Let E(3) be 3-dimensional Euclidean space. I have submitted the following question to Mathstackexchange and other mathematical websites, but have never received any responses-not even rejections on the grounds that it was inappropriate. If C is a subset of E(3) that is a simple closed curve, does C always contain 4 pairwise distinct points which are co-planar? Could the question-at this late date-still be open? It seems to me intuitively clear that the answer should be "yes" (even though I have not been able to prove it). I have seen theorems in the literature stating that if C is "smooth", then it always contains the vertices of a parallelogram or some other plane quadrilateral. But this is the closest I have found to the answer I am seeking.....Incidentally, I think I have a counter-example which proves that if C is homeomorphic to a straight line subset of E(3), then the answer to my question is negative. So I assume that this is well-known, although I have not run into any mention of it in the literature.



A question about simple closed curves in 3-dimensional Euclidean space
Take $C(0), C(1/4), C(1/2), C(3/4)$. If they are coplanar, then we are done. If they are not, they should form a tetrahedron, and we can choose a plane $X$ which completely divides the four points so that $C(0), C(1/2)$ lie on one side, and $C(1/4), C(3/4)$ lie on the other side. Then in each interval $(0, 1/4), (1/4, 1/2), (1/2, 3/4), (3/4, 1)$ there is a point on $X$. 


