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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 cannont find a "trivial" partition.

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.

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 cannont find a trivial partition.

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 cannont find a "trivial" partition.

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.

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 cannont find a trivial partition.