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

1$\begingroup$ I know how to do it with smooth circles. Maybe 15 years ago some Hopkins student (I forget his name, he is Canadian) mentioned he knew how to do it with round circles. But I don't think he ever described the construction to me. I'm curious why you're interested in this question? $\endgroup$ – Ryan Budney Jun 18 '10 at 17:46

2$\begingroup$ What's the difference between a smooth circle and a round circle? $\endgroup$ – Kiochi Jun 18 '10 at 18:13

2$\begingroup$ A smooth circle is just a smooth compact connected 1dimensional submanifold of $\mathbb R^3$, i.e. the image of a smooth nonconstant periodic function $f: \mathbb R \to \mathbb R^3$. Round means having constant curvature and zero torsion, also there's the equivalent definition of O'Rourke's, in the comments to his answer below. $\endgroup$ – Ryan Budney Jun 18 '10 at 18:16

1$\begingroup$ See related question: mathoverflow.net/questions/21327/… $\endgroup$ – Joel David Hamkins Jun 18 '10 at 18:39

4$\begingroup$ @Ryan I might be the Canadian you are remembering. The solution I know using round circles is the one described by Spencer below, and I learned it from Danny Arnon, who was a student of Mike Hopkins at the time. $\endgroup$ – Dan Christensen Jun 19 '13 at 10:35
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.

$\begingroup$ I understand your solution (provided I believe that there's such a well ordering). A little bit disappointed, but thank you! $\endgroup$ – Zarathustra Jun 18 '10 at 21:12

1$\begingroup$ I fixed some TeX. Namely, because Markdown gets to modify the text before jsMath does, it sees
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, and then jsMath doesn't process it. The solution is to put all TeX within backticks. $\endgroup$ – Theo JohnsonFreyd Jun 18 '10 at 21:49
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 lefthand 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 righthand figure. We can decompose the complement of the smaller Ushaped 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!
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 twopunctured 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.

3$\begingroup$ This is exactly the construction of Szulkin [MR0719756] (see my answer below). $\endgroup$ – Wlodek Kuperberg Jul 22 '13 at 16:15

$\begingroup$ But the question was about a partition into unit circles, and it's not obvious that an arbitrary twopunctured sphere can be covered with unit circles. $\endgroup$ – Timothy Chow Nov 6 '20 at 14:19
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.

2$\begingroup$ Is it known if the axiom of choice is required for congruent circles to partition? $\endgroup$ – Ryan Budney Jun 18 '10 at 19:28

1$\begingroup$ Ryan, I'm not sure. All the arguments I have heard use AC in this case, but I have never heard a proof that AC was required for such a thing (and it is difficult to see how such a proof would proceed). But perhas the geometers will come through with a positive construction... $\endgroup$ – Joel David Hamkins Jun 18 '10 at 19:40

$\begingroup$ The article is: mscand.dk/article/view/13850/11850 $\endgroup$ – Joel David Hamkins Apr 3 '14 at 14:44

$\begingroup$ Regarding whether AC is required, or whether there is a Borel partition of the various types, I asked the following question: mathoverflow.net/q/93601/1946 $\endgroup$ – Joel David Hamkins May 21 '15 at 12:39

1$\begingroup$ @JoelDavidHamkins : What paper is this? The links aren't working for me, and you didn't provide authors or a title or even a journal or year. $\endgroup$ – Timothy Chow Nov 4 '20 at 18:48
A very nice, explicit, elementary partition (without the Axiom of Choice) of $\mathbb{R}^3$ into geometric circles of variable radii is given in: [MR0719756] Szulkin, Andrzej. $\mathbb{R}^3$ is the union of disjoint circles. Amer. Math. Monthly 90 (1983), no. 9, 640–641.
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 threespace 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.

1$\begingroup$ Does "geometric" mean round with a fixed radius? $\endgroup$ – Ryan Budney Jun 18 '10 at 17:57

$\begingroup$ @Ryan: Yes. "A geometric circle in $R^3$ is the set of points in a fixed plane that lie a fixed positive distance from a center point that lies in the same plane." $\endgroup$ – Joseph O'Rourke Jun 18 '10 at 17:58

$\begingroup$ I mean, do all the circles have the same radius? Certainly their centres and axis can vary. $\endgroup$ – Ryan Budney Jun 18 '10 at 18:11

$\begingroup$ Ah, sorry. Somehow I completely misread your answer. It makes sense now and my comments were offtrack. $\endgroup$ – Ryan Budney Jun 18 '10 at 18:18

$\begingroup$ I don't understand, didn't your original answer say something about "using algebraic topology it's possible to show..." ? $\endgroup$ – Ryan Budney Jun 18 '10 at 18:23