Can a connected compact subset in $\mathbb{R} ^3$ always be enclosed in a closed ball of minimum radius?

Question

Given any connected compact set $A \subset \mathbb{R}^3$, does there exist a closed ball $B_r (x)$ of radius $r$ around $x\in A$ such that $r$ is minimum and $A\subset B_r (x)$ under usual topology ?

I tried considering a point $x'$ on the boundary of object first, then a ball $B_{(x',y')}(r')$ with centre as $y'$ and $x'$ as point on surface of ball so as to minimize distance of $x'$ from other points on boundary of A ($\partial A$), and then finding a ball $B_x(r)$ from {$B_{(x',y')}(r') | x' \in \partial A$} such that $r$ is minimum. But how do I show such $x$ and $r$ exist ? Also is the centre $x$ of such a ball necessarily the centre of gravity of$A$ ?

Answer 1

Even a compact, but not necessarily connected set $A$ in a Euclidean space can always be enclosed in closed ball in that space. But, no, the center of such a minimal ball is generally not the barycenter of $A$: for example, let $A$ be the union of the closed unit ball centered at the origin and a closed rectilinear segment of length 20 emanating from the origin. Since the segment has volume zero, it has no effect on the barycenter of $A$: this barycenter is the origin.

To show that there exists such a ball of minimal radius, consider a sequence of closed balls $B\left( \, \vec{c}_{k}, r_{k} \, \right)$, each enclosing $A$, and with the radii $r_{k}$ forming a strictly decreasing sequence that, furthermore, converges to: $$ r := \inf \; \{ \rho \; : \; \mbox{there exists a $\vec{c}_\rho$ such that $A \subset B\left(\vec{c}_\rho, \rho\right)$} \; \}. $$ Thus, if there exists a closed ball of radius $r$ and containing $A$, then such a ball is minimal such. The rest of this answer is aimed at proving that such a ball exists.

It is enough show that the sequence of the centers $\vec{c}_{k}$ (these are points in our Euclidean space, of whatever dimension) has a subsequence that converges to some limit $\vec{c}$. Once we show that, it will follow that $A \subset B\left( \, \vec{c}, r \, \right)$ (which can be proved by contradiction).

Since the set $A$ is bounded and since $$ \mbox{dist}\left(\vec{c}_{k}, A\right) := \inf \left\{ \left| \vec{x} - \vec{c}\right| \; : \; \vec{x} \in A \right\} < r_{k}, $$ it follows that the sequence $\vec{c}_{k}$ is bounded. Hence, it has a convergent subsequence.

Answer 2

Let $d$ be the usual (Cartesian) metric for the usual topology on $\Bbb R^3.$

Notation: $B_d(x,r)=\{y:d(x,y)<r\}$ is an $open$ ball.

Let $C\subset \Bbb R^3$ where $C$ is compact and has at least $2$ members. Since C is bounded there exists $$r=\sup\{d(x,y):x,y\in C\}\in \Bbb R.$$ And since $C$ has more than $1$ member, we have $r>0.$

(1).Note that $\overline {B_d(x,r)}\supset C$ for every $x\in C.$

(2).Let $D$ be the closure of $\cup \{\overline {B_d(x,2r)}:x\in C\}.$ Then $D$ is also bounded (because $C$ is bounded) so $D$ is compact. For $z\in D$ let $$f(z)=\inf \{s>0: \overline {B_d(z,s)}\supset C\}.$$

Now $f: D\to \Bbb R$ is continuous, in fact it is Lipschitz -continuous with respect to $d$:

If $z_1,z_2\in D$ then for all $x\in C$ and all $\epsilon >0$ we have $d(z_2,x)\le f(z_2)+\epsilon$ so by the $\triangle$ inequality we have $d(z_1,x)\le d(z_1,z_2)+f(z_2)+\epsilon.$ So $$\forall \epsilon >0\,(f(z_1\le d(z_1,z_2)+f(z_2)+\epsilon).$$ Interchanging the subscripts we also have $$\forall \epsilon >0\,(f(z_2)\le d(z_2,z_1)+f(z_1)+\epsilon).$$ And $d(z_1,z_2)=d(z_2,z_1)$ so we have $$\forall \epsilon >0\,(|f(z_1)-f(z_2)|\le d(z_1,z_2)+\epsilon.$$ Therefore $|f(z_1)-f(z_2)|\le d(z_1,z_2).$

(3).Since $D$ is compact and not empty, and $f:D\to \Bbb R$ is continuous, there exists $z_0\in D$ with $$f(z_0)=\min \{f(z):z\in D\}.$$ We have $C\subset \cap \{\overline {B_d(z_0,s)}:s>f(z_0)\}.$ And $f(x_0)>0$ because $C$ has at least two points.

(4). Note that $C\subset D$ so by Note (1) and the def'n of $r$ we have $f(z_0)\le \inf \{f(x):x\in C\}\le r.$

(5). We have $$C\subset \cap \{\overline {B_d(z_0,s)}:s>f(x_0)\}=\overline {B_d(z_0,f(z_0)}.$$ The "$=$" sign is justified by $f(z_0)>0.$

Finally:

(i).By def'n of $z_0$ and $f,$ we cannot have $C\subset \overline {B_d(z,s)}$ for any $s<f(z_0)$ and any $z\in D.$

(ii). And we cannot have $C\subset \overline {B_d(z,s)}$ for any $s<f(z_0)$ if $z\in \Bbb R^3 \setminus D$ because for any $x\in C$ we have $d(z,x)\ge 2r\ge 2f(z_0)>2s$ but any $w\in \overline {B_d(z,s)}$ must satisfy $d(z,w)\leq s.$