Let $X$ be a domain in the Riemann sphere $\widehat{\mathbb{C}}$. We say that $X$ is a circle domain if every connected component of its boundary is either a circle or a point.

It was conjectured by Koebe in 1909 that circle domains represent all planar domains up to conformal equivalence. This was proved by Koebe himself in the case of finitely many boundary components, and by Zheng-Xu He and Oded Schramm in 1993 in the countable case.

I'm trying to find examples of "bad" circle domains. In particular :

Question : Is there a circle domain whose boundary is not the union of countably many circles, countably many Cantor sets and countably many points?

I'm mostly interested in the case where the boundary has zero area.

Of course, the boundary of a circle domain contains at most countably many circles. The question is whether the set of point components can be written as a countable union of closed totally disconnected sets. The problem seems to be the set of point components that are accumulation points of circles...

Another question :

Question 2 : If $X$ is a circle domain, must there exist an element $z \in \partial X$ which is isolated from circles (in other words there is no sequence of circles in $\partial X$ converging to $z$)?

The answer is yes if the set of point components in closed, by Baire Category considerations...

Thank you, Malik

  • $\begingroup$ Just a question: is it true that if you take the quotient of $\hat{\mathbb{C}}$ by collapsing all circle component to a point (well, first you "fill the circles" to obtain discs, and then you collapse), the resulting space is still homeomorphic to $\hat{\mathbb{C}}$? Because then if I am not wrong the domain has only point boundary components and you need to look at totally disconnected closed subsets of the $2$-sphere. $\endgroup$ – Mathieu Baillif Sep 22 '15 at 14:55
  • $\begingroup$ (and a totally disconnected closed subset of the sphere is the union of countably many isolated points and a Cantor set.) $\endgroup$ – Mathieu Baillif Sep 22 '15 at 15:10
  • $\begingroup$ @MathieuBaillif Yes, it is true that the quotient space is homeomorphic to $\widehat{\mathbb{C}}$. It is a very special case of Moore's theorem on upper semi-continuous decompositions of the sphere. However, here I want the set of point components to be a countable union of Cantor sets and points, and one runs into trouble with this argument because the set of point components need not be closed... $\endgroup$ – Malik Younsi Sep 22 '15 at 16:49
  • $\begingroup$ In other words : If a circle in the boundary of $X$ is not isolated from point boundary components, then in the quotient space it will not correspond to an isolated point in the countable set. $\endgroup$ – Malik Younsi Sep 22 '15 at 17:43
  • $\begingroup$ True, but in the quotient space, the boundary is the union of a Cantor set $C$ and a countable subset $S$ (the points in $S$ may be non-isolated, of course, sorry for the mistake in my previous comment). In the original space, at most countably many of those points correspond to a circle. If you remove a point from a Cantor Set, you obtain a countable union of Cantor sets. Isn't is enough to answer your question 1 or I am missing something ? $\endgroup$ – Mathieu Baillif Sep 22 '15 at 18:35

Let $K$ be the boundary of your circle domain $\Omega$.

Let us suppose that every point of $K$ is accumulated on by a sequence of (pairwise different) circle components. Such an example is easy to construct (see e.g. the already existing answer to your Question 2, or simply add the circles inductively - see below for the details).

Claim. A countable union of closed and totally disconnected subsets of $K$ cannot contain all non-circle components of $K$.

Remark. This shows that the answer to your first question is negative.

Proof. First, suppose that $A$ is a totally disconnected closed subset of $K$, and that $U$ is some open set that intersects one of the circle components, say $C$. Then for any point $z$ on any arc of $C\setminus A$, there are arbitrarily small circle components accumulating on $z$. If they are small enough, then these components are themselves in the complement of $A$, and then there is also a small clopen (in $K$) neighbourhood of each of these components that is contained in the complement of $A$.

So, in summary: $U$ contains a clopen subset $X$ of $K$, which contains a circle component. Furthermore, the diameter of $X$ can be chosen as small as we wish.

Now if $(A_i)$ is a sequence of totally disconnected closed subsets of $K$, then we can proceed inductively: Find a nonempty clopen subset $X_1$ in the complement of $A_1$ as above, say having diameter less than $1$. Then find a nonempty clopen subset $X_2\subset X_1 \setminus A_1$ of $K$, having diameter less than $1/2$, etc.

Since each $X_k$ is clopen, and the diameters tend to zero, their (non-empty) intersection is disjoint from all circle components, and belongs to the complement of the union of all $A_i$. This completes the proof.

The proof in fact never used that the non-trivial sets in question are circles. Hence it shows:

Proposition. Let $K\subset\mathbb{C}$ be a compact set such that, for every $\newcommand{\eps}{\varepsilon}\eps>0$, there are only finitely many connected components of $K$ of diameter at least $\eps$. Assume furthermore that every point $z\in K$ is accumulated on by non-point components not containing $z$. Then any countable collection of closed and totally disconnected subsets of $K$ must omit some point components of $K$.

Remark. A slight adaptation of the proof shows furthermore that the set of omitted point components has the cardinality of the continuum.

EDIT. For completeness, let me outline the elementary construction of the domain in question, which provides more details about Misha's "Sierpinski carpet" suggestion. At each stage of the inductive construction, we have a collection $\mathcal{C}_k$ of finitely many pairwise disjoint circles, with $\mathcal{C}_{k+1}\supset \mathcal{C}_k$.

For each circle $C\in \mathcal{C}_k$, we also pick an open annulus $A_k(C)$ that surrounds $C$, separates $C$ from all other elements of $\mathcal{C}_k$, and whic becomes closer and closer to $C$ as $k\to\infty$.

Start with $\mathcal{C}_0$ having one circle $C$ in it, with some annulus $A(C)$ picked around it. Then, inductively, for every circle $C$ in $\mathcal{C}_k$, add a whole bunch of small circles to $\mathcal{C}_{k+1}$, within the inner curve of $A_k(C)$, such that every point of $C$ is close to one of these new circles. Then pick the annuli with the desired property, and so that they are pairwise disjoint.

Let $K$ be the closure of the union of all these circles. The construction ensures that this set is totally disconnected (any two points will be separated by one of the annuli), and clearly $K$ has all the desired properties).

  • $\begingroup$ Your argument is indeed quite reminiscent of the Baire theorem, but thank you for the details and also for the outline of the construction for Question 2. As usual, your answer is quite valuable! $\endgroup$ – Malik Younsi Sep 23 '15 at 11:36

For the 2nd part one can construct an example by imitating the Sierpinski Carpet construction. Or, you use Kleinian groups. Take two Fuchsian subgroups $F_1, F_2$ of $PSL(2,C)$ with disjoint limit circles. After replacing them with finite index subgroups if necessary, the subgroup $G$ of $PSL(2,C)$ that $F_1, F_2$ generate is a functional group, i.e. its domain of discontinuity contains a connected $G$-invariant component $\Omega$; $\Omega$ is easily seen to be a circular domain. At the same time, due to minimality of the action of $G$ on its limit set (which is the boundary of $\Omega$), each point of the limit set is the limit point of a sequence of circles whose radii tend to zero. You can find more details on such constructions (called "Klein combination") in Maskit's book "Kleinian Groups". He also gives a detailed background on Kleinian groups.

Edit: I have to think more about Part 1, I am not longer sure if it also has negative answer.

  • $\begingroup$ Thank you, I'll look into Klein combinations. Just to make sure : in your example, is any point on a circle also a limit point of infinitely many circles? $\endgroup$ – Malik Younsi Sep 22 '15 at 20:08
  • 3
    $\begingroup$ If yes, then your example of a circle domain $\Omega$ should also settle Question 1, unless I am missing something. Indeed, if $\partial \Omega$ were the union of countably many circles, countably many Cantor sets and countably many points, then one of these sets, say $A$, would have nonempty interior in $\partial \Omega$, by Baire. This means that there would exist some open set $U$ with $U \cap \partial \Omega \neq \emptyset$ and $U \cap \partial \Omega \subset A$. But then any point in $U \cap \partial \Omega$ would be isolated from circles... Am I missing something? $\endgroup$ – Malik Younsi Sep 22 '15 at 20:13
  • $\begingroup$ @MalikYounsi - just noticed your comment, which is a more elegant way of phrasing my answer - which in effect reproduces the proof of Baire's theorem. $\endgroup$ – Lasse Rempe-Gillen Sep 22 '15 at 21:59
  • $\begingroup$ @MalikYounsi: I still have no time to think about your question, but, yes, each point on each circle is an accumulation point of boundary circles of $\Omega$. $\endgroup$ – Misha Sep 23 '15 at 17:45

Ok, after writing many comments thinking I had a proof showing that Question 1 had a negative answer, I now believe that the following construction (inspired by Misha's answer) yields an example. I hope I am not mixing everything and that what I write makes sense.

Take the usual Cantor ternary set $K$ in $[0,1]$ (built by removing middle intervals) and embed it in the horizontal axis in $\mathbb{C}$. Say that $x\in K$ is an endpoint if it is the endpoint of an interval in some stage of the construction of $K$. Replace these endpoints by small circles that do not intersect or encircle any part of the rest of $K$. You can do that because there is some room on one side of any endpoint, which my be thought as the westmost or eastmost point in the added circle. The circles radii go to zero when you go "further down" in $K$, so you do not add new accumulation points, and the resulting subset $L\subset\mathbb{C}$ is compact. Then $L$ is the boundary of the domain $\mathbb{C}-L-\{\text{interior of the circles}\}$. If you remove the circle boundaries, you obtain the Cantor set minus its endpoints, which is homeomorphic to the irrationals (or the Baire space $\omega^\omega$, see for instance the answers to this question: https://math.stackexchange.com/questions/52073/two-questions-about-the-cantor-set-construction), which cannot be a countable union of Cantor sets.

  • $\begingroup$ This doesn't work because the boundary is still a union of countably many circles and a Cantor set (the Cantor set just happens to intersect all the circles). In fact, I had this very example in mind when I wrote my comment on the question about whether they had to be disjoint. However, it is still a nice counterexample to the second question. $\endgroup$ – Eric Wofsey Sep 22 '15 at 19:57
  • $\begingroup$ Ok, I thought they had to be disjoint, as I read your comment just after posting my answer. $\endgroup$ – Mathieu Baillif Sep 22 '15 at 19:58
  • $\begingroup$ Oh, this is not a counterexample to the second question unless you count points on a circle as being "limits of circles" (even if there are no other circles that approach the point). $\endgroup$ – Eric Wofsey Sep 22 '15 at 20:04
  • $\begingroup$ @EricWofsey I do not consider such points to be limits of circles. For me, limits of circles mean infinitely many circles approaching the point. $\endgroup$ – Malik Younsi Sep 22 '15 at 20:07

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.