---
tag: 0201Q4P2B5
kind: statement
role: theorem
name: Carathéodory's theorem
msc: [52A35]
uses: [0FQSDC2EQ2, 0KJH6SRFZK]
sources:
  - url: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)
    locator: lead section
  - url: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)
    locator: Proof
same_as:
  wikidata: Q2471737
  wikipedia: Carathéodory's theorem (convex hull)
---

## Statement

Let $P$ be a subset of $\mathbb{R}^d$, and let $x$ be a point of its [[convex hull|0FQSDC2EQ2]] $\operatorname{conv}(P)$.
Then there are at most $d + 1$ points $p_1, \dots, p_k$ of $P$, with $k \leq d + 1$, such that $x \in \operatorname{conv}\{p_1, \dots, p_k\}$.
Equivalently, $x$ is a convex combination of at most $d + 1$ points of $P$.

## Remarks

Thus $\operatorname{conv}(P)$ is the union of the simplices with vertices in $P$, of dimension at most $d$. Mathlib states this form as `convexHull_eq_union`.
The bound $d + 1$ is the best possible in general. Let $P$ be the three vertices of a triangle in $\mathbb{R}^2$. A point inside the triangle is not in the hull of two points of $P$.
The proof only uses the order and the field operations of $\mathbb{R}$. So it works over each ordered field.

## History

Carathéodory proved the theorem in 1911 for compact sets $P$. Steinitz extended it to arbitrary sets in 1914.

## Explanations

### L1 human

In the plane, a point in the convex hull of many points is always inside a triangle with corners from those points. You never need more than three corners. In space, four corners, the vertices of a tetrahedron, are always enough.