Distance between closed and open balls in a normed vector space

Question

Problem: Let $b\in E$(normed vector space), show that $d(b,B(a,r))=d(b,\overline{B}(a,r))$.

Notation: $d(a,C)=inf\{d(a,x):x\in C\}$

My attempt: Well, since $B(a,r)\subset \overline(B)(a,r)$, then $d(b,\overline{B}(a,r)) \leq d(b,B(a,r))$. So, I suppose $d(b,\overline{B}(a,r)) < d(b,B(a,r))$ to find a contradiction.

The real problem is that I have no idea how to continue nor even if im in the right path.

PD: Please do not use convergence :)

Thanks for the help!

Answer 1

Let $x \in \overset {-} B (a,r)$. Then there exists $\{x_n\} \subset B (a,r)$ such that $x_n \to x$. We have $d(b,B (a,r)) \leq d(b,x_n)$ for all $n$. If you let $n \to \infty$ you get $d(b,B (a,r)) \leq d(b,x)$. Taking inf over all $x$ we get $d(b,B (a,r)) \leq d(\overset {-} B (a,r))$. For a different proof not involving sequences let $x \in \overset {-} B (a,r)$ and $\epsilon >0$. There exists $y \in B(b,r)$ such that $d(x,y) <\epsilon$. We have $d(b,B(a,r))\leq d(b,y)<d(b,x)+\epsilon$. Take infimum over $x$ to get $d(b,B(a,r))\leq d(b,\overset {-} B (a,r))+\epsilon$. Since $\epsilon$ is arbitrary this gives $d(b,B(a,r))\leq d(b,\overset {-} B (a,r))$

Answer 2

A normed vector space $E$ is also a metric space with metric $d(x,y)=\|x-y\|.$ In a metric space $(E,d)$ let $\emptyset \ne C\subset E$ and $b\in E.$ We define $d(b,C)=\inf \{d(b,c):c\in C\}.$ We have $\{d(b,c'):c'\in \bar C\}\supset \{d(b,c):c\in C\}$ so $d(b, \bar C)\leq d(b,C).$

We show that $d(b,\bar C)<d(b,C)$ is untenable, and conclude that $d(b,\bar C)=d(b,C).$

For brevity of notation let $U=d(b,C)$ and $V=d(b,\bar C).$ Suppose $ U-V>0.$ Take $c'\in \bar C$ such that $d(b,c')\leq V+\frac {U-V}{2}.$ Now $\forall c\in C\;(d(c,b)\geq U).$ So for every $c\in C$ we have $$d(c,c')\geq d(c,b)-d(b,c')\geq$$ $$\geq U-(V+\frac {U-V}{2})=\frac {U-V}{2}. $$ So the open ball of positive radius $\frac {U-V}{2},$ centered at $c',$ is disjoint from $C.$ This contradicts $c'\in \bar C. $