Problem: Let $E\neq \{0\}$ a normed vector space and $B(a, r)$ the open ball of center $a\in E$ and radius $r > 0$. Given a $b\in E$, $d(b, B(a,r)) = 0$ if and only if $b\in \{x\in E : d(x, a)\leq r\}$.
Notation: $d(b,B) =\inf\{d(b,x):x\in B(a,r)\}$
My problem is showing $\leftarrow$. I know that if $b \in \overline{B}(a,r)$ I have 2 choices, $d(b,a)<r$ or $d(b,a)=r$. If $d(b,a)<r$ we are done, but how do i prove the second option? I guess we have to pick some 'smart' vectors, but I cant figure it out.
Thanks so much
Suppose you have $d(b,a)=r$. Consider the line connecting $a$ and $b$. It is parameterized by $t\mapsto (1-t)a+tb=:c_{t}$, with $t$ ranging from $0$ to $1$. You can check that for $0\leq t< 1$, the point $c_{t}$ lies in $B(a,r)$. You can also check that $d(c_{t},b)=(1-t)r$. Thus,
$$ d(b,B)=\inf_{e\in B}d(b,e)\leq\inf_{0<t<1}d(c_{t},b)=\inf_{0<t<1}(1-t)r=0. $$