# How bogus is the glitzy proof of Borsuk-Ulam?

Suppose $f: S^2 \rightarrow {\bf R}^2$ is continuous; let $A$ be the set of points $u \in S^2$ such that $f(u)-f(-u) \in {\bf R} \times \{0\}$ (where $-u$ denotes the antipode of $u$). Given $u,-u \in A$, must there exist a path in $A$ joining $u$ to $-u$?

The appealing argument presented at https://www.youtube.com/watch?v=csInNn6pfT4 (which I have also seen elsewhere), which purports to prove the existence of a $u \in S^2$ with $f(u)-f(-u) = (0,0)$, hinges on this unstated assumption.

Can anyone devise an $f$ for which $A$ is something like a topologist's sine curve?

ADDED LATER: I like Ilya Bogdanov's example, and I wonder whether we can improve it. Is there an $f$ such that no point in $A$ can be joined to its antipode by a continuous curve?

• He also states that stirring a cup of coffee must have a fixed point because it represents a continuous transformation, but that doesn't seem right either. – Jim Conant Oct 11 '16 at 21:57
• @Sorry for the stupid question, but why not? – Billy Rubina Oct 12 '16 at 5:51
• @OppaHilbertStyle: As a thought experiment: imagine inserting a divider into a coffee cup: it divides the coffee into two pieces, so can't be a continuous operation. There's really nothing conceptually different about inserting a spoon into the coffee, even if you can't prove it is a discontinuous operation using such a simple invariant as connectivity. – Jim Conant Oct 12 '16 at 21:44
• @OppaHilbertStyle: yes, forget the discreteness and just look at my thought experiment. – Jim Conant Oct 13 '16 at 3:55
• @OppaHilbertStyle: If you model the coffee as a $3$-ball and stirring as a continuous map $f:B^3\to B^3$, you also get some weird things. In particular (assuming none of the coffee sticks to the spoon), after stirring the coffee takes up the same space as before, so $f$ is a surjection. Also, no two points of the coffee get glued together, so $f$ is injective. Since $B^3$ is compact, $f$ is also closed, hence a homeomorphism. But then $f$ maps the boundary $S^2$ to $S^2$, which seems wrong. I think that when you stir the coffee, you expect the points at the top to mix with the interior points. – Dejan Govc Oct 13 '16 at 7:58

Let $S^2$ be the unit sphere in $\mathbb R^3$, $S^2_+=\{(x,y,z)\in S^2\mid z\geq0\}$ the upper hemisphere, $S^2_-$ the lower hemisphere and $B^2$ the closed unit disk in $\mathbb R^2$. Define $p:S^2\to B^2$ by $p(x,y,z)=(x,y)$. Then the restrictions of $p_+:S^2_+\to B^2$ and $p_-:S^2_-\to B^2$ of $p$ are homeomorphisms. Define a copy of the topologist's sine wave in $B^2$ by $$W=\overline{\{(x,y)\in B^2\mid x\neq 0\text{ and } y = \tfrac12\sin\tfrac{\pi}x\}}.$$ Then, $K=p_+^{-1}(W)\cup p_-^{-1}(W)$ is a closed subset of $S^2$ with the following properties:

1. $S^2\setminus K$ has two components $U_+$ and $U_-$ and the antipode of each point in $U_+$ lies in $U_-$ and vice versa.
2. Antipodes of points in $K$ lie in $K$, but there is no path in $K$ between any such pair of antipodes.

Now consider $f:S^2\to\mathbb R^2$ defined by $$f(x,y,z)=(0,\pm d((x,y,z),K)),$$ where $d$ is e.g. the Euclidean metric on the sphere and the sign is $+$ for $(x,y,z)\in U_+$ and $-$ for $(x,y,z)\in U_-$.

This function $f$ has all the desired properties: it is continuous and the set of points $u=(x,y,z)$ such that $f(u)-f(-u)\in\mathbb R\times\{0\}$ is precisely $K$, because for any $u\in S^2\setminus K$, the second components of $f(u)$ and $f(-u)$ have different signs.

Let $S^2=\{(x,y,z)\colon x^2+y^2+z^2=1\}$, and set $f(x,y,z)=(x,\sin 1000x)$. Then the set $A$ consists of many components, and only one of them contains opposite points.

You may choose whatever more interesing function instead of $\sin 1000x$...

• Nice! Can this example be modified so that the component that contains opposite points is not path-connected? – James Propp Oct 11 '16 at 22:09