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

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

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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? – Ryan Budney Jun 18 2010 at 17:46
What's the difference between a smooth circle and a round circle? – Kiochi Jun 18 2010 at 18:13
A smooth circle is just a smooth compact connected 1-dimensional submanifold of $\mathbb R^3$, i.e. the image of a smooth non-constant 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. – Ryan Budney Jun 18 2010 at 18:16
See related question: mathoverflow.net/questions/21327/… – Joel David Hamkins Jun 18 2010 at 18:39
Ryan, it was given to me as an exercise. I've spent many hours trying to solve it. Now I understand I was supposed to use the axiom of choice, which I actually tried; presumably not hard enough. – Zarathustra Jun 18 2010 at 19:29
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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.

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 I understand your solution (provided I believe that there's such a well ordering). A little bit disappointed, but thank you! – Zarathustra Jun 18 2010 at 21:12 I fixed some TeX. Namely, because Markdown gets to modify the text before jsMath does, it sees \{ and replaces it with {, and then jsMath doesn't process it. The solution is to put all TeX within backticks. – Theo Johnson-Freyd Jun 18 2010 at 21:49 Thank you, Theo! – Péter Komjáth Jun 19 2010 at 5:45

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.

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 Is it known if the axiom of choice is required for congruent circles to partition? – Ryan Budney Jun 18 2010 at 19:28 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... – Joel David Hamkins Jun 18 2010 at 19:40

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.

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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.

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Does "geometric" mean round with a fixed radius? – Ryan Budney Jun 18 2010 at 17:57
@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." – Joseph O'Rourke Jun 18 2010 at 17:58
I mean, do all the circles have the same radius? Certainly their centres and axis can vary. – Ryan Budney Jun 18 2010 at 18:11
Ah, sorry. Somehow I completely mis-read your answer. It makes sense now and my comments were off-track. – Ryan Budney Jun 18 2010 at 18:18
I don't understand, didn't your original answer say something about "using algebraic topology it's possible to show..." ? – Ryan Budney Jun 18 2010 at 18:23
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