Locally finite refinements of Euclidean open covers have order at most the dimension plus one

For $x\in \mathrm{\Omega }$ define $\delta (x):={\sup }_{U\in \mathcal{U}}\min \{1,\mathrm{dist}(x,{\mathbb{R}}^m\setminus U)\},\epsilon (x):=\min \{1,\mathrm{dist}(x,{\mathbb{R}}^m\setminus \mathrm{\Omega }),\delta (x)\}\mathrm{.}$ The set in the supremum is nonempty and contained in $[0,1]$, so [F3] makes $\delta (x)$ well defined; the empty-complement convention in [F2] covers $U={\mathbb{R}}^m$ or $\mathrm{\Omega }={\mathbb{R}}^m$. Since $\mathcal{U}$ covers $\mathrm{\Omega }$ and each $U$ is open, $\delta (x)>0$. For any nonempty closed set $E$, the triangle inequality gives $\mathrm{dist}(x,E)\le d_2(x,y)+\mathrm{dist}(y,E)$ by taking the infimum over $E$, and interchanging $x,y$ gives the reverse bound; capping at $1$ preserves this $1$-Lipschitz estimate. The supremum of a family of $1$-Lipschitz functions is $1$-Lipschitz because $f_U(x)\le f_U(y)+d_2(x,y)$ for every $U$, hence ${\sup }_U{f}_U(x)\le {\sup }_U{f}_U(y)+d_2(x,y)$, and likewise after interchanging $x,y$. The minimum of two $1$-Lipschitz functions satisfies the same estimate, so $\epsilon$ is $1$-Lipschitz. For each $x$, the definition of the supremum supplies $U_x\in \mathcal{U}$ with $\min \{1,\mathrm{dist}(x,{\mathbb{R}}^m\setminus U_x)\}>\delta (x)\mathrm{/}2$, hence $B(x,\delta (x)\mathrm{/}2)\subset U_x$. Also $\epsilon (x)\le \mathrm{dist}(x,{\mathbb{R}}^m\setminus \mathrm{\Omega })$.

Let $\mathcal{Q}$ be the family of all closed dyadic cubes $Q\subset \mathrm{\Omega }$ satisfying $\mathrm{diam}(Q)<\frac{1}{16}{\inf }_{y\in Q}\epsilon (y)\mathrm{.}$ The infimum exists by [F3], since $\epsilon \ge 0$ and $Q$ is nonempty. Every point of $\mathrm{\Omega }$ belongs to such a cube at sufficiently fine dyadic scale, because $\epsilon$ is continuous and positive there. Each qualifying cube has diameter $<1\mathrm{/}16$, so along the nested dyadic ancestors of any such cube only finitely many can qualify; choose a maximal qualifying ancestor. The maximal cubes cover $\mathrm{\Omega }$ and have pairwise disjoint interiors, since two dyadic cubes with intersecting interiors are nested and a contained qualifying cube cannot be maximal.

The maximal cubes are locally finite. Fix $x\in \mathrm{\Omega }$, put $a=\epsilon (x)>0$, and choose $r=a\mathrm{/}8$. If a maximal cube $Q$ meets $B(x,r)$ and has side length $s<a\mathrm{/}(512\sqrt{m})$, its dyadic parent $P$ has diameter $2\sqrt{m}s<a\mathrm{/}256$. Every $y\in P$ then satisfies $d_2(y,x)<a\mathrm{/}8+a\mathrm{/}256<a\mathrm{/}4$, so $P\subset \mathrm{\Omega }$ and the Lipschitz bound gives $\epsilon (y)>3a\mathrm{/}4$ on $P$. Thus $\mathrm{diam}(P)<a\mathrm{/}256<(3a\mathrm{/}4)\mathrm{/}16\le {\inf }_P\epsilon \mathrm{/}16$, so $P$ also qualifies, contradicting maximality of $Q$. Hence each maximal cube meeting $B(x,r)$ has side at least $a\mathrm{/}(512\sqrt{m})$. All such cubes lie in the bounded set $B(x,r+1\mathrm{/}16)$ and their centers are separated by at least this lower bound in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }}$: for two cubes with disjoint interiors, some coordinate intervals have disjoint interiors, so the centers differ in that coordinate by at least half the sum of the two side lengths, which is at least the lower bound. Partition a bounded box containing these centers into finitely many boxes of maximum-norm diameter smaller than that separation; each small box contains at most one center. Thus only finitely many such cubes meet $B(x,r)$.

A geometric $k$-simplex is the convex hull of vertices $v_0,\dots ,v_k$, where the convex hull is the set of sums ${\sum }_i{t}_i{v}_i$ with $t_i\ge 0$ and ${\sum }_i{t}_i=1$, and the vertices are affinely independent, meaning that ${\sum }_i{t}_i{v}_i=0$ and ${\sum }_i{t}_i=0$ imply $t_i=0$ for all $i$; its faces are the convex hulls of subsets of its vertices. A triangulation of $\mathrm{\Omega }$ is a locally finite family of closed simplices of dimensions at most $m$ covering $\mathrm{\Omega }$ such that two simplices meet only in a common face or not at all. A cubical cell here is a product of closed intervals, some possibly singletons, with dyadic endpoints; its boundary is the union of its proper faces, and its center is the coordinatewise midpoint. We construct a compatible triangulation of the maximal cubes and their shared face pieces. First, touching maximal cubes have side lengths differing by at most a factor of two. Indeed, let $z\in Q\cap R$, let their sides be $S\ge s$, and let $P$ be the dyadic parent of $R$. The qualification of $Q$ gives $\epsilon (z)>16\sqrt{m}S$. Since $\mathrm{diam}(P)=2\sqrt{m}s\le 2\sqrt{m}S<\epsilon (z)\le \mathrm{dist}(z,{\mathbb{R}}^m\setminus \mathrm{\Omega })$, every point of $P$ lies in $\mathrm{\Omega }$; if $\mathrm{\Omega }={\mathbb{R}}^m$, this containment is automatic. Thus $P$ is an admissible cube in the maximality comparison for $R$. Maximality of $R$ implies that $P$ fails the diameter inequality. The Lipschitz bound and $d_2(y,z)\le \mathrm{diam}(P)$ for $y\in P$ give ${\inf }_P\epsilon \ge \epsilon (z)-2\sqrt{m}s$, so $2\sqrt{m}s=\mathrm{diam}(P)\ge \frac{1}{16}{\inf }_P\epsilon \ge \frac{\epsilon (z)-2\sqrt{m}s}{16}>\sqrt{m}S-\frac{\sqrt{m}s}{8}$. Hence $S<17s\mathrm{/}8$, and since the side lengths are dyadic their ratio is at most two. Now fix a face $F$ of a maximal cube $Q$ of side $S$. Every maximal cube $R$ meeting $F$ touches $Q$, so the side of $R$ is at least $S\mathrm{/}2$. Their centers lie in a bounded box around $Q$, since each qualifying cube has diameter less than $1\mathrm{/}16$. Two distinct maximal cubes have disjoint interiors, so in some coordinate their intervals have disjoint interiors and their centers differ by at least half the sum of their side lengths, hence by at least $S\mathrm{/}2$ in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }}$. Partitioning a bounded box containing these centers into finitely many boxes of maximum-norm diameter less than $S\mathrm{/}2$ shows that only finitely many maximal cubes meet $F$. Their intersections with $F$ are products of dyadic intervals and cover $F$; collecting their endpoints gives a finite rectangular subdivision. If $F$ is shared by two maximal cubes, the same set of cubes meets $F$, so both sides induce the same subdivision. Repeat this construction on the lower-dimensional faces. Include the maximal cubes as top-dimensional cells and the rectangular pieces in their subdivided faces as lower-dimensional cells; a partitioned face is represented by its pieces, not also by the unsplit face. Include all faces of these pieces. The rectangular boundary pieces have common-face intersections after the indicated lower-dimensional subdivisions; an unsplit maximal cube can meet another cube along a union of such pieces, rather than along a single unsplit face. All pieces on a shared overlap use the same dyadic endpoint grid: touching cubes have sides $S$ or $S\mathrm{/}2$ relative to a largest incident cube of side $S$, so the only possible interior coordinate cut of one of its faces is its midpoint. Thus restricting that grid to a smaller incident face gives its induced subdivision as well. The cell family is locally finite because every neighborhood meeting only finitely many maximal cubes meets only finitely many of their finite subdivisions. Triangulate the boundary pieces inductively by dimension, choosing the center of each positive-dimensional cell and coning it to its already triangulated boundary. Here coning a boundary simplex to a point means taking their convex hull. In dimension one this subdivides each interval into the two segments from its center to its endpoints; in every higher dimension convexity makes the cones cover the cell, and distinct cones meet only along their common triangulated boundary cones. Use a single center and boundary triangulation for each shared cell. Finally cone each maximal cube from its center to its triangulated boundary. This gives a locally finite triangulation of $\mathrm{\Omega }$ of dimension at most $m$, with every simplex contained in a maximal cube. No infinite hyperplane is extended across unrelated cubes and no finite compact subcomplex is substituted for $\mathrm{\Omega }$.

For a simplex with vertices $v_0,\dots ,v_k$, its barycentric coordinates on that simplex are the unique coefficient functions ${\lambda }_0,\dots ,{\lambda }_k$ satisfying $y={\sum }_i{\lambda }_i(y)v_i$, ${\sum }_i{\lambda }_i(y)=1$, and ${\lambda }_i(y)\ge 0$; these functions are affine on the simplex; its relative interior is the set where every ${\lambda }_i>0$. For a vertex $v$ of the triangulation, define ${\lambda }_v$ on each simplex to be its barycentric coordinate when $v$ is a vertex of that simplex, and zero otherwise. The common-face condition makes these definitions agree on intersections, and local finiteness makes ${\lambda }_v$ continuous on $\mathrm{\Omega }$. Let $\mathrm{st}(v):=\{y\in \mathrm{\Omega }:{\lambda }_v(y)>0\}$, equivalently the union of the relative interiors of simplices containing $v$; it is open. Because intersections are common faces, each point lies in the relative interior of a unique simplex of the triangulation, its carrier. If a point lies in $\mathrm{st}(v_0),\dots ,\mathrm{st}(v_k)$, then $v_0,\dots ,v_k$ are vertices of that same simplex, whose dimension is at most $m$; hence $k+1\le m+1$. A neighborhood meeting only finitely many simplices meets stars of only their finitely many vertices; hence the star cover is locally finite and has order at most $m+1$.

If $y\in \mathrm{st}(v)$, then $y$ and $v$ lie in a simplex contained in a maximal cube $Q$ containing $v$. By step 2.1, $d_2(y,v)\le \mathrm{diam}(Q)<\epsilon (v)\mathrm{/}16\le \delta (v)\mathrm{/}16<\delta (v)\mathrm{/}2.$ Step 1.1 supplies $U_v\in \mathcal{U}$ with $B(v,\delta (v)\mathrm{/}2)\subset U_v$, so $\mathrm{st}(v)\subset U_v$. Choosing one such $U_v$ for each vertex defines a refinement map $\rho$; AC [A1] supplies these simultaneous choices. The star cover is therefore a locally finite open refinement of $\mathcal{U}$ with the required order.