- Let Ω be an open set in C and \(z \in \Omega\) . The connected component (or simply the component) of Ω containing z is the set \(\mathcal { C } _ { z }\) of all points w in Ω that can be joined to z by a curve entirely contained in Ω.
- Check first that \(\mathcal { C } _ { z }\) is open and connected. Then, show that \(w \in \mathcal { C } _ { z }\) defines an equivalence relation, that is: \(\mathrm { ( i ) } \ z \in \mathcal { C } _ { z } \ , \ \mathrm { ( i i ) }\) \(w \in \mathcal { C } _ { z }\) implies \(z \in \mathcal { C } _ { w }\) , and (iii) if \(w \in \mathcal { C } _ { z }\) and \(z \in { \mathcal { C } } _ { \zeta }\) , then \(w \in \mathcal { C } _ { \zeta }\)
Thus Ω is the union of all its connected components, and two components are either disjoint or coincide.
Show that Ω can have only countably many distinct connected components.
Prove that if Ω is the complement of a compact set, then Ω has only one unbounded component.
[Hint: For (b), one would otherwise obtain an uncountable number of disjoint open balls. Now, each ball contains a point with rational coordinates. For (c), note that the complement of a large disc containing the compact set is connected.]
Solution
Let \(\mathcal C_z\) be the set of points of \(\Omega\) that can be joined to \(z\) by a curve lying in \(\Omega\).
For openness, take \(w\in\mathcal C_z\). Since \(\Omega\) is open, some disc \(D(w,r)\) is contained in \(\Omega\). Every \(u\in D(w,r)\) can be joined to \(w\) by the line segment in that disc, and concatenating this segment with a curve from \(z\) to \(w\) shows \(u\in\mathcal C_z\). Hence \(\mathcal C_z\) is open. It is path connected by definition, hence connected.
The relation \(w\sim z\) defined by \(w\in\mathcal C_z\) is reflexive (use the constant curve), symmetric (reverse a joining curve), and transitive (concatenate joining curves). Hence it is an equivalence relation. Its equivalence classes are exactly the connected components \(\mathcal C_z\), so two components are either equal or disjoint and their union is \(\Omega\).
For countability, every component is a nonempty open subset of \(\mathbb C\cong\mathbb R^2\), hence contains a point of \(\mathbb Q^2\). Distinct components are disjoint, so choosing one rational point from each component gives an injection from the set of components into the countable set \(\mathbb Q^2\). Therefore there are at most countably many components.
Finally suppose \(\Omega=\mathbb C\setminus K\) with \(K\) compact. Choose \(R>0\) such that \(K\subset D(0,R)\). The exterior \[ E=\{z:|z|>R\} \] is connected and lies in \(\Omega\), so it is contained in a single component \(U\) of \(\Omega\). If \(V\) is any unbounded component, then \(V\) contains some point with modulus \(>R\), hence meets \(E\subset U\). Components that meet are equal, so \(V=U\). Thus \(\Omega\) has exactly one unbounded component.