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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.

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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.

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  • $\begingroup$ For the non projected version, let C and D be (singleton sets of) antipodal points, and then partition the (pairs of) great circle arcs between them into two sets. Gerhard "Wish I Thought Of That" Paseman, 2012.12.28 $\endgroup$
    – Gerhard Paseman
    Commented Dec 28, 2012 at 22:07
  • $\begingroup$ You can do the curly braces by \lbrace and \rbrace in math mode. I learned this by right-clicking math in others' posts. You can literally see how they typed their math this way. $\endgroup$
    – Yuichiro Fujiwara
    Commented Dec 28, 2012 at 22:09
  • $\begingroup$ Further, if C had more than one point, I suspect there is no nontrivial two coloring of the remainder that would avoid a circle with four colors. Gerhard "Not A Geometric Ramsey Theorist" Paseman, 2012.12.28 $\endgroup$
    – Gerhard Paseman
    Commented Dec 28, 2012 at 22:16
  • $\begingroup$ With Alexandre's example in mind, I think a painting argument shows that any such coloring must have two of the colors limited to singleton sets. Looking at the circles passing through D and C, they determine a pencil of pairs of arcs which must be monochromatic off of the C and D regions, and then D and C' (C' also having color of C) determines a distinct pencil from the first (excepting one pair of arcs). It then becomes clear that any connected open subset not colored by C or D is monochromatic. Gerhard "Some Details Remain, Of Course" Paseman, 2012.12.28 $\endgroup$
    – Gerhard Paseman
    Commented Dec 28, 2012 at 22:40
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    $\begingroup$ Time to eat my words. Pick a sphere colored A, then pick a circle on it and color it B. Then pick three points on the circle and color two of them C and one of them D. Any circle that has D, C, and B points must avoid the A color. This example can be generalized, but I don't see any set but A having measure greater than zero (with respect to the measure of spherical area). Gerhard "Please Pass The Salt Shaker" Paseman, 2012.12.28 $\endgroup$
    – Gerhard Paseman
    Commented Dec 28, 2012 at 22:56
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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.

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  • $\begingroup$ Maybe you didn't understand my question. Under your partition, every circle in the the complement of the equator (i.e. the fourth set) passes through the fourth set only. $\endgroup$
    – woodbass
    Commented Dec 28, 2012 at 18:31
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    $\begingroup$ I took it to mean a great circle. If not, do you mean round circles? Topologically embedded circles? $\endgroup$
    – Lee Mosher
    Commented Dec 28, 2012 at 19:24
  • $\begingroup$ Why not partition some circle S into 3 arbitrary non-empty sets, and the complement of S is the 4-th set. Let C be any round circle. It intersects S at at most 2 points, or coincides with S. So it intersects at most 3 sets. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 29, 2012 at 17:03
  • $\begingroup$ Indeed,as I observed also in a comment to your solution, Alexandre. Your solution has the feature that at least two colors occur on every circle though, which is not the case for this last suggestion. Gerhard "Ask Me About System Design" Paseman, 2012.12.29 $\endgroup$
    – Gerhard Paseman
    Commented Dec 30, 2012 at 6:28
  • $\begingroup$ @Alexandre Eremenko: There is a mistake in your construction. The line containing C and A passes through both B and D. $\endgroup$
    – woodbass
    Commented Jan 12, 2013 at 18:46

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