Is $R^3$ a union of disjoint closed curves? (Obviously $R^3$ minus a line is). Is this a classical problem?


Yes, $R^3$ is the union of disjoint circles. Not just homeomorphs of circles, but actual circles. The following simple construction is quoted from M. Jonsson and J. Wästlund, "Partitions of $R^3$ into curves", Math. Scand. 83 (1998), 192-204 (see Theorem 1.1), where it is attributed to A. Szulkin, "$R^3$ is the union of disjoint circles", Amer. Math. Monthly 90 (1983), 640-641. Here is a link to the Jonsson-Wästlund paper: http://www.mscand.dk/article.php?id=77 You can find out about other results like this, some of them depending on the axiom of choice, by reading that paper.

It is easy to see that it is possible to partition a two-punctured sphere into disjoint circles. Let $(x,y,z)$ be coordinates in $R^3$ and consider the union $C$ of the circles of radius $1$ in the $(x,y)$-plane centered at the points $(4k+1,0,0), k\in Z$. Then any sphere centered at the origin intersects $C$ in exactly two points. By covering each such sphere with disjoint circles, we obtain the desired partition.

Here is a picture: http://www.cut-the-knot.org/proofs/tessellation.shtml

I've been told that a partition of $R^3$ into circles was found independently and earlier, but not published, by William Thurston.


This is probably not what you wanted, but I can do it trivially: For every $r \in \mathbb{R}^3$, let $$C_{r}:I \to \mathbb{R}^3$$ be the map given by $C_{r}(x) = r$. That is, we let each curve be a trivial one point curve. These are all closed. Then we have that $$\mathbb{R}^3 = \bigcup_{r \in \mathbb{R}^3} C_r$$ and the union is disjoint.

  • $\begingroup$ Thank you. What if curves are homeomorphic to circles? $\endgroup$ – DVD May 15 '13 at 2:00
  • $\begingroup$ Yeah, I do not believe that is true, but have not yet been able to construct a proof. I will keep working on it. $\endgroup$ – Kris Williams May 15 '13 at 2:10

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