Whitney decomposition of a proper open subset of Euclidean space

The distance rule. For $x\in \mathrm{\Omega }$ put $d(x)=\mathrm{dist}(x,{\mathrm{\Omega }}^c)>0$; the positivity uses that $\mathrm{\Omega }$ is open and ${\mathrm{\Omega }}^c$ closed. Since the dyadic numbers $2^{-k}$, $k\in \mathbb{Z}$, partition $(0,\mathrm{\infty })$ into the intervals $(2^{-k-1},2^{-k}]$, there is exactly one $k(x)\in \mathbb{Z}$ with $2^{-k(x)-1}<d(x)\mathrm{/}(4\sqrt{n})\le 2^{-k(x)}$, and then $d(x)\mathrm{/}(4\sqrt{n})\le \mathrm{\ell }(x):=2^{-k(x)}<d(x)\mathrm{/}(2\sqrt{n})\mathrm{.}$ Let $Q(x)$ be the unique dyadic cube of generation $k(x)$ containing $x$, which exists because generation $k(x)$ partitions ${\mathbb{R}}^n$ by [L1]. Put $G=\{Q(x):x\in \mathrm{\Omega }\}$.

Cubes of the rule are contained in $\mathrm{\Omega }$ with controlled distance. If $y\in Q(x)$, then $\mathrm{\mid }y-x\mathrm{\mid }\le \sqrt{n}\mathrm{\ell }(x)<d(x)\mathrm{/}2$, so $d(y)\ge d(x)-\mathrm{\mid }y-x\mathrm{\mid }>d(x)\mathrm{/}2\ge \sqrt{n}\mathrm{\ell }(x)$ by [L2], and therefore $d(Q(x))={\inf }_{y\in Q(x)}d(y)\ge \sqrt{n}\mathrm{\ell }(x)>0$. In particular $Q(x)\cap {\mathrm{\Omega }}^c=\mathrm{\varnothing }$, that is, $Q(x)\subseteq \mathrm{\Omega }$, and $Q(x)\in G$ satisfies the lower bound of claim 1. Thus every point of $\mathrm{\Omega }$ lies in an element of $G$.

Maximal elements. For each fixed $Q\in G$, let $G_Q:=\{Q^{\mathrm{\prime }}\in G:Q\subseteq Q^{\mathrm{\prime }}\}$. Choose the witness $x\in \mathrm{\Omega }$ with $Q=Q(x)$ supplied by the definition of $G$. Every member $Q^{\mathrm{\prime }}$ of this particular set $G_Q$ contains $x$. By step 2.1, $\sqrt{n}\mathrm{\ell }(Q^{\mathrm{\prime }})\le d(Q^{\mathrm{\prime }})\le d(x)$, while dyadic nesting gives $\mathrm{\ell }(Q^{\mathrm{\prime }})\ge \mathrm{\ell }(Q)$. Thus its side length lies in the finite dyadic range $[\mathrm{\ell }(Q),d(x)\mathrm{/}\sqrt{n}]$; at each generation there is only one dyadic cube containing this fixed $x$. Therefore $G_Q$ is finite and nonempty. Its members are nested, so it has a unique largest member by side length, which is maximal in $G$. Let $\mathcal{W}$ be the set of all maximal elements of $G$. Every $Q\in G$ lies in one of them by the preceding finite-superset argument; the set of all dyadic cubes is countable, hence so is $\mathcal{W}$. This construction uses only $G$-supersets of each fixed cube, not a maximality principle over all cubes.

Covering, disjointness and the lower bound. Every $x\in \mathrm{\Omega }$ lies in some $Q(x)\in G$, hence in a maximal element of $G$ containing it, so $\mathrm{\Omega }={\bigcup }_{Q\in \mathcal{W}}Q$. Dyadic cubes are nested or disjoint by [L1], and maximality of the elements of $\mathcal{W}$ rules out proper inclusion, so distinct elements of $\mathcal{W}$ are disjoint (their interiors are disjoint, indeed the cubes themselves are disjoint as half-open sets). Each $Q\in \mathcal{W}$ belongs to $G$, so $\sqrt{n}\mathrm{\ell }(Q)\le d(Q)$ by step 2.1.

The upper bound is built into the rule. Fix $Q\in \mathcal{W}$ and $x\in \mathrm{\Omega }$ with $Q=Q(x)$, which exists because $\mathcal{W}\subseteq G$. By definition of $\mathrm{\ell }(x)$ one has $d(x)\le 4\sqrt{n}\mathrm{\ell }(x)$, and $d(Q)\le d(x)$ because $x\in Q$; with the lower bound of step 4.1 this gives $\sqrt{n}\mathrm{\ell }(Q)\le d(Q)\le d(x)\le 4\sqrt{n}\mathrm{\ell }(Q)$, which proves claim 1.

Touching cubes have comparable sizes. Let $Q,Q^{\mathrm{\prime }}\in \mathcal{W}$ have intersecting closures and let $z$ be a common point. Then $d(z)\ge d(Q)\ge \sqrt{n}\mathrm{\ell }(Q)$ by [L2] and step 4.1. Choose $y\in Q^{\mathrm{\prime }}$ with $d(y)<d(Q^{\mathrm{\prime }})+\epsilon$; since $z\in \overline{Q^{\mathrm{\prime }}}$ one has $\mathrm{\mid }z-y\mathrm{\mid }\le \mathrm{diam}(Q^{\mathrm{\prime }})=\sqrt{n}\mathrm{\ell }(Q^{\mathrm{\prime }})$, so $d(z)\le \mathrm{\mid }z-y\mathrm{\mid }+d(y)<\sqrt{n}\mathrm{\ell }(Q^{\mathrm{\prime }})+d(Q^{\mathrm{\prime }})+\epsilon \le \sqrt{n}\mathrm{\ell }(Q^{\mathrm{\prime }})+4\sqrt{n}\mathrm{\ell }(Q^{\mathrm{\prime }})+\epsilon =5\sqrt{n}\mathrm{\ell }(Q^{\mathrm{\prime }})+\epsilon ,$ the last inequality by step 5.1 applied to $Q^{\mathrm{\prime }}$. Letting $\epsilon \downarrow 0$ gives $\sqrt{n}\mathrm{\ell }(Q)\le 5\sqrt{n}\mathrm{\ell }(Q^{\mathrm{\prime }})$, hence $\mathrm{\ell }(Q)\le 5\mathrm{\ell }(Q^{\mathrm{\prime }})$; interchanging the roles of the two cubes gives $\mathrm{\ell }(Q^{\mathrm{\prime }})\le 5\mathrm{\ell }(Q)$. This proves claim 2 with the stated factor $5$.

Bounded overlap of small dilates. Fix $1\le R\le 2$ and $x\in {\mathbb{R}}^n$, and let $\mathcal{J}$ be the set of $j$ with $x\in R{Q}_j$; write $c_j$ for the centre, ${\mathrm{\ell }}_j=\mathrm{\ell }(Q_j)$ and $B_j=B(c_j,{\mathrm{\ell }}_j\mathrm{/}2)$. Distinct cubes of the family are disjoint and each contains $B_j$, so the balls $B_j$ are pairwise disjoint. If $x\in R{Q}_j$, then $x=c_j+R(y-c_j)$ for some $y\in Q_j$, so $\mathrm{dist}(x,Q_j)\le \mathrm{\mid }x-y\mathrm{\mid }=(R-1)\mathrm{\mid }y-c_j\mathrm{\mid }\le \frac{\sqrt{n}}{2}{\mathrm{\ell }}_j$. Since $d$ is 1-Lipschitz and $d(z)\ge d(Q_j)$ for $z\in Q_j$, this gives $d(x)\ge d(Q_j)-\mathrm{dist}(x,Q_j)\ge \frac{\sqrt{n}}{2}{\mathrm{\ell }}_j>0$. For the other direction, for each $\delta >0$ choose $u,v\in Q_j$ with $\mathrm{\mid }x-u\mathrm{\mid }<\mathrm{dist}(x,Q_j)+\delta$ and $d(v)<d(Q_j)+\delta$. Lipschitz continuity and $\mathrm{\mid }u-v\mathrm{\mid }\le \mathrm{diam}(Q_j)$ give $d(x)\le d(v)+\mathrm{\mid }x-v\mathrm{\mid }\le d(Q_j)+\mathrm{dist}(x,Q_j)+\mathrm{diam}(Q_j)+2\delta$; letting $\delta \downarrow 0$ yields $d(x)\le d(Q_j)+\mathrm{dist}(x,Q_j)+\mathrm{diam}(Q_j)$. By step 5.1 and $\mathrm{diam}(Q_j)=\sqrt{n}{\mathrm{\ell }}_j$, $d(x)\le 4\sqrt{n}{\mathrm{\ell }}_j+\sqrt{n}{\mathrm{\ell }}_j+\frac{\sqrt{n}}{2}{\mathrm{\ell }}_j=\frac{11}{2}\sqrt{n}{\mathrm{\ell }}_j,$ so ${\mathrm{\ell }}_j\ge 2d(x)\mathrm{/}(11\sqrt{n})$. The lower bound for $d(x)$ also gives ${\mathrm{\ell }}_j\le 2d(x)\mathrm{/}\sqrt{n}$, hence $\mathrm{\mid }x-c_j\mathrm{\mid }\le R\frac{\sqrt{n}}{2}{\mathrm{\ell }}_j\le \sqrt{n}{\mathrm{\ell }}_j\le 2d(x)$. The disjoint balls $B(c_j,d(x)\mathrm{/}(11\sqrt{n}))\subseteq B_j$ therefore have centres in $B(x,3d(x))$ and a common radius $d(x)\mathrm{/}(11\sqrt{n})$. The packing estimate [L3] bounds their number by $(66\sqrt{n}+1{)}^n$. Thus the dilated cubes have overlap at most $C_n=(66\sqrt{n}+1{)}^n$, uniformly for $1\le R\le 2$.

Counting touching cubes. Fix $Q\in \mathcal{W}$ with side $\mathrm{\ell }$ and generation $k$. By step 6.1, a touching cube has side in $[\mathrm{\ell }\mathrm{/}5,5\mathrm{\ell }]$, so its generation lies in $\{k-2,k-1,k,k+1,k+2\}$. At generations $k,k+1,k+2$, at most $3,4,6$ coordinate intervals, respectively, have closures meeting a given closed interval of length $\mathrm{\ell }$; hence the counts are at most $3^n,4^n,6^n$. At each of the two coarser generations, the interval of $Q$ lies inside a single dyadic interval of that generation. Its closure can meet at most that interval and one adjacent interval, because $\mathrm{\ell }$ is strictly less than the coarse side length and all endpoints lie on the fine grid. Thus each coarser generation contributes at most $2^n$ cubes. Summing proves claim 3 with $K(n)=2\cdot 2^n+3^n+4^n+6^n$.

Conclusion. Steps 4.1 and 5.1 provide a countable family of dyadic cubes with pairwise disjoint interiors, union $\mathrm{\Omega }$ and the two-sided distance estimate; step 6.1 gives the touching size comparison; step 7.1 gives the explicit touching count $K(n)=2\cdot 2^n+3^n+4^n+6^n$; step 6.2 gives the bounded overlap of the dilations with $1\le R\le 2$. This proves the lemma.