This is the kind of thing I don't actually believe in, but doesn't this "prove itself" by transfinite induction if you assume the axiom of choice? If you also assume the continuum hypothesis, you can choose a 1-1 correspondence between all points in R^3 union all directions and the set of all countable ordinals. Look inductively at the objects. If it's a point not yet included, there's at least one line whose direction isn't yet included and that doesn't pass through any points yet included, since you only have to avoid a countable set of points and directions. Choose some such line. Similarly, if its a direction not yet included, find some line in that direction not passing through any point yet included. This works much the same without assuming the continuum hypothesis, since at any stage in the induction the set of points or directions already used has cardinality less than the continuum, so there are some left over. Amusing: What happens when you look at Z^k or Q^k instead of R^3? Bill On Sep 18, 2004, at 5:00 PM, Daniel Asimov wrote:

Recently someone asked on sci.math.research whether R^3 could be foliated by straight lines, with exactly one line in each direction.  Someone else soon provided a proof that this is impossible (using homotopy theory).   If we drop the continuity requirement for a foliation, the problem becomes this one:   QUESTION:  Is R^3 the disjoint union of [bi-infinite] straight lines L_d such that each direction d occurs exactly once?    (I.e., each line through the origin of R^3 is parallel or equal to exactly one of the L_d's.)   My guess is yes, but it's just a guess.   --Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun