Is there a set of lines in $\mathbb{R}^3$ such that their union is equal to the set of points with not all coordinates rational? Does there exist a set L of pairwise disjoint lines, each line in an axis direction, in $\mathbb{R}^3$ such that their union is equal equal to $\{ x : x \text{ has at least one irrational coordinate}\}$?
 A: The answer is negative. Indeed, assume the converse. Let $P_0$ be a plane $z=\sqrt{2}$ and $l_0\in L$ be a line containing a point $(0,0,\sqrt{2})$. Then $l_0$ is contained in $P_0$. Each point $p\in P_0\setminus l_0$ is contained in some line $l_p\in L$. Since the lines $l_0$ and $l_p$ does not intersect, there are only two possibilities for direction of the line $l_p$: $l_p$ is parallel to the line $l_0$ or $l_p$ is perpendicular to the plane $L_0$. If for all points $p\in P_0\setminus l_0$ the first possibility holds then for all lines of the family $L$ the direction perpendicular to the plane $L_0$ is forbidden. If there exists a point $p_0\in P_0\setminus l_0$ for which holds the second possibility then this possibility also holds for all points from the line $l_p\subset P_0$, going through the point $p$ and parallel to the line $l_0$. These lines fill a plane $P_p$. Thus for all lines of the family $L$ the direction perpendicular to the plane $P_0$ is forbidden. Let $P$ be any plane perpendicular to the forbidden direction and containing a point which all coordinates are rational. The direction restriction implies that there is a direction $d_0$ such that any line $l\in L$, intersecting $P$, has direction $d_0$ too. So the set $(\bigcup L)\cap P$ is a union of parallel lines. But this set is a copy of a set $\Bbb R^2\setminus \Bbb Q^2$, which is not a union of parallel lines, a contradiction.
PS. Maybe the answer for a counterpart of this question for $\Bbb R^n$ is negative too.
