distance between two non-empty subsets, closed and compact, of a metric space

Question

$A,B\subseteq(X,d)$ metric space$,\;A\neq \emptyset,\;B\neq\emptyset,\;A\cap B=\emptyset,\;A$ is closed and $B$ is compact.

$d(x,A):=inf\{d(x,a)|\;a\in A\}$

$d(A,B):= inf\{d(b,A)|\;b\in B\}$

Prove that $d(A,B)>0$.

I tried to write definitions and properties (of closed and compact sets) and understood the problem "visually" but I can't write a rigorous proof. Help?

Answer 1

Is easier to prove that $A\cap B\neq\emptyset$ if and only if d$(A, B)=0$.

A direction is clear, namely, if $A\cap B\neq\emptyset$ then the distance between the 2 sets is zero.

For the other direction, build a sequence in $\lbrace x_{n}\rbrace_{n=1}^{\infty}$ in $B$ such that $d(x_{n}, A)\rightarrow 0$ as $n\rightarrow \infty$. Then choose a convergent subsequence in B of that sequence. Finally prove that that limit is an element of $A$ using the sequence $\lbrace d(x_{n}, A)\rbrace_{n=1}^{\infty}$ which converges to zero.

Answer 2

HINT: For each point $x\in B$ there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\cap A=\varnothing$. (Why?) These sets $B(x,\epsilon_x)$ cover $B$, and $B$ is compact, so ...

Note: $B(x,\epsilon)$ is the open ball of radius $\epsilon$ centred at $x$.