---
tag: 227ERQ43PY
kind: statement
role: theorem
name: Jung's theorem
msc: [52A20, 52A40]
uses: [0KW3PN547K, 0Q0H13G8ZB, F1W3QRBNGJ]
about: [0KW3PN547K]
sources:
  - url: https://en.wikipedia.org/wiki/Jung%27s_theorem
same_as:
  wikidata: Q1075398
  wikipedia: Jung's theorem
---

## Statement

Let $n \ge 1$, and let $S \subseteq \R^n$ be a bounded set with [[diameter|0KW3PN547K]] $d$.
Then $S$ is contained in a [[closed ball|0Q0H13G8ZB]] of radius
$$R = d \sqrt{\frac{n}{2(n+1)}}.$$

## Remarks

Take the smallest closed ball containing $S$; by a compactness argument its centre lies in the convex hull of the points of the closure of $S$ on its boundary sphere.
Since this centre is a convex combination of at most $n + 1$ such points, a computation with these points and the inequality $\abs{x_i - x_j} \le d$ gives the bound.
A classical alternative uses [[Helly's theorem|F1W3QRBNGJ]] to reduce to sets of $n + 1$ points.
The bound is sharp: the vertices of a regular $n$-simplex of edge $d$ have diameter $d$ and circumradius exactly $d \sqrt{n / (2(n+1))}$.
Example: for $n = 2$ the radius is $d / \sqrt{3}$, the circumradius of an equilateral triangle of side $d$.
For $n = 1$ the radius is $d/2$, as expected for a segment.

## Explanations

### L1 human

Any set of points whose pairwise distances are at most $d$ fits inside a ball of a fixed radius, a little larger than $d/2$.
In the plane the radius $d/\sqrt{3}$ suffices, and the equilateral triangle shows that nothing smaller always works.

### L2 human

The trivial bound is $R \le d$: a ball of radius $d$ centred at any point of $S$ contains $S$.
Jung's theorem gives the optimal constant, which grows from $1/2$ for $n = 1$ towards $1/\sqrt{2}$ as $n \to \infty$.
It is a typical result of convex geometry that relates two measures of size, the diameter and the circumradius.
It is often used together with the isodiametric inequality and in Borsuk-type partition problems.