Subset of $G_1(\mathbb{R}^n)$ having a line in common with every hyperplane of $G_{n-1}(\mathbb{R}^n)$ I am currently working on some problems related to Grassmann manifolds and eventually come to the following question.

Let $S$ be a subset of $G_1(\mathbb{R}^n)$ such that any element of $G_{n-1}:=G_{n-1}(\mathbb{R}^n)$ contains a line in $S.$ Is it true that there exists a hyperplane in $G_{n-1}$ that is completely covered by $S$ (that is, every line passing through the origin contained in that hyperplane belongs to $S$)?

Has this problem ever been discussed before? So far I could not come up with any good idea for it, at least not by "geometric arguments".
 A: There are many subsets $S$ with that property, and they need not cover any hyperplane.  
For instance, let $S\subset G_1(\mathbb{R}^n)$ be the set of lines $\{\mathbb{R}a\}_{a\in A}$ generated by a nonempty, symmetric (i.e. $A=-A$), connected subset $A$ of  $\mathbb{R}^n\setminus\{0\}$. For instance, a closed symmetric curve on the unit sphere (note that such a curve may meet any hyperplane in finitely many pairs of antipodal points).
The image of $A$ by any linear form on $\mathbb{R}^n$ is a non-empty symmetric interval, thus containing $0$. In other words, any hyperplane meets $A$, thus contains a line of $S$.  
A: The answer to the question is "no". It is probably best to think in terms of projective geometry: you are asking for a set $S$ of points in projective $n-1$ space such that every hyperplane contains a point in $S$, and $S$ does not contain a hyperplane. Let $S_0$ be the union of two hyperplanes $H_1$ and $H_2$ with two points $p_1\in H_1$ and $p_2\in H_2$ removed. Let $S$ be the union of $S_0$ with a point $q$ lying on the line between $p_1$ and $p_2$. Any hyperplane not meeting $S_0$ must meet $H_1$ only in $p_1$ and $H_2$ only in $p_2$, and so must contain the line between $p_1$ and $p_2$, and hence must contain $q$.
A: The answer is in the negative and a construction of a counterexample is provided here. I wrote this argument quickly, but I hope it is correct. Let me know if it is not.
We will construct a family of lines passing through the origin in $\mathbb{R}^3$ so that:
1. Every plane passing through the origin contains at least one of the lines from the family,
2. The union of the lines does not contain any plane passing through the origin.

The shortest way to do it is to use transfinite induction. While this is not an effective construction, it shows that you cannot prove your claim and this is what is important here. The argument used here is pretty similar to the one used in  
this post.

Order all planes by the initial ordinal of $2^{\aleph_0}$. Denote the planes by $P_\alpha$, where $\alpha$ is an ordinal. We can construct a corresponding family of lines $\ell_\alpha$ so that 

(a) $\ell_\alpha\subset P_\alpha$, 

(b) $\ell_\alpha\cap P_\beta=\{ 0\}$ for $\beta<\alpha$. 
Then (1) is obviously satisfied by (a). By (b), each plane $P_\beta$ can contain only lines $\ell_\alpha$, $\alpha\leq\beta$. The cardinality of such lines is less than $2^{\aleph_0}$ which is the cardinality of all lines in a plane so (2) follows.
 The construction of the family of lines $\ell_\alpha$ is pretty standard. Suppose we already have lines $\ell_\beta$, $\beta<\alpha$ satisfying (1) and (2). Consider the plane $P_\alpha$. $P_\alpha\cap P_\beta$, $\beta<\alpha$ is a line and the cardinality of such lines is less than $2^{\aleph_0}$ so there is a line $\ell_\alpha\subset P_\alpha$ which is different from all lines 
$P_\alpha\cap P_\beta$. Hence it is not contained in any of the planes $P_\beta$, $\beta<\alpha$ which is condition (2).
