TITLE: Finite sets of vectors closed under an orthogonality property
QUESTION [12 upvotes]: Say that a set $X \subset \mathbf{S}^{d-1}$ is ortho-closed if for any set $\{x_1,\dots,x_{d-1}\} \subset X$ there exists $x \in X$ such that $\langle x,x_i \rangle = 0$ for $i=1,\dots,d-1$. (Apologies for the terrible name - other suggestions are welcome.) Thus, as a silly example, the set of standard basis vectors is ortho-closed.
My rather vague question is, what (non-trivial) finite ortho-closed sets are there? Much more concretely (and actually what I'm most interested in), with $d=3$, is there a finite ortho-closed set containing the normalizations of the seven non-zero 0-1 vectors in $\mathbb{Z}^3$? (Or indeed is there a finite ortho-closed set containing an arbitrary fixed set?)
It seems to me that the answer ought to be `no', and that one ought to be able to construct an infinite sequence of vectors that would have to be in any such set. However I can't yet see how to do this, and it occurred to me that this is a fairly natural property which might already be known. Thus, any references would also be greatly appreciated.

REPLY [8 votes]: You can get a nice combinatorial proof of Ilya's result if you use the Friendship Theorem.
As Ilya did, we can identify opposite vectors. Your condition says that for any two distinct vectors $v_1$ and $v_2$ of $X$, there exists a vector $v \in X$ which is orthogonal to both $v_1$ and $v_2$. However, since you are in $\mathbb{R}^3$, the vector $v$ must be unique.
Now let us consider the vectors in our set $X$ to be vertices of a graph $G$, such that two vectors are adjacent if they are orthogonal. Then, by the above, this graph has the property that every pair of vertices has a unique common neighbor. The Friendship Theorem says that any such graph is a windmill graph, meaning that $G$ consists of a vertex $v$ adjacent to all other vertices of $G$, and the remaining vertices form a perfect matching.
Translating the orthogonalities encoded in the edges of $G$ back to the vector setting gives you exactly Ilya's result.