[proofplan] We prove the theorem by induction on the number $m$ of convex sets. The case $m \leq n+1$ is exactly the hypothesis. The essential step is the case $m=n+2$, where [Radon's theorem](/theorems/4086) produces a point lying in two convex hulls, and convexity forces that point to belong to every set. For larger $m$, we reduce the number of sets by replacing two sets with their intersection; the already proved $n+2$ case verifies the Helly hypothesis for the reduced family. [/proofplan]

[step:Reduce the theorem to an induction on the number of sets] Fix $n \in \mathbb{N}$. We prove the statement for every $m \in \mathbb{N}$ by induction on $m$. If $m \leq n+1$, then the index set $\{1,\dots,m\}$ has cardinality at most $n+1$, so the hypothesis applied to $I=\{1,\dots,m\}$ gives \begin{align*} \bigcap_{i=1}^m C_i \neq \varnothing. \end{align*} This proves the base range $m \leq n+1$. [/step]

[step:Prove the critical case of $n+2$ convex sets using Radon's theorem]Assume $m=n+2$. For each $i \in \{1,\dots,n+2\}$, define the index set \begin{align*} I_i := \{1,\dots,n+2\} \setminus \{i\}. \end{align*} Since $|I_i|=n+1$, the hypothesis gives \begin{align*} \bigcap_{j \in I_i} C_j \neq \varnothing. \end{align*} Choose a point $x_i \in \bigcap_{j \neq i} C_j$ for each $i \in \{1,\dots,n+2\}$. Apply Radon's theorem (citing a result not yet in the wiki: Radon's Theorem) to the finite set of points $\{x_1,\dots,x_{n+2}\} \subset \mathbb{R}^n$. There exist nonempty disjoint subsets $A,B \subset \{1,\dots,n+2\}$ with \begin{align*} A \cup B = \{1,\dots,n+2\} \end{align*} and a point $y \in \mathbb{R}^n$ such that \begin{align*} y \in \operatorname{conv}\{x_i : i \in A\} \cap \operatorname{conv}\{x_i : i \in B\}. \end{align*} We prove that $y \in C_k$ for every $k \in \{1,\dots,n+2\}$. Fix such a $k$. Since $A$ and $B$ are disjoint and cover $\{1,\dots,n+2\}$, either $k \notin A$ or $k \notin B$. If $k \notin A$, then for every $i \in A$ we have $i \neq k$, and the choice of $x_i$ gives $x_i \in C_k$. Since $C_k$ is convex, \begin{align*} \operatorname{conv}\{x_i : i \in A\} \subset C_k. \end{align*} Because $y \in \operatorname{conv}\{x_i : i \in A\}$, it follows that $y \in C_k$. If instead $k \notin B$, the same argument with $B$ in place of $A$ gives \begin{align*} \operatorname{conv}\{x_i : i \in B\} \subset C_k \end{align*} and hence $y \in C_k$. Thus $y \in C_k$ for every $k \in \{1,\dots,n+2\}$, so \begin{align*} y \in \bigcap_{k=1}^{n+2} C_k. \end{align*} Therefore the theorem holds when $m=n+2$.[/step]

[guided]The case of $n+2$ sets is the only genuinely geometric part of the proof. For each index $i \in \{1,\dots,n+2\}$, define \begin{align*} I_i := \{1,\dots,n+2\} \setminus \{i\}. \end{align*} This is the subfamily obtained by deleting the single set $C_i$. Since $|I_i|=n+1$, the hypothesis applies and gives \begin{align*} \bigcap_{j \in I_i} C_j \neq \varnothing. \end{align*} Choose \begin{align*} x_i \in \bigcap_{j \neq i} C_j. \end{align*} Thus $x_i$ is a point that lies in every set except possibly $C_i$. Now apply Radon's theorem (citing a result not yet in the wiki: Radon's Theorem) to the $n+2$ points $x_1,\dots,x_{n+2}$ in $\mathbb{R}^n$. The theorem gives a partition of the index set into two nonempty disjoint subsets $A$ and $B$ with \begin{align*} A \cup B = \{1,\dots,n+2\} \end{align*} and a point $y \in \mathbb{R}^n$ satisfying \begin{align*} y \in \operatorname{conv}\{x_i : i \in A\} \cap \operatorname{conv}\{x_i : i \in B\}. \end{align*} The point $y$ is the candidate for a common point of all the sets. Fix an arbitrary $k \in \{1,\dots,n+2\}$. Because $A$ and $B$ are disjoint, $k$ cannot belong to both of them. Hence either $k \notin A$ or $k \notin B$. Suppose first that $k \notin A$. Then every $i \in A$ satisfies $i \neq k$, so the defining property of $x_i$ gives $x_i \in C_k$. Since $C_k$ is convex and contains each point $x_i$ with $i \in A$, it contains their convex hull: \begin{align*} \operatorname{conv}\{x_i : i \in A\} \subset C_k. \end{align*} Since $y$ belongs to this convex hull, $y \in C_k$. If instead $k \notin B$, the identical argument using the set $B$ gives \begin{align*} \operatorname{conv}\{x_i : i \in B\} \subset C_k, \end{align*} and since $y \in \operatorname{conv}\{x_i : i \in B\}$, again $y \in C_k$. The index $k$ was arbitrary, so $y \in C_k$ for all $k \in \{1,\dots,n+2\}$. Therefore \begin{align*} y \in \bigcap_{k=1}^{n+2} C_k, \end{align*} which proves the theorem for $n+2$ convex sets.[/guided]

[step:Reduce a larger family by intersecting the last two sets]Assume $m>n+2$ and assume the theorem has been proved for all families of $m-1$ convex subsets of $\mathbb{R}^n$. Let $C_1,\dots,C_m \subset \mathbb{R}^n$ satisfy the stated Helly hypothesis. Define a new family $D_1,\dots,D_{m-1}$ by \begin{align*} D_i &:= C_i \quad \text{for } 1 \leq i \leq m-2,\\ D_{m-1} &:= C_{m-1} \cap C_m. \end{align*} Each $D_i$ is convex: this is immediate for $1 \leq i \leq m-2$, and $D_{m-1}$ is convex because it is the intersection of the convex sets $C_{m-1}$ and $C_m$. We verify the Helly hypothesis for the family $D_1,\dots,D_{m-1}$. Let $J \subset \{1,\dots,m-1\}$ be nonempty with $|J| \leq n+1$. If $m-1 \notin J$, then \begin{align*} \bigcap_{j \in J} D_j = \bigcap_{j \in J} C_j \neq \varnothing \end{align*} by the original hypothesis. If $m-1 \in J$, define \begin{align*} K := (J \setminus \{m-1\}) \cup \{m-1,m\}. \end{align*} Then \begin{align*} \bigcap_{j \in J} D_j = \bigcap_{k \in K} C_k. \end{align*} Moreover $|K|=|J|+1 \leq n+2$. If $|K| \leq n+1$, the original hypothesis gives $\bigcap_{k \in K} C_k \neq \varnothing$. If $|K|=n+2$, then every subfamily of $\{C_k : k \in K\}$ with at most $n+1$ members has nonempty intersection by the original hypothesis, so the already proved $n+2$ case gives \begin{align*} \bigcap_{k \in K} C_k \neq \varnothing. \end{align*} Thus in all cases, \begin{align*} \bigcap_{j \in J} D_j \neq \varnothing. \end{align*}[/step]

[guided]The induction step reduces the number of sets by one, but it must preserve the hypothesis. The natural way to do this is to merge the last two sets. Define \begin{align*} D_i &:= C_i \quad \text{for } 1 \leq i \leq m-2,\\ D_{m-1} &:= C_{m-1} \cap C_m. \end{align*} The sets $D_1,\dots,D_{m-1}$ are convex: the first $m-2$ are the original convex sets, and $D_{m-1}$ is convex because an intersection of convex sets is convex. We now check the hypothesis for this reduced family. Let $J \subset \{1,\dots,m-1\}$ be nonempty with $|J| \leq n+1$. We must show \begin{align*} \bigcap_{j \in J} D_j \neq \varnothing. \end{align*} If $m-1 \notin J$, then no merged set appears. Therefore the intersection is just an intersection of the corresponding original sets: \begin{align*} \bigcap_{j \in J} D_j = \bigcap_{j \in J} C_j. \end{align*} Since $|J| \leq n+1$, the original Helly hypothesis gives this intersection is nonempty. Now suppose $m-1 \in J$. Then the intersection over the $D_j$ includes the merged set $D_{m-1}=C_{m-1}\cap C_m$. Define the corresponding original index set \begin{align*} K := (J \setminus \{m-1\}) \cup \{m-1,m\}. \end{align*} With this definition, \begin{align*} \bigcap_{j \in J} D_j = \bigcap_{k \in K} C_k. \end{align*} The set $K$ has one more element than $J$, so \begin{align*} |K| = |J|+1 \leq n+2. \end{align*} If $|K| \leq n+1$, the original hypothesis directly gives \begin{align*} \bigcap_{k \in K} C_k \neq \varnothing. \end{align*} If $|K|=n+2$, then the original hypothesis says that every subfamily of $\{C_k : k \in K\}$ with at most $n+1$ members has nonempty intersection. Therefore the $n+2$ case already proved applies to this subfamily and gives \begin{align*} \bigcap_{k \in K} C_k \neq \varnothing. \end{align*} Hence every subfamily of at most $n+1$ members of the reduced family $D_1,\dots,D_{m-1}$ has nonempty intersection.[/guided]

[step:Apply the induction hypothesis to the reduced family] By the previous step, the family $D_1,\dots,D_{m-1}$ is a family of $m-1$ convex subsets of $\mathbb{R}^n$ satisfying the Helly hypothesis. The induction hypothesis therefore gives \begin{align*} \bigcap_{i=1}^{m-1} D_i \neq \varnothing. \end{align*} Using the definition of the $D_i$, this intersection is \begin{align*} \bigcap_{i=1}^{m-1} D_i = \left(\bigcap_{i=1}^{m-2} C_i\right) \cap C_{m-1} \cap C_m = \bigcap_{i=1}^m C_i. \end{align*} Hence \begin{align*} \bigcap_{i=1}^m C_i \neq \varnothing. \end{align*} This completes the induction on $m$, and therefore proves Helly's theorem. [/step]