Convention 2.7. We will be interested in several families $\mathcal {H} \subseteq 2^X$ of subsets of the vertex set X of a graph (and later, more generally an arbitrary set X), which have the property of being closed under the complement operation $\neg : 2^X \to 2^X$ . For example, $\mathcal {H}_{\mathrm {cvx}}$ will denote the family of convex sets whose complement is also convex; see Definition 2.22.

For such families $\mathcal {H}$ , we will always use the plain symbol $\mathcal {H}$ to denote the family including the two trivial subsets $\varnothing , X \in \mathcal {H}$ , while $\mathcal {H}^{*} := \mathcal {H} \setminus \{\varnothing , X\}$ will denote the nontrivial elements of $\mathcal {H}$ . We also write $\mathcal {H}^G(X) = \mathcal {H}(X) = \mathcal {H}$ (e.g., $\mathcal {H}_{\mathrm {cvx}}^G(X)$ ) when the vertex/edge set needs to be specified.

Proof. $[x,y] \cap [x,z] \supseteq [x,{\langle {x,y,z} \rangle }]$ follows from (2.5).

Now let $w \in [x,y] \cap [x,z]$ . Then ${\langle {w,y,z} \rangle } \in [w,y] \cap [w,z] \cap [y,z] \subseteq [x,y] \cap [x,z] \cap [y,z]$ again by (2.5), whence ${\langle {w,y,z} \rangle } = {\langle {x,y,z} \rangle }$ .

Since $w \in [x,y]$ and ${\langle {x,y,z} \rangle } = {\langle {w,y,z} \rangle } \in [w,y]$ , again by (2.5), $w \in [x,{\langle {x,y,z} \rangle }]$ .

Proof. Let $a, b \in \mathrm {cone}_{x}(y)$ and $c \in [a,b]$ ; we will show

which by (2.5) along with gives , as desired. By (2.5), it suffices to show the following:

• • by definition of ${\langle {a,b,x}\rangle }$ .

• • , i.e., $y \in [x, {\langle {a,b,x}\rangle }]$ , by Lemma 2.11.

• • again by Lemma 2.11, because ${\langle {a,c,x} \rangle } \in [a,x]$ and ${\langle {a,c,x} \rangle } \in [a,b]$ , the latter because ${\langle {a,c,x} \rangle } \in [a,c] \subseteq [a,b]$ by (2.5).

Proof. Pick any $a_0 \in A$ , and inductively given $a_n$ , if there is $a \in A$ such that $a_n \not \in [x,a]$ , then put $a_{n+1} := {\langle {x,a,a_n} \rangle }$ . Then by repeated application of (2.5), so this sequence must terminate in at most $d(a_0,x)$ steps at a point in ${\mathrm {cvx}}(A)$ which is between x and every point in A; this proves existence.

$\operatorname {\mathrm {proj}}_A(x)$ is also between x and every point in ${\mathrm {cvx}}(A)$ , since $\mathrm {cone}_{x}(\operatorname {\mathrm {proj}}_A(x))$ is convex and contains A. Now for uniqueness: if there were two such points $a, b \in {\mathrm {cvx}}(A)$ , then we would have and , whence $a = b$ by (2.4).

Proof. First we check $\operatorname {\mathrm {proj}}_A$ is a median homomorphism. Let $x, y, z \in X$ with . Then

$$ \begin{align*} w := {\langle {\operatorname{\mathrm{proj}}_A(y),\operatorname{\mathrm{proj}}_A(x),\operatorname{\mathrm{proj}}_A(z)}\rangle} \in [\operatorname{\mathrm{proj}}_A(y),x] \cap [\operatorname{\mathrm{proj}}_A(y),z] = [\operatorname{\mathrm{proj}}_A(y),{\langle {\operatorname{\mathrm{proj}}_A(y),x,z}\rangle} ] \end{align*} $$

since , and similarly for z; and

$$ \begin{align*} [\operatorname{\mathrm{proj}}_A(y),x] \cap [\operatorname{\mathrm{proj}}_A(y),z] = [\operatorname{\mathrm{proj}}_A(y),\operatorname{\mathrm{proj}}_{[x,z]}(\operatorname{\mathrm{proj}}_A(y))] \subseteq [\operatorname{\mathrm{proj}}_A(y),y] \end{align*} $$

by Lemmas 2.11 and 2.16. Thus, $w \in [y, \operatorname {\mathrm {proj}}_A(y)]$ , so by definition of $\operatorname {\mathrm {proj}}_A(y)$ , we have $w = \operatorname {\mathrm {proj}}_A(y)$ , which means that .

For uniqueness, note that if $f : X \to {\mathrm {cvx}}(A)$ is a median homomorphism fixing ${\mathrm {cvx}}(A)$ , then for $x \in X$ , for every $a \in A$ , we have ${\langle {x,f(x),a}\rangle } \in {\mathrm {cvx}}(A)$ , whence ${\langle {x,f(x),a}\rangle } = f({\langle {x,f(x),a}\rangle }) = {\langle {f(x),f(f(x)),f(a)}\rangle } = {\langle {f(x),f(x),a}\rangle } = f(x)$ , i.e., $f(x) \in [x,a]$ ; thus, $f(x) = \operatorname {\mathrm {proj}}_A(x)$ .

Proof. By Remark 2.17, we have $\operatorname {\mathrm {proj}}_A \circ \operatorname {\mathrm {proj}}_B = \operatorname {\mathrm {proj}}_{\operatorname {\mathrm {proj}}_A(B)} \circ \operatorname {\mathrm {proj}}_A$ ; thus, composing on the right with another $\operatorname {\mathrm {proj}}_A$ or $\operatorname {\mathrm {proj}}_B$ yields the same thing. In particular, $\operatorname {\mathrm {proj}}_A \circ \operatorname {\mathrm {proj}}_B$ is an idempotent median homomorphism. Its image is clearly $\operatorname {\mathrm {proj}}_A(B)$ , which is convex by Remark 2.19; thus in fact, the map is $\operatorname {\mathrm {proj}}_{\operatorname {\mathrm {proj}}_A(B)}$ .

If $A \cap B \ne \varnothing $ , then $A \cap B$ is fixed by both $\operatorname {\mathrm {proj}}_A$ and $\operatorname {\mathrm {proj}}_B$ , and thus contained in $\operatorname {\mathrm {im}}(\operatorname {\mathrm {proj}}_A \circ \operatorname {\mathrm {proj}}_B) = \operatorname {\mathrm {proj}}_A(B)$ , which is clearly contained in A. To see that it is also contained in B: let $c \in A \cap B$ ; then for any $b \in B$ , we have $\operatorname {\mathrm {proj}}_A(b) \in [c,b] \subseteq B$ .

Proof. All of the projections $\operatorname {\mathrm {proj}}_{A_0}, \dotsc , \operatorname {\mathrm {proj}}_{A_{n-1}}$ commute; thus, applying the composite of all of them in any order to any point yields a point in the intersection (since each $\operatorname {\mathrm {proj}}_{A_i}$ could have been applied last).

Proof. For $(x,y) \in G$ , $H := \operatorname {\mathrm {proj}}_{\{x,y\}}^{-1}(y)$ is indeed a half-space, being a median preimage of a half-space; and clearly, $(x,y) \in \partial _{\mathsf {ie}} H$ . Conversely, for any nontrivial half-space $H \subseteq X$ and $(x,y) \in \partial _{\mathsf {ie}} H$ , that H is a half-space means precisely that the indicator function $X \twoheadrightarrow \{x,y\}$ taking H to y is a median homomorphism, which clearly fixes the convex set $\{x,y\}$ , thus is equal to $\operatorname {\mathrm {proj}}_{\{x,y\}}$ .

Proof. Given a square, each vertex is between its neighbors, from which it is easily seen that a hyperplane containing one edge must also contain the parallel edge; thus, hyperplane-equivalence contains said equivalence relation. Conversely, if $(a,b), (c,d) \in \partial _{\mathsf {ie}} H$ for some $H \in \mathcal {H}_{\mathrm {cvx}}^{*}(X)$ , find a geodesic between $a,c$ , which must lie in the convex set $\partial _{\mathsf {ov}} H$ and hence be matched via $\partial _{\mathsf {ie}} H$ to a geodesic between $b,d$ , which together with the matching forms the desired strip of squares.

Proof. Given such an n-cube, the n hyperplanes cutting it are clearly pairwise non-nested. Conversely, given pairwise non-nested $H_0, \dotsc , H_{n-1} \in \mathcal {H}_{\mathrm {cvx}}(X)$ , pick , let , and let for each $\vec {a} \in \{0,1\}^n$ ; using Proposition 2.20, this is easily seen to be an n-cube.

Proof. $\Longrightarrow $ : Two non-nested half-spaces yield a 4-cycle by the preceding lemma.

$\Longleftarrow $ : A cycle has two distinct edges on a hyperplane, whence there is a 4-cycle by Lemma 2.25.

Proof. $\Longrightarrow $ is clear. Conversely, suppose $H \supseteq K$ and $x \in \partial _{\mathsf {iv}} H \cap \partial _{\mathsf {ov}} K$ ; then clearly $H \supsetneq K$ , and any $H \supseteq L \supseteq K$ must either contain x but not $\operatorname {\mathrm {proj}}_{\neg H}(x) \in \partial _{\mathsf {ov}} H$ , hence equal H, or contain $\operatorname {\mathrm {proj}}_K(x) \in \partial _{\mathsf {iv}} K$ but not x, hence equal K.

Proof. By flipping $H, K$ if necessary, we may assume there is $x \in \partial _{\mathsf {iv}} H \cap \partial _{\mathsf {ov}} K \subseteq H \cap \neg K$ . If $\operatorname {\mathrm {proj}}_K(x) \in \neg H$ or $\operatorname {\mathrm {proj}}_{\neg H}(x) \in K$ , then these points are equal and $H = \neg K = \mathrm {cone}_{\operatorname {\mathrm {proj}}_K(x)}(x)$ . Otherwise, $H \cap K \ne \varnothing $ and $\neg H \cap \neg K \ne \varnothing $ . If also $\neg H \cap K \ne \varnothing $ , then $H, K$ are non-nested. Otherwise, $K \subseteq H$ and $x \in \partial _{\mathsf {iv}} H \cap \partial _{\mathsf {ov}} K$ , so H is a successor of K.

Proof. Let be any geodesic, so $n = d(x,y)$ . Then each half-space $x \not \in H \ni y$ has $x_i \not \in H \ni x_{i+1}$ for a unique $i < n$ , so we get a bijection between the set of such H and n.

Proof. If A or B is empty, this is trivial; so suppose not. Pick a geodesic , where $n = d(A, B)$ . Then $H := \mathrm {cone}_{x_1}(x_0)$ works. Indeed, note that $x_0 = \operatorname {\mathrm {proj}}_A(x_n)$ (being the closest point in A to $x_n$ ), whence $A \subseteq \mathrm {cone}_{x_n}(x_0) \subseteq \mathrm {cone}_{x_1}(x_0)$ by (2.5), and similarly $B \subseteq \mathrm {cone}_{x_0}(x_n) \subseteq \mathrm {cone}_{x_0}(x_1)$ .

Proof. By Lemma 2.32, there are finitely many half-spaces $H \subseteq [x,y]$ ; by Lemma 2.33, a vertex ${z \in [x,y]}$ is determined by the half-spaces containing it, whence $[x,y]$ is finite. By the proof of Proposition 2.14, ${\mathrm {cvx}}(\{x_0,\dotsc ,x_n\})$ consists of points of the form which are in an interval between $x_n$ and a point in an interval between $x_{n-1}$ and ….

Proof. $\Longrightarrow $ follows from Remark 2.19. Conversely, suppose $f^{-1}$ preserves half-spaces, and let $x, y, z \in X$ with $f(x) \not \in [f(y),f(z)]$ ; then there is a half-space $H \subseteq Y$ containing $f(x)$ but not $f(y), f(z)$ , whence $x \in f^{-1}(H) \not \ni y, z$ , whence $x \not \in [y,z]$ .

Proof. For any $(x,y) \in \partial _{\mathsf {ie}} H$ , by Lemma 2.23, H is the unique half-space containing y but not x.

Proof. We must show that every clopen orientation $U \subseteq \mathcal {H}_{\mathrm {cvx}}(X)$ is $\hat {x}$ for some $x \in X$ . Since U is clopen, there is a finite $A \subseteq X$ , which we may assume to be convex by Corollary 2.34, such that for every $H \in \mathcal {H}_{\mathrm {cvx}}(X)$ , $H \in U$ iff there is $K \in U$ with $H \cap A = K \cap A$ . If $K \cap A = \varnothing $ for some $K \in U$ , then $\varnothing \in U$ , a contradiction; thus, A intersects every element of U. Also, every $H, K \in U$ have $H \cap K \ne \varnothing $ , or else $H \subseteq \neg K \implies \neg K \in U$ . Thus, by Corollary 2.21, there is $x \in \bigcap U \cap A$ , whence $U \subseteq \hat {x}$ , whence $U = \hat {x}$ since both are orientations.

Proof. First, note that compactness together with the Priestley separation axiom implies that a closed upward-closed $F \subseteq P$ disjoint from a closed downward-closed $D \subseteq P$ may be separated by a clopen upward-closed $F \subseteq U \subseteq \neg D$ , by a standard finite covering argument.

Now for a closed partial orientation $F \subseteq P$ , we have $F \subseteq \neg \neg (F)$ , whence there is a clopen upward-closed $F \subseteq U \subseteq \neg \neg (F)$ , whence also $F = \neg (\neg (F)) \subseteq \neg (\neg U) = \neg \neg (U)$ , whence F is contained in the clopen partial orientation $U \cap \neg \neg (U)$ . Thus, we may assume to begin with that F is clopen.

Taking $F = \{q \in P \mid q \ge p\}$ shows that each $0 < p \in P$ is in a clopen partial orientation.

Thus, we may assume $\neg 0 \in F$ , that is, $0 \not \in \neg \neg (F)$ . So each $p \in \neg \neg (F)$ is in a clopen partial orientation, and so by compactness $\neg \neg (F)$ is contained in a finite union of such sets $G_0, \dotsc , G_{n-1} \subseteq P$ . Now put $F_0 := F$ and $F_{i+1} := F_i \cup (G_i \cap \neg \neg (F_i))$ ; then $U := F_n$ works.

Proof. There is at least one vertex, by Lemma 2.43 applied to $\{\neg 0\}$ .

By definition, $U, V$ are neighbors iff $U \setminus V$ is a single vertex, which must then be minimal in U since V was upward-closed. Conversely, for minimal $p \in U \setminus \{\neg 0\}$ , $V := U \triangle \{p, \neg p\}$ is still an orientation (since $p \in U$ is minimal and $\neg p \in \neg (U) = \neg U$ is maximal), and clopen since p is isolated.

It follows that $d(U,V) = {\lvert {U \setminus V}\rvert }$ : $\ge $ because a path may modify one element at a time; $\le $ because we may modify U to V in ${\lvert {U \setminus V}\rvert }$ steps by flipping minimal elements. In particular, note that $d(U,V) < \infty $ , that is, the graph is connected, since $U \setminus V$ is a compact set of isolated points.

The conditions $U \cap V \subseteq W$ and $W \subseteq U \cup V$ are equivalent since $U, V, W$ are orientations (take image under $\neg : P \to P$ followed by complement). The condition $(U \setminus W) \cup (W \setminus V) \subseteq U \setminus V$ is equivalent to $U \setminus W \subseteq \neg V$ and $W \setminus V \subseteq U$ , which are equivalent to the two former conditions. This third condition is equivalent to $d(U,W) + d(W,V) = d(U,V)$ , that is, $W \in [U,V]$ .

Thus, $M \in [U,V] \cap [V,W] \cap [W,U]$ iff $(U \cap V) \cup (V \cap W) \cup (W \cap U) \subseteq M \subseteq (U \cup V) \cap (V \cup W) \cap (W \cup U)$ iff $M = \{p \in P \mid p \text { belongs to at least two of } U, V, W\}$ ; and this set is easily seen to still be a clopen orientation if $U, V, W$ are.

Proof. It is an order-embedding, since for $p \not \le q \in P$ , $F := \{r \in P \mid p \le r \mathrel {\,\text {or}\,} \neg q \le r\}$ is a closed partial orientation, and hence extends by Lemma 2.43 to $U \in \mathcal {U}^\circ (P)$ such that $U \in \hat {p} \setminus \hat {q}$ . To see that it is surjective: take a nontrivial half-space $H \subseteq \mathcal {U}^\circ (P)$ and an edge $(U,V) \in \partial _{\mathsf {ie}} H$ ; then by Proposition 2.44, $V \setminus U = \{p\}$ is a singleton, and so by Lemma 2.23, $H = \mathrm {cone}_{U}(V)$ , which by Proposition 2.44 is the set of $W \in \mathcal {H}_{\mathrm {cvx}}(\mathcal {U}^\circ (P))$ such that $V \setminus U \subseteq W$ , which means that $H = \hat {p}$ .

Proof. By Corollary 2.35, this is equivalent to $(g^{*})^{-1} : 2^{\mathcal {U}^\circ (P)} \to 2^{\mathcal {U}^\circ (Q)}$ mapping half-spaces to half-spaces. Indeed, the composite of $(g^{*})^{-1}$ with the canonical isomorphism $P \cong \mathcal {H}_{\mathrm {cvx}}(\mathcal {U}^\circ (P)) \subseteq 2^{\mathcal {U}^\circ (P)}$ is easily seen to be g composed with $Q \cong \mathcal {H}_{\mathrm {cvx}}(\mathcal {U}^\circ (Q))$ .

Proof. $\Longrightarrow $ is obvious; $\Longleftarrow $ by Proposition 2.42.

Proof. Half-spaces non-nested with $H \in \mathcal {H}_{\mathrm {cvx}}(X)$ correspond to nontrivial half-spaces of $\partial _{\mathsf {iv}} H$ , by Lemma 2.27.

Proof. We easily have (i) $\iff $ (ii) (using that H is a successor of K iff $\neg K$ is a successor of $\neg H$ ). By the preceding corollary, it thus remains to show, assuming G has finite hyperplanes, that (i) iff G is locally finite.

Suppose (i) holds. Then G is locally finite since the neighbors of x correspond to half-spaces containing x on the inner boundary, which correspond to a clique of adjacent hyperplanes by Corollary 2.31.

Now suppose (i) fails; let $\partial _{\mathsf {ie}} H$ for $H \in \mathcal {H}_{\mathrm {cvx}}^{*}(X)$ be adjacent to infinitely many other hyperplanes. Then some belongs to infinitely many other hyperplanes, whence x has infinite degree.

Proof. (i) $\implies $ (ii) follows from Proposition 2.44 applied to the principal orientations $\hat {x}, \hat {y} \in \mathcal {U}^\circ (\mathcal {H})$ . Conversely, assume (ii), let $A \in 2^X \setminus \{\varnothing , X\}$ , and let $x \in A \not \ni y$ . For every $H \in \mathcal {H} \setminus \{A\}$ such that $x \in H \not \ni y$ , we have either some $x_H \in A \setminus H$ or some $y_H \in H \setminus A$ . Then the set of all $B \in 2^X$ containing x and each $x_H$ but not y or any $y_H$ is a clopen neighborhood of A disjoint from $\mathcal {H} \setminus \{A\}$ . Thus, the limit points of $\mathcal {H}$ are $\subseteq \{\varnothing , X\}$ , proving (i).

Proof. Since $\mathcal {H}_{\mathrm {cvx}}(X) \setminus \mathcal {H}$ consists only of isolated points, $\mathcal {H} \subseteq \mathcal {H}_{\mathrm {cvx}}(X)$ is closed. So given $U \in \mathcal {U}^\circ (\mathcal {H})$ , the upward-closure of U in $\mathcal {H}_{\mathrm {cvx}}(X)$ is closed by compactness of $\{(H,K) \mid H \subseteq K\} \subseteq \mathcal {H}_{\mathrm {cvx}}(X)^2$ , and a partial orientation since U is, and hence extends by Lemma 2.43 to a clopen orientation $U \subseteq V \subseteq \mathcal {H}_{\mathrm {cvx}}(X)$ , so that $U \subseteq V \cap \mathcal {H}$ , whence $U = V \cap \mathcal {H}$ since both are orientations. This shows that $\mathcal {U}^\circ (\mathcal {H}_{\mathrm {cvx}}(X)) \to \mathcal {U}^\circ (\mathcal {H})$ is surjective.

The composite $X \cong \mathcal {U}^\circ (\mathcal {H}_{\mathrm {cvx}}(X)) \twoheadrightarrow \mathcal {U}^\circ (\mathcal {H})$ is given by $x \mapsto \hat {x} \cap \mathcal {H}$ , which identifies two vertices iff they belong to the same half-spaces in $\mathcal {H}$ , that is, the same $\mathcal {H}$ -block. If two $\mathcal {H}$ -blocks $[x]_{\mathcal {H}}, [y]_{\mathcal {H}}$ are joined by an edge, then that means the interval $[[x]_{\mathcal {H}}, [y]_{\mathcal {H}}]$ consists of only $\{[x]_{\mathcal {H}}, [y]_{\mathcal {H}}\}$ , whence $[x,y] \subseteq [x]_{\mathcal {H}} \cup [y]_{\mathcal {H}}$ , so any geodesic from x to y must contain an edge between $[x]_{\mathcal {H}}, [y]_{\mathcal {H}}$ .

Proof. Let $A \in \mathcal {H}^{*}_{\partial <\infty }$ ; we must show that A is a finite Boolean combination of half-spaces. By Corollary 2.34, is finite; thus (by Lemma 2.33), $A \cap Y$ is a finite Boolean combination of half-spaces in Y, whence $\operatorname {\mathrm {proj}}_Y^{-1}(A \cap Y)$ is a finite Boolean combination of half-spaces in X. We claim that $A = \operatorname {\mathrm {proj}}_Y^{-1}(A \cap Y)$ . The intersections of both with Y are clearly the same; thus, it suffices to show that for $x \in X \setminus Y$ , $x \in A \iff x \in \operatorname {\mathrm {proj}}_Y^{-1}(A \cap Y)$ . Indeed, if $x \in A$ , then $\operatorname {\mathrm {proj}}_Y(x) \in A \cap Y$ , or else the last vertex in A on a geodesic from x to $\operatorname {\mathrm {proj}}_Y(x)$ would be on the inner boundary of A but not in Y, contradicting the definition of Y. Similarly if $x \not \in A$ .

Proof. Injectivity: for $U, V \in \widehat {X}$ , the set of $A \in \mathcal {H}_{\partial <\infty }(X)$ such that $A \in U \iff A \in V$ is a Boolean algebra, and hence if it contains $\mathcal {H}_{\mathrm {cvx}}(X)$ , must be all of $\mathcal {H}_{\partial <\infty }(X)$ by the above lemma. Surjectivity: $V \in \mathcal {U}(\mathcal {H}_{\mathrm {cvx}}(X))$ has the finite intersection property by Corollary 2.21, and hence extends to an ultrafilter $V \subseteq U \subseteq \mathcal {H}_{\partial <\infty }(X)$ such that $V = U \cap \mathcal {H}_{\mathrm {cvx}}(X)$ .

Proof. Let $U \in \widehat {X} \setminus X$ be an end, $\widehat {A} \ni U$ be a clopen neighborhood where $A \in \mathcal {H}_{\partial <\infty }(X)$ . By Corollary 2.34, ${\mathrm {cvx}}(\partial _{\mathsf {ov}} A) \subseteq X$ is finite. By Lemma 2.58 (applied to it and each point on its outer boundary), it is a finite intersection of half-spaces. One such half-space H must not contain U since $U \not \in {\mathrm {cvx}}(\partial _{\mathsf {ov}} A)$ . Then $U \in \neg \widehat {H}$ , whence $A \cap \neg H \ne \varnothing $ , whence $\neg H \subseteq A$ since $\partial _{\mathsf {ov}} A \cap \neg H = \varnothing $ .

Proof. The half-space neighborhoods of U form a filter base in X converging to U.

Proof. Let $f : X \to Y$ be such a map, and $R < \infty $ with $d(f(x), f(x')) \le 0 \implies d(x, x') \le R$ .

Proof. (a) Let $S < \infty $ such that $d(x, x') \le 1 \implies d(f(x), f(x')) \le S$ . For $A \subseteq Y$ and $(x,x') \in \partial _{\mathsf {ie}} f^{-1}(A)$ , we have a path of length $\le S$ between $f(x) \not \in A$ and $f(x') \in A$ , whence , that is, , whence . Since f is a coarse embedding, this is enough.

(b) For $x \in f^{-1}(A) \setminus g^{-1}(A)$ , we have a path of length $\le d(f, g)$ between $f(x) \in A$ and $g(x) \not \in A$ , whence . Swapping $f, g$ , the same holds for $x \in g^{-1}(A) \setminus f^{-1}(A)$ . Thus, , which similarly to (a) implies the claim.

(c) follows by taking a quasi-inverse g, so that $d(1_X, g \circ f), d(1_Y, f \circ g) < \infty $ .

Proof. If 5.2 holds, then each $x \in X$ is separated from each of its finitely many neighbors by only finitely many $H \in \mathcal {H}$ , proving 5.2. Conversely, if 5.2 holds, then for $x, y \in X$ , pick any path between them; $H \in \mathcal {H}$ separating $x, y$ must separate some edge along this path, and there are only finitely many such H for each edge, proving 5.2.

Proof. A has a clopen neighborhood of all $B \subseteq X$ containing $\partial _{\mathsf {iv}} A$ and disjoint from $\partial _{\mathsf {ov}} A$ , which is disjoint from $\mathcal {H}_{\mathrm {conn}}(X) \setminus \{A\}$ .

Proof. Let $H \in \mathcal {H}_{\partial <\infty }$ be non-nested with $K \in \mathcal {H}$ . There is a path in K between $H \cap K, \neg H \cap K$ , whence ; similarly, . Now any path between must contain a vertex on . Thus, fixing for each a path $p_{xy}$ between them, we have that any K non-nested with H must contain some z on some $p_{xy}$ on its boundary . So given H, there are finitely many , for each of which there are finitely many z on $p_{xy}$ , for each of which there are finitely many $K \in \mathcal {H}$ with by Lemma 5.2.

Proof. (ii) $\implies $ (iii): An end cannot map to a vertex, since it is non-isolated, and hence does not have a minimal neighborhood in $\mathcal {H}$ by (ii). And two ends cannot map to the same end, since they are separated by some $H \in \mathcal {H}$ by (ii).

(iii) $\implies $ (ii): By Lemma 5.9, $\mathcal {U}^\circ (\mathcal {H})$ has finite hyperplanes. Thus, by Lemma 2.61, half-spaces form a neighborhood basis for ends in $\widehat {\mathcal {U}^\circ (\mathcal {H})}$ . Now given an end $U \in \widehat {X} \setminus X$ and a clopen neighborhood $\widehat {A} \ni U$ , by (iii), the image of U under the induced map is not in the image of $\neg \widehat {A}$ , and the latter is closed by compactness, and hence is disjoint from some half-space $\widehat {H}$ containing U.

(ii) + (iii) $\implies $ (i): By Lemma 5.9, it remains to check 4.1(ii) that each $\mathcal {H}$ -block is finite and 4.1(iv) each $H \in \mathcal {H}^{*}$ has only finitely many successors. The former follows from the fact that no end in $\widehat {X}$ maps to a vertex in $\mathcal {U}^\circ (\mathcal {H})$ . For the latter, let $H \in \mathcal {H}^{*}$ . For any end $U \in \neg \widehat {H}$ , since $\neg \widehat {H}$ is not a minimal neighborhood of U, there is some $H \subsetneq K \in \mathcal {H}^{*}$ with $U \in \neg \widehat {K}$ ; by taking K minimal (which is possible since H is isolated), we get $U \in \neg \widehat {K}$ for some successor K of H. Thus, by compactness of $\neg \widehat {H} \setminus X$ , there are finitely many successors $H \subsetneq K_1, \dotsc , K_n \in \mathcal {H}^{*}$ such that every end in $\neg \widehat {H}$ is in some $\neg \widehat {K}_i$ . It follows by compactness that must be finite. Now let $H \subsetneq K \in \mathcal {H}^{*}$ be another successor not equal to any $K_i$ ; thus, for each i, we have $K \setminus K_i, K_i \setminus K, K \cap K_i \ne \varnothing $ (the last because $H \subseteq K \cap K_i$ ). If $\neg K \cap \neg K_i \ne \varnothing $ for some i, then $K, K_i$ are non-nested; by Lemma 5.9, there are only finitely many possibilities for K for each i. Otherwise, we have , so there are again only finitely many such K.

(i) $\implies $ (iii): Since each $\mathcal {H}$ -block is finite by 4.1(ii), no end in $\widehat {X}$ maps to a vertex in $\mathcal {U}^\circ (\mathcal {H})$ . Suppose two distinct ends $U, V \in \widehat {X}$ map to the same end $W \in \widehat {\mathcal {U}^\circ (\mathcal {H})}$ . Since half-spaces $\widehat {H} \ni W$ form a neighborhood basis for W by Lemma 2.61, it follows that the collection of $H \in \mathcal {H}$ containing both $U, V$ (as opposed to neither) is an ultrafilter basis. But then the intersection of all such H is a set of vertices in X not separated by any element of $\mathcal {H}$ , which is infinite because, for example, for any $K \in \mathcal {H}$ separating U from V, each such H must meet which is a finite set, and we can find infinitely many such K with pairwise disjoint boundaries. So 4.1(ii) fails.

Proof. Let

$$ \begin{align*} \mathcal{H}' := \{\neg D \mid H \in \mathcal{H} \mathrel{\,\text{and}\,} C \in H/G \mathrel{\,\text{and}\,} D \in \neg C/G\} \end{align*} $$

(where $H/G$ denotes the set of G-components of H). Then every such $\neg D \in \mathcal {H}'$ has $\partial _{\mathsf {ie}} \neg D \subseteq \partial _{\mathsf {ie}} C \subseteq \partial _{\mathsf {ie}} H$ . Now fix an end $U \in \widehat {X} \setminus X$ and a neighborhood $A \in \mathcal {H}_{\partial <\infty }$ of U; we must find a smaller neighborhood in $\mathcal {H}'$ . Let be finite connected. Then $\neg B$ is a neighborhood of U, so there is $\neg B \supseteq H \in \mathcal {H}$ such that $U \in \widehat {H}$ . Let $C \subseteq H$ be the component containing U, and $D \subseteq \neg C$ be the component containing B. Then $\neg D \in \mathcal {H}'$ , with $U \in \widehat {C} \subseteq \neg \widehat {D}$ , and $\neg D \subseteq A$ since $\neg D$ is connected, disjoint from , and contains $U \in \widehat {A}$ . (See Figure 3.)

Proof. Let $(X,G), (Y,T)$ be connected locally finite graphs, $f : X \to Y$ be a coarse equivalence with quasi-inverse $g : Y \to X$ , and suppose $\mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le S}}(Y)$ is dense towards ends for $S < \infty $ . By Lemma 2.67, pick $R < \infty $ so that for any $H \in \mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le S}}(Y)$ , we have $f^{-1}(H) \in \mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le R}}(X)$ . Then for any $U \in \widehat {X} \setminus X$ and $A \in \mathcal {H}_{\partial <\infty }(X)$ with $U \in \widehat {A}$ , letting $B := \neg \operatorname {\mathrm {Ball}}_{d(1_X, g \circ f)}(\neg A)$ , we have $f^{-1}(g^{-1}(B)) \subseteq \operatorname {\mathrm {Ball}}_{d(1_X, g \circ f)}(B) \subseteq A$ , and $A \triangle B, B \triangle f^{-1}(g^{-1}(B))$ are finite, so $U \in {f^{-1}(g^{-1}(B))}^\wedge $ , so $\hat {f}(U) \in \widehat {g^{-1}(B)}$ , so there is $g^{-1}(B) \supseteq H \in \mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le S}}(Y)$ with $\hat {f}(U) \in H$ , so $f^{-1}(H) \in \mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le R}}(X)$ with $U \in \widehat {f^{-1}(H)}$ and $f^{-1}(H) \subseteq f^{-1}(g^{-1}(B)) \subseteq A$ .

Proof. By Lemma 5.18 and that $\mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le 1}}$ is clearly dense towards ends in a tree.

Proof. By Corollary 5.20, Corollary 5.17, and Remark 4.7.

Proof. Take R satisfying Theorem 5.23. Let $H \subsetneq K \in \mathcal {H}^{*}_{{\operatorname {\mathrm {diam}}(\partial ) \le R}} \cap \mathcal {H}_{\mathrm {conn}}$ with ; thus, $\operatorname {\mathrm {Ball}}_S(H) \subseteq K$ . Pick any . Then $\operatorname {\mathrm {Ball}}_R(x)$ contains and is contained in $\operatorname {\mathrm {Ball}}_{S-1}(H)$ , so $\operatorname {\mathrm {Ball}}_1(\operatorname {\mathrm {Ball}}_R(x)) \subseteq K$ , that is, $\operatorname {\mathrm {Ball}}_R(x) \cap \operatorname {\mathrm {Ball}}_1(\neg K) = \varnothing $ . Thus, $\operatorname {\mathrm {Ball}}_1(\neg K)$ is contained in a single component L of $\neg \operatorname {\mathrm {Ball}}_R(x)$ , with $H \cap L = \varnothing $ since $H \cap \neg K = \varnothing $ and whence every component of $\neg \operatorname {\mathrm {Ball}}_R(x)$ is contained in or disjoint from H. Then $\neg L \in \mathcal {H}^{*}_{{\operatorname {\mathrm {diam}}(\partial ) \le R}} \cap \mathcal {H}_{\mathrm {conn}}$ by Theorem 5.23 with $H \subsetneq H \cup \operatorname {\mathrm {Ball}}_R(x) \subseteq \neg L \subseteq \neg \operatorname {\mathrm {Ball}}_1(\neg K) \subsetneq K$ , so K is not a successor of H.

Proof. The above lemma shows that for $H, K \in \mathcal {H}^{*}_{{\operatorname {\mathrm {diam}}(\partial ) \le R}} \cap \mathcal {H}_{\mathrm {conn}}$ , if one is a successor of the other, then their boundaries are at bounded distance apart. Also, the proof of Lemma 5.9 shows that for such $H, K$ which are non-nested, there is a vertex in which lies on a geodesic $p_{xy}$ between two points , and hence, $H, K$ have boundaries at distance $\le R$ apart. Thus, $H, K$ which are adjacent in the pocset $\mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le R}} \cap \mathcal {H}_{\mathrm {conn}}$ (in the sense of Corollary 2.31) have boundaries at bounded distance apart. Since these boundaries also have diameter $\le R$ , and G has bounded degree, the adjacency graph on $\mathcal {H}^{*}_{{\operatorname {\mathrm {diam}}(\partial ) \le R}} \cap \mathcal {H}_{\mathrm {conn}}$ has bounded degree. From the proof of Corollary 2.50, we see that $\mathcal {U}^\circ (\mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le R}} \cap \mathcal {H}_{\mathrm {conn}})$ has bounded degree, and also from Corollary 2.49 that for each of its half-spaces H, the median subgraph $\partial _{\mathsf {iv}} H$ has boundedly many half-spaces, which implies bounded size of $\partial _{\mathsf {iv}} H$ since $\partial _{\mathsf {iv}} H \hookrightarrow 2^{\mathcal {H}_{\mathrm {cvx}}(\partial _{\mathsf {iv}} H)}$ (Lemma 2.33).

Finally, to show (c): for $x, y \in X$ , we have

$$ \begin{align*} d(\hat{x}, \hat{y}) = {\lvert {\{H \in \mathcal{H}_{{\operatorname{\mathrm{diam}}(\partial) \le R}} \cap \mathcal{H}_{\mathrm{conn}} \mid x \in H \not\ni y\}}\rvert} \end{align*} $$

by Lemma 2.32. We may upper-bound this for $x \mathrel {G} y$ since any such H must have , whence $d(\hat {x}, \hat {y})$ is upper-bounded for all $x, y$ by a multiple of $d(x, y)$ . And we may lower-bound $d(\hat {x}, \hat {y})$ by a multiple of $d(x, y)$ using Lemma 5.24, by letting $\neg H \subseteq \neg \{x\}$ be the component (radial cut) containing y, so that is bounded by the degree bound of G (which we may assume to be $\le R$ ); then letting $H \subseteq K \subseteq \neg \{y\}$ be the component containing ; and then using Lemma 5.24 to find a sequence in $\mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le R}} \cap \mathcal {H}_{\mathrm {conn}}$ with consecutive boundaries at bounded distance, thus with n at least a multiple of $d(x, y)$ .

Proof. The Borel reduction to a treeable CBER constructed in Corollary 5.21, ultimately Theorem 4.6, is componentwise given by the quasi-isometry $x \mapsto \hat {x}$ to a median graph of the above result, followed by the identity map to the subforest given by Theorem 3.5.

Proof. Let $g : Y \to X$ be a quasi-inverse of f, and let $S < \infty $ be such that $d(y,y') \le 1 \implies d(g(y),g(y')) \le S$ . Let $R := \max (d_X(1_X, g \circ f), S)$ , and let G be the distance $\le R$ graph on X. Then $g : (Y,T) \to (X,G)$ is $1$ -Lipschitz, whence $g \circ f : (X,d) \to (X,G)$ is bornologous, whence so is $1_X$ since $d_G(1_X, g \circ f) \le 1$ . And $1_X : (X,G) \to (X,d)$ is clearly R-Lipschitz.

Proof. By the above, for each class, there is some R such that the distance $\le R$ graph has $\mathcal {H}_{{\operatorname {\mathrm {diam}}(\partial ) \le S}}$ dense towards ends for some S; apply Remark 4.7. The last statement follows from Corollary 5.26.

Proof. Let $X = \{x_0, x_1, \dotsc \}$ be a bijective enumeration. Let $F_0 := F$ , and inductively for each n, let $F_{n+1} \subseteq F_n$ be given by $F_{n+1}(x) := F_n(x)$ for $x \ne x_n$ , while $F_{n+1}(x_n) \subseteq F_n(x_n)$ is given as follows: for each $y \in \operatorname {\mathrm {Ball}}_{\le 1}(x_n)$ , pick some $z_y \in F_n(x_n) \cap F_n(y)$ ; then let $F_{n+1}(x_n) \subseteq Y$ be the convex hull of these $z_y$ ’s. Then each $F_n$ is a tree decomposition, whence so is , since the definition of tree decomposition only requires checking $F'(x), F'(y)$ for each edge $x \mathrel {G} y$ .

Proof. For each $y \in Y$ , since F is a tree decomposition, we have a partition

$$ \begin{align*} X \setminus F^{-1}(y) = \bigsqcup_{y' \mathrel{T} y} (F^{-1}(\mathrm{cone}_{y}(y')) \setminus F^{-1}(y)), \end{align*} $$

which is invariant with respect to the induced subgraph on $X \setminus F^{-1}(y)$ , whence only finitely many pieces are nonempty. Since F has finite width, it follows that each $F^{-1}(\mathrm {cone}_{y}(y')) \subseteq X$ has finite boundary. This implies that $\widehat {F}$ is a function on ends: it cannot map an end to a vertex, since each $F^{-1}(y)$ is finite; and it cannot map an end $U \in \widehat {X} \setminus X$ to two distinct ends in $\widehat {Y} \setminus Y$ , since they are separated by some half-space $\mathrm {cone}_{y}(y') \subseteq Y$ whence U belongs to either $F^{-1}(\mathrm {cone}_{y}(y'))$ or its complement. Now if the $y'$ th piece of the above partition is empty, that means $F^{-1}(\mathrm {cone}_{y}(y')) \subseteq F^{-1}(y)$ , whence we may remove $\mathrm {cone}_{y}(y')$ from the tree Y and still maintain that we have a tree decomposition $F \cap (X \times (Y \setminus \mathrm {cone}_{y}(y'))) = {\operatorname {\mathrm {proj}}_{Y \setminus \mathrm {cone}_{y}(y')}} \circ F$ ; moreover, $\widehat {F}$ did not map any end of X into $\widehat {\mathrm {cone}_{y}(y')}$ (again since $F^{-1}(y)$ is finite). Let $Y' \subseteq Y$ be the result of removing all such $\mathrm {cone}_{y}(y')$ with $F^{-1}(\mathrm {cone}_{y}(y')) \setminus F^{-1}(y) = \varnothing $ .

Proof. By the preceding lemmas, let $F \subseteq X \times Y$ be a tree decomposition with Y locally finite, each $F(x) \subseteq Y$ finite, and width $\le N-1$ . Let $U \in \widehat {X} \setminus X$ and $A \in \mathcal {H}_{\partial <\infty }(X)$ with $U \in \widehat {A}$ . Then the end $\widehat {F}(U) \in \widehat {Y}$ and the finite set are separated by some half-space $\mathrm {cone}_{y}(y') \subseteq Y$ such that $\widehat {F}(U) \in \widehat {\mathrm {cone}_{y}(y')}$ and . It follows that $U \in \widehat {F^{-1}(\mathrm {cone}_{y}(y'))} \setminus F^{-1}(y)$ , which is a union of some of the components of $X \setminus F^{-1}(y)$ (as in the proof of the preceding lemma), while . Thus, the component H of $X \setminus F^{-1}(y)$ containing U is contained in A, since it contains $U \in \widehat {A}$ and is disjoint from ; and $H \in \mathcal {H}_{{\min \partial \le N}}$ since $\partial _{\mathsf {ov}} H \subseteq F^{-1}(y)$ which has size $\le N$ . (See Figure 4.)