As title says, can a union of lines that are not paralell or perpendicular to each other be $\mathbb{R}^3$? The number of lines does not matter. It may be countable or uncountable.
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2What do you mean when you say "fill up"? – ymbirtt Sep 23 '13 at 11:19
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union, I am saying. Edited my question. – Euclid Sep 23 '13 at 11:22
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1It may not be countable. – njguliyev Sep 23 '13 at 11:24
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OK, so what about possibility of union being $\mathbb{R}^3$? – Euclid Sep 23 '13 at 11:36
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Does 'perpendicular' imply that the lines intersect? Or does it mean perpendicular after translation? – Simon Markett Sep 23 '13 at 11:40
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I meant intersect. Sorry for causing confusion. – Euclid Sep 23 '13 at 11:49
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Can anyone provide vector calculus reasoning? – Euclid Sep 23 '13 at 12:26
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1@Euclid: Your original -- perhaps unclear -- question appears to have been adequately answered by (another) Arthur below. To preserve this answer I have reverted the question back to its previous version. Feel free to ask your indended question as a new question. (Incidentally, now by "intersecting" to you mean just intersecting, or orthogonal and intersecting? Be careful when phrasing your next question. Cheers!) – user642796 Sep 23 '13 at 17:03
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There is an injection from the set of points in $\Bbb R^3$ to the set of directions contained in the (open) first octant, for instance, via space-filling curves. (Fill the space with a curve, this gives a bijection from $\Bbb R^3$ to $\Bbb R$. Now inject that real line into the space $\{(\theta, \phi) | 0 < \theta, \phi < \pi/2\}$, consisting of directions for the lines in spherical coordinates.)
No two such directions are orthogonal, and every point gets a line through it with a unique direction, so it turns out you can fill the space with lines neither parallel nor perpendicular.
Arthur
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