[proofplan] We first reduce the open cover to explicit continuous bump functions built from distances to the complements of the covering sets. These functions form a [partition of unity](/page/Partition%20of%20Unity) on $S^d$, so they define a continuous map from the sphere into the standard simplex in $\mathbb{R}^{d+1}$. After subtracting the barycenter of the simplex, the map lands in a $d$-dimensional hyperplane, so Borsuk-Ulam gives a point where the shifted map has the same value at $x$ and $-x$. Equality of all partition functions at this antipodal pair forces one positive coordinate at both points, and subordination puts both points in the same member of the cover. [/proofplan]

[step:Handle the zero-dimensional sphere directly] If $d = 0$, then $S^0 = \{-1,1\} \subset \mathbb{R}$. The cover consists of the single [open set](/page/Open%20Set) $U_1$, and the hypothesis gives $S^0 = U_1$. Taking $x = 1 \in S^0$, we have $x = 1 \in U_1$ and $-x = -1 \in U_1$. Hence the conclusion holds for $d = 0$. For the rest of the proof, assume $d \geq 1$. [/step]

[step:Construct a partition of unity subordinate to the open cover]If some $U_i$ equals $S^d$, then any $x \in S^d$ satisfies $x,-x \in U_i$, and the conclusion follows. Hence assume that each complement $F_i := S^d \setminus U_i$ is nonempty. For each $i \in \{1,\dots,d+1\}$, define the distance function \begin{align*} \rho_i:S^d \to [0,\infty), \qquad x \mapsto \operatorname{dist}(x,F_i) := \inf_{y \in F_i} |x-y|. \end{align*} Because $F_i$ is closed in the compact [metric space](/page/Metric%20Space) $S^d$, the function $\rho_i$ is continuous. Since the sets $U_i$ cover $S^d$, for every $x \in S^d$ there is an index $i$ with $x \in U_i$, equivalently $x \notin F_i$. As $F_i$ is closed, this gives $\rho_i(x) > 0$ for at least one $i$. Therefore the function \begin{align*} R:S^d \to (0,\infty), \qquad x \mapsto \sum_{j=1}^{d+1} \rho_j(x) \end{align*} is continuous and strictly positive. For each $i \in \{1,\dots,d+1\}$, define \begin{align*} \lambda_i:S^d \to [0,1], \qquad x \mapsto \frac{\rho_i(x)}{R(x)}. \end{align*} Then each $\lambda_i$ is continuous, and for every $x \in S^d$, \begin{align*} \sum_{i=1}^{d+1} \lambda_i(x) = 1. \end{align*} Moreover, if $\lambda_i(x) > 0$, then $\rho_i(x) > 0$, so $x \notin F_i$ and therefore $x \in U_i$.[/step]

[guided]The goal of this step is to replace the open cover by numerical coordinates that remember which open sets contain a point. We do this without invoking an abstract partition-of-unity theorem, using only the metric on the sphere. Assume first that no $U_i$ is already all of $S^d$; otherwise the theorem is immediate because every antipodal pair lies in that $U_i$. For each index $i \in \{1,\dots,d+1\}$, define the closed complement \begin{align*} F_i := S^d \setminus U_i. \end{align*} Since $U_i$ is open in the [subspace topology](/page/Subspace%20Topology) of $S^d$, the set $F_i$ is closed in $S^d$. Define \begin{align*} \rho_i:S^d \to [0,\infty), \qquad x \mapsto \operatorname{dist}(x,F_i) := \inf_{y \in F_i} |x-y|. \end{align*} This function is continuous because distance to a fixed nonempty subset of a metric space is $1$-Lipschitz with respect to the ambient Euclidean distance restricted to $S^d$: for $x,z \in S^d$ and every $y \in F_i$, the triangle inequality gives $|x-y| \leq |x-z| + |z-y|$, hence $\rho_i(x) \leq |x-z|+\rho_i(z)$ after taking the infimum over $y \in F_i$. Interchanging $x$ and $z$ gives $|\rho_i(x)-\rho_i(z)| \leq |x-z|$. Now fix $x \in S^d$. Since the sets $U_1,\dots,U_{d+1}$ cover $S^d$, there is at least one index $i$ such that $x \in U_i$. This means $x \notin F_i$. Because $F_i$ is closed in $S^d$, a point outside $F_i$ has positive distance from $F_i$ inside the compact metric space $S^d$, so $\rho_i(x)>0$. Thus the sum \begin{align*} R:S^d \to (0,\infty), \qquad x \mapsto \sum_{j=1}^{d+1} \rho_j(x) \end{align*} is everywhere positive and continuous. We can therefore normalize the distance functions by defining, for each $i$, \begin{align*} \lambda_i:S^d \to [0,1], \qquad x \mapsto \frac{\rho_i(x)}{R(x)}. \end{align*} The quotient is continuous because $R(x)>0$ for every $x \in S^d$. The normalization gives \begin{align*} \sum_{i=1}^{d+1} \lambda_i(x) = \frac{\sum_{i=1}^{d+1}\rho_i(x)}{R(x)} = 1. \end{align*} Finally, the subordination property is exactly encoded by positivity: if $\lambda_i(x)>0$, then $\rho_i(x)>0$, so $x$ has positive distance from $F_i$ and in particular $x \notin F_i$. Since $F_i = S^d \setminus U_i$, this proves $x \in U_i$.[/guided]

[step:Shift the simplex-valued map into a $d$-dimensional hyperplane] Define the continuous map \begin{align*} f:S^d \to \mathbb{R}^{d+1}, \qquad x \mapsto (\lambda_1(x),\dots,\lambda_{d+1}(x)). \end{align*} Let \begin{align*} c := \left(\frac{1}{d+1},\dots,\frac{1}{d+1}\right) \in \mathbb{R}^{d+1}. \end{align*} Define \begin{align*} \bar{f}:S^d \to \mathbb{R}^{d+1}, \qquad x \mapsto f(x)-c. \end{align*} Let \begin{align*} H := \left\{y=(y_1,\dots,y_{d+1}) \in \mathbb{R}^{d+1} : \sum_{i=1}^{d+1} y_i = 0\right\}. \end{align*} For every $x \in S^d$, \begin{align*} \sum_{i=1}^{d+1} \bar{f}_i(x) = \sum_{i=1}^{d+1} \lambda_i(x) - \sum_{i=1}^{d+1} \frac{1}{d+1} = 1-1 = 0. \end{align*} Thus $\bar{f}$ takes values in the $d$-dimensional real [vector space](/page/Vector%20Space) $H$. [/step]

[step:Apply Borsuk-Ulam to obtain an antipodal equality of weights]Choose a linear isomorphism \begin{align*} A:H \to \mathbb{R}^d. \end{align*} Define \begin{align*} g:S^d \to \mathbb{R}^d, \qquad x \mapsto A(\bar{f}(x)). \end{align*} The map $g$ is continuous because $\bar{f}$ and $A$ are continuous. By the [Borsuk-Ulam Theorem](/theorems/6462) (citing a result not yet in the wiki: Borsuk-Ulam Theorem), applied to the continuous map $g:S^d \to \mathbb{R}^d$, there exists $x \in S^d$ such that \begin{align*} g(x) = g(-x). \end{align*} Since $A$ is injective, this implies \begin{align*} \bar{f}(x) = \bar{f}(-x). \end{align*} Adding $c$ to both sides gives \begin{align*} f(x) = f(-x). \end{align*} Therefore, for every $i \in \{1,\dots,d+1\}$, \begin{align*} \lambda_i(x) = \lambda_i(-x). \end{align*}[/step]

[guided]The partition functions give a map into $\mathbb{R}^{d+1}$, but Borsuk-Ulam applies to maps from $S^d$ into a $d$-dimensional target. The reason for subtracting the barycenter $c$ is that the coordinates of $f(x)$ always sum to $1$, so after subtracting $c$ they sum to $0$ and lie in the $d$-dimensional hyperplane \begin{align*} H := \left\{y=(y_1,\dots,y_{d+1}) \in \mathbb{R}^{d+1} : \sum_{i=1}^{d+1} y_i = 0\right\}. \end{align*} To put this in the standard form of Borsuk-Ulam, choose a linear isomorphism \begin{align*} A:H \to \mathbb{R}^d. \end{align*} Such an isomorphism exists because $H$ is a $d$-dimensional real vector space. Define \begin{align*} g:S^d \to \mathbb{R}^d, \qquad x \mapsto A(\bar{f}(x)). \end{align*} The map $g$ is continuous: $\bar{f}$ is continuous by construction from the continuous functions $\lambda_i$, and $A$ is a [linear map](/page/Linear%20Map) between finite-dimensional normed vector spaces. Now apply the Borsuk-Ulam Theorem (citing a result not yet in the wiki: Borsuk-Ulam Theorem) to this continuous map $g:S^d \to \mathbb{R}^d$. Its conclusion gives a point $x \in S^d$ with \begin{align*} g(x) = g(-x). \end{align*} By the definition of $g$, this means \begin{align*} A(\bar{f}(x)) = A(\bar{f}(-x)). \end{align*} Because $A$ is injective, the equality after applying $A$ forces equality before applying $A$: \begin{align*} \bar{f}(x) = \bar{f}(-x). \end{align*} Finally, $\bar{f}=f-c$, so adding the same vector $c$ to both sides gives \begin{align*} f(x) = f(-x). \end{align*} Since $f$ has coordinates $(\lambda_1,\dots,\lambda_{d+1})$, this is exactly the coordinatewise equality \begin{align*} \lambda_i(x) = \lambda_i(-x) \end{align*} for every $i \in \{1,\dots,d+1\}$.[/guided]

[step:Use a positive coordinate to find one set containing both antipodal points] For the point $x \in S^d$ found above, the partition identity gives \begin{align*} \sum_{i=1}^{d+1} \lambda_i(x) = 1. \end{align*} Since each $\lambda_i(x) \geq 0$, at least one index $i \in \{1,\dots,d+1\}$ satisfies $\lambda_i(x)>0$. By the subordination property proved above, $\lambda_i(x)>0$ implies $x \in U_i$. For the same index $i$, the antipodal equality of weights gives \begin{align*} \lambda_i(-x) = \lambda_i(x) > 0. \end{align*} Again by subordination, $\lambda_i(-x)>0$ implies $-x \in U_i$. Thus $x \in U_i$ and $-x \in U_i$ for this index $i$, which is the required antipodal pair in a single member of the cover. [/step]