Let a covering \(S^d = U_1 \cup \dots \cup U_d \cup U_{d+1}\) be given as specified, and assume that there are no antipodal points in any of the sets \(U_i\). We define a map \(f : S^d \to \mathbb {R}^d\) by

Here \(\delta (x, U_i)\) denotes the distance of \(x\) from \(U_i\). Since this is a continuous function in \(x\), the map \(f\) is continuous. Thus the Borsuk–Ulam theorem tells us that there are antipodal points \(x^*, -x^*\) with \(f(x^*) = f(-x^*)\). Since \(U_{d+1}\) does not contain antipodes, we get that at least one of \(x^*\) and \(-x^*\) must be contained in one of the sets \(U_i\), say in \(U_k\) (\(k \leq d\)). After exchanging \(x^*\) with \(-x^*\) if necessary, we may assume that \(x^* \in U_k\). In particular this yields \(\delta (x^*, U_k) = 0\), and from \(f(x^*) = f(-x^*)\) we get that \(\delta (-x^*, U_k) = 0\) as well.

If \(U_k\) is closed, then \(\delta (-x^*, U_k) = 0\) implies that \(-x^* \in U_k\), and we arrive at the contradiction that \(U_k\) contains a pair of antipodal points.

If \(U_k\) is open, then \(\delta (-x^*, U_k) = 0\) implies that \(-x^*\) lies in \(\overline{U_k}\), the closure of \(U_k\). The set \(U_k\), in turn, is contained in \(S^d \setminus (\overline{U_k})\), since this is a closed subset of \(S^d\) that contains \(U_k\). But this means that \(-x^*\) lies in \(S^d \setminus (\overline{U_k})\), so it cannot lie in \(-U_k\), and \(x^*\) cannot lie in \(U_k\), a contradiction.

For our ground set let us take \(2k + d\) points in general position on the sphere \(S^{d+1}\). Suppose the set \(V(n, k)\) of all \(k\)-subsets of this set is partitioned into \(d + 1\) classes, \(V(n, k) = V_1 \, \dot{\cup } \, \dots \, \dot{\cup } \, V_{d+1}\). We have to find a pair of disjoint \(k\)-sets \(A\) and \(B\) that belong to the same class \(V_i\).

For \(i = 1, \dots , d + 1\) we set

Clearly, each \(O_i\) is an open set. Together, the open sets \(O_i\) and the closed set \(C = S^{d+1} \setminus (O_1 \cup \dots \cup O_{d+1})\) cover \(S^{d+1}\). Invoking Lyusternik–Shnirel’man (43.1) we know that one of these sets contains antipodal points \(x^*\) and \(-x^*\). This set cannot be \(C\)! Indeed, if \(x^*, -x^* \in C\), then by the definition of the \(O_i\)’s, the hemispheres \(H_{x^*}\) and \(H_{-x^*}\) would contain fewer than \(k\) points. This means that at least \(d + 2\) points would be on the equator \(H_{x^*} \cap H_{-x^*}\) with respect to the north pole \(x^*\), that is, on a hyperplane through the origin. But this cannot be since the points are in general position. Hence some \(O_i\) contains a pair \(x^*, -x^*\), so there exist \(k\)-sets \(A\) and \(B\) both in class \(V_i\), with \(A \subset H_{x^*}\) and \(B \subset H_{-x^*}\).

But since we are talking about open hemispheres, \(H_{x^*}\) and \(H_{-x^*}\) are disjoint, hence \(A\) and \(B\) are disjoint, and this is the whole proof.

TODO