Hatcher Exercise 9, Chapter 1, using Borsuk-Ulam's theorem 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.  
 A: Yeah I have been trying at this problem for a while. Let me tell you my progress. Basically, I can prove it for one subset.
Define a map from $S^2 \to \mathbb{R}^2$ as follows. For each point $p$ in the sphere draw its tangent plane through the origin. Then let $x$ be the measure of the subset $A$ above the plane (where above is given by the direction of the normal vector) and $y$ be the measure of $A$ below the plane. Then the map sends $p$ to $(x,y)$. Borsuk-Ulam implies that $x=y$ (if you believe this map is continuous, which is not hard to prove anyway).
Notice we have a degre of freedom to play with because we imposed the condition that the plane goes through the origin. Hope it helps! Please let me know if you make any progress. Cheers!
A: 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 $A_3$.
A: BW's got the correct way, and here I add some details and mathematical rigor. $m$ is the Lebesgue measure on $\mathbb R^3$.
We can assume that $m(A_i)$ are all positive for $i=1,\,2,\,3$. If, say, $A_1$ and $A_2$ have measures $>0$ and $m(A_3)=0$, then we solve the problem for the three sets $A_1$, $A_1$ and $A_2$. We get a plane equally bisecting the measures of $A_1$ and $A_2$. And indeed this plane bisects the measure of $A_3$ equally: zero and zero.
So here's the outline (same as BW's but notations):
Let $v$ be an element of $S^2$, the unit sphere in $\mathbb R^3$. Let $t$ be a real number. Let $H^v_t$ be the set of all $x \in \mathbb R^3$ such that $x\cdot v < t$, where the dot means the Euclidean inner product.
Define $g_{i,\,v}(t)$ to be $m(H^v_t \cap A_i)/m(A)$. Then it is clear that $g_{i,\,v}$ is a monotonically increasing function $\mathbb R \to \mathbb R$. The compactness of $A_i$ ensures $g_{i,\,v}$ is real-valued. We will prove soon that $g_{i,\,v}$ is continuous, and that the set $g_{i,\,v}^{-1}(1/2)$ is a closed, bounded interval.
We define $f_i(v)$ to be the minimum of this interval $g_{i,\,v}^{-1}(1/2)$. Let $\hat f_i(v)$ denote the maximum of $g_{i,\,v}^{-1}(1/2)$. Generally, $f_i(v)$ and $\hat f_i(v)$ does not equal; think of the case that $A_1$ is a union of two disjoint closed balls. Anyway, $f_i$ and $\hat f_i$ are in fact continuous functions of $S^2$ into $\mathbb R$, which will be proved soon. Furthermore, we have $f_i(-v) = -\hat f_i (v)$ for all $v \in S^2$.
Finally, consider a function $h:S^2 \to \mathbb R^2$ given by $h = (f_2 - f_1 ,\, f_3 - f_1 )$. Use Borsuk-Ulam theorem with $h$ to find $v\in S^2$ for which $h(v)=h(-v)$, that is,
$$ f_2(v) - f_1(v) = f_2(-v) - f_1(-v) \,\,\,\, \text{and} \,\,\,\, f_3(v) - f_1(v) = f_3(-v) - f_1(-v). $$
By the relation between $f_i$ and $\hat f_i$, we know that
$$ f_1(v)+\hat f_1(v) = f_2(v)+\hat f_2(v) = f_3(v)+\hat f_3(v). $$
Let $c$ be the common value above. Then $c \in g_{1,\,v}^{-1}(1/2)\cap g_{2,\,v}^{-1}(1/2)\cap g_{3,\,v}^{-1}(1/2)$. Hence we conclude that the wanted plane is $x\cdot v = c$.
Now we dig into the details. For brevity, we drop the index $i$ from all expressions. Fix $v \in S^2$, and we show first that $g_v$ is continuous. The compact set $A$ is bounded in $\mathbb R^3$, so it is contained in a ball $B$ centered at $0\in \mathbb R^3$. Let $r$ be the radius of $B$. Let $C_v$ be the solid cylinder which is parallel to $v$ and tangent to $B$. Then $A$ is contained in $C_v$.
Let $\epsilon$ be a positive number. Choose a positive number $\delta$ less than $m(A)\epsilon / \pi r^2$. If two real numbers $t_1$ and $t_2$ satisfy $t_1 \le t_2$ and $t_2 -t_1 < \delta$, then we have
$$ g_v(t_2) -g_v(t_1) = m((H^v_{t_2} - H^v_{t_1}) \cap C_v \cap A) / m(A) \le \pi r^2 \delta / m(A) < \epsilon. $$
This shows that $g_v$ is uniformly continuous.
$A$ is compact, so if we choose a very large positive number $t$, then $g_v(-t)=0$ and $g_v(t)=1$. By the monotonicity and continuity of $g_v$, the set $g_v^{-1}(1/2)$ is closed, bounded, connected subset of $\mathbb R$: a closed interval.
Now we show that $f$ is continuous. Let $\theta$ be a real number between $0$ and $\pi/2$. Let $v=(v_1,\,v_2,\,v_3)$ and $w=(w_1,\,w_2,\,w_3)$ be elements of $S^2$ which satisfy $\cos ^{-1} (v\cdot w) = \theta$. We prove that
$$ f(v) \cos \theta - r \sin \theta \le f(w) \le f(v) \cos \theta + r \sin \theta, $$
which implies that if $w \to v$, then $f(w) \to f(v)$. Note that if $w$ tends to $v$, then $\theta$ tends to $0^+$.
We need to show the only case when $v$ is $(0,\,0,\,1)$. What is left is done by isometry of $\mathbb R^3$. Anyway in this case, $w_3 = \cos \theta$ and $\sin \theta = \sqrt{w_1^2 + w_2^2}$. Denote by $x=(x_1,\,x_2,\,x_3)$ an arbitrary point in $\mathbb R^3$. We have to show, in the cylindrical domain $x_1^2 + x_2^2 \le r^2$, that the plane $x_3 = f(v)$ lies between the planes $P^+_w$ and $P^-_w$, where
$$ P^\pm_w : \, w_1 x_1 + w_2 x_2 + w_3 x_3 = f(v) \cos \theta \pm r \sin \theta. $$
By the Cauchy-Schwarz inequality we have
$$ | w_1 x_1 + w_2 x_2 | \le \sqrt{w_1^2 + w_2^2}\sqrt{x_1^2 + x_2^2} \le r \sin \theta. $$
So the equation for $P^+_w$ implies $w_3 x_3 \ge f(v) \cos \theta$. $P^-_w$ implies the direction of inequality inverted. But by the range of $\theta$, the number $w_3 = \cos \theta$ is positive, therefore the desired result.
The continuity of $\hat f$ comes from the equation $f(-v) = -\hat f(v)$.
