Question

Is it possible to partition $\mathbb R^3$ into unit circles?

Answer 1

The construction is based on a well ordering of $R^3$ into the least ordinal of cardinality continuum. Let $\phi$ be that ordinal and let $R^3=\{p_\alpha:\alpha<\phi\}$ be an enumeration of the points of space. We define a unit circle $C_\alpha$ containing $p_\alpha$ by transfinite recursion on $\alpha$, for some $\alpha$ we do nothing. Here is the recursion step. Assume we have reached step $\alpha$ and some circles $\{C_\beta:\beta<\alpha\}$ have been determined. If some of them contains (=covers) $p_\alpha$, we do nothing. Otherwise, we choose a unit circle containing $p_\alpha$ that misses all the earlier circles. For that, we first choose a plane throu $p_\alpha$ that is distinct from the planes of the earlier circles. This is possible, as there are continuum many planes throu $p_\alpha$ and less than continuum many planes which are the planes of those earlier circles. Let $K$ be the plane chosen. The earlier circles intersect $K$ in less than continuum many points, so it suffices to find, in $K$, a unit circle going throu $p_\alpha$ which misses certain less than continuum many points. That is easy: there are continuum many unit circles in $K$ taht pass throu $p_\alpha$ and each of the bad points disqualifies only 2 of them.

Answer 2

Evelyn Sander says here, "Geometric circles of unit radius are called hoops. Using the Axiom of Choice, J.H. Conway and H.T. Croft showed that it is nevertheless possible to discontinuously fill three-space using disjoint hoops." The "nevertheless" was to contrast with filling continuously. This was a report on a talk by Daniel Asimov in 1994, who showed that it is not possible to fill continuously with hoops.

Answer 3

In this article, the authors prove that not only can you partition $R^3$ into congruent circles, but you can do so into unlinked congruent circles. They also prove a variety of other similar results: $R^3$ can be partitioned into isometric copies of any family of continuum many real analytic curves. And they consider the question in higher dimensions, and also the role of AC in the proofs: for example, in $R^3$ no AC is needed for circles, if different sizes are allowed.

Answer 4

Péter's proof is very clever and, while there is no real need to resurrect this thread, the following is quite straightforward in case one is not inclined to hunt for it in the literature on this subject:

Observe that you can cover a two-punctured sphere with circles. Now consider a family of circles lying in the $xy$ plane, radii 1, centred at the points $(4k+1,0,0)$ for $k \in \mathbb{Z}$. Each sphere about the origin intersects this family in exactly two places.

Answer 5

Even though this question is old, I'd like to give what I regard as a very beautiful solution. It is different from the others in that the circles used are not round (but they are unlinked). First observe that the circles $x^2 + y^2 = r^2$, $z = c$, for $r \geq 1$ and $c$ any real number, decompose all of $\Bbb{R}^3$ except an open cylinder into circles. At first glance, this seems to have accomplished nothing, since the open cylinder is homeomorphic to $\Bbb{R}^3$, so we have reduced the original problem to an equivalent problem. However, look at the left-hand figure of the included image, which shows an open cylinder embedded as a U shape, with the ends going to infinity in the same direction. Since this is just a deformation of the original embedding, we can decompose the complement into circles. To handle the interior, embed an open cylinder into it, as shown in the right-hand figure. We can decompose the complement of the smaller U-shaped cylinder into circles. We continue in this way, making sure that the embedded cylinders go off to infinity, so that every point of $\Bbb{R}^3$ is included at some finite stage.

It seems like we have never really solved the problem, but instead have just pushed it away so much that it vanishes into thin air!