I heard that the following are equivalent:
(1) (Borsuk-Ulam) If $f:S^n\to\Bbb R^n$, there is $x\in S^n$ such that $f(x)=f(-x)$.
(2) If $A_{i}\subset S^n$ is closed for $i=1,2,…,n+1$ such that $A_1\cup\cdots A_{n+1} = S^n$, then at least one of these sets must contain a pair of antipodal points
(3) If $A_{i}\subset S^n$ is open for $i=1,2,…,n+1$ such that $A_1\cup\cdots A_{n+1} = S^n$, then at least one of these sets must contain a pair of antipodal points
Well I know $(1)\Rightarrow(2)$. But don't know the above is equivalent. How can I show? By the way, are those equivalent?
Best Answer
$(2)\Rightarrow(3)$ This is just topology. Suppose first that $S^n=A_1\cup A_2$ with each $A_i$ open. We have $(S^n\setminus A_1)\cap (S^n\setminus A_2)=S^n\setminus(A_1\cup A_2)=\emptyset$, so since each $S^n\setminus A_i$ is closed we can use normality to find an open set $U$ with $S^n\setminus A_1\subseteq U\subseteq \overline U\subseteq A_2$. Putting $B_1=S^n\setminus U$ and $B_2=\overline U$ we have $S^n=B_1\cup B_2$ with each $B_i$ a closed set satisfying $B_i\subseteq A_i$.
Now suppose $S^n=A_1\cup\dots\cup A_{n+1}$ with each $A_i$ open. Inductively repeating the above procedure we find closed $B_1,\dots,B_{n+1}\subseteq S^n$ with $B_i\subseteq A_i$ and $S^n=B_1\cup\dots\cup B_{n+1}$. By hypothesis some $B_i$ contains an antipodal point, and hence so does some $A_i$. $\square$
$(3)\Rightarrow(1)$ Let $f:S^n\rightarrow \mathbb{R}^n$ be given and put $g(x)=f(x)-f(-x)$ so that $g(x)=0$ if and only if $f(x)=-f(x)$. Note that $g(-x)=-g(x)$ for all $x\in S^n$. To establish a contradiction we suppose that $0\not\in g(S^n)$.
Take the $n$ open sets $U_i=\{x_i>0\}\subseteq\mathbb{R}^n$, $i=1,\dots,n$, and add to this collection the set $U_{n+1}=\{x_i<0\;\text{for all}\;i=1,\dots,n\}$. Because of the supposition we have $g(S^n)\subseteq\bigcup^{n=1}_{i=1}U_i$.
Letting $A_i=g^{-1}(U_i)$ we can now write $S^n=A_1\cup\dots\cup A_{n+1}$ as a union of $n+1$ open sets. By assumption one of the sets contains an antipodal pair. But this means that there is $x\in S^n$ such that $g(x),g(-x)=-g(x)\in U_i$ for some $i$, and this is absurd. Thus we have a contradiction to the assumption that $g$ is nonvanishing. $\square$.