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 A}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.
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!
