We can express $\mathbb{R}^3$ as a disjoint union of circles. There are some constructive ways of doing this, although it's easier to construct them sequentially by transfinite induction, applying the following step to each point $P_{\alpha}$ in the well-ordering of $\mathbb{R}^3$ by the initial ordinal of $2^{\aleph_0}$:

  • Suppose the point $P_{\alpha}$ has not already been covered (otherwise, move on to the next point);
  • Choose a plane passing through that point, such that the direction of the normal vector is distinct from all previous planes;
  • This plane intersects the union of all existing circles in fewer than $2^{\aleph_0}$ points, so we can draw a circle on that plane passing through $P_{\alpha}$ and disjoint from all existing circles.

We can also express $\mathbb{R}^3 - \ell$ as a disjoint union of pairwise linked circles, where $\ell$ is an arbitrary line. Specifically, we take the stereographic projection of the Hopf fibration, and remove the `circle' passing through the point at infinity.

This suggests that there may be a mutual generalisation of both results, namely that $\mathbb{R}^3$ can be expressed as a disjoint union of pairwise linked circles. Can anyone prove or disprove this conjecture?

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    $\begingroup$ The question mathoverflow.net/questions/28647 discusses several different constructions without the linking condition. $\endgroup$ Jul 5, 2014 at 20:01
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    $\begingroup$ See also mathoverflow.net/q/93601/1946, which asks whether for any of the various cases of surprising geometrical decompositions where we don't have an explicit concrete construction, whether there is any proof that there is no Borel decomposition. This seems to be open. But if one entertains the possibility that there is no such decomposition, then it would seem to be more approachable to refute Borel decompositions, as this would allow one to use measure-theoretic or descriptive set-theoretic concepts in the argument. $\endgroup$ Jul 6, 2014 at 2:47

1 Answer 1


Whilst this doesn't answer the original question, I shall show that it's possible to partition $\mathbb{R}^3 \setminus \{ 0 \}$ into pairwise-linked disjoint circles.

I'll begin with a couple of observations:

Observation 1: Note that when we apply the transfinite induction to the weaker version of the problem (which doesn't require the circles to be linked), we actually have an extra degree of freedom. For example, it's possible to assert that all circles have unit radius, and the proof still works since, even after choosing a point $P$, a plane $\Pi$ incident with $P$, and a radius $R > 0$, there are $2^{\aleph_0}$ circles of radius $R$ lying on $\Pi$ and passing through $P$.

Observation 1 is quite well known and is included in the literature on the subject. We won't use it directly, but the idea of being able to choose the 'sizes' of circles is key to our proof.

Observation 2: If we have two disjoint great circles on $S^3$, they must necessarily be Hopf-linked. To see this, note that great circles are intersections of $S^3$ with planes, and disjointness of the circles imply that the corresponding planes are complementary (their direct sum is $\mathbb{R}^4$), in which case we can apply a linear transformation to map them to our 'standard planes' $x = y = 0$ and $w = z = 0$. This is, for example, why all of the circles in the Hopf fibration are linked.

So, we would like to be able to express $\mathbb{R}^3 \setminus \{ 0 \}$ as a disjoint union of (stereographic projections of) great circles, since then we are done by Observation 2. This is indeed possible by the usual transfinite induction proof, since the appropriate analogue of Observation 1 still applies (namely that after choosing a point $P \in \mathbb{R}^3 \setminus \{ 0 \}$ and plane $\Pi$ incident with $P$, there are $2^{\aleph_0}$ (stereographic projections of) great circles lying on $\Pi$ and passing through $P$).

This proof doesn't work for $\mathbb{R}^3$ since every great circle passing through $0$ is a line and therefore invalid.

  • $\begingroup$ I think a slight modification of your argument should do the trick. Take a circle going through the origin. Then for every other point in $\mathbb{R}^3$, there is an open set of stereographic projections of great circles which link this circle (since the set of plane through a point on $S^3$ cover the unit ball in $\mathbb{R}^4$). I think then your argument proceeds as before, being careful to choose a plane through each point which intersects the open set linking the circle through the origin. $\endgroup$
    – Ian Agol
    Mar 10, 2018 at 21:25
  • $\begingroup$ I don't see how any modification of my argument can work, because the great circles are invariant under the antipodal map, whereas the space we're covering ($\mathbb{R}^3 = S^3 \setminus \{ 0 \}$) is not. $\endgroup$ Mar 11, 2018 at 0:07
  • $\begingroup$ Oh, of course, I missed that the antipodal points can't be covered. $\endgroup$
    – Ian Agol
    Mar 11, 2018 at 2:26

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