How does one prove that union of skew lines can be $\mathbb{R}^3$ using vector calculus? Space-filling curve methods are available, but i would like to know the method using vector calculus, as I heard that the method exists.
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Could you be more specific? $\mathbb R^3$ is, clearly, the union of all lines. – azarel Sep 24 '13 at 01:54
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I just want to show that $\mathbb{R}^3$ can be a union of some skew lines. Isn't this all that can be provided? – Skew Sep 24 '13 at 01:59
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That's clearly true. For each $x\in\mathbb R^3$ pick a skew line $\ell_x$ containing $x$ then $\mathbb R^3=\bigcup{\ell_x: x\in\mathbb R^3}$ – azarel Sep 24 '13 at 02:06
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@azarel What guarantees the lines are pairwise skew? Such a set can be constructed using the axiom of choice but that's likely not what the OP is looking for. – user7530 Sep 24 '13 at 02:07
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@user7530 I don't see any restriction imposed on the set of lines. That's why I asked for clarification. – azarel Sep 24 '13 at 02:10
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An explicit construction has been given in this answer. It uses complex numbers to help labeling the lines. – achille hui Sep 24 '13 at 02:19
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The key idea is to foliate $\mathbb{R}^3$ with (one-sheet) hyperboloids. Each hyperboloid is ruled with skew lines, and by construction rulings of different hyperboloids aren't parallel and don't intersect.
See here for discussion of this problem: https://mathoverflow.net/questions/1194/how-to-partition-r3-into-pairwise-non-parallel-lines/3714#3714
