Let A and B be disjoint closed subsets of Rn. Define d(A,B)=inf{∥a−b∥:a∈A and b∈B}. Show that if A={a} is a singleton, then d(A,B)>0.

Question

Let $A$ and $B$ be disjoint closed subsets of $\mathbb{R}^n$. Define $d(A,B)=\inf \{||a-b||: a \in A, b \in B\}$.

I have to show that if $A=\{a\}$ is a singleton set, then $d(A,B)>0$ and I have no idea how to do this.

Answer 1

HInts:

If $d(A,B)=0$, then by the definition of $d(A,B)$, there exists a sequence $\{x_n\} \in B$ such that converges to $a \in A$ (Can you construct it by yourself?). Since $B$ is closed, then $a \in B$, and hence $A\cap B=\{a\}\not=\emptyset $. It is a contradiction!

Answer 2

Since $B$ is closed and $a\notin B$ there is an $\epsilon>0$ such that $U_\epsilon(a)$ does not intersect $B$. It follows that $|x-y|\geq\epsilon$ for all $x\in\{a\}$, $y\in B$, and this proves $d\bigl(\{a\},B)\geq\epsilon$.