Let $S$ be a nonempty bounded subset of a metric space $(X,d)$. The main question is simply:

Question 0.Does there exist a smallest closed ball $B(x,r):=\{y \in X: d(x,y) \le r\}$, with $x \in X$ and $r\ge 0$, containing $S$?

(As usual, smallest here is meant with partial order of set inclusion.) As it is, the question is "too general": pick $X=\mathbf{R}\setminus \{0\}$ with its relative topology. Then $\{-1,1\}$ does not have a smallest ball containing it.

This is, of course, not a new question. As we can read in the abstract here, it goes back at least to Sylvester (1857) for the case of the Euclidean space $\mathbf{R}^2$.

I discovered that there are a lot of works on the finite dimensional Euclidean spaces in search of efficient alghorithms to find its solution, with applications to statistics, operation research, data analysis, etc.: see the introduction on these two Wikipedia pages, namely, the smallest circle problem in dimension $d=2$, and the smallest bounding sphere in dimension $d\ge 1$.

I came across this problem reading the first page of this article of Lev. At his first definition, he writes: *The limit circle of a bounded sequence of complex elements is the (unique) circle of least radius which contains within or on its boundary the limit points of the sequence.* (Is its existence obvious?)

Looking at the preceeding Wikipedia page, you can find a "Proof that the minimum covering disk is unique". However, what it really proves is: "*If* there exists a ball with smallest radius (hence, with respect to the total order of $\le$) of balls containing $S$ then it is unique. However, it seems to miss:

- the existence part (more explicitly, our Question 0 above);
- why two minimal balls have necessarily the same radius.

**First (incomplete) attempt to Question 0, with the additional hypothesis that $X$ is complete.** If $|S|=1$ the claim is trivial, let us assume that $|S|\ge 2$ or equivalently $d:=\mathrm{diam}(S)>0$. Let $\mathscr{B}$ be the family of closed balls containing $S$, which is nonempty since $S$ is bounded. If $(B(x_n,r_n): n\ge 1)$ is a decreasing chain in $\mathscr{B}$ (in particular, each $r_n \ge d$), then the set $F:=\bigcap_n B(x_n,r_n)$ is closed and contains $S$ (however, potentially $F$ is not a ball). Now, we have that if $n>k\ge 1$ then $d(x_n,x_k)\le r_k-r_n$. Also, the sequence $(r_n)$ is lower bounded by $d$ and monotonically descreasing, hence $r_\star:=\lim_nr_n$ exists and $(x_n)$ is Cauchy. Since $X$ is complete there exists also $x_\star:=\lim_n x_n$. Now, fix $\varepsilon>0$, pick an arbitrary point $s \in S$, and note that $d(s,x_\star) \le d(s,x_n)+d(x_n,x_\star)$, which is smaller than $r_\star+\varepsilon$ whenever $n$ is sufficiently large. Therefore $S\subseteq B(x_\star,r_\star) \subseteq \bigcap_n B(x_n,r_n)$, hence every (countable) chain in $\mathscr{B}$ has a lower bound in $\mathscr{B}$. Hence there exists a *minimal* element in $\mathscr{B}$ by Zorn's lemma.

If $X$ is, additionally, a Hilbert space then the same Wikipedia's argument linked above proves that: if $B(x_1,r_1),B(x_2,r_2)$ are two minimal elements of $\mathscr{B}$, then there exists $B(x_3,r_3) \in \mathscr{B}$ with radius $r_3\le \min\{r_1,r_2\}$. Wlog $B_3(x_3,r_3)$ is minimal. Hence, if $X$ is Hilbert then the family $\mathscr{B}_{\mathrm{minimal}}$ of minimal balls containing $S$ is downward directed. How can we conclude that $\mathscr{B}_{\mathrm{minimal}}$ has a minimum?

Note: the answer to Question 0 (with respect to set inclusion) is "no" even if $X$ is complete. A counterexample is: let $X$ be the unit sphere on $\mathbf{R}^2$ and pick $S:=\{(1,0), (-1,0)\}$. Then there are two minimal (with respect to set inclusion) bounding spheres with intersection $S$, but no minimum.

**Second attempt to Question 0, with the additional hypothesis that $X$ metric space with the Heine--Borel property.** (In particular, all finite dimensional spaces, but not infinite dimensional normed vector spaces.) Pick a distinghished point $0 \in X$. Let us say $S\subseteq B(0,M)$, with $M>0$. Without loss of generality, $S$ is nonempty compact (by the Heine--Borel property). For each $x \in X$, the function $S\to \mathbf{R}: s\mapsto d(s,x)$ admits a maximum, let us say $m_x$, and satisfies $d(0,x)-M \le m_x\le M+d(0,x)$. We would like to minime $m_x$. If this problem has a solution $x$, then $x \in B(0,3M)$, which is compact. Hence, define the function
$$
f: B(0,3M) \to \mathbf{R}: x \mapsto m_x.
$$
Thanks to Weiestrass' theorem, it would be sufficient to show that $f$ is continuous. Hence: is $f$ continuous? Fix $\varepsilon>0$ and $x_0 \in B(0,3M)$ and $s_0 \in S$ with $\|x_0-s_0\|=m_{x_0}$. If $\|x-x_0\|\le \varepsilon$ then $m_x=\max_{s \in S}\|x-s\|\le \|x-x_0\|+\max_{s \in S}\|x_0-s\|\le \varepsilon+m_{x_0}$ and viceversa $m_{x_0}\le \varepsilon+m_x$. Therefore, if $\|x-x_0\|\le \varepsilon$ then $\|m_x-m_{x_0}\|\le \varepsilon$.

**Third attempt to Question 0, with the additional hypothesis that $X$ is Hilbert.** If $|S|=1$ the claim is obvious. Otherwise, suppose that $|S|\ge 2$ and define
$$
R:=\{r \in \mathbf{R}: S\subseteq B_X(\eta,r) \text{ for some }\eta \in X\}.
$$
Note that $R\neq \emptyset$ since $S$ is bounded. We claim that $r_\star:=\inf R$ is actually its minimum.

Now, for each $n \in \omega$, we can fix $\eta_n \in X$ such that $S\subseteq B_X(\eta_n,r_n)$ with $r_n:=r_\star+2^{-n}$. We claim that the sequence $(\eta_n)$ is Cauchy. For, fix $n_1,n_2 \in \omega$ with $n_1>n_2$. If $B_X(\eta_{n_1},r_{n_1}) \subseteq B_X(\eta_{n_2}, r_{n_2})$ then $\|\eta_{n_1}-\eta_{n_2}\|\le r_{n_2}-r_{n_1}\le 2^{-n_2}$. Otherwise, fix a vector $v \in V:= S_X(\eta_{n_1},r_{n_1}) \cap S_X(\eta_{n_2}, r_{n_2})$ and let $\eta$ be its projection on the line passing through $\eta_{n_1}$ and $\eta_{n_2}$. It follows that $$ r_{n_1}^2 \le \|v-\eta\|^2 =\|v-\eta_{n_2}\|^2-\|\eta-\eta_{n_2}\|^2\le r_{n_2}^2-\frac{1}{4}\|\eta_{n_1}-\eta_{n_2}\|^2. $$ Therefore $$ \|\eta_{n_1}-\eta_{n_2}\| \le \sqrt{r_{n_2}^2-r_{n_1}^2}\le \sqrt{r_{n_2}^2-r_\star^2}=\sqrt{2^{1-n_2}r_\star+2^{-2n_2}}, $$ which can be made arbitrarily small if $n_2=\min\{n_1,n_2\}$ is sufficiently large.

Hence $(\eta_n)$ is convergent to some $\eta_\star \in X$. Then, let us show that $S\subseteq B_X(\eta_\star, r_\star)$. Pick $x \in X$ and, for each $n \in \omega$, fix an integer $k_n\ge n$ such that $\|\eta_{k_n}-\eta_\star\|<2^{-n}$. It follows that $$ \forall n \in\omega, \quad \|x-\eta_\star\|\le \|x-\eta_{k_n}\|+\|\eta_{k_n}-\eta_\star\|\le r_\star+2^{-k_n}+2^{-n}\le r_\star+2^{-1-n}. $$ Therefore $x \in B_X(\eta_\star, r_\star)$, so that $r_\star=\min R$.

Lastly, we claim that $\eta_\star$ is unique. For, suppose that $S\subseteq B_X(\eta_1, r_\star)$ and $S\subseteq B_X(\eta_2, r_\star)$, for some distinct $\eta_1,\eta_2 \in X$. Then $S$ would be contained in their intersection. But this implies that $ S\subseteq B_X\left(\frac{1}{2}(\eta_1+\eta_2), r_{\star\star} \right)$ where $r_{\star\star}:=\sqrt{r_\star^2-\frac{1}{4}\|\eta_1-\eta_2\|^2}$. However, this would imply that $r_{\star\star}<r_\star$ and $r_{\star\star} \in R$, contradicting that $r_\star=\min R$.

To conclude, my questions are:

Question 1.Am I missing something obvious? How can we conclude attempt 1? Does Question 0 have a positive answer in every Banach space?

Question 2.Is there a "good" reference on this problem for the existence of the smallest bounding ball? E.g., I didn't find the above arguments anywhere..