Let n be a positive integer not less than 2. Does anyone know of a theorem stating that- for each n- there exists a simple closed curve c(n), which (1) is a subset of n-dimensional Euclidean space E(n) and (2) does not contain n+1 pairwise distinct points all belonging to the same (n-1)-dimensional Hyperplane of E(n)? For n=2 there is such a theorem. This theorem is almost "trivial". We have merely to picture the circumference of a circle. I am asking whether, and if so, to what extent this "trivial" theorem has been generalized. What is the situation when n is greater than 2?



A question about simple closed curves in finite dimensional Euclidean spaces
Extending my comment, I can give an almost complete answer. If $n$ is even, an example is an appropriate normal rational curve, e.g., $t\mapsto(t/p,t^2/p,\ldots,t^n/p)$, where $p(t)$ is any real polynomial of degree $n$ and without real roots. I conjecture that, for $n$ odd, such a curve doesn't exist. Here is a proof for $n=3$. The curve is obviously not planar. Consider the pencil of planes through a generic tangent. Apart from the tangency point, a generic plane intersects the curve at an even number of points, hence, one can find one with at least two points of intersection. Now, move this plane slightly off the tangency point to get $4$ coplanar points on the curve. I think that this proof can be generalized to any $n$ by considering a pencil of sufficiently osculating hyperplanes.


**Added:** My proof for $n=3$ was too complicated. Here is a cleaner proof for any odd $n$, not using any smoothness. Take $n$ generic points on the curve and consider the hyperplane $H$ through these points. For homological reason, the intersection of $H$ and the curve must be $0\bmod2$; hence, there is one more point.


