Existence of a finite subset of the unit sphere with a particular orthogonality property

Let $$\mathbb{S}^2$$ be the unit sphere in $$\mathbb{R}^3$$. Claim: There exists no finite subset $$A$$ of $$\mathbb{S}^2$$ such that for every $$x \in A$$ there are at least eight elements in $$A$$ that are orthogonal to $$x$$. Question: Is this claim correct, and how can it be proved?

(I have tried many different constructions of such an A, both analytically and numerically, and they all failed, so I am confident that the claim of non-existence is correct, but I am not able to prove this. Even a pointer to relevant related literature would be appreciated.)

• What about $\{e_1,e_2,e_3,-e_1,-e_2,-e_3\}$ where $e_1,e_2,e_3$ is an orthonormal basis? Oct 2 '20 at 16:56
• Oh, damn, I simplified my problem too much. I am not allowed to distinguish between x and -x. I need to edit my question to explain that. Thank you very much! Oct 2 '20 at 17:00
• Ok, so instead of explaining that I am working with a projective space (where $x$ is identified with $-x$) I decided to just change "at least four elements" into "at least eight elements". Thank you, again, Berci, for immediately picking up on that. Oct 2 '20 at 17:02
• +1 Very interesting question! Some simple observations: If $A$ and $B$ are such subsets of $\mathbb{S}^2$, then so is $A\cup B$, and so is $O(A)$ for any orthogonal transformation $O$. In particular $-A$ is such a subset, so without loss of generality we may assume that $A=-A$, i.e. that for every $x\in A$ also $-x\in A$. This brings the problem back to projective space... Oct 7 '20 at 22:23
• Moreover, if $A$ is such a subset satisfying $A=-A$, let $B=A/\{\pm1\}$ be the set of antipodal pairs in $A$, and $G$ the graph on $B$ whose edges are the orthogonal pairs in $B$. Then $\deg v\geq4$ for all $v\in G$, and any two vertices in $G$ have at most one common neighbour, i.e. $G$ contains no $4$-cycles. Oct 7 '20 at 22:28