- Show that every open \(U \subseteq \RR^n\) is a countable union of almost disjoint closed cubes.
Solution
A dyadic cube of level \(k \geq 0\) is a cube \(\prod_{i=1}^n [m_i 2^{-k}, (m_i+1)2^{-k}]\) with \(m_i \in \ZZ\). Its parent is the unique dyadic cube of level \(k-1\) containing it, for \(k \geq 1\). Closed cubes are almost disjoint if their interiors are pairwise disjoint. Let \(U \subseteq \RR^n\) be open.
1. The dyadic cubes of a fixed level are countably many, are almost disjoint, and cover \(\RR^n\). Two dyadic cubes are either almost disjoint or one contains the other.
The cubes of level \(k\) are indexed by \(\ZZ^n\) and form the grid of side \(2^{-k}\). If \(Q\) has level \(k\) and \(Q'\) has level \(k' \leq k\), then \(Q\) lies in exactly one cube of level \(k'\), its ancestor; either that ancestor is \(Q'\), or it is almost disjoint from \(Q'\) and hence so is \(Q\).
2. If \(U = \RR^n\), the level-\(0\) dyadic cubes give the required union.
Step 1.
3. Assume \(U \neq \RR^n\), and let \(\mathcal F\) be the set of dyadic cubes \(Q \subseteq U\) that are of level \(0\) or whose parent is not contained in \(U\). Then \(U = \bigcup_{Q \in \mathcal F} Q\).
3.1. Every \(x \in U\) lies in some dyadic cube contained in \(U\).
Choose \(r > 0\) with \(B(x,r) \subseteq U\). A dyadic cube containing \(x\) of side \(2^{-k} < r/\sqrt n\) has diameter less than \(r\), so it lies in \(B(x,r)\).
3.2. Every \(x \in U\) lies in some \(Q \in \mathcal F\).
For each \(k\) choose a dyadic cube \(Q_k(x) \ni x\) of level \(k\) with \(Q_{k+1}(x) \subseteq Q_k(x)\), starting from a level-\(0\) cube containing \(x\) and descending by children. By step 1 some \(Q_k(x) \subseteq U\) for large \(k\), since every level-\(k\) cube containing \(x\) has diameter \(\sqrt n\,2^{-k}\). Let \(k_0\) be the least \(k\) with \(Q_k(x) \subseteq U\). Then either \(k_0 = 0\), or the parent \(Q_{k_0-1}(x)\) is not contained in \(U\); in both cases \(Q_{k_0}(x) \in \mathcal F\).
3.3. Q.E.D.
Each \(Q \in \mathcal F\) lies in \(U\), and step 2 gives the reverse inclusion.
4. \(\mathcal F\) is countable and almost disjoint.
1. The parent is not contained in \(U\) because \(Q \in \mathcal F\), so \(Q' \not\subseteq U\), a contradiction.
5. Q.E.D.
Step 2 handles \(U = \RR^n\); otherwise steps 3 and 4 write \(U\) as the countable almost disjoint union of the cubes in \(\mathcal F\).