Is there a covering of $\mathbb{R}^3$ with pairwise skew lines?
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Ehsan M. Kermani
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3This is probably ignorant, but my gut reaction would be to define the (uncountable) set: ${\ell_{x_0y_0}=(x_0,y_0,z), z\in\mathbb R:x_0,y_0\in\mathbb R}$. Just take the line parallel to the $z$-axis at every point in the $xy$-plane. – Ian Coley Apr 24 '13 at 08:20
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@FrankMcGovern skew lines not parallel lines, http://en.wikipedia.org/wiki/Skew_lines – Ehsan M. Kermani Apr 24 '13 at 08:23
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1See http://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines – Mariano Suárez-Álvarez Apr 24 '13 at 08:32
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Ah! thank you @MarianoSuárez-Alvarez. I'm going to delete this question. – Ehsan M. Kermani Apr 24 '13 at 08:43
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Well, don't. The question I linked to does not directly answer your question; but the method given in its answer does apply to it, I think. If you manage to adapt it to your question, then add an answer. – Mariano Suárez-Álvarez Apr 24 '13 at 08:46
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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.
achille hui
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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.
David Bevan
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