TITLE: Open covering of $S^n$ by sets not containing antipodal points
QUESTION [7 upvotes]: Given an $n$-dimensional sphere $S^n$ and an open cover such that none of the open sets contain antipodal points, does there exist a point on $S^n$ that belongs to at least $n+1$ open sets from the open cover?
This is true for $S^1$, and it seems to be true for $S^2$. I am not sure if the Lusternik–Schnirelmann theorem would be useful here. I do not know too much about this field, so perhaps this problem is trivial. Any help would be appreciated!

REPLY [5 votes]: This is not true for $\mathbb{S}^3$.
Start with the central projection of 4-cube to $\mathbb{S}^3$.
Consider a covering with one set $U$ that contains all vertices and edges + one open set that is very close to the remaining 2- and 3-cells.
The multiplicity of this covering can be made to be 3.
All open sets except $U$ do not contain opposite points.
After a small perturbation, opposite points disappear in $U$.
Since the perturbation is small, the property remains valid for the remaining open set.