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!
 A: 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.
