I am currently working on some problems related to Grassmann manifolds and eventually come to the following question.



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> 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 $S$)?
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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".



Subset of $G_1(\mathbb{R}^n)$ having a line in common with every hyperplane of $G_{n-1}(\mathbb{R}^n)$
**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](https://mathoverflow.net/questions/173387/can-r3-be-expressed-as-a-disjoint-union-of-pairwise-linked-circles) 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).














