3
$\begingroup$

Suppose we have a domain $U$ in $\mathbb{C}$ (an open connected set). Denote the union of all bounded connected components of $\mathbb{C} - U$ by $K$ and the union of all unbounded connected components of $\mathbb{C} - U$ by $H$. It seems natural to be that $U \cup K$ should be open, or equivalently, that $H$ be closed.

However, I have not been able to prove this and I started doubting it. All connected components of $H$ are closed since they are connected components of the complement of $U$, an open set. Also, it's not very hard to see that any point on the boundary of $K$ is also on the boundary of $U$ (any ball around it contains points in $K \subset \mathbb{C} - U$ and supposing some ball does not contain points from $U$ means that the ball - a connected set - is contained in $K \cup H$ and contains points from 2 different connected components of it).

This is not enough though: if we consider the unit open disc and cut out a solid closed triangle from it and also remove a sequence of points converging to one of the vertices, when we put the traingle back in, the resulting set is not open even though all points on the boundary of the traingle are boundary points of the cut-out disk.

We need to show that no sequence of points from $H$ converges to a point in $K$. I don't know if this is true, but it seems like it should be. Also, I'm not sure whether the connectivity of $U$ would play any role at all here, but for my purposes I can just assume that $U$ is also connected, not just open.

Please let me know if anything is unclear.

Edit: As seen in the answers, this has been confirmed to be true, but if anyone could find a proof that avoid the use of the Riemann sphere, that would be what I need. I'm trying to prove some other result in the Euclidean metric and this is an intermediate step that I've been stuck on.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, this is true. I will add a proof when I have more time. $\endgroup$ Nov 13, 2021 at 3:49

1 Answer 1

1
$\begingroup$

Suppose that $U\subset {\mathbb C}$ is a domain. I will work with the 1-point compactification of the complex plane, the Riemann sphere $S^2= {\mathbb C}\cup \{\infty\}$ that I will identify with the unit sphere with the (angular) distance $d$. Each component $C$ of $U^c= {\mathbb C} \setminus U$ is either closed in $S^2$ (precisely when $C$ is bounded in the complex plane) or not closed and its closure $\bar{C}$ equals $C\cup \{\infty\}$. Suppose that $C_k$ is a sequence of unbounded complementary components. We have the corresponding sequence $\bar{C}_k$ of compacts in $S^2$.

The subset ${\mathcal K}(S^2)$ of nonempty compacts in $S^2$ is equipped with the Hausdorff distance $d_H$. The metric space $({\mathcal K}(S^2), d_H)$ is compact. Hence, WLOG, we can assume that the sequence of compacts $\bar{C}_k$ converges to some compact $K$. Since $\infty\in \bar{C}_k$ for each $k$, $\infty\in K$ as well.

Moreover, since all compacts $\bar{C}_k$ are disjoint from the open set $U$, so is their limit $K$. It is a nice exercise to check that the Hausdorff-limit of a sequence of connected compacts is again connected (may fail to be path-connected though). It does not necessarily follow that $K':=K\setminus \{\infty\}$ is connected, but it does follow that $K'$ contains no compact components. From the viewpoint of the Euclidean metric on the complex plane, this simply means that each component of $K'$ is unbounded.

Suppose now that $z\in U^c$ is the limit of a sequence $z_k\in C_k$. Then $z\in K'$. Let $C$ denote the component of $U^c$ containing $z$. Let $K'_z$ denote the component of $K'$ containing $z$. As noted above, $K'_z$ is noncompact. Since $K'_z\subset U^c$ is connected, we get $K'_z\subset C$. Since $K'_z$ is unbounded as a subset of the complex plane, so is $C$. Thus, we obtain:

Proposition. The union of unbounded components of $U^c$ is a closed subset of the complex plane.

Remark. 1. There is nothing special here about the complex plane, the same result holds for domains in ${\mathbb R}^n$.

  1. In my proof I never used connectedness of $U$, so the result holds for all open subsets of ${\mathbb R}^n$.
$\endgroup$
3
  • $\begingroup$ This is a nice proof, but I was looking for something that avoids the use of the Riemann sphere. This proof might be able to be "translated" in the Euclidean metric, though it's definitely not easy to do so. $\endgroup$ Nov 15, 2021 at 19:39
  • $\begingroup$ @StefanOctavian: One can easily adopt the same proof to the Euclidean metric, even though it would be artificial. Namely, you introduce a Euclidean disk $B$ of sufficiently large radius $R$ that contains the entire sequence $z_k$ and its limit $z$. Then replace all the components $C_k$ by components of their intersections with $B$ containing both $z_k$ and a point in $\partial B$. Then conclude that the component $C\subset U^c$ containing $z$, has nonempty intersection with the boundary of $B$. Since this holds for all $R\ge R_0$, you conclude that $C$ is unbounded. $\endgroup$ Nov 15, 2021 at 19:45
  • $\begingroup$ In any case, if you want to study complex analysis (I think, that's the origin of the question), using the Riemann sphere should be one of the standard tools in your skill-set. Doing otherwise, will seriously handicap you. $\endgroup$ Nov 15, 2021 at 19:48

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.