[proofplan] We construct the desired points as a normalized Gale transform of the vertices of a cyclic polytope. The key input is that the cyclic polytope in dimension $2k-2$ is $(k-1)$-neighborly, so every set of at most $k-1$ vertices is a face. Gale duality converts this face condition into the statement that the complementary Gale vectors contain the origin as a full-dimensional relative interior point of their convex hull. If some open hemisphere contained fewer than $k$ Gale points, its complement would lie in a closed half-space through the origin, contradicting that full-dimensional interior condition. [/proofplan]

[step:Handle the case $k=1$ by a simplex construction] Assume first that $k=1$. Then $n \geq 2$, and the target sphere is $S^{n-2} \subset \mathbb{R}^{n-1}$. Choose points $p_1,\dots,p_n \in S^{n-2}$ forming the vertices of a regular simplex centered at the origin in $\mathbb{R}^{n-1}$. Thus the affine span of the points is all of $\mathbb{R}^{n-1}$ and there are positive coefficients $\alpha_1,\dots,\alpha_n \in (0,\infty)$ such that \begin{align*} \sum_{i=1}^n \alpha_i p_i = 0. \end{align*} Let $u \in S^{n-2}$. If $u \cdot p_i \leq 0$ for every $i \in \{1,\dots,n\}$, then taking the Euclidean [inner product](/page/Inner%20Product) of the displayed identity with $u$ gives \begin{align*} 0 = \sum_{i=1}^n \alpha_i (u \cdot p_i). \end{align*} Since every $\alpha_i$ is positive and every summand is nonpositive, each term satisfies $u \cdot p_i = 0$. Hence $u$ is orthogonal to every $p_i$. Because the simplex vertices affinely span $\mathbb{R}^{n-1}$ and are centered at the origin, their linear span is $\mathbb{R}^{n-1}$, forcing $u=0$, which contradicts $u \in S^{n-2}$. Therefore at least one $p_i$ lies in the open hemisphere determined by $u$, which is the required conclusion when $k=1$. [/step]

[step:Choose a cyclic polytope whose Gale space has the target dimension] Assume from now on that $k \geq 2$. Define the integers \begin{align*} r := 2k-2 \end{align*} and \begin{align*} d := n-r-1 = n-2k+1. \end{align*} Since $n \geq 2k$, we have $d \geq 1$, and the desired sphere is $S^{d-1} = S^{n-2k}$. Choose [real numbers](/page/Real%20Numbers) $t_1<\cdots<t_n$. Define the moment-curve map $\gamma: \mathbb{R} \to \mathbb{R}^r$ by \begin{align*} \gamma(t) := (t,t^2,\dots,t^r). \end{align*} For each $i \in \{1,\dots,n\}$, set $q_i := \gamma(t_i) \in \mathbb{R}^r$. The polytope \begin{align*} P := \operatorname{conv}\{q_1,\dots,q_n\} \subset \mathbb{R}^r \end{align*} is the cyclic polytope with these vertices. We use two standard prerequisites about cyclic polytopes for this proof. The affine general position theorem for the moment curve says that the points $q_1,\dots,q_n$ are in affine general position, meaning every subset of at most $r+1$ of them is affinely independent. The neighborliness theorem for cyclic polytopes says that the cyclic polytope in dimension $r=2k-2$ is $(k-1)$-neighborly, meaning every subset of at most $k-1$ vertices is the vertex set of a face of $P$. [/step]

[step:Form a Gale transform and normalize its vectors onto the sphere] Let $e_1,\dots,e_n$ denote the standard basis of $\mathbb{R}^n$. Define the homogenization map $A:\mathbb{R}^n \to \mathbb{R}^{r+1}$ by specifying \begin{align*} A e_i := (q_i,1) \end{align*} for each $i \in \{1,\dots,n\}$ and extending linearly. Since $q_1,\dots,q_n$ affinely span $\mathbb{R}^r$, the [linear map](/page/Linear%20Map) $A$ has rank $r+1$. Therefore its kernel \begin{align*} K := \ker A \subset \mathbb{R}^n \end{align*} has dimension \begin{align*} \dim K = n-(r+1)=d. \end{align*} Choose a linear isomorphism $\Phi:\mathbb{R}^d \to K$. For each $i \in \{1,\dots,n\}$, define the linear functional $\ell_i:\mathbb{R}^d \to \mathbb{R}$ by \begin{align*} \ell_i(a) := (\Phi(a))_i, \end{align*} where $(\Phi(a))_i$ denotes the $i$-th coordinate of the vector $\Phi(a) \in \mathbb{R}^n$. By the Euclidean representation of linear functionals on $\mathbb{R}^d$, there is a unique vector $g_i \in \mathbb{R}^d$ such that \begin{align*} \ell_i(a)=g_i \cdot a \end{align*} for every $a \in \mathbb{R}^d$. The ordered family $(g_1,\dots,g_n)$ is a Gale transform of the affine configuration $(q_1,\dots,q_n)$. We claim that no Gale vector $g_i$ is zero. Fix $i \in \{1,\dots,n\}$. If $g_i=0$, then the coordinate functional $e_i^*:\mathbb{R}^n\to\mathbb{R}$, $x\mapsto x_i$, vanishes on $K=\ker A$. Thus every linear relation among the columns $(q_1,1),\dots,(q_n,1)$ has coefficient $0$ on the $i$-th column. Equivalently, the column $(q_i,1)$ is not in the linear span of the remaining columns, so deleting the $i$-th column lowers the rank of $A$. On the other hand, since $n-1\geq 2k-1=r+1$, we may choose $r+1$ indices $m\neq i$. By affine general position of the moment-curve vertices, the corresponding points $q_m$ are affinely independent. Therefore their homogenizations $(q_m,1)$ are linearly independent in $\mathbb{R}^{r+1}$, so they span $\mathbb{R}^{r+1}$. Hence the columns with $m\neq i$ already have rank $r+1$, contradicting the conclusion that deleting the $i$-th column lowers the rank. Thus $g_i\neq 0$ for every $i$. Define points $p_i \in S^{d-1}$ by \begin{align*} p_i := \frac{g_i}{|g_i|} \end{align*} for $i \in \{1,\dots,n\}$. [/step]

[step:Translate small vertex sets into full-dimensional convex hulls of complementary Gale points]Let $I \subset \{1,\dots,n\}$ satisfy $\# I \leq k-1$, and define its complement \begin{align*} J := \{1,\dots,n\}\setminus I. \end{align*} By the neighborliness of $P$, the set $\{q_i : i \in I\}$ is the vertex set of a face of $P$. We use the following strengthened Gale face criterion as the exact Gale-duality prerequisite needed here. For an affinely spanning configuration $q_1,\dots,q_n\in\mathbb{R}^r$ with homogenization kernel $K\subset\mathbb{R}^n$ and Gale transform $g_1,\dots,g_n\in\mathbb{R}^d$ obtained from an isomorphism $\mathbb{R}^d\to K$, a subset $I\subset\{1,\dots,n\}$ is the vertex set of a face of $\operatorname{conv}\{q_1,\dots,q_n\}$ if and only if \begin{align*} 0 \in \operatorname{relint}\operatorname{conv}\{g_j : j \in \{1,\dots,n\}\setminus I\}. \end{align*} The strengthened form also asserts that, when $\{q_i:i\in I\}$ is affinely independent, the complementary Gale vectors $\{g_j:j\in \{1,\dots,n\}\setminus I\}$ affinely span $\mathbb{R}^d$; hence the relative interior above is ordinary Euclidean interior in $\mathbb{R}^d$. The hypotheses of this criterion are satisfied because $q_1,\dots,q_n$ affinely span $\mathbb{R}^r$, the map $A$ is the standard homogenization of this configuration, and $(g_1,\dots,g_n)$ was constructed from a linear isomorphism $\mathbb{R}^d\to\ker A$. Since $\#I\leq k-1\leq r+1$ and the moment-curve vertices are in affine general position, the set $\{q_i:i\in I\}$ is affinely independent. Therefore \begin{align*} 0 \in \operatorname{int}_{\mathbb{R}^d}\operatorname{conv}\{g_j : j \in J\}. \end{align*} It remains to justify that normalization preserves this interior containment. Suppose, to the contrary, that \begin{align*} 0 \notin \operatorname{int}_{\mathbb{R}^d}\operatorname{conv}\{p_j : j \in J\}. \end{align*} Since $\operatorname{conv}\{p_j:j\in J\}$ is a compact convex subset of $\mathbb{R}^d$, the separating hyperplane theorem for a point outside the interior of a compact convex set gives a nonzero vector $v\in\mathbb{R}^d$ such that \begin{align*} v\cdot p_j \leq 0 \end{align*} for every $j\in J$. Because $g_j=|g_j|p_j$ and $|g_j|>0$, this implies \begin{align*} v\cdot g_j \leq 0 \end{align*} for every $j\in J$. Hence $\operatorname{conv}\{g_j:j\in J\}$ is contained in the closed half-space $\{x\in\mathbb{R}^d:v\cdot x\leq 0\}$, which cannot contain $0$ as an interior point because $v\neq 0$. This contradicts the preceding interior containment for the $g_j$. Therefore \begin{align*} 0 \in \operatorname{int}_{\mathbb{R}^d}\operatorname{conv}\{p_j : j \in J\}. \end{align*}[/step]

[guided]Fix a subset $I \subset \{1,\dots,n\}$ with $\#I \leq k-1$, and let \begin{align*} J := \{1,\dots,n\}\setminus I. \end{align*} The reason for studying such an $I$ is that an alleged bad hemisphere will later give exactly such a set: the labels of the points lying inside the hemisphere. The cyclic polytope input says that every subset of at most $k-1$ vertices is a face. Since $\#I \leq k-1$, the set $\{q_i : i \in I\}$ is therefore the vertex set of a face of the cyclic polytope \begin{align*} P=\operatorname{conv}\{q_1,\dots,q_n\}. \end{align*} We now use the strengthened Gale face criterion in its full-dimensional form. The accepted prerequisite is this: for an affinely spanning configuration $q_1,\dots,q_n\in\mathbb{R}^r$ with homogenization kernel $K\subset\mathbb{R}^n$ and Gale transform $g_1,\dots,g_n\in\mathbb{R}^d$ obtained from an isomorphism $\mathbb{R}^d\to K$, a subset $I$ is the vertex set of a face of $\operatorname{conv}\{q_1,\dots,q_n\}$ exactly when the complementary Gale vectors have the origin in the relative interior of their convex hull. The strengthened part needed here says that if $\{q_i:i\in I\}$ is affinely independent, then the complementary Gale vectors affinely span all of $\mathbb{R}^d$, so relative interior becomes ordinary Euclidean interior. The criterion applies here because $q_1,\dots,q_n$ affinely span $\mathbb{R}^r$, the linear map $A:\mathbb{R}^n \to \mathbb{R}^{r+1}$ is their homogenization, and the vectors $g_i \in \mathbb{R}^d$ were constructed from the kernel of $A$. Since $\#I\leq k-1\leq r+1$ and the moment-curve vertices are in affine general position, $\{q_i:i\in I\}$ is affinely independent. Thus the complementary Gale vectors affinely span $\mathbb{R}^d$, and the relative interior conclusion from the face criterion is ordinary Euclidean interior: \begin{align*} 0 \in \operatorname{int}_{\mathbb{R}^d}\operatorname{conv}\{g_j : j \in J\}. \end{align*} Finally, we must check that replacing $g_j$ by the normalized point $p_j$ does not lose the interior condition. The right way to see this is by half-spaces. Suppose instead that \begin{align*} 0 \notin \operatorname{int}_{\mathbb{R}^d}\operatorname{conv}\{p_j : j \in J\}. \end{align*} Because $\operatorname{conv}\{p_j:j\in J\}$ is compact and convex, the separating hyperplane theorem for a point outside the interior of a compact convex set gives a nonzero vector $v\in\mathbb{R}^d$ such that \begin{align*} v\cdot p_j \leq 0 \end{align*} for every $j\in J$. Since $g_j=|g_j|p_j$ with $|g_j|>0$, the same inequalities hold for the unnormalized Gale vectors: \begin{align*} v\cdot g_j \leq 0 \end{align*} for every $j\in J$. Therefore $\operatorname{conv}\{g_j:j\in J\}$ lies in the closed half-space $\{x\in\mathbb{R}^d:v\cdot x\leq 0\}$. A closed half-space through the origin with nonzero normal vector cannot contain a Euclidean neighbourhood of $0$, so this contradicts \begin{align*} 0 \in \operatorname{int}_{\mathbb{R}^d}\operatorname{conv}\{g_j : j \in J\}. \end{align*} Hence the normalized vectors also satisfy \begin{align*} 0 \in \operatorname{int}_{\mathbb{R}^d}\operatorname{conv}\{p_j : j \in J\}. \end{align*}[/guided]

[step:Rule out an open hemisphere containing fewer than $k$ points] Let $u \in S^{d-1}$, and define the open hemisphere \begin{align*} H_u := \{x \in S^{d-1} : u \cdot x > 0\}. \end{align*} Suppose, toward a contradiction, that $H_u$ contains fewer than $k$ of the points $p_1,\dots,p_n$. Define \begin{align*} I := \{i \in \{1,\dots,n\} : u \cdot p_i > 0\}. \end{align*} Then $\#I \leq k-1$. Let \begin{align*} J := \{1,\dots,n\}\setminus I. \end{align*} By the previous step, \begin{align*} 0 \in \operatorname{int}_{\mathbb{R}^d}\operatorname{conv}\{p_j : j \in J\}. \end{align*} For every $j \in J$, the definition of $J$ gives $u \cdot p_j \leq 0$. Since the map $L_u:\mathbb{R}^d \to \mathbb{R}$ defined by \begin{align*} L_u(x) := u \cdot x \end{align*} is linear, it follows that \begin{align*} L_u(x) \leq 0 \end{align*} for every \begin{align*} x \in \operatorname{conv}\{p_j : j \in J\}. \end{align*} Thus this convex hull is contained in the closed half-space \begin{align*} \{x \in \mathbb{R}^d : u \cdot x \leq 0\}. \end{align*} For $\varepsilon>0$, let $B(0,\varepsilon):=\{x\in\mathbb{R}^d:|x|<\varepsilon\}$ denote the open Euclidean ball centered at $0$ with radius $\varepsilon$. A subset of this closed half-space cannot contain $0$ as an interior point in $\mathbb{R}^d$: since $u \neq 0$, every such ball contains the point $\varepsilon u/(2|u|)$, and \begin{align*} u \cdot \frac{\varepsilon u}{2|u|} = \frac{\varepsilon}{2}|u| > 0. \end{align*} Therefore no Euclidean ball centered at $0$ is contained in the half-space $\{x:u\cdot x\leq 0\}$. This contradicts \begin{align*} 0 \in \operatorname{int}_{\mathbb{R}^d}\operatorname{conv}\{p_j : j \in J\}. \end{align*} Hence every open hemisphere of $S^{d-1}=S^{n-2k}$ contains at least $k$ of the points $p_1,\dots,p_n$. This proves the theorem. [/step]