let $(X,d)$ be a metric space and let $A,B\subseteq X$. we define the distance between $A$ and $B$ as:
$$\operatorname{dist}(A,B)=\inf\{d(a,b):a \in A,b \in B\}$$
1
show that for any $x \in X$, we have $\operatorname{dist}(A,B)\le \operatorname{dist}(x,A)+\operatorname{dist}(x,B)$. (Hint please)
2
if $A \subseteq B$ and $x \in X$, prove that $\operatorname{dist}(x,A)\le\operatorname{dist}(x,B)+\operatorname{diam}(B)$?
where: $\operatorname{diam}(B)=\sup\{d(y,z):y,z \in B\}$.
Best Answer
HINTS:
(1) If $\operatorname{dist}(A,B)>\operatorname{dist}(x,A)+\operatorname{dist}(x,B)$, it follows from the definition of supremum that there are $a\in A$ and $b\in B$ such that $d(a,b)>\operatorname{dist}(x,A)+\operatorname{dist}(x,B)$. Now apply the triangle inequality.
(2) Let $b\in B$; then $d(x,a)\le d(x,b)+d(b,a)$.