Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
I don't think so. If you add a point at $\infty,$ then yes, this is the Hopf fibration of $\mathbb S^3.$ en.wikipedia.org/wiki/Hopf_fibration –  Will Jagy May 14 '13 at 4:00
add comment

2 Answers

up vote 5 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
    
Thank you. What if curves are homeomorphic to circles? –  Daved May 15 '13 at 2:00
    
Yeah, I do not believe that is true, but have not yet been able to construct a proof. I will keep working on it. –  Kris Williams May 15 '13 at 2:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.