Bounded sets in $\Bbb{R}^n$ Let $S\subset \Bbb{R}^n$ is a bounded set. I want to show that $\underline{\underline{the}}$ sphere(closed ball) of $\textbf{smallest radius}$ which contains S is $\underline{unique}$.  the assumption of 
having two such spheres leads to an immediate contradiction. Why?
 A: Part 1.  There is a smallest radius of (closed) ball containing $S$.
Define $f\colon\mathbb R^n\to\mathbb R$ by $f(x) = \sup_{a\in S} \|x-a\|$.  Note that: $f$ is well-defined because $S$ is bounded; $f$ is continuous ($1$-Lipschitz, even); and $S$ is contained in a ball centred at $x$ and of radius $r$ if and only if $f(x)\le r$.  Now, let $B$ be the ball centred at the origin and of radius $2f(0)$.  $B$ is compact, so $f$ has a minimum on this set; this minimum is $\le f(0)$ since $0\in B$; and $f(x)\ge f(0)$ for all $x\notin B$, since every point of $S$ is at least $f(0)$ away from every point of $B^C$.  So the minimum on $B$ is a global minimum.
Part 2.  If $S$ is contained in a ball of radius $r$ centred at $x$, and also contained in a ball of radius $r$ centred at $y$, with $x\ne y$, then $S$ contained in a ball of radius strictly less than $r$ centred at $\frac12(x+y)$.
Proof: Let $a\in S$.  By assumption, $\|x-a\|\le r$ and $\|y-a\|\le r$; by the parallelogram identity,
$$
\|\tfrac12(x+y)-a\|^2
= \tfrac12\|x-a\|^2 + \tfrac12\|y-a\|^2 - \|\tfrac12(x-y)\|^2
\le r^2 - \|\tfrac12(x-y)\|^2
$$
So $S$ is contained in a ball centred at $\frac12(x+y)$ and of radius $\sqrt{r^2 - \|\frac12(x-y)\|^2}$, which is strictly less than $r$, as desired.
A: LEt $B^n( r ) $ and $B^n(r')$ be two closed balls with the smallest radius. Therefore, by definition we must have $ r \leq r' $ and $r' \leq r $. But this implies that $r = r'$. And hence the result follows.
