# Filling $\mathbb{R}^3$ with skew lines

I would like to know if it is possible to fill $\mathbb{R}^3$ with lines with the following two properties:

(1) Every point $x \in \mathbb{R}^3$ is contained in precisely one line.

(2) Every neighborhood of every point is pierced by lines whose directions fill out the sphere of possible line orientations, in this sense: For every point $x$ and every $\epsilon > 0$, the lines that pass through a point in the ball $B_\epsilon(x)$ of radius $\epsilon$ centered on $x$ have the property that, were they all translated to pass through the origin, the closure of the set of points that constitutes their intersection with an origin-centered sphere $S$, fills out $S$ completely. This image below is meant to suggest the idea:

I am sure there is a more concise way to phrase the second condition; apologies for my ungainly formulation. I want to be able to find every line orientation within a neighborhood of every point.

Perhaps condition (2) is not possible to achieve in conjunction with (1). But I don't see an argument. Any ideas/insights/pointers would be appreciated—Thanks!

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The same construction as in the following question should give a YES answer. mathoverflow.net/questions/28647/… – Anton Petrunin Apr 2 '12 at 20:01
And I even supplied one of the answers to that question... Thanks, Anton! This accords with Andreas's sketch. – Joseph O'Rourke Apr 2 '12 at 20:21
Let's try to prove that there can be no Borel such partition of space into lines... – Joel David Hamkins Apr 2 '12 at 20:25
I would appreciate a definition of a Borel partition (or, a pointer to where I can learn). Thanks! – Joseph O'Rourke Apr 2 '12 at 20:29
One way to formalize what it means is that the relation of "being on the same line" for the partition, which is a binary relation on points in $\mathbb{R}^3$, would be a Borel subset of $\mathbb{R}^6$. – Joel David Hamkins Apr 2 '12 at 20:32

I have to give a lecture in a few minutes, so this will be just a quick sketch. List, in a well-ordered sequence of length $\mathfrak c$ (the initial ordinal of cardinality continuum) the requirements that (1) some line passes through $x$ (one requirement for each $x\in\mathbb R^3$) and (2) some line passes through $B$ in direction $d$ (one requirement for each open ball $B$ and direction $d$). Now go through the requirements, one at a time, and choose, for each one, a line fulfilling that requirement and disjoint from previously chosen lines. (Exception: If you get to a requirement (1) and the relevant $x$ is on a previously chosen line, skip that requirement since it's already satisfied.) I claim it's easy to check that you never get stuck, i.e., at any stage, the previously chosen, strictly fewer than $\mathfrak c$ lines, cannot block all the lines that would satisfy your current requirement.
Since this "construction" depends on well-ordering a set of the cardinality of the continuum, it will give a horrible decomposition of $\mathbb R^3$. I don't see at the moment whether this can be done "nicely", for example with a Borel partition.