2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.