# $S^2$ covered by 6 sets - Borsuk-Ulam

Let $A, B, C \subset S^2$ be 3 disjoint closed sets. Each of these sets should not contain antipodal points. I want to prove, that the six sets $A, B, C, -A, -B, -C$ don't cover $S^2$.

(with $-M = \{ -x \; | \; x \in M \}$)

If I had only the 3 sets $A, B$ and $C$ it wouldn't be a problem (with Borsuk-Ulam), but I can't extend it to this case. Clearly the sets $-A, -B$ and $-C$ are disjoint. But $A$ could be the same as $-B$ for example.

• How does the proof go when you've only got three sets? – Michael Albanese Jan 19 '16 at 2:56
• Assume A,B,C cover $S^2$. Create the continuous map $f:S^2 \rightarrow \mathbb{R}^2 : x \mapsto (d(x,A),d(x,B))$. Then Borsuk-Ulam gives a point $x$ with $f(x) = f(-x)$. If $f(x) = (0,0)$ then $x, -x$ lie in A (and B). If not they lie in C. – Boris Jan 19 '16 at 3:02
• @MichaelAlbanese: Or simpler: $S^2$ is connected so it cannot be the disjoint union of finitely many closed sets unless one of those sets is the entire space. But then this set must obviously contain antipodes. – Henning Makholm Jan 19 '16 at 3:08

In the comments, you've already provided a proof via Borsuk-Ulam that three antipode-free closed sets cannot cover $S^2$.

Now note that $A \cup -B$ is antipode-free, since $A$, and $B$ are each antipode-free and disjoint. Similarly, $B \cup -C$ and $C \cup -A$ are each antipode-free. Thus the union of these three sets, which is the union of the six sets in question, cannot be $S^2$ by Borsuk-Ulam.

I'll prove something stronger. $A$, $B$, and $C$ don't need to be pairwise disjoint; we only need $A\cap B\cap C$ to be $\emptyset$.

Consider the map: \begin{align} f:S^2&\to\Bbb R^2\\ x&\mapsto\left(\frac{d(x,A)}{d(x,A)+d(x,B)+d(x,C)},\frac{d(x,B)}{d(x,A)+d(x,B)+d(x,C)}\right) \end{align} This is defined, since the denominators are never zero; remember that $A\cap B\cap C=\emptyset$.

By Borsuk-Ulam, there is a point $x\in S^2$ such that $f(x)=f(-x)$.

If the first (or second) coordinate of $f(x)=f(-x)$ is zero, then $x$ and $-x$ are both in $A$ (or $B$). This is impossible, since $A$ and $B$ don't contain antipodal points.

If the coordinates of $f(x)$ add up to $1$, then $d(\pm x,C)=0$, and both $x$ and $-x$ are in $C$. This is impossible, since $C$ doesn't contain antipodal points.

The only possibility is that the coordinates are nonzero and don't add up to $1$. This means that $x$ and $-x$ are both not in any of the three sets. That would imply that $x$ is not in $A$, $B$, $C$, $-A$, $-B$, or $-C$, which means that they do not cover $S^2$. QED

Note that we need the condition $A\cap B\cap C=\emptyset$. If the triple intersection is allowed to be nonempty, the theorem is false. For example, consider dividing the northern hemisphere into three equal parts, letting three lines of longitude (the vertical ones) be the borders between the sets. Then the three sets intersect at the north pole, and $A$, $B$, $C$ and their opposites cover the sphere.

• (You could try using lines of latitude to divide the northern hemisphere into three sets, but then the lowermost set has antipodal points along the equator.) – Akiva Weinberger Jan 21 '16 at 21:34