Suppose $A_1,A_2,A_3$ are compact sets in $\mathbb{R}^3$, use Borsuk-Ulam theorem to show that there is one plane $P$ in $\mathbb{R}^3$ that simultaneously divides each $A_i$ into two pieces of equal measures.
I believe we should use the corollary of the theorem that says "whenever $S^2$ is expressed as the union of three closed sets $A_1,A_2,A_3$ then at least one of these sets must contain a pair of antipodal points," but I have no idea how to do it, any help is appreciated.
Best Answer
Here's a bit of a rough sketch I hope works, at least in $\mathbb{R}^3$:
Each $s\in S^2$ can be identified with a unit vector in $\mathbb{R}^3$. For each $s\in S^2$, you can define $P_i^s$ to be the unique (edit: the plane actually need not be unique, but can by chosen in a systematic way as pointed out in the comments below) plane with normal vector $s$ which divides $A_i$ into two pieces of equal measure. At least in $\mathbb{R}^3$, I think it's intuitively clear such $P_i^s$ exist. Since the $A_i$ are compact, they are bounded, so by continuously sliding an affine plane along the line determined by $s$, at some point none of $A_i$ on one side of the plane, but as we slide along, the amount of measure of $A_i$ on that side of the plane increases from none to all. So by the intermediate value theorem, at some point the plane must divide the measure of $A_i$ equally.
So let $d_1\colon S^2\to\mathbb{R}$ be the continuous function for which $d_1(s)$ which measures the distance from $P_3^s$ to $P_1^s$, where the distance is positive (nonnegative, I guess) if we travel in the $s$ direction to get from $P_3^s$ to $P_1^s$, and negative (nonpositive, I guess) if we travel in the $-s$ direction to get from $P_3^s$ to $P_1^s$. Likewise define $d_2\colon S^2\to\mathbb{R}$ for the distance from $P_3^s$ to $P_2^s$.
Then define $$ \varphi\colon S^2\to\mathbb{R}^2\colon s\mapsto (d_1(s),d_2(s)). $$ This is a continuous map, and $\varphi(-s)=-\varphi(s)$ since changing the direction of $s$ changes the sign of the distance between the planes found above. By Borsuk-Ulam, there exists $s_0\in S^2$ such that $\varphi(s_0)=\varphi(-s_0)$, which means $d_1(s_0)=-d_1(s_0)$, so $d_1(s_0)=0$, and likewise $d_2(s_0)=0$. This means that the distances from $P_1^{s_0}$ and $P_2^{s_0}$ to $P_3^{s_0}$ are both $0$, which means all three planes are the same. So there is a single plane which bisects $A_1$, $A_2$, and $A_3$.