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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    $\begingroup$ The same construction as in the following question should give a YES answer. mathoverflow.net/questions/28647/… $\endgroup$ Apr 2, 2012 at 20:01
  • $\begingroup$ And I even supplied one of the answers to that question... Thanks, Anton! This accords with Andreas's sketch. $\endgroup$ Apr 2, 2012 at 20:21
  • $\begingroup$ Let's try to prove that there can be no Borel such partition of space into lines... $\endgroup$ Apr 2, 2012 at 20:25
  • $\begingroup$ I would appreciate a definition of a Borel partition (or, a pointer to where I can learn). Thanks! $\endgroup$ Apr 2, 2012 at 20:29
  • $\begingroup$ 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 $\mathbb{R}^6$. $\endgroup$ Apr 2, 2012 at 20:32

1 Answer 1


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.

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    $\begingroup$ Thank you, Andreas, and I hope your lecture went well! :-) $\endgroup$ Apr 2, 2012 at 20:20
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    $\begingroup$ Despite that the title of this question contains "skew lines", the question itself does not require that the lines be skew to each other. But unless I'm mistaken, this answer can be trivially sharpened to satisfy that by requiring any chosen line to be not parallel to any previously chosen line. $\endgroup$
    – Dale
    Aug 9, 2020 at 14:38

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