Prove Proposition 1.4: Every open set \(G\subset\mathbb R\) can be written as a finite or countable union of disjoint open intervals \((a_j,b_j)\) with at most one \(a_j=-\infty\) and at most one \(b_j=\infty\).
Solution
1. Components of \(G\) are open intervals.Proof: for \(x \in G\), let \(C_x\) be the connected component of \(x\) in \(G\).Since \(G\) is open, \(C_x\) is open (the components of an open set in \(\mathbb{R}\) are open: if \(y \in C_x\), then \((y-\delta, y+\delta) \subseteq G\) for some \(\delta\), and \((y-\delta,y+\delta)\cup C_x\) is connected, so it equals \(C_x\)). Being a connected subset of \(\mathbb{R}\), \(C_x\) is an interval; an open interval \((a_x, b_x)\) with \(a_x, b_x \in [-\infty,\infty]\).Distinct components are disjoint, and \(G = \bigcup_{x\in G} C_x\).
2. The collection of components is countable.Proof: each component \((a_x, b_x)\) is a nonempty open interval, so it contains a rational \(q\).Distinct components are disjoint, hence contain distinct rationals; choosing one rational per component gives an injection from the set of components into \(\mathbb{Q}\), which is countable.
3. Endpoint conditions.Proof: at most one component has \(a_j = -\infty\): two distinct components are disjoint, and two intervals \((a_1, b_1)\), \((a_2, b_2)\) both unbounded below would both contain \((\min(a_1,a_2), \min(b_1,b_2))\) — they would intersect.Similarly, at most one component has \(b_j = +\infty\).
4. Q.E.D.