# Covering the space by disjoint unit circles

Sierpinski has proved the following two interesting theorems.

Theorem 1. The Euclidean plane $\mathbb{R}^2$ is not a union of nondegenerate disjoint circles.

Theorem 2. The Euclidean space $\mathbb{R}^3$ is a union of nondegenerate disjoint circles.

In the second of the two main parts of his proof, Sierpinski blows a circle until it fits his needs. Thinking about this, it seems that if one uses rotations instead, one obtains the following theorem.

Theorem 3. The Euclidean space $\mathbb{R}^3$ is a union of disjoint unit circles.

The reason is that fixing a point on the unit circle, the intersection of every three rotations about this point contains just this point.

I think this version is much more cute. E.g., note that the analogous version of Theorem 1 becomes much more obvious when restricting to unit circles (one can have at most countably many disjoint unit circles in the plane!). Also, unit circles are not special. One only needs the following property of the desired curve.

Free Rotations Property. There exists a point on the curve such that, considering rotations of the curve about this point, every uncountable intersection of such rotations contains only the rotation point.

Thus, the theorem applies, e.g., to any regular polygon of perimeter 1, or even to a fixed "8" shape (a nice name for Theorem 3 may thus be Room for 8 :) ). In the plane, it is known that even if we allow deforming the 8 figures, we cannot have more than countably many of disjoint 8's.

In fact, I do not know a curve not having the Free Rotations Property. Maybe even for each "nice enough" curve there is a finite number $k$ and a rotation point on it such that every $k$ rotations intersect in one point only.

My questions ( Assuming that my analysis is correct ): Is the above known? If yes, could you provide a reference? What is the largest known class of curves with the mentioned Free Rotations Property?

• It was pointed out to me that one must avoid things like the Peano curve. So perhaps "nice enough" should be interpreted as "at least simple (i.e., injective)". – Boaz Tsaban Apr 3 '14 at 13:58
• Embarrasingly, I am unable to find the exact reference to Sierpinski. Maybe he already answered my questions there, at least to some extent. I would appreciate a pointer. – Boaz Tsaban Apr 3 '14 at 14:04
• The question and other variations have come up on MO several times. See, for example, mathoverflow.net/a/28655/1946, and also mathoverflow.net/q/93601/1946. – Joel David Hamkins Apr 3 '14 at 14:10

In their article Partitions of $\mathbb{R}^3$ into curves, Mathematica Scandinavica 1998, the authors M. Jonsson and J. Wästlund show that space can be partitioned in unlinked unit circles or other kinds of curves.

Abstract. A general technique for obtaining partitions of $\mathbb{R}^3$ into curves satisfying various properties is presented. The method can be used to partition $\mathbb{R}^3$ into unlinked circles of radius one, which answers a question posed by Wilker [7], or into arbitrary collections of real analytic curves. We also apply the method to study the set of bijections of the open unit disk.

• Thanks Joel, for the answer and for the pointers to the earlier discussions. In fact I searched MO before posting my question (and looked at the automatic suggestions while composing my questions), but it did not work out for me. So yes, the question is closed. – Boaz Tsaban Apr 3 '14 at 15:17
• I think your question isn't technically closed, and since it is a little different than the others, since you are more explicitly focused on which kind of curves the argument works with, I guess we might just as well leave it open. – Joel David Hamkins Apr 3 '14 at 15:20

Here is an answer to your question in another post. Is it possible to partition $\mathbb R^3$ into unit circles? Unfortunately it's not a constructive solution, so we're still missing that.

I do know a nice constructive solution using circles of unbounded radii.

Consider the points on the $x$ axis $(4n + 1, 0, 0)$ for $n \in Z$. Consider the circles of radius $1$ on the $x-y$ plane with center at these points.

Now consider all spheres with radius $r>0$ and center $(0,0,0)$. Each such sphere intersects the set of these circles at exactly two points. Remove these two points from the sphere. What remains of the sphere can certainly be decomposed into disjoint circles. Altogether we have a partition of $R^3$ into a disjoint union of circles.

This solution was pointed out to me many years ago by Yuval Peres.