Bounded set in finite normed vector space

Question

Let $E$ be a finite-dimensional normed vector space and $A$ a non-empty and bounded subset of $E$. Let $I = \{R\geq 0: \exists \mathcal{B}_{F}(x, R), A \subset \mathcal{B}_{F}(x, R)\}$, and $r=\inf I$.

Let $R_n = r+\frac{1}{n}$. Show that $\forall n \in \mathbb{N}^*$, there exists $x_n \in E$ such that $A \subset \mathcal{B}_F(x_n, R_n)$.

Answer 1

$r+\frac1n > r = \inf I$ we have that $r+\frac1n$ is not a lower bound for $I$, or $\inf I$ would not be the largest lower bound of $I$. So some $R_n \in I$ obeys $R_n < r+\frac1n$, and by the definition of $I$ this means there is some $x_n$ such that $A \subseteq B(x_n, R_n)\subseteq B(x_n, r + \frac1n)$ for this $R_n$. Done.

Answer 2

Your $I$ has the following property: if $s\in I$ and $t>s$ then $t\in I$. Therefore $I$ is an interval, more precisely either $(r,\infty)$ or $[r,\infty)$. Regardless, $r+\epsilon$ is always in $I$ for any $\epsilon>0$. Which, by the definition of $I$, means that there is $x\in e$ such that $A\subseteq B(x,r+\epsilon)$.