A Problem  about partitioning $S^2$ Question: Can the 2-dimensional sphere $S^2$ be partitioned into four nonempty sets such that every circle in $S^2$  passes through just three  of these four sets?
Here, "just three" means "exactly three", circle is in the ordinary sense, i.e. round circle (or say 1-sphere), not necessarily great circle. If "just three" means "at most three" as Alexandre Eremenko supposed, then the answer is yes (there are a lot of examples). For example, see the example just given by Lee Mosher.
EDIT. I prove that if such partitioning exists, then these four subsets are all dense in $S^2$. So, it seems that we cannot find a "trivial" partition.
 A: I suppose you mean circles in the literal sense, round circles.
I suppose "just three" means "at most three". The answer is "yes".
Let us identify your sphere with the extended
complex plane via stereographic projection, so that your
circles are straight lines or circles. Consider the following sets $A,B,C,D$.
Let $D=\{ \infty  \}$, $C=\{ 0 \}$, $A$ the set of points with argument
commensurable with $\pi$,
and $B$ the set of points with argument non-commensurable with $\pi$.
These sets partition the extended plane.
If a circle does not contain $D$, it intersects at most 3 sets.
If it contains $D$, it is a straight line. If this straight line
does not contain $C$, it intersects at most 3 sets.
If it contains $C$, it intersects either $C,A,D$ or $C,B,D$ but not all 4.
Sorry, do not know how to make curly braces {} with Mathjack.
A: First partition the equator into three nonempty sets such that every antipodal pair intersects exactly two of them. Then take the fourth set to be the complement of the equator.
