For the record, I misstated the proof, although the idea is correct. For the inductive step: some collection of lines have already been chosen. If you want to now choose a line through a point, the previous choices rule out lines that lie in some collection of planes, of lower cardinality than the continuum, so you can choose a plane not in that collection. The plane intersects the collection of lines in a set of points of cardinality less than the continuum, so you can find a line through that point not intersecting the previously chosen line. If you want to choose a line in a given direction, again some finite collection of planes is ruled out. The same proof works for Q^3, using ordinary induction. In this context, I'm confident it could be done constructively (although I haven't thought it through)--- so how about looking for a nice system for doing it in Q^3. It's possible it could be then generalized to a constructive method for R^3. Bill On Sep 19, 2004, at 4:13 AM, Dean Hickerson wrote:

...

Recently someone asked on sci.math.research whether R^3 could be foliated by straight lines, with exactly one line in each direction.

That was me. Unfortunately, I borrowed the word "foliate" from an earlier posting, unaware that it implied some sort of continuity.

...

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?

That's the question I was trying to ask.

Thanks to William Thurston for the proof that this follows from the axiom of choice. I wonder if there's a constructive proof.

Dean Hickerson dean@math.ucdavis.edu

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