Combinatorial Generalizations of Jung’s Theorem

Abstract

We consider combinatorial generalizations of Jung’s theorem on covering a set by a ball. We prove the “fractional” and “colorful” versions of the theorem.

1 Introduction

The famous theorem of Jung states that any set with diameter $1$ in $\mathbb R ^d$ can be covered by the ball of radius $R_d=\sqrt{\tfrac{d}{2(d+1)}}$ (see [3]).

The proof of this theorem is based on Helly’s theorem:

Theorem 1

(Helly’s theorem) Let $\fancyscript{P}$ be a family of convex compact sets in $\mathbb R ^d$ such that an intersection of any $d+1$ of them is not empty. Then the intersection of all of the sets from $\fancyscript{P}$ is not empty.

Helly’s theorem has many generalizations. Katchalski and Liu in 1979 [8] proved a “fractional” version of Helly’s theorem and Kalai in 1984 [7] gave the strongest version of it. In 1979, Lovász suggested a “colorful” version of Helly’s theorem. In this paper, we give analogous generalizations of Jung’s theorem.

2 The Fractinal Version of Jung’s Theorem

Let us recall the fractional version of Helly’s theorem.

Theorem 2

(Katchalski and Liu [8]) For every $d \ge 1$ and every $\alpha \in (0,1]$ there exists a $\beta = \beta (d,\alpha ) > 0$ with the following property. Let $\mathcal X _1,\mathcal X _2,\ldots ,\mathcal X _n$ be convex sets in $\mathbb R ^d$ such that $\bigcap \mathcal X _i \ne \emptyset$ for at least $\alpha C_n^{d+1}$ index sets $I \subseteq [n]$ of size $(d + 1)$. Then there exists a point contained in at least $\beta n$ sets among $\mathcal X _i$.

The best possible value of $\beta (d,\alpha )$ is $1 - (1 - \alpha )^{1/(d+1)}$ [7] and, in particular, $\beta \rightarrow 1$ as $\alpha \rightarrow 1$.

Using this we can prove the fractional version of Jung’s theorem.

Theorem 3

For every $d \ge 1$ and every $\alpha \in (0,1]$ there exists a $\beta = \beta (d,\alpha ) > 0$ with the following property. Let $\mathcal V$ be an $n$-point set in $\mathbb R ^d$ such that for at least $\alpha C^2_n$ of pairs $\{x,y\}$ ($x, y \in \mathcal V$) the distance between $x$ and $y$ is less than $1$. Then there exists a ball of radius $R_d$, which covers $\beta n$ points of $\mathcal V$. Moreover, $\beta \rightarrow 1$ as $\alpha \rightarrow 1$.

Proof

Let us show that $\beta >0$ exists.

We construct a graph $G$ on points of $V$ as vertices. Two vertices of $G$ are connected if and only if the distance between the points is not greater than $1$.

Then the degree of some vertex $v$ of $G$ is not less than $\alpha (n-1)$. The ball with center at the point corresponding to $v$ and radius $1$ contains at least $\alpha (n-1)+1$ points of $\mathcal V$. Any ball of radius $1$ can be covered by $c_d$ balls of radius $R_d$, where $c_d$ is a constant depending only on the dimension $d$. Then one of these balls should cover more than $\frac{\alpha (n-1)}{c_d}$ points of the set $\mathcal V$. Therefore the statement of the theorem holds for $\beta ={\alpha }/{2c_d}$, because

$$\begin{aligned} \frac{\alpha }{2c_d}n<\frac{\alpha (n-1)}{c_d}. \end{aligned}$$

Note that this argument works for any radius $R$ not necessarily equal to $R_d$.

Let us show that $\beta \rightarrow 1$ as $\alpha \rightarrow 1$. Any pair of vertices belongs to $C_{n-2}^{d-1}$ subsets of $d+1$ vertices. Therefore, if there are at most $(1-\alpha )C^2_n$ empty edges in $G$, then the number of incomplete subgraphs on $d+1$ vertices is not greater than

$$\begin{aligned} (1-\alpha )C^2_nC_{n-2}^{d-1}=(1-\alpha )C_n^{d+1}C^2_{d+1}. \end{aligned}$$

We see that among the $C_n^{d+1}$ subgraphs of $d+1$ vertices at least $\alpha ^{\prime }C_n^{d+1}$ are full, where $\alpha ^{\prime }=1-(1-\alpha )C^2_{d+1}$.

Consider the family $\fancyscript{B}$ of $n$ balls of radius $R_d$ with centers at the points of $\mathcal V$. If the distances between the centers of some set of balls is no greater than $1$, then these balls have nonempty intersection (because Jung’s theorem implies that these centers can be covered by a ball of radius $R_d$). Therefore, among $C^{d+1}_n$ subsystems of $d+1$ balls from $\fancyscript{B}$, at least $\alpha ^{\prime }C_n^{d+1}$ have nonempty intersection. From fractional Helly’s theorem, it follows that there is a point which belongs to $\beta (d,\alpha ^{\prime })n$ balls. The ball with the center at this point and radius $R_d$ covers $\beta n$ points of $\mathcal V$. To conclude the proof, we note that $\alpha ^{\prime }$ tends to $1$, as $\alpha \rightarrow 1$. The result of Kalai now implies that $\beta (d,\alpha ^{\prime })$ also tends to $1$. $\square$

Note that using an approximation arguments it is possible to prove the same theorem for a measure.

3 Close Sets

We will use the following definition.

Definition 1

We call two nonempty sets $\mathcal V _1$ and $\mathcal V _2$ close if for any points $x\in \mathcal V _1$ and $y\in \mathcal V _2$, the distance between $x$ and $y$ is not greater than $1$.

It is easy to see that if two close sets $\mathcal V _1$ and $\mathcal V _2$ are given, the diameter of each of them is not greater than $2$. Moreover, the following theorem holds.

Theorem 4

The union of several pairwise close sets in  $\mathbb R ^d$  can be covered by a ball of radius $1$.

Proof

Denote by $\mathcal X$ one of the sets and by $\mathcal Y$ union of other sets. Without loss of generality, we may assume that $\mathcal X$ and $\mathcal Y$ are convex closed sets, because the condition of the theorem also holds for $\mathrm{cl}\,(\mathrm{conv}\mathcal X )$ and $\mathrm{cl}\,(\mathrm{conv}\mathcal Y )$. If $\mathcal X$ and $\mathcal Y$ have nonempty intersection, then a unit ball with center at any point from the intersection covers $\mathcal X$ and $\mathcal Y$.

Suppose they do not intersect. Choose points $x\in \mathcal X$ and $y \in \mathcal Y$ so that the length of $[x, y]$ is minimal. Let $m$ be the midpoint of the segment $[x,y]$. We will show that the unit ball with center at $m$ covers $\mathcal X$ and $\mathcal Y$. It is known that the hyperplane perpendicular to $[x,y]$ and passing through $m$ separates $\mathcal X$ and $\mathcal Y$ (see [6]). Take any point $z$ from $\mathcal X$. Note that the angle $zmy$ is obtuse (Fig. 1). Therefore the segment $[m, z]$ is shorter than $[y, z]$, which is no longer than $1$. The same argument works for any point $z$ from $\mathcal Y$. $\square$

Fig. 1

Covering of two bodies

It is clear that the diameter of the covering ball in this theorem could not be decreased. The following two questions arise naturally.

Suppose that a family of pairwise close sets $\mathcal V _1,\,\mathcal V _2$, ..., $\mathcal V _n$ in $\mathbb R ^d$ is given.

• 1.  What is the minimal  $R$  so that at least one of the sets  $\mathcal V _i$  can be covered by a ball of radius $R$?

• 2.  What is the minimal $D$  so that at least one of the sets  $\mathcal V _i$ has diameter no greater than $D$?

For $n=3$ and $d=2$, the author learned the answer to the first question from Vladimir Dol’nikov. His arguments work well for all $n>d$. In Theorem 5, we find the exact value of $R$ for all pairs $d$ and $n$ in the first question. In Theorem 7, we show that the second question is equivalent to the well-known problem about spherical antipodal codes.

4 Colorful Jung’s Theorem

Theorem 5

Let $\mathcal V _1,\,\mathcal V _2$, ..., $\mathcal V _n$ be pairwise close sets in $\mathbb R ^d$. Then one of the sets $\mathcal V _i$ can be covered by a ball of radius $R$.

$$\begin{aligned} R&= \tfrac{1}{\sqrt{2}}\quad {\textit{if}}\; n\le d;\\ R&= R_d=\sqrt{\tfrac{d}{2(d+1)}}\quad {\textit{if}} \;n > d. \end{aligned}$$

Proof

First, let us show that $R$ from the statement of the Theorem is minimal. Suppose $n \le d$. Consider a crosspolytope with $2n$ vertices and with edge length $1$. Let the sets $\mathcal V _i$ be the pairs of opposite vertices of the crosspolytope. Then the distance between any two points from different sets is equal to $1$, and distance between two points from the same set is exactly $\sqrt{2}$. Therefore the radius of the minimal cover ball for each of the sets equals $\tfrac{1}{\sqrt{2}}$.

For $n>d$, let the sets $\mathcal V _i$ coincide with regular simplices with edge length $1$. In this case, the sets $\mathcal V _i$ are close to each other and the radius of the minimal cover ball for each set equals $R_d$.

Now let us show that one of the sets $\mathcal V _i$ can be covered by a ball of radius $R$. Without loss of generality, we may assume that the sets $\mathcal V _i$ are all closed, since the condition of the theorem holds for the closure of these sets.

Suppose $n \le d$. For any set $\mathcal V _i$, consider the minimal ball $B(o_i, r_i)$ covering this set. Let $r_1$ be the minimal radius among $r_i$. Note that there exists a point $x$ of the set $\mathcal V _2$ which does not belong to the interior of the ball $B(o_1, r_1)$. Indeed, if all the points of the set $\mathcal V _2$ belong to the interior of the ball $B(o_1, r_1)$, then $\mathcal V _2$ can be covered by a ball of radius less than $r_1$. This contradicts the assumption that $r_1$ is minimal among $r_i$. Let $s$ be a hyperplane passing through the point $o_1$ and perpendicular to the segment $[o_1,x]$ (Fig. 2). This hyperplane divides the sphere $S(o_1, r_1)$ onto two hemispheres. Denote by $S^{\prime }(o_1, r_1)$ the hemisphere, which lies in the halfspace generated by $s$ not containing $x$. We need the following lemma (see [3] statements 2.6 and 6.1 or [5] Lemma 2).

Fig. 2

The minimal covering ball and a point outside

Lemma 1

If a sphere $S$ is the boundary of the minimal ball which covers a closed set $\mathcal V$, then the center of $S$ belongs to $\mathrm{conv}(\mathrm{cl}\,\mathcal V \cap S)$.

Therefore the hemisphere $S^{\prime }(o_1, r_1)$ contains at least one point $y$ from the set $\mathcal V _1$. Note that distance from any point of the hemisphere $S^{\prime }(o_1, r_1)$ to $x$ is no less than $\sqrt{2}r_1$. Thus the distance between the points $x$ and $y$ is no less than $\sqrt{2}r_1$. Since $\mathcal V _1$ and $\mathcal V _2$ are close and the distance between $x$ and $y$ is no greater than $1$, we have $r_1\le \frac{1}{\sqrt{2}}$.

Consider the second case: $n\ge d+1$. This argument is due to V. L. Dol’nikov.

We will use the following theorem of Lovász [1].

Theorem 6

(Colorful Helly’s theorem) Let $\fancyscript{F}_1,\fancyscript{F}_2, \ldots ,\fancyscript{F}_{d+1}$ be $d+1$ finite families of convex sets in $\mathbb R ^d$. If $\cap _{i=1}^{d+1}\mathcal X _i \ne \emptyset$ for all choices of $\mathcal X _1 \in \fancyscript{F}_1,\,\mathcal X _2 \in \fancyscript{F}_2$, ..., $\mathcal X _{d+1} \in \fancyscript{F}_{d+1}$, then for some $i$ the tersection of all sets from $\fancyscript{F}_i$ is not empty.

Let $\fancyscript{B}_i$ be a family of balls of radius $R_d$ with centers at the points of the set $\mathcal V _i$. Note that this set of families satisfies condition of colorful Helly’s theorem. Indeed, if we choose one ball from each of the sets $\mathcal B _i\in \fancyscript{B}_i$, then the distance between its centers will not be greater than $1$. From Jung’s theorem, it follows that they can be covered by a ball $\mathcal B$ of radius $R_d$. This means that the center of $\mathcal B$ is contained in all sets $\mathcal B _i,i=1,2, \ldots ,n$. Thus they have nonempty intersection.

Applying colorful Helly’s theorem to the families $\fancyscript{B}_i$, we obtain that for some $i$, the balls of the family $\fancyscript{B}_i$ have nonempty intersection. Therefore, all points of the set $\mathcal V _i$ can be covered by a ball of radius $R_d$ centered at any point of intersection of all balls from the family $\fancyscript{B}_i$. $\square$

Remark 1

Note that, by the same arguments, it can be shown that for $n \le d$ all sets except one can be covered by a ball of radius $1/\sqrt{2}$. For the proof, one considers the minimal ball which covers all sets except one.

If $n>d$, then all but $d$ sets can be covered by a ball of radius $R_d$. Here one should use the modified version of colorful Helly’s theorem: Suppose a collection of families $\{\fancyscript{F}_i\}$ of convex sets is given and the intersection of any $d+1$ sets from different families is not empty. Then there is a point which pierces all but $d$ sets from the families $\{\fancyscript{F}_i\}$.

The proof is the same as the classical proof of colorful Helly’s theorem (see [9]): consider $d$ sets from different families whose intersection has the lowest higher point among all intersections of $d$ sets from different families. This point should be contained in all but the chosen $d$ sets of the families $\{\fancyscript{F}_i\}$.

5 The Bound on the Diameter of a Set

In this section, we study a bound on the diameter of $\mathcal V _i$.

Definition 2

A spherical code is a finite set of points in $\mathbb S ^d$. A spherical code is called antipodal if it is symmetric with respect to the center of the sphere.

A spherical code $\mathcal W$ of $n$ points is called optimal if it has a maximal minimal diameter ($\min |x-y|$, where $x,\,y \in \mathcal W$ and $x\ne y$) among all spherical antipodal codes of the cardinality $n$.

By $D_d(n)$, we denote the minimal diameter of the optimal spherical antipodal code of cardinality $2n$ on the unit sphere $\mathbb S ^{d-1}$.

Theorem 7

Let $\mathcal V _1,\,\mathcal V _2$, ..., $\mathcal V _n$ be pairwise close sets in $\mathbb R ^d$. Then one of the sets $\mathcal V _i$ has diameter no greater than

$$\begin{aligned} D=\frac{2}{\sqrt{4-D_d(n)^2}}. \end{aligned}$$

Proof

Suppose otherwise. Then the diameter of each $\mathcal V _i$ is no less than $D^{\prime }>D$. Again, we may assume that all sets $\mathcal V _i$ are closed convex sets. Then in each set $\mathcal V _i$, it is possible to choose two points $a_i$ and $b_i$ with distance between them equal to $D^{\prime }$. So, we can assume that each set $\mathcal V _i$ is a two-point set: $\mathcal V _i=\{a_i, b_i\}$.

Let us show, without loss of generality, we may assume that the midpoints of the segments $[a_i,b_i]$ coincide.

Indeed, denote the origin by $o$.

Suppose

$$\begin{aligned} a_i^{\prime }=o+\frac{\overrightarrow{b_ia_i}}{2}, \quad b_i^{\prime }=o+\frac{\overrightarrow{a_ib_i}}{2}. \end{aligned}$$

Then

$$\begin{aligned}&|a_i^{\prime }-a_j^{\prime }|=|\overrightarrow{a_i^{\prime }a_j^{\prime }}|\\&\qquad \qquad \qquad =\frac{|\overrightarrow{a_ib_i}+\overrightarrow{b_ja_j}|}{2}=\frac{|\overrightarrow{a_ib_j}+\overrightarrow{a_jb_i}|}{2}\\&\qquad \qquad \qquad \le \tfrac{1}{2}(|a_i-b_j|+|a_j-b_i|) \le 1. \end{aligned}$$

The same inequality holds for all other pairs of points $(a_i^{\prime }, b_j^{\prime })$ and $(b_i^{\prime }, b_j^{\prime })$.

We may assume that the points of the sets $\mathcal V _i$ lie on the sphere of radius $D^{\prime }/2$ and form the antipodal code. Since the distance between the points, say, $a_i$ and $a_j$ is not greater than $1$, the distance between $a_i$ and $b_j$ should be greater than $\sqrt{D^{\prime 2}-1}$.

Therefore,

$$\begin{aligned} \begin{array}{l} \displaystyle \frac{2\sqrt{D^{\prime 2}-1}}{D^{\prime }}<D_d(n) \\ \quad \Leftrightarrow 4D^{\prime 2}-4<D_d(n)^2{D^{\prime 2}} \\ \quad \Leftrightarrow 4D^{\prime 2}-D_d(n)^2{D^{\prime 2}}<4 \\ \quad \Leftrightarrow D^{\prime 2}\le \displaystyle \frac{4}{4-D_d(n)^2}\\ \quad \Leftrightarrow D^{\prime }\le \displaystyle \frac{2}{\sqrt{4-D_d(n)^2}}=D. \end{array} \end{aligned}$$

This contradicts the assumption $D^{\prime }>D$ and concludes the proof. $\square$

It is clear that the corresponding optimal antipodal code yields equality for $D$.

Unfortunately, the precise value of $D_d(n)$ is known only for a few cases. In particular, the following numbers are known (see [2, 4]):

$$\begin{aligned} \begin{array}{l} \displaystyle D_2(n)=2\sin \displaystyle \frac{\pi }{2n};\\ D_d(n)=\sqrt{2} \text{ for} n\le d;\\ D_3(6)=\displaystyle \frac{2}{\sqrt{10 +2\sqrt{5}}};\\ D_4(12)=D_8(120)=D_{24}(98280)=1. \end{array} \end{aligned}$$