A$ and $B$ are disjoint, we have:

$$d(A, B) = \lim_{k \to \infty} |x_k - y_k| = |x_0 - y_0| > 0$$

Answer 3

Define, for $x\in B$ the function $f(x)= \inf_{a \in A} ||x-a||_2$

Then $f(x) > 0$ for all $x \in B$ since $A \cap B = \emptyset$.

Moreover, $f(x)$ is continuous so it attains its infimum on the compact set $B$ and this infimum is strictly positive. That is, there exist $x_0 \in B$ such that $f(x_0) = \delta > 0$.

Hence, $\inf_{a \in A, b \in B} ||a-b||_2 \geq f(x_0) = \delta > 0$ as needed.