Let \(d=n-2k+1\) and suppose, to the contrary, that there is a partition of \(\binom{[n]}{k}\) into intersecting families \(\mathcal{F}_1,\dots,\mathcal{F}_{d}\text{.}\)

Let \(x_1,\dots,x_n\) be uniformly random points in \(\mathbb{S}_d\) chosen independently of one another. For each \(x\in\mathbb{S}_d\text{,}\) define \(H(x)\) to be the open hemisphere centred at \(x\text{.}\) For each \(A\in \binom{[n]}{k}\text{,}\) define

\begin{equation*}
H(A) := \bigcap_{j\in F}H(x_j)\text{.}
\end{equation*}

The set \(H(A)\) is open since it is the intersection of finitely many open sets. Also, for each \(1\leq i\leq d\text{,}\) define

\begin{equation*}
U_i:=\bigcup_{A\in \mathcal{F}_i}H(A)\text{.}
\end{equation*}

Then

\(U_i\) is open as well. Define

\(U_0:=\mathbb{S}_d\setminus\bigcup_{i=1}^{d}U_i\text{.}\) Then the sets

\(U_0,U_1,\dots,U_d\) cover

\(\mathbb{S}_d\text{.}\) By

Theorem 1.5.5, there exists an index

\(i\) and

\(x\in \mathbb{S}_d\) such that

\(\{x,-x\}\subseteq U_i\text{.}\) We divide the proof into cases.

### Case.

\(i=0\text{.}\)

Consider the set \(T:=\{j\in [n]: x_j\in H(x)\}\text{.}\) We claim that \(|T|\leq k-1\text{.}\) If not, then let \(A\) be a subset of \(T\) of cardinality \(k\text{.}\) Note that, for any two points \(y,z\) in the sphere, we have \(z\in H(y)\) if and only if \(y\in H(z)\text{.}\) So, since \(x_j\in H(x)\) for all \(j\in A\text{,}\) we must have that \(x\in H(A)\text{.}\) Since \(\mathcal{F}_1,\dots,\mathcal{F}_d\) partition \(\binom{[n]}{k}\text{,}\) there must be some \(i\in [d]\) such that \(A\in\mathcal{F}_i\text{.}\) But then \(x\in U_i\) which contradicts the fact that \(x\in U_0\text{.}\)

Therefore, \(|T|\leq k-1\text{.}\) Analogously, the set \(T':=\{j\in [n]: x_j\in H(-x)\}\) also has cardinality at most \(k-1\text{.}\) Thus, among the points \(x_1,\dots,x_n\text{,}\) there are at least \(n-2k+2 = d+1\) points that are in \(\mathbb{S}_d\setminus (H(x)\cup H(-x))\text{.}\) In other words, there are at least \(d+1\) such points that are contained in the plane \(\{z: \langle z,x\rangle=0\}\subseteq \mathbb{R}^{d+1}\text{.}\) However, \(x_1,\dots,x_n\) were chosen at random, and so the probability that any plane contains more than \(d\) of them is zero. So, with probability \(1\text{,}\) the set \(U_0\) cannot contain two antipodal points \(x\) and \(-x\text{.}\)

### Case.

\(1\leq i\leq d\text{.}\)

Since \(x\in U_i\text{,}\) there must exist \(A\in \mathcal{F}_i\) such that \(x\in H(A)\text{.}\) This implies that \(x_j\in H(x)\) for all \(j\in A\text{.}\) Similarly, there exists \(A'\in \mathcal{F}_i\) such that \(x_j\in H(-x)\) for all \(j\in A'\text{.}\) Since the hemispheres \(H(x)\) and \(H(-x)\) are disjoint, we have that \(\{x_j: j\in A\}\cap \{x_j:
j\in A'\}=\emptyset\text{.}\) However, this implies that \(A\cap A'=\emptyset\text{,}\) which contradicts the assumption that \(\mathcal{F}_i\) is intersecting and completes the proof!