Maximal dyadic cubes covering a proper open set

$\mathcal{F}$ is nonempty and covers $\mathrm{\Omega }$ locally: fix $x\in \mathrm{\Omega }$; by [F2] there is $r>0$ with $B_2(x,r)\subseteq \mathrm{\Omega }$. Choose $k$ with $3n\sqrt{n}{2}^{-k}<r$, let $Q$ be the generation-$k$ dyadic cube containing $x$, and let $y\in 5\sqrt{n}Q$. Since $x,y\in 5\sqrt{n}Q$ and $5\sqrt{n}Q$ has side $5\sqrt{n}{2}^{-k}$, [F1] gives $d_{\mathrm{\infty }}(x,c(Q))\le \frac{1}{2}{2}^{-k}$ and $d_{\mathrm{\infty }}(c(Q),y)\le \frac{5}{2}\sqrt{n}{2}^{-k}$, so by [F2] $d_2(x,y)\le d_2(x,c(Q))+d_2(c(Q),y)\le n(\frac{1}{2}+\frac{5}{2}\sqrt{n})2^{-k}\le 3n\sqrt{n}{2}^{-k}<r$; hence $y\in B_2(x,r)\subseteq \mathrm{\Omega }$. Thus $5\sqrt{n}Q\subseteq \mathrm{\Omega }$ and $Q\in \mathcal{F}$ contains $x$.

Maximal elements exist: let $Q\in \mathcal{F}$ and $w\mathrm{\notin }\mathrm{\Omega }$, which exists because $\mathrm{\Omega }$ is proper, and fix $x\in Q$. If an ancestor $A$ of $Q$ of side $s$ lies in $\mathcal{F}$, then $5\sqrt{n}A\subseteq \mathrm{\Omega }$, so $w\mathrm{\notin }5\sqrt{n}A$; now $x\in Q\subseteq A$ gives $d_{\mathrm{\infty }}(x,c(A))\le s\mathrm{/}2$ and every point of $5\sqrt{n}A$ is within ${\mathrm{\ell }}^{\mathrm{\infty }}$-distance $(5\sqrt{n}\mathrm{/}2)s$ of $c(A)$, so [F2] gives $d_2(x,c(A))\le ns\mathrm{/}2$ and, for $z=w$ first, $d_2(w,x)\ge d_{\mathrm{\infty }}(w,x)\ge \frac{1}{2}(5\sqrt{n}-1)s$, by the reverse triangle inequality in $d_{\mathrm{\infty }}$; that is, $s\le 2d_2(w,x)\mathrm{/}(5\sqrt{n}-1)=:M$. The ancestors of $Q$ have side lengths $2^{-j}$ with $j\le k$ increasing as $j$ decreases, so by [F4] only finitely many of them have side $s\le M$; hence only finitely many ancestors of $Q$ lie in $\mathcal{F}$, and among those finitely many there is one of least generation, which is a maximal element of $\mathcal{F}$ containing $Q$. Taking $Q$ arbitrary shows that every cube of $\mathcal{F}$ lies below a maximal element, and in particular maximal elements exist.

The maximal elements are pairwise disjoint: if $Q,Q^{\mathrm{\prime }}$ are maximal and meet, then by [F1] one contains the other, and maximality forces $Q=Q^{\mathrm{\prime }}$. They are at most countable: the map sending a dyadic cube to its centre is injective on any family of pairwise disjoint cubes (a cube contains its own centre), its values are points of ${\mathbb{Q}}^n$ because $c(Q_{k,m}{)}_i=(m_i+\frac{1}{2})2^{-k}\in \mathbb{Q}$, and ${\mathbb{Q}}^n$ is at most countable, so [F3] makes the family at most countable.

Their union is exactly $\mathrm{\Omega }$: each maximal element lies in $\mathcal{F}$, hence is contained in $\mathrm{\Omega }$, so the union is a subset of $\mathrm{\Omega }$; conversely, for $x\in \mathrm{\Omega }$ step 1.1 supplies $Q\in \mathcal{F}$ with $x\in Q$, and step 2.1 supplies a maximal element containing $Q$, hence containing $x$. Thus $\mathrm{\Omega }=\bigcup \{Q:Q\text{ maximal in }\mathcal{F}\}$.

Let $Q$ be maximal in $\mathcal{F}$ and let $P$ be its dyadic parent: $P$ has side $2\mathrm{\ell }(Q)$, its centre differs from $c(Q)$ by at most $\frac{1}{2}\mathrm{\ell }(Q)$ in each coordinate, so $P\subseteq 3Q$. Maximality gives $P\mathrm{\notin }\mathcal{F}$, that is, $5\sqrt{n}P\subseteq ̸\mathrm{\Omega }$, so there is $y\in 5\sqrt{n}P$ with $y\mathrm{\notin }\mathrm{\Omega }$; every point of $P$ is within ${\mathrm{\ell }}^{\mathrm{\infty }}$-distance $\mathrm{\ell }(Q)$ and every point of $5\sqrt{n}P$ within ${\mathrm{\ell }}^{\mathrm{\infty }}$-distance $5\sqrt{n}\mathrm{\ell }(Q)$ of the centre of $P$, so [F2] bounds the $d_2$-distance between any $x\in Q\subseteq P$ and $y$ by $n\mathrm{\ell }(Q)+5n\sqrt{n}\mathrm{\ell }(Q)\le 6n\sqrt{n}\mathrm{\ell }(Q)$; in particular $\mathrm{dist}(Q,{\mathrm{\Omega }}^c)\le 6n\sqrt{n}\mathrm{\ell }(Q)$ while $5\sqrt{n}Q\subseteq \mathrm{\Omega }$.