Let $U$ be an open set in $\mathbb{C}$. I would like to prove the following result:

There exists a sequence of compact sets $\{K_n\}$ with the following properties:

  1. Each $K_n$ is a subset of $U$.
  2. $K_n \subset \mathrm{int}(K_{n+1}) \ \forall n$, where $\mathrm{int}()$ denotes interior.
  3. $\bigcup_{n} K_n = U$.
  4. each bounded component of the complement of $K_n$ meets the complement of $U$

(This result is used without proof in Theorem 1.4.3 of "Several Complex Variables with Connections to Algebraic Geometry and Lie Groups" by Joseph L. Taylor.)

Construction of a sequence with the first 3 properties has already been answered here. How do I ensure that the fourth property is also satisfied?

  • $\begingroup$ Why not just take many small closed disks $\subset U$, why do you think it may fail. $\endgroup$ – reuns Feb 6 at 9:24
  • $\begingroup$ @reuns $U$ need not be a connected set. $\endgroup$ – rationalbeing Feb 6 at 9:33

The construction in the linked answer does satisfy condition (4)!

To explain why, I'm going to write out the construction in a slightly different way.

First, I'm going to extend the complex plane to include the point at infinity. The resulting space is the Riemann sphere, $S^2 := \mathbb C \cup \{ \infty \}$.

Now for each $n \in \mathbb N$, we define the open set, $$ V_n = B(\infty , n ) \cup \bigcup_{a \in \mathbb C - U} B (a, \tfrac 1 n ) \ \ \ \ (\star),$$

where $B(a , \tfrac 1 n )$ is the disk $ \{ z \in S^2 : |z - a | < \tfrac 1 n \}$, and $B(\infty , n )$ is the "disk at $\infty$", $\{ z \in S^2 : |z| > n \}$.

Then we define

$$ K_n = \mathbb C - V_n.$$

It's clear that these $K_n$'s obey (1), (2) and (3). Your question is asking why these $K_n$'s obey (4).

I claim that for each $n$, every connected component $C$ of $V_n$ meets $S^2 - U$. (In other words, $K_n$ doesn't have more holes than $U$.) This is sufficient to imply your property (4).

To prove this, pick a $z \in C$. Notice that $z$, being an element of $V_n$, must be contained in $B(a, \tfrac 1 n)$ for some $a \in \mathbb C - U$ (or in $B(\infty, n)$ for that matter, but the argument in this case is the same, so I won't bother writing it out). Then observe that $C \cup B(a, \tfrac 1 n )$ is a connected subset of $V_n$, since it is the union of two connected subsets of $V_n$ with non-empty intersection. But we know that $C$ is a maximal connected subset of $V_n$ (this is what is means for $C$ to be a connected component of $V_n$), so $B(a, \tfrac 1 n) $ must actually be contained in $C$. Hence $C$ contains $a$, which is in $S^2 - U$.

In fact, we can prove a stronger result. Notice that no connected component of $S^2 - U$ can intersect two different connected components of $V_n$. (Since $S^2 - U \subset V_n$, each connected component $D \subset S^2 - U$ is also a connected set in $V_n$. So if $C_1$ and $C_2$ are two different connected components of $V_n$ that both intersect $D$, then $C_1 \cup D \cup C_2$ would be a connected subset of $V_n$ that is strictly larger than $C_1$ and $C_2$, contradicting the assumption that $C_1$ and $C_2$ are maximal connected subsets in $V_n$.) The conclusion is that each connected component of $S^2 - U$ is contained in a single connected component of $V_n$. Thus, having previously established that each connected component $C$ of $V_n$ contains a point $a \in S^2 - U$, and having now established that the connected component $D$ of $S^2 - U$ that contains $a$ must in fact be wholly contained in $C$, we arrive at the following conclusion:

Property (4a): For each $n \in \mathbb N$, every connected component of $S^2 - K_n$ contains a connected component of $S^2 - U$.

P.S. My answer is based on Rudin, Theorem 13.3.

  • $\begingroup$ Thanks! I had seen the proof in Rudin but it didn't make much sense to me. Now it is much better :) But why can't we prove this directly, without going for one point compactification of the complex plane? $\endgroup$ – rationalbeing Feb 5 at 4:19
  • 1
    $\begingroup$ @rationalbeing It just seems easier this way! You handle the "unbounded" region using the "disk at infinity". $\endgroup$ – Kenny Wong Feb 5 at 8:08
  • $\begingroup$ So by taking the compactification of space, we get rid of the problem that might occur when $U$ is unbounded. Like when $U=\mathbb{C}$. As pointed here. With your definition of $B(\infty, n)$ it makes sense to consider $K_n = \mathbb{C}\setminus V_n$. Thanks! $\endgroup$ – rationalbeing Feb 5 at 10:31

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