Is it possible to represent $\mathbb{R^{3}}$ as a union of countable non-intersecting and non-coplanar lines?
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1See here: http://math.stackexchange.com/questions/371302/cover-mathbbr3-with-skew-lines – hot_queen Jan 23 '14 at 16:23
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No, they have measure 0, thus so does their countable union. Whereas $\mathbb R^3$ has infinite measure.
Also by the baire category theorem a complete metric space cannot be written as the countable union of closed, nowhere-dense sets and lines are nowhere dense in $\mathbb R^3$.
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5How about if I don't put the restriction to the union to be countable? – Konrad Szałwiński Jan 22 '14 at 23:59
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Unsure. My intuition says no but no proof comes to mind. Perhaps it is possible. – Sempliner Jan 23 '14 at 10:36