I've been writing some linear algebra problems with colleagues, and the following question occurred to us:
Let $S^2$ denote the unit sphere in $\mathbb{R}^3$. Does there exist a partition $S^2=\bigsqcup_\alpha B_\alpha$ such that each $B_\alpha$ is an orthonormal basis for $\mathbb{R}^3$?
The analogous question for the unit circle in the plane is easily affirmative, but that approach doesn't seem to work here.
Using the Axiom of Choice, yes you can.
To get such a partition, start by enumerating all the points of the sphere with order type $\mathfrak{c}$ (the least ordinal number with the same cardinality as the sphere): say $\langle p_\alpha :\, \alpha < \mathfrak{c} \rangle$ is such an enumeration. Now we define our partition elements one at a time, via a transfinite recursion of length $\mathfrak{c}$. At stage $\alpha$ of the recursion, suppose we've already selected, at previous stages of the recursion, some bases $\{ B_\xi :\, \xi < \alpha \}$ that will be in our partition. Now consider the point $p_\alpha$. There are two possibilities: either we already put $p_\alpha$ into one of the $B_\xi$'s for some $\xi < \alpha$, or we didn't. In the first case, we do nothing at stage $\alpha$ of the recursion: formally, we could define $B_\alpha = \emptyset$ in this case. In the second case, we choose an orthonormal basis $B_\alpha$ that contains $p_\alpha$, and that is disjoint from everything we put into our partition at an earlier stage. This is possible there is a $\mathfrak{c}$-sized collection $\mathcal C$ of orthonormal bases, any two of which intersect only in $p_\alpha$ (just rotate a basis a bit around $p_\alpha$); because $\bigcup_{\xi < \alpha}B_\xi$ contains $\leq 3 \cdot |\alpha| < \mathfrak{c}$ points, one of the bases in $\mathcal C$ contains no points from $\bigcup_{\xi < \alpha}B_\xi$. We choose some such basis to be $B_\alpha$. In the end, $\{ B_\alpha :\, \alpha < \mathfrak{c} \} \setminus \{\emptyset\}$ is the desired partition.
