Suppose I have a Banach space $V$ and a set $A \subseteq V$ such that for all $\epsilon > 0$ there exists $v$ such that $A \subseteq \overline{B}(v, r + \epsilon)$. Does there exist $c$ such that $A \subseteq \overline{B}(c, r)$?


The answer is clearly yes for finite dimensional normed spaces: Define $T\_\epsilon = \bigcap\_{a \in A} \overline{B}(a, r + \epsilon)$. The $T\_\epsilon$ form a chain of closed sets and for $\epsilon > 0$ are non-empty, so have the finite intersection property. Thus when $V$ is finite dimensional they have non-empty intersection, and any element of the intersection works as $c$.


For more general Banach spaces I feel like you should be able to choose a cauchy sequence $x\_n$ such that $x\_n \in T\_{\epsilon\_n}$ with $\epsilon\_n \to 0$, but I can't seem to make it work.


Note that an arbitrary choice of $x\_n \in T\_{\epsilon\_n}$ can't be guaranteed to be Cauchy: If $V$ is $l^\infty$ and $A = \{ x : x\_0 = 0, ||x|| \leq 1 \}$ then diam$(T\_\epsilon) \geq 2$ because you can choose $c\_0$ arbitrarily in $[-1, 1]$


Note also that the assumption of $V$ a Banach space is essential: If $V$ is not Banach and $c$ is an element of the completion which is not in $V$ then $A = \overline{B}(c, 1) \cap V$ has no center. 



Radii and centers in Banach spaces
I believe that the property does not hold for all Banach spaces, but my counterexample is a little involved. If you've the patience then follow me through...


Let $V=\bigoplus\_{n=1}^\infty \ell^n\_2$ where $\ell^p\_2$ is $\mathbb{R}^2$ with
norm $\lVert\cdot\rVert\_p$ (Note: $n$ is taking the role of $p$). For $i\geq1$
and $j\in\{0,1\}$ we have $e\_{i,j}$, the $j^{th}$ standard basis vector of
$\ell^i\_2$ in $V$.


Give $V$ the norm $\lVert v\rVert=\sup\_n\lVert v\_n\rVert\_n$.


Let $W=\{v\in V:\lVert v\_n\rVert\_n\to 0\}$. I assert that $W$ is a Banach space.
Certainly every $e\_{i,j}\in W$.


Let $A=\{e\_{k,0}+e\_{k,1}, e\_{k,0}-e\_{k,1}:k\geq 1\}$.


Fact: Let $r(A)$ be the infimum of radii of balls containing $A$. Then $r(A)\leq1$


Proof:


Let $c\_N=\sum\_{i=1}^n e\_{i,0}$. We wish to compute the distance of each point
of $A$ from $c\_N$.


For $k\leq N$ we have
$\lVert c\_N-e\_{k,0}-e\_{k,1}\rVert$
$=\lVert\sum\_{i=1\ (i\not=k)}^Ne\_{i,0}-e\_{k,1}\rVert$
$=\sup\{\lVert e\_{i,0}\rVert\_i:i\leq N,i\not=k\}\cup\{\lVert-e\_{k,1}\rVert\_k\}$
$=1$
and similarly for $\lVert c\_N-e\_{k,0}+e\_{k,1}\rVert$.


For $k>N$ we have
$\lVert c\_N-e\_{k,0}-e\_{k,1}\rVert$
$=\max(\lVert c\_N\rVert,\lVert e\_{k,0}+e\_{k,1}\rVert\_k)$
$= \max(1,(1+1)^\frac{1}{k})$
$= 2^\frac{1}{k}$
$\leq 2^\frac{1}{N}$
and similarly for $\lVert c\_N-e\_{k,0}+e\_{k,1}\rVert$.


Thus $A\subseteq \overline{B}(c\_N,2^\frac{1}{N})$ and so $r(A)\leq2^\frac{1}{N}$. Letting
$N\to\infty$ we have $r(A)\leq 1$.


QED


Fact: $A$ is not contained in a ball of radius $1$.


Proof:


Suppose $A\subseteq \overline{B}(c,1)$. Then in particular for every $n$ we have
$\lVert c-e\_{n,0}-e\_{n,1}\rVert\leq 1$ and thus $\lVert c\_n-e\_{n,0}-e\_{n,1}\rVert\_n\leq 1$. Similarly $\lVert c\_n-e\_{n,0}+e\_{n,1}\rVert\_n\leq 1$.


Simple consideration of $\ell^n\_2$ shows that this implies $c\_n=e\_{n,0}$.
Thus $\lVert c\_n\rVert=1\not\to0$ and $c\not\in W$, contradicting the assumption.


QED


