Cover $\mathbb{R}^3$ with skew lines 
Is there a covering of $\mathbb{R}^3$ with pairwise skew lines?

 A: For each $\omega = x_0 +  y_0 i\in \mathbb{C} \simeq \mathbb{R}^{2}$. Consider the set $l_{\omega} \subset \mathbb{R}^3$ defined by:
$$l_{\omega} = \left\{ (x,y,z) \in \mathbb{R}^{3} : \frac{x + iy}{1 + iz} = \omega \right\}$$
Since the map $(x,y,z) \mapsto \frac{x+iy}{1+iz}$ is defined over all $\mathbb{R}^3$, we have: 
$$\mathbb{R}^3 = \cup_{\omega \in \mathbb{C}}\;l_{\omega}\quad\text{ and }\quad
l_{\omega_1} \cap l_{\omega_2} = \emptyset \;\text{ for distinct }\;\omega_1, \omega_2 \in \mathbb{C}$$ 
Furthermore, each $l_{\omega}$ is a straight line as it admits a linear parametrization:
   $$\mathbb{R} \ni t \mapsto ( x_0 - y_0 t,\, y_0 + x_0 t,\, t ) \in l_{\omega}$$
With this parametrization, one can see that its tangent vector is along the direction
proportional to $(-y_0, x_0, 1)$.
This means $l_{\omega_1} \not\parallel l_{\omega_2}$ for distinct $\omega_1, \omega_2 \in \mathbb{C}$.
In short, $\{ l_{\omega} : \omega \in \mathbb{C} \}$ is a partition of $\mathbb{R}^3$ into a family of lines in which all pairs are skew.
A: Take some injection $\varphi:\mathbb{R}^3\rightarrow\mathbb{R}$ (e.g. interleaving of binary expansions). Now orient a line through each point $(x,y,z)$ with direction $(\cos\theta,\sin\theta,0)$ where $\theta=\tan^{-1}\varphi(x,y,z)$.
This gives a covering of $\mathbb{R}^3$. The answer to this mathoverflow question can easily be modified to give a partition.
