Let \(C_1\) and \(C_2\) be disjoint simple closed curves in \(S^2\).
Show that \(S^2 - C_1 - C_2\) has precisely three components. [Hint: If \(W_1\) is the component of \(S^2 - C_1\) disjoint from \(C_2\), and if \(W_2\) is the component of \(S^2 - C_2\) disjoint from \(C_1\), show that \(\overline{W}_1 \cup \overline{W}_2\) does not separate \(S^2\).]
Show that these three components have boundaries \(C_1\) and \(C_2\) and \(C_1 \cup C_2\), respectively.
Solution
By the Jordan curve theorem, each \(C_i\) separates \(S^2\) into exactly two components, each having boundary \(C_i\).
Since \(C_2\) is connected and disjoint from \(C_1\), it lies wholly in one component of \(S^2-C_1\). Let \(W_1\) be the other component. Similarly, let \(W_2\) be the component of \(S^2-C_2\) disjoint from \(C_1\). Then \(W_1\) and \(W_2\) are disjoint: if a point lay in both, the connected sets \(C_1\) and \(C_2\) would lie on opposite sides of each other in both Jordan decompositions, which is impossible; equivalently, by the Jordan-Schoenflies theorem the closures \(\overline W_1\) and \(\overline W_2\) are disjoint closed disks.
The set \[ K=\overline W_1\cup\overline W_2 \] is the union of two disjoint closed disks. Its complement is an open annulus and is therefore connected. Hence \[ R=S^2-K \] is connected. Moreover \[ S^2-(C_1\cup C_2)=W_1\ \sqcup\ R\ \sqcup\ W_2. \] All three sets are nonempty, open, and connected, so these are precisely the three components. This proves (a).
For (b), the Jordan curve theorem gives \[ \partial W_1=C_1,\qquad \partial W_2=C_2. \] The remaining component \(R\) lies between the two curves. Every point of \(C_1\) is approached from \(W_1\) on one side and from \(R\) on the other, so \(C_1\subset\partial R\); similarly \(C_2\subset\partial R\). Conversely, a boundary point of \(R\) cannot lie in the open set \(S^2-(C_1\cup C_2)\), since there it would belong to one of the three open components. Thus \[ \partial R\subset C_1\cup C_2. \] Therefore \[ \partial R=C_1\cup C_2. \] So the three component boundaries are \(C_1\), \(C_2\), and \(C_1\cup C_2\), respectively.