Is it possible to partition $\mathbb R^3$ into unit circles? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:54:38Z http://mathoverflow.net/feeds/question/28647 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles Is it possible to partition $\mathbb R^3$ into unit circles? Zarathustra 2010-06-18T17:39:20Z 2010-11-27T14:56:50Z <p>Is it possible to partition $\mathbb R^3$ into unit circles?</p> http://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles/28648#28648 Answer by Joseph O'Rourke for Is it possible to partition $\mathbb R^3$ into unit circles? Joseph O'Rourke 2010-06-18T17:55:24Z 2010-06-18T17:55:24Z <p>Evelyn Sander says <a href="http://www.geom.uiuc.edu/docs/forum/hoops_links/khoops.html" rel="nofollow">here</a>, "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.</p> http://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles/28650#28650 Answer by Péter Komjáth for Is it possible to partition $\mathbb R^3$ into unit circles? Péter Komjáth 2010-06-18T18:32:07Z 2010-06-18T21:48:01Z <p>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 <code>$R^3=\{p_\alpha:\alpha&lt;\phi\}$</code> 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 <code>$\{C_\beta:\beta&lt;\alpha\}$</code> have been determined. If some of them contains (=covers) <code>$p_\alpha$</code>, we do nothing. Otherwise, we choose a unit circle containing <code>$p_\alpha$</code> that misses all the earlier circles. For that, we first choose a plane throu <code>$p_\alpha$</code> that is distinct from the planes of the earlier circles. This is possible, as there are continuum many planes throu <code>$p_\alpha$</code> 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 <code>$p_\alpha$</code> which misses certain less than continuum many points. That is easy: there are continuum many unit circles in $K$ taht pass throu <code>$p_\alpha$</code> and each of the bad points disqualifies only 2 of them. </p> http://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles/28655#28655 Answer by Joel David Hamkins for Is it possible to partition $\mathbb R^3$ into unit circles? Joel David Hamkins 2010-06-18T19:07:17Z 2010-06-18T19:07:17Z <p>In <a href="http://www.mscand.dk/article.php?id=77" rel="nofollow">this article</a>, the authors prove that not only can you partition $R^3$ into congruent circles, but you can do so into <em>unlinked</em> 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. </p> http://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles/47507#47507 Answer by Spencer for Is it possible to partition $\mathbb R^3$ into unit circles? Spencer 2010-11-27T14:56:50Z 2010-11-27T14:56:50Z <p>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:</p> <p>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.</p>