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.



Finite sets of vectors closed under an orthogonality property
We identify opposite vectors (or we may factor $S^{2}$ by $\{\pm 1\}$). 


We claim that all finite ortho-closed sets on $S^2$ are coplanar after removing one vector. (I think the same argument works in higher dimension as well.)


Firstly, one useful property:  *If $a,b,c\in X$, where $c$ is not orthogonal to $a$, then the (normalized) projection of $c$ onto the plane spanned by $a$ and $b$ also lies in $X$.* Indeed, $X$ contains the vector $x\perp a,b$; next, it contains the vector $y\perp x,c$; finally, it contains $z\perp x,y$, and $z$ is the required projection.


Now, starting from any two vectors $a,x\in X$, we find in $X$ the vectors $b\perp a,x$ and $c\perp a,b$, thus $X$ contains an orthogonal triple $a,b,c$. If all vectors from $X$ distinct from $a,b,c$ lie in exactly one of the planes $\langle a,b\rangle$, $\langle a,c\rangle$, and $\langle b,c\rangle$, then our claim is true. Otherwise, using the property, we find two such planes containing some other vectors from $X$. We ay assume that these vectors are $u\_0\in\langle a,b\rangle$ and $v\_0\in\langle a,c\rangle$. Notice that $a$ forms acute angles with $u\_0$ and $v\_0$.


Now, let $a\_0$ be the projection of $a$ onto $\langle u\_0,v\_0\rangle$, and let $u\_1$ and $v\_1$ be the projections of $a\_0$ onto $\langle a,b\rangle$ and $\langle a,c\rangle$, respectively. Then $\angle (a,u\_1)<\angle (a,u\_0)$ and $\angle(a,v\_1)<\angle(a,v\_0)$. Repeating the process we find an infinite number offvectors in $X$.


