Theorem. Let \(U\) be a simply connected open set in \(\mathbb{R}^2\). If \(C\) is a simple closed curve lying in \(U\), then each bounded component of \(\mathbb{R}^2 - C\) also lies in \(U\).
(This condition actually characterizes the simply connected open sets of \(\mathbb{R}^2\). See [RW]. The space \(\mathbb{R}^2 - C\) has, of course, only one bounded component, as we shall prove in the next section.)
Solution
Goal: Prove that if \(U \subseteq \mathbb{R}^2\) is a simply connected open set and \(C \subseteq U\) is a simple closed curve, then every bounded component of \(\mathbb{R}^2 \setminus C\) is contained in \(U\).
1. Setting up proof by contradiction:Proof:
1.1. Let \(B\) be a bounded component of \(\mathbb{R}^2 \setminus C\).
1.2. Suppose for contradiction that \(B \not\subseteq U\).
1.3. Then there exists a point \(p \in B\) such that \(p \notin U\).
2. Fundamental group of the punctured plane and winding number:Proof:
2.1. Parameterize the simple closed curve \(C\) as a continuous loop \(\gamma: S^1 \to C \subseteq U\) based at some point \(x_0 \in C\).
2.2. Since \(p \in B\) lies in a bounded component of \(\mathbb{R}^2 \setminus C\), the Jordan curve theorem / winding number theorem (Theorem 61.1) implies that the loop \(\gamma\) generates a non-trivial homology/homotopy class in the punctured plane \(\mathbb{R}^2 \setminus \{p\}\):\[[\gamma] \neq 1 \in \pi_1(\mathbb{R}^2 \setminus \{p\}, x_0) \cong \mathbb{Z}.\]
3. Nullhomotopy in \(U\) and induced homomorphism:Proof:
3.1. Since \(p \notin U\), we have the topological inclusion \(i: U \hookrightarrow \mathbb{R}^2 \setminus \{p\}\).
3.2. The loop \(\gamma\) lies entirely inside the simply connected open set \(U\).
3.3. Because \(\pi_1(U, x_0) = \{1\}\), the loop \(\gamma\) is nullhomotopic in \(U\): \([\gamma]_U = 1 \in \pi_1(U, x_0)\).
3.4. The continuous inclusion map \(i: U \to \mathbb{R}^2 \setminus \{p\}\) induces a group homomorphism:\[i_*: \pi_1(U, x_0) \to \pi_1(\mathbb{R}^2 \setminus \{p\}, x_0).\]
3.5. Applying \(i_*\) gives:\[[\gamma]_{\mathbb{R}^2 \setminus \{p\}} = i_*([\gamma]_U) = i_*(1) = 1 \in \pi_1(\mathbb{R}^2 \setminus \{p\}, x_0).\]
4. Contradiction and conclusion:Proof:
4.1. Step 3 proves \([\gamma] = 1\) in \(\pi_1(\mathbb{R}^2 \setminus \{p\}, x_0)\), directly contradicting step 2 which established \([\gamma] \neq 1\).
4.2. Hence no such point \(p \in B \setminus U\) exists, which forces \(B \subseteq U\). Q.E.D.