[proofplan] We argue by contradiction. If no $F_i$ contains an antipodal pair, we first thicken each $F_i$ to an [open set](/page/Open%20Set) $U_i$ that is still disjoint from its antipodal image. From the open cover $(U_i)_{i=0}^n$ we build an explicit continuous [partition of unity](/page/Partition%20of%20Unity) by distance-to-the-complement functions, avoiding any appeal to a separate partition-of-unity theorem. The coordinate functions of this partition produce a continuous odd map from $S^n$ to the unit sphere in the hyperplane $\sum_i y_i=0$, which is an $(n-1)$-sphere; this contradicts the no-map form of the [Borsuk-Ulam theorem](/theorems/6462). [/proofplan] [step:Handle the zero-dimensional sphere directly] If $n=0$, then $S^0=\{-1,1\}\subset \mathbb{R}$. Since the only member of the cover is $F_0$ and $S^0=F_0$, both $1$ and $-1$ lie in $F_0$. Thus $F_0$ contains an antipodal pair. Henceforth assume $n\geq 1$. [/step] [step:Thicken each closed set without creating antipodal intersections] Assume, for contradiction, that no $F_i$ contains an antipodal pair. Define the antipodal map \begin{align*} A:S^n \to S^n,\quad A(x)=-x. \end{align*} For each $i\in\{0,\dots,n\}$, let $-F_i:=A(F_i)=\{-x:x\in F_i\}$. If $F_i=\varnothing$, set $U_i:=\varnothing$. If $F_i\neq\varnothing$, then $F_i$ and $-F_i$ are disjoint compact subsets of the [metric space](/page/Metric%20Space) $(S^n,d)$, where $d:S^n\times S^n\to[0,\infty)$ is given by $d(x,y)=|x-y|$. Define \begin{align*} \delta_i:=\inf\{|x-y|:x\in F_i,\ y\in -F_i\}. \end{align*} Compactness and disjointness give $\delta_i>0$. Define \begin{align*} U_i:=\{x\in S^n:\operatorname{dist}(x,F_i)<\delta_i/3\}, \end{align*} where $\operatorname{dist}(x,F_i):=\inf\{|x-y|:y\in F_i\}$. Then $U_i$ is open in $S^n$ and contains $F_i$. We claim that $U_i\cap A(U_i)=\varnothing$. If $x\in U_i\cap A(U_i)$, then $x\in U_i$ and $A(x)=-x\in U_i$. Hence \begin{align*} \operatorname{dist}(x,F_i)<\delta_i/3 \end{align*} and \begin{align*} \operatorname{dist}(-x,F_i)<\delta_i/3. \end{align*} Because $\operatorname{dist}(x,-F_i)=\operatorname{dist}(-x,F_i)$, we also have \begin{align*} \operatorname{dist}(x,-F_i)<\delta_i/3. \end{align*} Thus there are points $a\in F_i$ and $b\in -F_i$ such that $|x-a|<\delta_i/3$ and $|x-b|<\delta_i/3$. By the triangle inequality, \begin{align*} |a-b|\leq |a-x|+|x-b|<2\delta_i/3. \end{align*} This contradicts the definition of $\delta_i$ as the distance between $F_i$ and $-F_i$. Therefore $U_i\cap A(U_i)=\varnothing$. Since $F_i\subset U_i$ for every $i$ and the $F_i$ cover $S^n$, the family $(U_i)_{i=0}^n$ is an open cover of $S^n$. [/step] [step:Build a continuous partition of unity from the open cover] For each $i\in\{0,\dots,n\}$, define a [continuous function](/page/Continuous%20Function) \begin{align*} g_i:S^n\to[0,\infty),\quad g_i(x)=\operatorname{dist}(x,S^n\setminus U_i). \end{align*} Here $S^n\setminus U_i$ is closed in $S^n$, so the distance function is continuous. Since $(U_i)_{i=0}^n$ covers $S^n$, for every $x\in S^n$ there is an index $i$ such that $x\in U_i$. Because $U_i$ is open in the metric space $S^n$, this implies $g_i(x)>0$. Therefore the function \begin{align*} G_0:S^n\to(0,\infty),\quad G_0(x)=\sum_{j=0}^{n} g_j(x) \end{align*} is continuous and strictly positive. For each $i\in\{0,\dots,n\}$, define \begin{align*} \varphi_i:S^n\to[0,1],\quad \varphi_i(x)=\frac{g_i(x)}{G_0(x)}. \end{align*} Then each $\varphi_i$ is continuous, $\varphi_i(x)=0$ for $x\notin U_i$, and \begin{align*} \sum_{i=0}^{n}\varphi_i(x)=1 \end{align*} for every $x\in S^n$. [guided] The purpose of this step is to replace the closed cover by continuous coordinates. For each open set $U_i$, we measure how far a point is from leaving $U_i$. This is the function \begin{align*} g_i:S^n\to[0,\infty),\quad g_i(x)=\operatorname{dist}(x,S^n\setminus U_i). \end{align*} The function $g_i$ is continuous because distance to a fixed closed subset of a metric space is continuous. It also has the exact support behavior we need: if $x\notin U_i$, then $x\in S^n\setminus U_i$, so $g_i(x)=0$; if $x\in U_i$, then openness of $U_i$ gives a radius $r>0$ such that the relative open ball $B_{S^n}(x,r):=\{z\in S^n:|z-x|<r\}$ satisfies $B_{S^n}(x,r)\subset U_i$, hence every point of $S^n\setminus U_i$ is at distance at least $r$ from $x$, so $g_i(x)>0$. Now define the sum \begin{align*} G_0:S^n\to(0,\infty),\quad G_0(x)=\sum_{j=0}^{n}g_j(x). \end{align*} Why is $G_0(x)$ never zero? Since the open sets $U_0,\dots,U_n$ cover $S^n$, every $x\in S^n$ lies in at least one $U_i$, and for that index $i$ we just proved $g_i(x)>0$. Thus the sum is strictly positive. We may therefore normalize: \begin{align*} \varphi_i:S^n\to[0,1],\quad \varphi_i(x)=\frac{g_i(x)}{G_0(x)}. \end{align*} Each $\varphi_i$ is continuous as a quotient of continuous functions with strictly positive denominator. Also $\varphi_i(x)=0$ whenever $x\notin U_i$, because then $g_i(x)=0$. Finally, \begin{align*} \sum_{i=0}^{n}\varphi_i(x)=\sum_{i=0}^{n}\frac{g_i(x)}{G_0(x)}=\frac{G_0(x)}{G_0(x)}=1. \end{align*} Thus the functions $\varphi_0,\dots,\varphi_n$ form the continuous coordinate system subordinate to the open cover that we will use to construct the odd map. [/guided] [/step] [step:Construct a nonvanishing odd map into the hyperplane] Define the continuous map \begin{align*} \Phi:S^n\to\mathbb{R}^{n+1},\quad \Phi(x)=(\varphi_0(x),\dots,\varphi_n(x)). \end{align*} Let \begin{align*} H:=\{y=(y_0,\dots,y_n)\in\mathbb{R}^{n+1}:\sum_{i=0}^{n}y_i=0\} \end{align*} be the $n$-dimensional linear hyperplane in $\mathbb{R}^{n+1}$. Define \begin{align*} \Psi:S^n\to H,\quad \Psi(x)=\Phi(x)-\Phi(-x). \end{align*} The image lies in $H$ because \begin{align*} \sum_{i=0}^{n}\bigl(\varphi_i(x)-\varphi_i(-x)\bigr)=1-1=0. \end{align*} Moreover $\Psi$ is odd, since \begin{align*} \Psi(-x)=\Phi(-x)-\Phi(x)=-\Psi(x). \end{align*} We claim that $\Psi(x)\neq 0$ for every $x\in S^n$. If $\Psi(x)=0$, then $\varphi_i(x)=\varphi_i(-x)$ for every $i$. Since $\sum_{i=0}^{n}\varphi_i(x)=1$, there exists an index $i$ such that $\varphi_i(x)>0$. Then also $\varphi_i(-x)>0$. By construction of $\varphi_i$, positivity implies $x\in U_i$ and $-x\in U_i$, contradicting $U_i\cap A(U_i)=\varnothing$. Hence $\Psi$ is nowhere zero. [/step] [step:Normalize the odd map and contradict Borsuk-Ulam] Let \begin{align*} S(H):=\{y\in H:|y|=1\} \end{align*} be the unit sphere in the Euclidean [vector space](/page/Vector%20Space) $H$. Since $\Psi(x)\neq 0$ for every $x\in S^n$, define \begin{align*} \Theta:S^n\to S(H),\quad \Theta(x)=\frac{\Psi(x)}{|\Psi(x)|}. \end{align*} The map $\Theta$ is continuous because $\Psi$ is continuous and nowhere zero. It is odd because $\Psi$ is odd and $|\Psi(-x)|=|\Psi(x)|$: \begin{align*} \Theta(-x)=\frac{\Psi(-x)}{|\Psi(-x)|}=\frac{-\Psi(x)}{|\Psi(x)|}=-\Theta(x). \end{align*} The hyperplane $H$ has dimension $n$, so $S(H)$ is an $(n-1)$-sphere. Choose a linear isometry $L:H\to\mathbb{R}^n$ and define $\widetilde{\Theta}:S^n\to S^{n-1}$ by $\widetilde{\Theta}(x)=L(\Theta(x))$. Since $L$ is linear and continuous, $\widetilde{\Theta}$ is continuous and odd. This contradicts the no-map form of the Borsuk-Ulam theorem, which states that for $n\geq 1$ there is no continuous odd map $S^n\to S^{n-1}$. [guided] We now turn the nowhere-zero odd map $\Psi:S^n\to H$ into a map whose codomain is a sphere. Since $\Psi(x)\neq 0$ for every $x\in S^n$, the Euclidean norm $|\Psi(x)|$ is strictly positive. Therefore the normalization \begin{align*} \Theta:S^n\to S(H) \end{align*} defined by $\Theta(x)=\Psi(x)/|\Psi(x)|$ is a well-defined continuous map into the unit sphere $S(H)=\{y\in H:|y|=1\}$. The normalization preserves oddness because the Euclidean norm is invariant under multiplication by $-1$. For every $x\in S^n$, \begin{align*} \Theta(-x)=\frac{\Psi(-x)}{|\Psi(-x)|}=\frac{-\Psi(x)}{|\Psi(x)|}=-\Theta(x). \end{align*} Thus $\Theta$ is a continuous odd map from $S^n$ to the unit sphere in $H$. The target $S(H)$ is an $(n-1)$-sphere because $H=\{y\in\mathbb{R}^{n+1}:\sum_{i=0}^{n}y_i=0\}$ is an $n$-dimensional Euclidean vector space with the [inner product](/page/Inner%20Product) inherited from $\mathbb{R}^{n+1}$. Choose a linear isometry $L:H\to\mathbb{R}^n$. Since $L$ preserves Euclidean norms, it maps $S(H)$ onto $S^{n-1}$. Define $\widetilde{\Theta}:S^n\to S^{n-1}$ by $\widetilde{\Theta}(x)=L(\Theta(x))$. The map $\widetilde{\Theta}$ is continuous because it is a [composition of continuous maps](/theorems/4960), and it is odd because $L$ is linear: \begin{align*} \widetilde{\Theta}(-x)=L(\Theta(-x))=L(-\Theta(x))=-L(\Theta(x))=-\widetilde{\Theta}(x). \end{align*} This is exactly what the no-map form of the Borsuk-Ulam theorem forbids: for $n\geq 1$, there is no continuous odd map $S^n\to S^{n-1}$. Since we have constructed such a map from the assumption that no $F_i$ contains an antipodal pair, that assumption is impossible. [/guided] The contradiction shows that the assumption was false. Therefore at least one $F_i$ contains an antipodal pair. [/step]