- Show that every open \(U \subseteq \RR\) is a countable union of disjoint open intervals.
Solution
Let \(U \subseteq \RR\) be open.
1. Each connected component \(C\) of \(U\) is an open interval \((a,b)\) with \(-\infty \leq a < b \leq \infty\).
Components of an open subset of a locally connected space are open, so \(C\) is open in \(\RR\). A nonempty connected subset of \(\RR\) is an interval, and an open interval of \(\RR\) has the form \((a,b)\) with \(a,b \in [-\infty, \infty]\).
2. Distinct components are disjoint, and \(U\) is their union.
The components are the equivalence classes of the relation \(x \sim y\) if and only if \(x\) and \(y\) lie in a common connected subset of \(U\), and equivalence classes partition \(U\).
3. There are at most countably many components.
Each component is a nonempty open interval, so it contains a rational number; choose one, \(q_C\), for each component \(C\). By step 2 distinct components are disjoint, so \(C \mapsto q_C\) is injective into the countable set \(\QQ\).
4. Q.E.D.
Steps 1 and 2 express \(U\) as a disjoint union of open intervals, and step 3 shows the family is countable.