# Nested sequence of compact subsets covering an open set in $\mathbb{C}$

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?

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

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.

• 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? – rationalbeing Feb 5 at 4:19
• @rationalbeing It just seems easier this way! You handle the "unbounded" region using the "disk at infinity". – Kenny Wong Feb 5 at 8:08
• 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! – rationalbeing Feb 5 at 10:31