Proof. Assertions (i) and (ii) already have been observed in [ Reference Bürger and Kurkofka6 , Lemma 2·13] and [ Reference Pitz39 , Lemma 4·3]. For convenience of the reader, we give a self-contained argument for all three properties (i)–(iii) at once. For this, suppose $T\subseteq G$ is a rooted tree that contains U cofinally.

To see the claim, suppose for a contradiction that for some finite set of vertices X and some collection $\mathscr{C}$ of components of $G-X$ we have that $\bigcup \mathscr{C}$ meets U finitely, but V(T) infinitely. By increasing X, we may assume that $\bigcup \mathscr{C} \cap U = \emptyset$ . Since V(T) is infinite but $\lceil{X}\rceil$ is finite, there exists $t \in (T \setminus \lceil{X}\rceil) \cap \bigcup \mathscr{C}$ . Then $\lfloor{t}\rfloor \subseteq \bigcup \mathscr{C}$ avoids U, a contradiction to the assumption that U is cofinal in T.

Now to see (i), consider any end $\varepsilon \notin \partial_{\Omega} {U}$ . Then there is a finite set of vertices X such that $C(X,\varepsilon)$ avoids U. By Claim 3·2, also V(T) intersects $C(X,\varepsilon)$ finitely, witnessing $\varepsilon \notin \partial_{\Omega} {T}$ . The argument for (iii) is similar. Finally, for (ii), consider a component C ^′ of $G-T$ . Then there is a component C of $G-U$ with $C' \subseteq C$ . Assuming that U has finite adhesion, $X=N(C)$ is finite. Claim 3·2 applied to X and C yields that V(T) intersects C finitely. Then $N(C') \subseteq X \cup (V(T) \cap V(C))$ is finite, so T has finite adhesion.

Proof. Let U be a given set of vertices in a connected graph G. By Zorn’s lemma, there is an inclusionwise maximal set of combs attached to U with pairwise disjoint spines. Write $\mathcal{R}$ for this collection of spines. Let S be the set of all centres of (infinite) stars attached to U. We will show that

$$U^* \;:\!=\; U \cup \bigcup_{R \in \mathcal{R}} V(R) \cup S$$

is an envelope for U. The verification relies on the following claim:

To see the claim, consider some finite set of vertices X, and assume that $\mathscr{C}$ is a collection of components of $G-X$ such that U meets $\bigcup \mathscr{C}$ finitely. Then S avoids $\bigcup \mathscr{C}$ . Consider next a spine $R \in \mathcal{R}$ that meets $\bigcup \mathscr{C}$ . Since R eventually lies in a component $C \notin \mathscr{C}$ , we have that R meets $\bigcup \mathscr{C}$ finitely. Moreover, R also meets X, and since the spines in $\mathcal{R}$ are pairwise disjoint, there are at most $|X|$ spines from $\mathcal{R}$ that meet $\bigcup\mathscr{C}$ , and each does so finitely. Hence $\bigcup_{R \in \mathcal{R}} V(R)$ meets $\bigcup \mathscr{C}$ finitely, too, and the claim follows.

Now to see $\partial_{\Omega} {U^*} = \partial_{\Omega} {U}$ , consider any end $\varepsilon \notin \partial_{\Omega} {U}$ . Then there is a finite set of vertices X such that $C(X,\varepsilon)$ avoids U. By Claim 3·4, also $U^*$ intersects $C(X,\varepsilon)$ finitely, witnessing $\varepsilon \notin \partial_{\Omega} {U^*}$ . The argument that if U is concentrated, then so is $U^*$ , follows similarly from Claim 3·4.

To see that $U^*$ has finite adhesion, suppose for a contradiction that there is a component C of $G-U^*$ with infinite neighbourhood. Then by a routine application of Theorem 2·1, we either find a star or a comb attached to $U^*$ whose centre v or spine R is contained in C. Then for all finite sets of vertices X (chosen disjoint from v in the star case), the centre v or a tail of R lives in a component of $G-X$ that meets $U^*$ infinitely, and hence also U by Claim 3·4. But then it is straightforward to inductively construct a star with centre v or a comb with spine R attached to U, violating the choice of S or $\mathcal{R}$ respectively.

To get a connected envelope for U that also satisfies the moreover-part, note that there are inclusionwise minimal subtrees of G containing $U^*$ which satisfy that $T \cap C$ is connected for every component C of $G-U$ : Take a spanning tree $T_C$ for every component C of $G-U$ , and extend the forest $\bigcup_C T_C$ to a rooted spanning tree T of G. Then the down-closure of $U^*$ in T can serve as the desired envelope by Theorem 3·1.

Proof. First, suppose that for some finite $X \subseteq U$ there are infinitely many $n\in \mathbb{N} $ with $N_G(D_n) = X$ . Write $\mathscr{C}$ for the collection of components of $G-X$ containing an end from $\partial_{\Omega} {U}$ . Then $\Omega(X,\mathscr{C})$ is an open neighbourhood around $\partial_{\Omega} {U}$ avoiding infinitely many $\varepsilon_n$ , witnessing $\varepsilon_n \not\to \partial_{\Omega} {U}$ .

Now assume that the map $ \mathbb{N} \ni n\mapsto N_G(D_n)$ is finite-to-one, and let us suppose for a contradiction that the sequence $\varepsilon_0,\varepsilon_1,\ldots$ does not converge to $\partial_{\Omega} {U}$ in the end space of G. Then there exists an open set around $\partial_{\Omega} {U}$ avoiding infinitely many $\varepsilon_n$ , and by compactness of $\partial_{\Omega} {U}$ we may find a finite vertex set $X\subseteq V(G)$ such that infinitely many $\varepsilon_n$ live outside of $\bigcup_{\varepsilon \in \partial_{\Omega} {U}} C(X,\varepsilon)$ . Since the map $n\mapsto D_n$ has finite fibres, we may assume without loss of generality that the finite vertex set X meets none of the components $D_n$ in which these ends live. Hence, infinitely many neighbourhoods $N_G(D_n)$ avoid $\bigcup_{\varepsilon \in \partial_{\Omega} {U}} C(X,\varepsilon)$ . But each of these neighbourhoods consists of vertices in U, and since the fibres of the map $n\mapsto N_G(D_n)$ are finite, the union of these neighbourhoods is an infinite subset of U that lies outside of $\bigcup_{\varepsilon \in \partial_{\Omega} {U}} C(X,\varepsilon)$ , contradicting the assumption that U is concentrated.

Proof. (i) $\Rightarrow$ (ii). For limits $\sigma$ , this holds by assumption, so suppose that $\sigma$ is a non-limit and consider any component C of $G-T^{\leqslant\sigma}$ . Let us assume for a contradiction that C has infinitely many neighbours. Using that these form a well-ordered chain and that $\sigma$ is a non-limit, we find an $\omega$ -chain of infinitely many neighbours $t_0\lt t_1\lt\cdots$ of C and a limit $t\in T^{\lt\sigma}$ such that $t_n\lt t$ for all $n\lt\omega$ . Let $\mu$ denote the height of t and note that $\mu$ is a limit. Then the component of $G-T^{\leqslant\mu}$ which contains C has infinite neighbourhood, contradicting our assumption that G has finite adhesion.

(ii) $\Rightarrow$ (iii). Suppose t is a successor, and let $\sigma$ be the height of the predecessor of t. By Theorem 2·3 (iii), the unique component of $G-T^{\leqslant \sigma}$ containing t is spanned by $\lfloor{t}\rfloor$ , so has finite neighbourhood.

(iii) $\Rightarrow$ (i). Let $\sigma$ be a limit ordinal. By Theorem 2·3 (iii), the components of $G-T^{\leqslant \sigma}$ are spanned by $\lfloor{t}\rfloor$ for $t \in T^{\sigma+1}$ , so have finite neighbourhoods by assumption (iii).

Proof. (i) $\Rightarrow$ (ii). If G has uniformly finite adhesion, then every vertex set of the form for a limit $t\in T$ has all its neighbours in the finite set $S_t\cup\{t\}$ .

(ii) $\Rightarrow$ (iii). Let us assume for a contradiction that some has infinitely many neighbours. Then we find an $\omega$ -chain of infinitely many neighbours $t_0\lt t_1\lt\cdots$ of and a limit $\ell\leqslant t$ such that $t_n\lt\ell$ for all $n\lt\omega$ . But then also has infinite neighbourhood, a contradiction.

(iii) $\Rightarrow$ (i). Let t be a limit node of T. Then is as desired.

Proof. The down-neighbours of t form an infinite chain $\mathscr{C}\subseteq\mathring{\lceil t\rceil}$ . Since G has finite adhesion, below every successor in $\mathring{\lceil t\rceil}$ there are only finitely many elements of $\mathscr{C}$ . It follows that $\mathscr{C}$ must be an $\omega$ -chain.

Proof. (i) Suppose first that T is special with antichain partition $\{U_n\colon n \in \mathbb{N} \}$ . Following Diestel, Leader and Todorcevic [ Reference Diestel and Leader20 ], we construct a T-graph as follows: successors are connected to their predecessors, and given a limit $t \in T$ , pick down-neighbours $t_0 \lt_T t_1 \lt_T t_2 \lt_T \cdots\lt_T t$ with $t_i \in U_{n_i}$ recursively such that $t_{i-1} \lt t_i \lt t$ and each $n_{i}$ is smallest possible. We claim that the resulting T-graph G has uniformly finite adhesion. Indeed, $S_t$ may be chosen to consist of all elements of $\mathring{\lceil t \rceil}$ contained in an antichain with smaller index than the antichain containing t.

Conversely, suppose that there is a T-graph G of uniformly finite adhesion with root r, say. We argue that there exists a regressive function $f \colon (T - r) \to T$ such that each fibre can be covered by countably many antichains (then T itself is special by a well-known observation of Todorcevic [ Reference Todorcevic53 , Theorem 2·4]). Indeed, map all successors to their unique predecessor, and map a limit t to any of its neighbours in $(\textrm{max}\, S_t,t)$ . This regressive map is ${\leqslant}2-1$ on branches of T: Indeed, suppose for a contradiction that there are two limits $t\lt t'$ with $f(t) = f(t')$ . Since $t' \gt t \gt f(t) = f(t')$ we have $f(t') \in S_t$ . But then $f(t) \in S_t$ , a contradiction.

(ii) Suppose that the successors of T admit an antichain partition $(U_n)_{n \in \mathbb{N} }$ . Following the same procedure as above, we construct a T-graph as follows: successors are connected to their predecessors, and given a limit $t \in T$ , pick down-neighbours $t_0 \lt_T t_1 \lt_T t_2 \lt_T \cdots\lt_T t$ with $t_i \in U_{n_i}$ recursively such that $t_{i-1} \lt t_i \lt t$ and each $n_{i}$ is smallest possible. We claim that the resulting T-graph G has finite adhesion. Indeed, for any limit ordinal $\sigma$ and any component C of $G-T^{\leqslant\sigma}$ we have $C = \lfloor{s}\rfloor$ for a unique successor node s of some limit $t \in T^\sigma$ by Theorem 2·3 (iii). Hence, N(C) is finite, as all elements of $N(C) \cap \mathring{\lceil t\rceil}$ belong to antichains with smaller indices than the antichain containing s.

Conversely, suppose that there is a T-graph G of finite adhesion. Let $L\subseteq T$ consist of all limits of T that have successors in T, and for every limit $\ell\in L$ let the set $S(\ell)$ consist of all the successors of $\ell$ in T. We obtain a new order tree T ^′ from T as follows: First, add for each $\ell\in L$ and $s\in S(\ell)$ a new node $v(\ell,s)$ that we declare to be a successor of $\ell$ and a predecessor of s. Then delete L. Note that there is a natural epimorphism $\varphi\colon T' \to T$ which is the identity on $T\setminus L$ and which sends each $v(\ell,s)$ to $\ell$ . Clearly, this epimorphism has the property that any two distinct elements of $\varphi^{-1}(\ell)$ are mutually incomparable.

Let G ^′ be the T ^′-graph with $V(G')\;:\!=\;T'$ and

$$E(G') \;:\!=\; \{tt'\;:\;t \lt t' \in T' \; \wedge \; \varphi(t)\varphi(t') \in E(G)\}.$$

Consider a limit $t=v(\ell,s)$ of T ^′. Since G has finite adhesion, $\lfloor{s}\rfloor_T$ has finite neighbourhood say $S_s$ in G. Since every $t' \gt t$ in T ^′ satisfies $\varphi(t') \in \lfloor{s}\rfloor_T$ , the set $S_t=\varphi^{-1}(S_s) \cap \lceil t\rceil_{T'}$ witnesses that G ^′ has uniformly finite adhesion. (To see that $S_t$ is finite, note that since the elements of $\varphi^{-1}(\ell)$ are mutually incomparable, we have $|\varphi^{-1}(\ell) \cap \lceil t\rceil_{T'}| \leqslant 1$ for each $\ell \in S_s$ .) It follows from (i) that T ^′ is special, so admits a countable antichain partition. This induces an antichain partition on the successors of T ^′.

Proof. If G is a T-graph of finite adhesion, then combining the previous theorems shows that T embeds into $( \mathbb{R} ,\leqslant)$ . In particular, for every branch B of T there is an order-preserving injection from $(B,\leqslant)$ into $( \mathbb{R} ,\leqslant)$ . Since B is well-ordered but $ \mathbb{R} $ contains no uncountable well-ordered subsets, it follows that B is countable.

Proof. (i) We show the forward implication indirectly. For this, suppose that there is a top t of $\varrho$ such that $t\in\varrho_n$ for infinitely many $n\in A$ . Then the finite vertex set $S_t\cup\{t\}$ separates $\varepsilon$ from $\varepsilon_n$ for infinitely many $n\in A$ at once, a contradiction.

For the backward implication, let $X\subseteq V(G)$ be any finite vertex set; we have to find a number $N\in \mathbb{N} $ such that $\varepsilon_n$ lives in $C(X,\varepsilon)$ for all $n\in A$ with $n\geqslant N$ . Let Y be the set of all tops t of $\varrho$ for which $\lfloor t\rfloor$ meets X. By assumption, we find a large enough $N\in \mathbb{N} $ such that all high-rays $\varrho_n$ with $n\in A$ and $n\geqslant N$ avoid Y. We claim that all corresponding ends $\varepsilon_n$ live in $C(X,\varepsilon)$ . For this, let any $n\in A$ with $n\geqslant N$ be given. Let $t'_0\lt t'_1\lt\cdots$ be a cofinal $\omega$ -chain in the high-ray $\varrho_n$ such that $t'_0$ is a top of $\varrho$ . Let $t_0\lt t_1\lt\cdots$ be a cofinal $\omega$ -chain in the high-ray $\varrho$ such that X has no vertex on $\varrho$ above or equal to $t_0$ and $t_0$ is a neighbour of $t'_0$ . Using that the intervals $[t_k,t_{k+1}]$ and $[t'_k,t'_{k+1}]$ of the T-graph G are connected for all $k\in \mathbb{N} $ by Theorem 2·3 (iv), we find rays $R\in\varepsilon$ and $R_n\in\varepsilon_n$ which start in $t_0$ and $t'_0$ , respectively, and avoid X. Since $t_0 t'_0$ is an edge of G, it follows that $\varepsilon_n$ lives in $C(X,\varepsilon)$ .

(ii) We show the forward implication indirectly. For this, suppose that there is a successor node $t\in \varrho$ such that $\varrho\cap\varrho_n\subseteq\mathring{\lceil t\rceil}$ for infinitely many $n\in B$ . Then $N_G(\lfloor t\rfloor)$ is a finite vertex set by Theorem 4·2, and it separates infinitely many $\varepsilon_n$ from $\varepsilon$ simultaneously.

For the backward implication, let any finite vertex set $X\subseteq V(G)$ be given. Let $t\in\varrho$ be any successor such that $X\subseteq V_{\mathring{\lceil t\rceil}}$ . By assumption, we find a large enough number $N\in B$ such that $t\in\varrho_n$ for all $n\geqslant N$ (with $n\in B$ ). Let $t_0\lt t_1\lt\cdots$ be a cofinal $\omega$ -chain in the high-ray $\varrho$ with $t_0\;:\!=\;t$ . Since the intervals $[t_n,t_{n+1}]\subseteq\varrho$ are connected in G by Theorem 2·3 (iv), we find a ray $R\in\varepsilon$ in $G-X$ which starts in t. Similarly, we find a ray $R_n\in\varepsilon_n$ in $G-X$ which starts in t, for every $n\geqslant N$ . Hence $\varepsilon_n\in\Omega(X,\varepsilon)$ for all $n\geqslant N$ .

Proof. By definition of our subbase, every open set U with $x \in U \subseteq \mathcal{R}(T)$ contains a basic set B of the form

\begin{align*} x \in B = \bigcap_{t \in E} [t] \cap \bigcap_{s \in F'} [s]^\complement= \bigcap_{t \in E} [t] \setminus \bigcup_{s \in F'} [s] \subseteq U\end{align*}

for some finite sets $E,F' \subseteq T$ . Pick B with $|E| + |F'|$ of minimal size. Since $x \in B$ implies $E \subseteq x$ , we could replace E by its maximum. Hence, $|E| \leqslant 1$ . Next, we claim that for every $s\in F'$ there is a top $\tilde s$ of x with $\tilde s \leqslant s$ . Indeed, if say $\lceil s_0\rceil \cap x \subsetneq x$ , pick $t \in x \setminus (E \cup \lceil s_0\rceil)$ to obtain

\begin{align*} x \in [t] \setminus \bigcup_{s \in F'_0} [s] \subseteq B\end{align*}

for $F'_0 = F' \setminus {\left\lbrace {s_0} \right\rbrace}$ , contradicting the minimality of $|E|+|F'|$ . Hence, for every $s\in F'$ there is a top $\tilde s$ of x with $\tilde s \leqslant s$ , and by setting $F = \{\tilde s \;:\; s \in F'\}$ , we obtain

\begin{align*} x \in [t,F] \subseteq B \subseteq U \subseteq \mathcal{R}(T)\end{align*}

with $t \in x$ and F a finite set of tops of x as desired.

Proof. Let G be a uniform T-graph. Then T is special by Theorem 4·6. By Theorem 5·2, the end space $\Omega(G)$ is naturally homeomorphic to the topology $\tau_{seq}$ on $\mathcal{R}(T)$ that is given by describing the convergent sequences as follows: Let x and $x_n$ ( $n \in \mathbb{N} $ ) be high-rays in a special order tree T. Let $A\subseteq \mathbb{N} $ consist of all numbers n for which $x \subsetneq x_n$ , and let $B\;:\!=\; \mathbb{N} \setminus A$ .

1. (i) We have convergence $x_n \to x$ for $n \in A$ and $n\to\infty$ if and only if A is infinite and for every top t of x there are only finitely many $n \in A$ with $t \in x_n$ .

2. (ii) We have convergence $x_n \to x$ for $n \in B$ and $n\to\infty$ if and only if B is infinite and for every node $t \in x$ there are only finitely many $n \in B$ with $x \cap x_n \subseteq \mathring{\lceil t\rceil}$ .

Write $\tau$ for the standard topology on the ray space $\mathcal{R}(T) \subseteq {\left\lbrace {0,1} \right\rbrace}^T$ . In order to verify that $\tau$ and $\tau_{seq}$ induce the same closed sets, given $X \subseteq \mathcal{R}(T)$ we need to show that $x \in \overline{X}$ with respect to $\tau$ if and only if there is a sequence $(x_n) \subseteq X$ such that $x_n \to x$ in $\tau_{seq}$ .

For the forwards implication, suppose that $x \in \overline{X} \setminus X$ . If there are infinitely many tops $s_n$ ( $n \in \mathbb{N} $ ) of x such that $[s_n] \cap X \neq \emptyset$ then select $x_n \in [s_n] \in X$ and note that $x_n \to x$ in $\tau_{seq}$ according to (i). Hence, we may suppose that there are only finitely many tops $F = {\left\lbrace {s_1,\ldots, s_k} \right\rbrace}$ of x such that $X \cap [s_i] \neq \emptyset$ . Let $X' = X \setminus [F]$ . Since $[F] \not\ni x$ is clopen, it follows that $x \in \overline{X'}$ . Let $t_n$ be an increasing, cofinal sequence in x (which exists since T is special). Since $\bigcap_{n \in \mathbb{N} }[t_n] \cap X' = \emptyset$ but $[t_n] \cap X' \neq \emptyset$ , we can choose pairwise distinct $x_n \in X' \cap [t_n]$ , and we have convergence $x_n \to x$ in $\tau_{seq}$ according to (ii).

For the backwards implication, suppose $x_n \to x$ in $\tau_{seq}$ . We show that $x \in \overline{X}$ with respect to $\tau$ . Let [t, F] be a standard basic open neighbourhood of x where $t \in x$ and F a finite set of tops of x. If $x_n \to x$ according to (i), then only finitely many members of $\{x_n \;:\; n \in \mathbb{N} \}$ lie above F, so infinitely many $x_n$ belong to [t, F]. If $x_n \to x$ according to (ii), then only finitely many members of $x_n$ do not contain t, so infinitely many $x_n$ belong to [t, F].

Proof. Since $(T,\mathcal{V})$ has finite adhesion, so does the T-graph $\dot{G}$ , and we may apply Theorem 4·7 to $\dot{G}$ .

Proof. Since t is contained in $O(\varepsilon)$ , the end $\varepsilon$ does not lie in the closure of $V_{T\setminus\lfloor t\rfloor}$ , i.e., every ray in $\varepsilon$ has a tail in $G[V_{\lfloor t\rfloor}]$ . Let $s_i$ ( $i\in I$ ) be the successors of t in T. Since no $s_i$ is contained in $O(\varepsilon)$ and $(T,\mathcal{V})$ has finite adhesion, every ray in $\varepsilon$ meets each vertex set $V_{\lfloor s_i\rfloor}$ only finitely often. As $V_t$ pairwise separates all vertex sets $V_{\lfloor s_i\rfloor}$ in the subgraph $G[V_{\lfloor t\rfloor}]$ which contains a tail of every ray in $\varepsilon$ , it follows that every ray in $\varepsilon$ meets $V_t$ infinitely often. In particular, $V_t$ is infinite, so t must be a limit.

Proof. Since $\dot{G}$ is a T-graph, each interval $[t_n,t_{n+1}]$ is connected in it by Theorem 2·3 (iv), so in particular it contains a ray R with $V(R)\subseteq\varrho$ that traverses all $t_n$ in the correct order. Using that the induced subgraphs $G[V_t]$ with $t\in T$ are connected, it is straightforward to construct the desired comb from R.

Proof. First, we show that $\varepsilon$ is contained in the intersection on the left-hand side of the equation. For this, let any $t\in\varrho$ be given and put $X\;:\!=\;\bigcup\,\{V_s\mid s\in\varrho\text{ and }s\geqslant t\}$ . Assume for a contradiction that $\varepsilon$ is not contained in $\partial_{\Omega} {X}$ . Then $\varepsilon$ contains a ray R which avoids X. Since $t \in \varrho$ implies $\varepsilon \notin \partial_{\Omega} {V_{T\setminus\lfloor t\rfloor}}$ , a tail (so without loss of generality all) of R avoids $V_\varrho$ . By Theorem 2·3 (iii), the components of $G-V_{\varrho}$ are of the form $V_{\lfloor{r}\rfloor}$ for r minimal in $T-\varrho$ , and since $\varepsilon$ corresponds to $\varrho$ , the component of $G-V_{\varrho}$ containing R has to be of the form $G[V_{\lfloor{r}\rfloor}]$ for r a top of $\varrho$ .

This last statement implies $\varepsilon\in\partial_{\Omega} {V_{\lfloor{r}\rfloor}}$ . By definition of $O(\varepsilon) = \varrho$ , the fact $r\notin\varrho$ implies $\varepsilon\in\partial_{\Omega} {V_{T\setminus \lfloor{r}\rfloor}}$ , and so we can extend R to a comb in G which is attached to $V_{T\setminus \lfloor{r}\rfloor}$ . However, since $\dot{G}$ is a T-graph and $(T,\mathcal{V})$ has finite adhesion, it follows that all but finitely many of the paths that have been added to R in order to obtain this comb must pass through X, which gives $\varepsilon\in\partial_{\Omega} {X}$ , a contradiction.

It remains to show the forward inclusion of the equation. For this, let $\delta$ be an element of the intersection on the left. We have just shown that $\varepsilon$ is an element of this intersection as well. Pick rays $R\in\varepsilon$ and $S\in\delta$ . We show that R and S are equivalent in G. Let $t_0\lt t_1\lt\cdots$ be a cofinal $\omega$ -chain in the high-ray $\varrho$ , and put $X_n\;:\!=\;\bigcup\,\{V_s\mid s\in\varrho,s\geqslant t_n\}$ . We extend both R and S to combs $C_R$ and $C_S$ in G with teeth $u_R^n$ and $u_S^n$ in $X_n$ for each $n\in \mathbb{N} $ , respectively. Using that all intervals $[t_n,t_{n+1}]$ of the T-graph $\dot{G}$ are connected by Theorem 2·3 (iv), we find infinitely many pairwise disjoint $\{u_R^n\mid n\in \mathbb{N} \}$ – $\{u_S^n\mid n\in \mathbb{N} \}$ paths in G, which shows that R and S are equivalent as desired.

Proof. Every high-ray of T has some ends of G corresponding to it by Theorem 6·4. Every high-ray of T has at most one end of G corresponding to it by Theorem 6·5.

Proof. Using Theorem 6·6 we may let $\varepsilon$ be the unique end of G that corresponds to the high-ray $\mathring{\lceil t\rceil}$ . To show that $N(V_{\lfloor t\rfloor})$ is concentrated in $\varepsilon$ , it suffices to show that for every infinite set U of neighbours of $V_{\lfloor t\rfloor}$ there exists a comb in G attached to U with its spine in $\varepsilon$ . For this, let U be any infinite set of neighbours of $V_{\lfloor t\rfloor}$ . Since $\mathring{\lceil t\rceil}$ is a high-ray, it contains a cofinal $\omega$ -chain $t_0\lt t_1\lt\cdots$ , and since $(T,\mathcal{V})$ has finite adhesion, by Theorem 4·4 all but finitely many of the vertices of U are contained in $V\restriction\{t'\in T\mid t_n\leqslant t'\lt t\}$ for each $n\in \mathbb{N} $ . Hence we find a cofinal $\omega$ -chain $t'_0\lt t'_1\lt\cdots$ in $\mathring{\lceil t\rceil}$ such that $V_{t'_n}$ contains a vertex $u_n$ of U for all $n\in \mathbb{N} $ . Applying Theorem 6·4 to $\{u_n\mid n\in \mathbb{N} \}$ yields the desired comb.

Proof. (i) The forward implication is evident. The backward implication follows from the assumption that $(T,\mathcal{V})$ is sequentially faithful.

(ii) We prove the forward implication by contradiction. So assume for a contradiction that $\varepsilon_n\to\varepsilon$ for $n\in B$ and $n\to\infty$ while there are a successor node $t\in\varrho$ and an infinite subset $B'\subseteq B$ such that $\varrho\cap\varrho_n\subseteq\mathring{\lceil t\rceil}$ . Then $X\;:\!=\;N_G(V_{\lfloor t\rfloor})$ is finite by (PT2). For every $n\in B'$ , the end $\varepsilon_n$ lives outside of $C(X,\varepsilon)$ as $\varrho\cap\varrho_n\subseteq\mathring{\lceil t\rceil}$ . As B ^′ is infinite by assumption, this contradicts that $\varepsilon_n\to\varepsilon$ for $n\in B$ and $n\to\infty$ .

For the backward implication, we have to prove that $\varepsilon_n\to\varepsilon$ for $n\in B$ and $n\to\infty$ . For this, let $X\subseteq V(G)$ be any finite vertex set. Let $t\in\varrho$ be any successor node such that $X\subseteq V_{\mathring{\lceil t\rceil}}$ . By assumption, we find a large enough number $N\in \mathbb{N} $ such that $t\in\varrho_n$ for all $n\in B$ with $n\geqslant N$ . We claim that $\varepsilon_n\in\Omega(X,\varepsilon)$ for all $n\in B$ with $n\geqslant N$ . Indeed, consider any such n, and pick any vertex $v\in V_{t}$ . Theorem 6·4 yields a ray $R_n\in\varepsilon_n$ that is included in $G[\,V\restriction\{t'\in\varrho_n\colon t'\geqslant t\}\,]$ and starts at v. Similarly, Theorem 6·4 yields a ray $R\in\varepsilon$ that is included in $G[\,V\restriction\{t'\in\varrho\colon t'\geqslant t\}\,]$ and starts at v. Then $R_n\cup R$ is a double ray in G which avoids X with one tail in $\varepsilon_n$ and another in $\varepsilon$ , yielding $\varepsilon_n\in\Omega(X,\varepsilon)$ .

Proof. Let $G_C$ be the graph that is obtained from $G[C\cup N(C)]$ by removing all the edges that run between any two vertices in N(C). By the star-comb lemma (2·1), $G_C$ contains either a star or a comb Z attached to N(C). Let $G^+_C$ be the supergraph of $G_C$ in which we make N(C) complete. Let $\varepsilon^+$ be the end of $G^+_C$ containing the rays from the clique induced by N(C), and note that $N(C)\cup Z$ is concentrated in $\varepsilon^+$ in the graph $G^+_C$ . Now apply Theorem 3·3 inside $G^+_C$ to find a connected envelope $U^*$ for $N(C) \cup Z$ that is concentrated in $\varepsilon^+$ and so that $U\;:\!=\; U^* \cap C$ is connected. We verify that U satisfies (i)–(iv).

(i) Suppose that U is infinite. We need to show that for any finite set X of vertices of G, almost all vertices of U are contained in $C(X,\varepsilon)$ . By assumption, almost all vertices of N(C) are contained in $C(X,\varepsilon)$ , and by increasing X if necessary, we may assume that $C(X,\varepsilon)$ is the only component of $G-X$ containing vertices of N(C). Now since $U^*$ is concentrated in $\varepsilon^+$ , it follows that almost all vertices of $U^*$ can be connected to N(C) in $G^+_C - X$ . Then the same is true in $G - X$ , and so all these vertices belong to $C(X,\varepsilon)$ . Thus, we have shown that almost all vertices of $U^*$ (and hence of U) belong to $C(X,\varepsilon)$ , as desired.

(ii) Every component of $C - U$ is also a component of $G_C^+ -U^\ast$ , and so has finite neighbourhood by the definition of an envelope. Note that it does not matter here whether we take the neighbourhood in G or in $G_C^+$ , because $N_G(C)$ is included in $U^\ast$ .

(iii) Given a sequence $(\varepsilon_n)_{n\in \mathbb{N} }$ as in the statement of (iii), note first that U must be infinite. Hence U is concentrated in $\varepsilon$ by (i), and so the claim follows by Theorem 3·3.

(iv) $N_G(U)\cap N_G(C)$ is infinite by the choice of $Z \subseteq U^*$ .

Proof. Let G be any connected graph. We will use the following notation. Suppose that $(T_1,\mathcal{V}_1)$ and $(T_2,\mathcal{V}_2)$ are two partition trees of G and that $T_1$ has a final level. Furthermore, suppose that $\mathcal{V}_1=(V_t^1\mid t\in T_1)$ and that $\mathcal{V}_2=(V^2_t\mid t\in T_2)$ . We write $(T_1,\mathcal{V}_1)\leqslant (T_2,\mathcal{V}_2)$ if the following two conditions are met:

1. (i) $T_2$ extends $T_1$ so that all the points of $T_2\setminus T_1$ lie above the final level of $T_1$ ;

2. (ii) $V^1_t=V^2_t$ for all $t\in T_1$ below the final level, and $V_t^1=V^2_{\lfloor t\rfloor}$ for all $t\in T_1$ in the final level.

Case 1. In the first case, i is a successor ordinal $i=j+1$ . Then $T=T_j$ has a final level $F_j$ of limit height. Consider any point $C\in F_j$ . The point C is a top of a high-ray $\varrho$ of T, and for the high-ray $\varrho$ there is a unique end $\varepsilon$ of G with $\tau(\varepsilon)=\varrho$ by Theorem 6·6. Furthermore, C is a connected subgraph of G whose neighbourhood is concentrated in $\varepsilon$ by Theorem 6·7. We apply Theorem 7·2 to $C\subseteq G$ and $\varepsilon$ to find a non-empty connected vertex set $U_C\subseteq V(C)$ that satisfies the following conditions:

1. (i) $U_C$ is either finite or concentrated in $\varepsilon$ ;

2. (ii) every component of $C-U_C$ has finite neighbourhood in G;

3. (iii) every sequence $(\varepsilon_n)_{n\in \mathbb{N} }$ of ends of G, where each $\varepsilon_n$ lives in a component $D_n$ of $C-U$ so that the map $ \mathbb{N} \ni n\mapsto N_G(D_n)$ is finite-to-one, converges to $\varepsilon$ ;

4. (iv) $N_G(U_C)\cap N_G(C)$ is infinite.

In each component D of $C-U_C$ we use Theorem 2·2 to pick an inclusionwise maximal normal tree T(C, D) rooted in a vertex that sends an edge to $U_C$ . Then the neighbourhood in G of every component K of $D-T(C,D)$ is included in $U_C\cup T(C,D)$ , where the proportion $N(K)\cap U_C$ is finite (because N(D) is finite by (ii)) while $N(K)\cap T(C,D)$ is infinite; and in fact the infinitely many neighbours of K in T(C, D) determine a normal ray R(K) of T(C, D).

We obtain the tree $T_i$ from T in two steps, as follows. First, we add for each C in the final level $F_j$ of $T=T_j$ and every component D of $C-U_C$ the order tree defined by T(C, D) directly above the point C so that the root of T(C, D) becomes a successor of C. Second, we add for each C and D every component K of $D-T(C,D)$ as a top of the high-ray $\lceil R(K)\rceil_{T_i}$ . The family $\mathcal{V}_i$ is defined as follows.

1. (i) For each $t\in T-F_j$ we let $V^i_t\;:\!=\;V^j_t$ .

2. (ii) For each $C\in F_j$ we let $V^i_C\;:\!=\;U_C$ .

3. (iii) For each $t\in T(C,D)$ we let $V^i_t\;:\!=\;\{t\}$ .

4. (iv) For each K in the final level of $T^i$ we let $V^i_K\;:\!=\;V(K)$ .

Clearly, $(T_j,\mathcal{V}_j)\leqslant (T_i,\mathcal{V}_i)$ if $(T_i,\mathcal{V}_i)$ is a partition tree. We claim that $(T_i,\mathcal{V}_i)$ is a partition tree of G. (PT1) follows with (iv). Condition (ii) ensures that $(T_i,\mathcal{V}_i)$ has finite adhesion (PT2). (PT3) holds by construction.

To see that the partition tree $(T_i,\mathcal{V}_i)$ is sequentially faithful, let any end $\varepsilon$ of G be given that corresponds to a high-ray $\tau(\varepsilon)$ of $T_i$ . Here, $\tau$ is defined with regard to $(T_i,\mathcal{V}_i)$ . If the order-type of the high-ray $\tau(\varepsilon)$ is equal to the height of $F_i$ , then $(T_i,\mathcal{V}_i)$ is sequentially faithful at $\varepsilon$ for the trivial reason that all tops of $\tau(\varepsilon)$ lie in the final level $F_i$ and therefore have no successors. Hence we may assume that $\tau(\varepsilon)$ is included in $T_j\subseteq T_i$ . If the order-type of $\tau(\varepsilon)$ is less than the height of $F_j$ , then all successors of the tops of $\tau(\varepsilon)\subseteq T_i$ are present in $T_j$ . Hence $(T_i,\mathcal{V}_i)$ is sequentially faithful at $\varepsilon$ because $(T_j,\mathcal{V}_j)\leqslant (T_i,\mathcal{V}_i)$ is. Otherwise, the order-type of $\tau(\varepsilon)$ is equal to the height of $F_j$ . Then the successors of the tops of $\tau(\varepsilon)\subseteq T_i$ are missing in $T_j$ . To see that $(T_i,\mathcal{V}_i)$ is sequentially faithful at $\varepsilon$ , let any sequence $(\varepsilon_n)_{n\in \mathbb{N} }$ of ends of G be given that live in the up-closures $\lfloor s_n\rfloor$ of successors of tops of $\tau(\varepsilon)$ such that the map $ \mathbb{N} \ni n\mapsto N_G(V_{\lfloor s_n\rfloor})$ is finite-to-one. By construction, each successor $s_n$ is the root of a normal tree $T(C_n,D_n)$ . Let $X\subseteq V(G)$ be any finite vertex set. We have to find a number $N\in \mathbb{N} $ such that all ends $\varepsilon_n$ with $n\geqslant N$ live in the component $C(X,\varepsilon)$ . Only the vertex sets of finitely many tops C of $\tau(\varepsilon)$ meet X, all others must be included in $C(X,\varepsilon)$ . This partitions the successors $s_n$ , and hence the sequence $\varepsilon_n$ , into finitely many subsequences: the first subsequence is formed by all ends $\varepsilon_n$ that already live in $C(X,\varepsilon)$ , and the other subsequences are formed by all ends $\varepsilon_n$ that live in C for a common top C of $\tau(\varepsilon)$ . Applying (iii) to all these subsequences except the first yields N.

To show that $(T_i,\mathcal{V}_i)$ displays all the ends of G that do not live at points in $F_i$ , it suffices by Theorem 6·6 to show that every end $\varepsilon$ of G with $\tau(\varepsilon)\in T_i$ lives at some $K\in F_i$ (i.e. satisfies $\tau(\varepsilon)=K$ ). Here, $\tau$ is defined with regard to $(T_i,\mathcal{V}_i)$ . Let any end $\varepsilon$ of G be given with $\tau(\varepsilon)\in T_i$ , and recall that $\tau(\varepsilon)$ must be a limit. Then $\tau(\varepsilon)$ cannot be a point of $T-F_j$ , because $(T_j,\mathcal{V}_j)$ displays all the ends of G that do not live at points in $F_j$ . And $\tau(\varepsilon)$ cannot be a point $C\in F_j\subseteq T_i$ , because this would imply $\varepsilon\in\partial_{\Omega} {U_C}$ (by Theorem 6·3) where by (i) the end $\varepsilon$ would then be sent to the high-ray $\mathring{\lceil C\rceil}_{T_i}$ instead of C. Therefore, $\tau(\varepsilon)\in F_i$ is the only possibility, as desired.

Case 2. In the second case, i is a limit, so T has no final level (or else the construction would have terminated for some $j\lt i$ ). Then for every $t\in T$ there is a least ordinal $j(t)\lt i$ for which t is contained in $T_{j(t)}$ but not in the final level $F_{j(t)}$ , so $V_t^j=V_t^{j(i)}$ for all $j\in [j(t),i)$ . We let $V_t^i\;:\!=\;V_t^{j(t)}$ for all $t\in T$ . If C is any component of $G-\bigcup_{t\in T}V_t^i$ , then for each $j\lt i$ there is a unique point $C_j\in F_j$ with $C\subseteq C_j$ , and $B(C)\;:\!=\;\{C_j\mid j\lt i\}$ is a branch T. We obtain $T_i$ from T by adding each component C as a top of the branch B(C) of T. Letting $V_C\;:\!=\;V(C)$ for all these tops completes the definition of $(T_i,\mathcal{V}_i)$ .

To ensure that $(T_i,\mathcal{V}_i)$ is a partition tree, we have to show that each C has cofinal down-neighbourhood in $G/\mathcal{V}_i$ . Since $(T_i,\mathcal{V}_i)$ has finite adhesion, it suffices to show that the collection $T_C$ of all points of T whose parts contain neighbours of C is an infinite subset of B(C). The inclusion $T_C\subseteq B(C)$ is immediate from the fact that each $G/\mathcal{V}_j$ with $j\lt i$ is a $T_j$ -graph in which $C_j$ is a maximal vertex. So assume for a contradiction that $T_C$ is finite and consider any $j\lt i$ with $T_C\subseteq T_j-F_j$ . Then $C=C_j$ since otherwise $C\subsetneq C_j$ has a neighbour in $C_j$ that is contained in no part $V_t$ with $t\in T_C$ , contradicting the definition of $T_C$ . But then $C_j$ has finite neighbourhood, contradicting the fact that $C_j\in F_j$ is a vertex at limit height in the $T_j$ -graph $G/\mathcal{V}_j$ .

To see that $(T_i,\mathcal{V}_i)$ is sequentially faithful, consider any end $\varepsilon$ of G for which $\tau(\varepsilon)$ is a high-ray of $T_i$ . If $\tau(\varepsilon)$ is a high-ray of any $T_j$ with $j\lt i$ , then we are done because $(T_j,\mathcal{V}_j)$ is sequentially faithful at $\varepsilon$ by assumption. Otherwise $\tau(\varepsilon)$ is a branch of T with all tops in $F_i$ , so $(T_i,\mathcal{V}_i)$ is sequentially faithful at $\varepsilon$ for the trivial reason that these tops have no successors in $T_i$ .

To show that $(T_i,\mathcal{V}_i)$ displays all the ends of G that do not live at points in $F_i$ , it suffices by Theorem 6·6 to show that every end $\varepsilon$ of G with $\tau(\varepsilon)\in T_i$ satisfies $\tau(\varepsilon)\in F_i$ . Let us assume for a contradiction that G has an end $\varepsilon$ such that $\tau(\varepsilon)\;=\!:\;t\in T_i$ is a point below $F_i$ . Then t lies below $F_j$ for $j\;:\!=\;j(t)\lt i$ , contradicting our assumption that $(T_j,\mathcal{V}_j)$ displays all the ends of G that do not live at points in $F_j$ .

We terminate the construction at the first ordinal $\kappa$ with $T_{\kappa+1}=T_{\kappa}$ . Then $\kappa\leqslant\omega_1$ because each $(T_\alpha,\mathcal{V}_\alpha)$ defines a $T_\alpha$ -graph $G/\mathcal{V}_\alpha$ of finite adhesion that has a final level $F_\alpha$ of height at least $\alpha$ , and these $T_\alpha$ -graphs cannot have uncountable branches by Theorem 4·7. By assumption, $(T_\kappa,\mathcal{V}_\kappa)$ is a sequentially faithful partition tree of G that displays all the ends of G which do not live at points in $F_\kappa$ . We claim that $(T_\kappa,\mathcal{V}_\kappa)$ displays all the ends of G. For this, it suffices to show that no end $\varepsilon$ of G lives at a point in $F_\kappa$ . And indeed, no end of G can live at a point in $F_\kappa$ , because otherwise the construction would not have terminated. Therefore, $(T_\kappa,\mathcal{V}_\kappa)$ is the desired partition tree.