It is an easy exercise to show that the Euclidean plane cannot be partitioned into round circles (note however that it is possible to do so for $\mathbb{R}^3$). It seems almost obvious that it is not possible to partition the plane into Jordan curves either.

However, I am not able to design a proof that does not use the choice axiom. With choice, assume you have such a partition $J$ and define on $J$ a partial ordering: $j<k$ if the curve $j$ is contained in the interior of $k$. Any decreasing chain $j_n$ has a lower bound: if $K_n$ is the closure of the interior of $j_n$, then $K_n$ is a decreasing sequence of compact sets, thus there is some point $x$ in the intersection. Then the curve of $J$ that contains $x$ is a lower bound of $(j_n)$. Now Zorn Lemma ensures that there is a minimal element $j$ in $J$. But this is obviously impossible since $j$ would have a non-empty interior, therefore containing another curve of $J$.

The question is therefore the following: can we prove that there exist no partition of the plane into Jordan curves without assuming the Choice axiom?

  • 6
    $\begingroup$ Instead of using the axiom of choice, choose a dense countable set of points in the plane. Associated to the first point in the set, there is a circle. Inside this circle, find the point in your dense countable set with least index and repeat. This might "guide" you through the circles, without using the axiom of choice. $\endgroup$ – damiano Apr 14 '10 at 12:04
  • 2
    $\begingroup$ For the R^3 case, here is an interesting article: mscand.dk/article.php?id=77. They partition R^3 into unlinked congruent circles, and also consider other more arbitrary families of curves. Partitioning R^3 into circles is constructive, but for the more exotic partitions, the Axiom of Choice is used. $\endgroup$ – Joel David Hamkins Apr 14 '10 at 12:38
  • $\begingroup$ Somewhat off-topic question: a point is not a Jordan curve? Wikipedia's definitions seem to allow this (see articles on "Jordan Curve" and "Curve"). $\endgroup$ – Nate Eldredge Apr 14 '10 at 15:52
  • 1
    $\begingroup$ @Nate Elredge: here I consider non constant curves only. It seems the most reasonnable choice: if you want the Jordan theorem to apply to all Jordan curves, then certainly you don't want to call a point a Jordan curve. $\endgroup$ – Benoît Kloeckner Apr 14 '10 at 16:42

It is possible to change your argument so that the choice is over countable set; hope this is good enough. Namely, topology on the plane has countable base (say, circles at rational points with rational radii); let's index this base as $U_1,\dots,U_n,\dots$; your argument can be used to construct a sequence of Jordan curves $C_1,\dots,C_n,\dots$ such that $C_{i+1}$ is contained in the interior of $C_i$, and $U_i$ is not contained in the interior of $C_i$.

| cite | improve this answer | |
  • 1
    $\begingroup$ damiano typed his comment as I was typing this... $\endgroup$ – t3suji Apr 14 '10 at 12:15
  • 5
    $\begingroup$ I think it is not quite good enough, but Damiano's idea works here. The issue is that there could be many possible choices for C_i+1 satisfying your condition of being contained in the interior of C_i and not having U_i in its interior. But if you enumerate a countable dense subset of the plane, then each point is contained on exactly one curve, so the choice is completely determined by the well-ordering of the countable dense set: you pick the first point lying on a curve in the interior of the previous curve. $\endgroup$ – Joel David Hamkins Apr 14 '10 at 12:30
  • $\begingroup$ Thanks for the comment, I completely agree it is better to have a determined construction. It is somehow my sloppy attitude that countable choice is acceptable even if continuum choice is not, because so many construction depend on it... Out of curiosity: we are using the Jordan Curve Theorem; does its proof rely on choice? In fact, don't we have to use that a descending chain of compact sets $K_n$ has non-empty intersection? How would one prove this without countable choice? $\endgroup$ – t3suji Apr 14 '10 at 14:38
  • $\begingroup$ @t3suji: this is a very reasonable curiosity, given what the question asks! You can prove the Jordan Curve Theorem easily using homology. I am not sure that this does not use the axiom of choice, but it seems that you can do everything quite explicitly in the various arguments. As for your question about compact sets: if you already know that the sets are compact and it is a descending chain, they are required to have a common point by compactness! $\endgroup$ – damiano Apr 14 '10 at 15:21
  • $\begingroup$ I accept this answer, and want also to thank damiano. Finally, I take the opportunity to give a question that is not a true one, in the sense that I saw a proof some years ago. Can you prove that one cannot partition the plane by Y shaped subsets? $\endgroup$ – Benoît Kloeckner Apr 14 '10 at 16:51

Isn't the "proper" 3D analogue of filling the plane with closed curves to fill space with surfaces?

I mean, the decomposition of the 3 sphere into two tori (Hopf fibration) is nice and all, but you're using a result involving link theory as the analogue of a result from measure theory...

What I would consider the proper analogue of partitioning the plane with Jordan curves is to partition ${\mathbb R}^3$ with closed, orientable surfaces. And what makes that question superficially a lot more interesting than the ${\mathbb R}^2$ case is that all Jordan curves are homeomorphic to a circle, but closed orientable surfaces can have any genus. However, the proof you give still goes through in that case because a closed orientable surface has nonempty interior...

| cite | improve this answer | |
  • 1
    $\begingroup$ It's also more interesting because of wild embeddings like the <a href="mathworld.wolfram.com/… horned sphere.</a> $\endgroup$ – Allen Knutson Apr 14 '10 at 17:20
  • $\begingroup$ That's analogous to considering closed space-filling curves in ${\mathbb R}^2$. $\endgroup$ – Miguel Apr 14 '10 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.