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

2012-04-02T19:45:34Z

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!

Answer by Andreas Blass for Filling $\mathbb{R}^3$ with skew lines

2012-04-02T20:05:04Z

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.

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

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

Answer by Andreas Blass for Filling $\mathbb{R}^3$ with skew lines