Proof. Suppose that $I_1\cap I_2$ is a proper intersection. That is, $I_1\cap I_2 \neq \emptyset$ , $I_1\setminus I_2\neq \emptyset$ and $I_2\setminus I_1\neq \emptyset$ . Write $X=\{x_1,\ldots,x_t\}$ and $I_1=\{x_{p_1},x_{p_1+1},\ldots,x_{q_1}\}$ , $I_2=\{x_{p_2},x_{p_2+1},\ldots,x_{q_2}\}$ for $1\leqslant p_1\lt p_2\leqslant q_1\lt q_2\leqslant t$ . Let $u=a(x_{p_1},x_{q_1})$ and $v=a(x_{p_2},x_{q_2})$ . We claim that either $u$ is an ancestor of $v$ or $v$ is an ancestor of $u$ . Let $z\in I_1\cap I_2$ . By definition, both $u$ and $v$ are ancestors of $z$ . This means that there exists descending paths connecting $z$ to $u$ and $z$ to $v$ in $T_X$ with vertices in different levels. However, every vertex in $T_X$ has at most one father. Therefore, either the path $z$ to $u$ contains the path $z$ to $v$ or vice-versa. If the path $z$ to $u$ contains the path $z$ to $v$ , then $u$ is an ancestor of $v$ . Hence $u$ is an ancestor of $I_2\setminus I_1$ , which contradicts the fact that $I_1$ is closed (Condition ( $\star \star$ ) of Definition 2.2). The other case is analogous.

Proof. By Proposition 2.3 we obtain that either $I_1\cap I_2=\emptyset$ or $I_1\subseteq I_2$ . If the first case happens, then $I_1$ and $I_2$ satisfy condition (1) and we are done. Hence, we may assume that $I_1\subseteq I_2$ . If $I_2$ is a broken comb, then by definition $A_2=I_2$ . Thus in this case $I_1\subseteq A_2$ , satsifying condition (2). Now suppose without loss of generality that $I_2=A_2\cup B_2$ is a left maximal comb and write $A_2=\{x_1,\ldots, x_{\ell }\}$ , $B_2=\{y_1,\ldots, y_s\}$ . If $I_1\cap B_2=\emptyset$ , then $I_1\subseteq A_2$ and again condition (2) holds.

At last, it remains to deal with the case that $I_1\cap B_2\neq \emptyset$ . Since $I_1$ is an interval of $X$ and $I_1\subseteq I_2$ , then in particular $I_1$ is an interval of $I_2$ . Write $I_1=\{x_p,\ldots,x_\ell \}\cup \{y_1,\ldots, y_q\}$ for $1\leqslant p \leqslant \ell$ and $1\leqslant q \leqslant s$ . By condition (a3*) of Definition 2.4, the set $A_2\cup \{y_1,\ldots,y_{q-1}\}$ is closed. Therefore for any $z \in A_2\cup \{y_1,\ldots,y_{q-1}\}$ the greatest ancestor $a(z,y_q)$ of $z$ and $y_q$ is the same as the greatest ancestor of $a(x_1,y_q)$ . In particular, this implies that $a(x_p,y_q)$ is an ancestor for the entire set $A_2$ . Hence $A_2\subseteq I_1$ and consequently $I_1=A_1\cup B_1$ is a left comb satisfying condition (3), because $A_1=A_2$ and $B_1$ is an initial segment of $B_2$ . Note that $I_1$ is not maximal in this case, since the set $I_1\cup \{y_{q+1}\}$ is also a left comb. Thus if $I_1$ is a maximal comb, then it either satisfies (1) or (2).

Proof. The proof basically consists of checking the consistency of our definition. If $F(I)=F(I')=\emptyset$ , then $I$ and $I'$ are either of type $2$ or $6$ . In both cases $\chi _0(I)=\chi _0(I')=0$ .

If $I=\{x_1,\ldots,x_s\}$ and $I'=\{x'_1,\ldots,x'_{s'}\}$ are of type $1$ , $3$ or $4$ , then since $a(I)=F(I)=F(I')=a(I')$ we obtain by Fact 2.1 that $s=s'$ and $a(x_i,x_{i+1})=a(x'_i,x'_{i+1})$ for every $1\leqslant i \leqslant s$ . Therefore $\delta (x_i,x_{i+1})=\delta (x'_i,x'_{i+1})$ for every $1\leqslant i \leqslant s$ and by the colouring defined in Section 3.3, it follows that $\chi _0(I)=\chi _0(I')$ .

The last case that we need to check is when $I=A\cup B=\{x_1,\ldots,x_{s}\}$ and $I'=A'\cup B'=\{x'_1,\ldots,x'_{s'}\}$ are of type $5$ , where $|A|=\ell$ and $|A'|=\ell '$ . As usual, we assume that $I$ and $I'$ are left combs. Since $a(\{x_\ell,\ldots,x_s\})=F(I)=F(I')=a(\{x'_{\ell '},\ldots,x'_{s'}\})$ , it follows again by Fact 2.1 that $s-\ell =|B|=|B'|=s'-\ell '$ and $a(x_i,x_{i+1})=a(x'_i,x'_{i+1})$ for $\ell \leqslant i\leqslant s-1$ . Thus $\delta (x_i,x_{i+1})=\delta (x'_i,x'_{i+1})$ for $\ell \leqslant i \leqslant s-1$ and by the colouring of type 5 we obtain that $\chi _0(I)=\chi _0(I')$ .

Proof. Suppose without loss of generality that $|I|\leqslant |I'|$ . By Proposition 2.7 either $I\cap I'=\emptyset$ or $I\subseteq A'$ . If $I\cap I'=\emptyset$ , then by the fact that $I, I'$ are closed we obtain that $a(I)\cap a(I')=\emptyset$ . Since $F(I)\subseteq a(I)$ by definition, it follows that $F(I)\cap F(I')=\emptyset$ .

Now suppose that $I\subseteq A'$ . We may assume that $F(I), F(I')\neq \emptyset$ and consequently that $I,I'$ are of type $1$ , $3$ , $4$ or $5$ . Since maximal combs of types $1$ , $3$ , $4$ and $5$ have size at least $r$ , our assumption implies that $|I'|\geqslant |I|\geqslant r$ .

We claim that $|A'|\gt r$ . Suppose that $|A'|=r$ . Since $r\leqslant |I|\leqslant |A'|$ we obtain that $I=A'$ and $|I|=r$ . The maximality of $I$ implies that it is either a left or right maximal comb (otherwise we could extend the comb to $I\cup B$ ). However, in this case $I$ is of type $2$ . Thus $F(I)=\emptyset$ , which contradicts our assumption on $I$ . Therefore, $|A'|\gt r$ and consequently $I'$ is of type $5$ . Write $I'=\{x'_1,\ldots,x'_{s'}\}$ with $|A'|=\ell '$ and assume that $I'$ is a left comb. Then by definition

\begin{align*} F(I')=\{a(x'_i,x'_{i+1})\}_{\ell '\leqslant i \leqslant s'-1}. \end{align*}

Since $I\subseteq A'$ we obtain that $F(I)\subseteq a(A')$ . By the structure of a left comb (see Figure 10) we have that

\begin{align*} \delta (x'_1,x'_{\ell '})\gt \delta (x'_{\ell '},x'_{\ell '+1})\gt \delta (x'_{\ell '+1},x'_{\ell '+2})\gt \ldots \gt \delta (x'_{s'-1},x'_{s'}). \end{align*}

Figure 10. A picture of $I$ and $\{a(x_i,x_{i+1})\}_{\ell '\leqslant i \leqslant s-1}$ .

Thus $a(I)\cap \{a(x'_i,x'_{i+1})\}_{\ell '\leqslant i \leqslant s'-1}=\emptyset$ and consequently $F(I)\cap F(I')=\emptyset$ .

Proof. Let $V\;:\!=\;V(H)$ . Given a closed interval $I\subseteq V$ , by condition (ii) of Definition 2.2 there exists a vertex $u \in a(I)$ such that $I=V(u)$ . Consider the partition of $I$ given by $I=I^L\cup I^R$ , where $I^L=V_L(u)$ are the left descendants of $u$ and $I^R=V_R(u)$ are the right descendants of $u$ . Let $u^L$ be the left child of $u$ and $u^R$ be the right child. Hence, $I^L=V(u^L)$ and $I^R=V(u^R)$ and consequently $I=I^L \cup I^R$ is a partition of a closed interval in $V$ into two non empty closed intervals in $V$ (see Figure 12).

Figure 12. Partition of a closed interval into two other closed intervals.

We will construct our set $M'$ iteratively. This is done in two stages. In the first stage we start with the closed interval $Y_0=V$ and proceed recursively as follows: For a closed interval $Y_i\subseteq V$ , let $Y_i=Y_i^L\cup Y_i^R$ be the partition described above in two closed intervals. The choice of $Y_{i+1}$ is determined by the conditions below

1. (P1) Set $Y_{i+1}\;:\!=\;Y_i^L$ if $|Y_i^L\cap M|\geqslant |Y_i^R\cap M|$ .

2. (P2) Set $Y_{i+1}\;:\!=\;Y_i^R$ if $|Y_i^L\cap M|\lt |Y_i^R\cap M|$ .

We stop the process whenever $Y_{i}\cap K_0=\emptyset$ or $Y_{i}\cap K_1=\emptyset$ . Note that since $Y_i^L$ and $Y_i^R$ are non empty, at each iteration of the process the size of $|(K_0\cup K_1)\cap Y_i|$ reduces at least by one. Thus, in a finite amount of time the process terminates. Let $Y$ be the closed interval obtained in the end. We may assume without loss of generality that $Y\cap K_1=\emptyset$ . Write $Y=K_Y\cup M_Y$ , where $K_Y\subseteq K_0$ and $M_Y\subseteq M$ . It is not hard to check by the construction that $|M_Y|\geqslant m/2$ .

For the second stage, let $Z_0=Y$ . Given a closed interval $Z_i \subseteq V$ , let $Z_i=Z_i^L\cup Z_i^R$ be the partition into two non empty closed intervals. By definition we have that $Z_i^L\lt Z_i^R$ . We say that a partition $Z_i^L\cup Z_i^R$ is of type $A$ if $Z_i^R\cap K_0=\emptyset$ and of type $B$ if $Z_i^R\cap K_0\neq \emptyset$ . The choice of $Z_{i+1}$ will depend on the type of partition as follows:

Type A: $Z_i^R\cap K_0 = \emptyset$ .

1. (A1) Set $Z_{i+1}\;:\!=\;Z_i^L$ if $|Z_i^R|\lt \frac{1}{2}k^{-1/2}m^{1/2}$ .

2. (A2) Set $Z_{i+1}\;:\!=\;Z_I^R$ if $|Z_i^R|\geqslant \frac{1}{2}k^{-1/2}m^{1/2}$ and stop the process.

Type B: $Z_i^R\cap K_0 \neq \emptyset$ .

1. (B) Set $Z_{i+1}\;:\!=\;Z_i^R$ .

We terminate the process if we either reach condition (A2) or if $Z_{i+1}$ is a singleton. Since $|Z_{i+1}|\lt |Z_i|$ , the process is finite. Let $Z$ be the closed interval obtained at the end. We split into two cases.

If the process terminates after some instance of condition (A2), then it means that $Z=Z_i^R$ is a closed interval in $V$ with $|Z|\geqslant \frac{1}{2}k^{-1/2}m^{1/2}$ for some index $i$ . Because we are in a partition of type $A$ we also obtain that $Z\subseteq M$ . Thus, if we set $M'=Z$ the simple subdaisy $H[K_0\cup M'\cup K_1]$ satisfies condition (1) of the statement.

Now suppose that the process terminates with $|Z|=1$ . Then it means that for every partition of type A we had an instance of condition (A1). If $Z_{i+1}$ is a set obtained after condition (A1), then $|Z_{i+1}\cap M|\gt |Z_i\cap M|-\frac{1}{2}k^{-1/2}m^{1/2}$ and $|Z_{i+1}\cap K_0|=|Z_{i+1}\cap K_0|$ . That is, condition (A1) removes less than $\frac{1}{2}k^{-1/2}m^{1/2}$ element of $M$ from $Z_i$ and no elements of $K_0$ from it. Moreover, if $Z_{i+1}$ is obtained after condition (B), then $|Z_{i+1}\cap M|=|Z_i\cap M|$ and $|Z_{i+1}\cap K_0|\lt |Z_i\cap K_0|$ . That is, $M$ remains unaffected, but $K_0$ loses at least one element from $K_i$ to $K_{i+1}$ .

Consider the sequence of operations applied to $Z_0$ in order to obtain $Z$ . Since we start with a set $Z_0=Y$ with $|Z_0\cap M|=|M_Y|\geqslant m/2$ , we obtain that during our process we had at least

\begin{align*} \frac{\frac{m}{2}}{\frac{1}{2}k^{-1/2}m^{1/2}}=k^{1/2}m^{1/2} \end{align*}

instances of condition (A1) in the sequence. Similarly, since $|Z_0\cap K_0|=|K_Y|\leqslant k$ , we obtain that we had at most $k$ instances of condition (B) in the sequence. Hence, by the pigeonhole principle there exists a sequence of consecutive applications of condition (A1) of length at least $m^{\prime}=k^{1/2}m^{1/2}/(k+1)\geqslant \frac{1}{2}k^{-1/2}m^{1/2}$ .

Figure 13. Sequence of closed intervals $Z_j^R,\ldots, Z_{j+m^{\prime}-1}^R$ .

Let $Z_j, Z_{j+1},\ldots, Z_{j+m^{\prime}}$ be the closed intervals involved in the sequence. That is, $Z_{i+1}$ is obtained from $Z_i$ by a condition (A1) for every $j\leqslant i \leqslant j+m^{\prime}-1$ . By the algorithm, we obtain closed intervals $Z_j^R,\ldots,Z_{j+m^{\prime}-1}^R \subseteq M$ all of them with size less than $\frac{1}{2}k^{-1/2}m^{1/2}$ (Figure 13). For every $j\leqslant i \leqslant j+m^{\prime}-1$ , choose a point $z_i\in Z_i^R$ .

Set $M'=\{z_j,\ldots,z_{j+m^{\prime}-1}\}$ . We claim that $M'$ is a set satisfying condition (2) of the statement. Let $H'=H[K_0\cup M'\cup K_1]$ and $V'=V(H')$ . To see that condition (2) is satisfied we just need to find a maximal comb $I=A\cup B \subseteq V'$ such that $M'\subseteq B$ . Let $K'=K_0\cap Z_{j+m^{\prime}}$ and consider the interval $I'=K' \cup M'$ in $V'$ . By construction, the intervals $K'$ and $K'\cup \{z_{j+i},\ldots,z_{j+m^{\prime}-1}\}$ are closed in $V'$ for every $0\leqslant i\leqslant m^{\prime}-1$ . Therefore, by condition (a3*) of Definition 2.4, the interval $I'=A'\cup B'$ is a left comb and $M' \subseteq B'$ . Since every comb can be extended to a maximal one, there exists a maximal left comb $I=A\cup B$ with $A=A'$ and $B'\subseteq B$ such that $M'\subseteq B$ and we are done.

Proof. If $I\in{\mathcal{C}}_{V',M'}$ is such that $I\cap M'=\emptyset$ , then either $I\subseteq K_0$ or $I\subseteq K_1$ . Since $X=K_0\cup P\cup K_1$ for some $P\in M'^{(r)}$ , we obtain that $\Psi (I)=I\cap X=I$ . This shows that $\Psi$ is a bijection from the intervals of $V'$ disjoint of $M'$ to the intervals of $X$ disjoint of $P$ .

Now suppose that $I \in{\mathcal{C}}_{V',M'}$ is such that $M'\subseteq I$ . Then $I$ can be written as $I=K_I\cup M'$ with $K_I\subseteq K_0\cup K_1$ . Thus $\Psi (I)=I\cap X=K_I \cup P$ . Since $K_I\neq K_{I'}$ for $I\neq I'$ , we obtain that $\Psi$ is an injection from the intervals of $V'$ containing $M'$ to the intervals of $X$ containing $P$ . To check surjectivity, just notice that $K_I\cup P$ is an interval if and only if $K_I\cup M'$ is an interval.

It remains to prove that $I$ is closed if and only if $\Psi (I)$ is closed. Throughout the rest of the proof, for a set $S\subseteq V$ we define

\begin{align*} x_S=\min (S), \quad y_S=\max (S), \quad u_S=a(x_S,y_S). \end{align*}

Note that the backwards direction is straightforward from the definition of being closed.

The following observation will be useful for the rest of the proof.

In particular, Fact 5.4 applied to $W=M'$ says that an element $y\notin M'$ have the same common ancestor with any $x\in M'$ . We split the proof of the forward implication depending on the structure of $H'$ given by Lemma 5.1.

Case 1: $M'$ is a closed interval in $V'$ .

The proof of Case 1 is slightly different depending on the location of the interval $I$ in $V'$ .

Figure 14. A picture of Fact 5.4.

Case 1.1: $I \in{\mathcal{C}}_{V',M'}$ such that $I\cap M'=\emptyset$ .

As seen before, we have that $\Psi (I)=I$ . By condition ( $\star \star$ ) of Definition 2.2 there is no vertex $x\in X\setminus I$ such that $u_I$ is an ancestor of $x$ . If there is a descendant of $u_I$ in $V'\setminus I$ , then the descendant is in the set $V'\setminus X=M'\setminus P$ . Since $I\cap M'=\emptyset$ , Fact 5.4, applied to the closed interval $M'$ , implies that for every $y\in I'$ and $x\in M'$ we have $a(x,y)=a(y,u_{M'})$ . Thus, if $u_I$ is an ancestor of some $x \in M'$ , then $u_I$ is an ancestor of $u_{M'}$ . This implies that $u_I$ is an ancestor for the entire set $M'$ and in particular of $P$ , which contradicts the fact that $I$ is closed in $X$ . Therefore, $I$ is a closed interval in $V'$ .

Case 1.2: $I \in{\mathcal{C}}_{V',M'}$ such that $M'\subseteq I$ .

Suppose that $I=K_I\cup M'$ is a interval in $V'$ containing $M'$ . We need to prove that $I=K_I\cup M'$ is closed in $V'$ if $\Psi (I)=I\cap X=K_I\cup P$ is closed in $X$ . If $K_I=\emptyset$ , then $I=M'$ which is by assumption closed in $V'$ . Otherwise, we claim that $u_I=u_{\Psi (I)}$ . That is $I$ and $\Psi (I)$ have the same common ancestor.

The assumption that $K_I\neq \emptyset$ gives us that either $x_I\lt \min (M')$ or $y_I\gt \max (M')$ . Assume without loss of generality that $y_I\gt \max (M')$ . Thus, $y_I\in K_I$ and we have that $y_I=y_{\Psi (I)}=\max (K_I)$ . If $x_I \notin M'$ , then similarly we have $x_I=x_{\Psi (I)}$ and consequently $u_I=a(x_I,y_I)=a(x_{\Psi (I)},y_{\Psi (I)})=u_{\Psi (I)}$ . Now if $x_I \in M'$ , then $x_I\notin K_I$ . This implies that $x_{\Psi (I)} \in P\subseteq M'$ . Since both $x_I,x_{\Psi (I)} \in M'$ and $y_I=y_{\Psi (I)}\notin M'$ , by Fact 5.4 we obtain that $u_I=a(x_I,y_I)=a(u_{M'},y_I)=a(u_{M'},y_{\Psi (I)})=a(x_{\Psi (I)},y_{\Psi (I)})=u_{\Psi (I)}$ . Hence, $I=K_I\cup M'$ and $\Psi (I)=K_I\cup P$ have the same common ancestor.

To finish the proof note that by condition ( $\star \star$ ) of Definition 2.2 there are no descendants of $u_{\Psi (I)}$ in $X\setminus \Psi (I)$ . Since $u_I=u_{\Psi (I)}$ and $V\setminus I'=K\setminus K_I=X\setminus \Psi (I)$ , we conclude that there are no descendants of $u_I$ in $V'\setminus I$ and consequently $I$ is closed in $V'$ .

Case 2: $M'\subseteq Q$ for some maximal comb $Q=A^Q\cup B^Q$ in $V'$ with $M'\subseteq B^Q$ .

We may assume without loss of generality that $Q$ is a maximal left comb. Let $K_0^Q=K_0\cap Q$ and $K_1^Q=K_1\cap Q$ . Clearly, $Q=K_0^Q\cup M'\cup K_1^Q$ with $K_0^Q\lt M'\lt K_1^Q$ . Moreover, $Q=A^Q\cup B^Q$ with $A^Q\lt B^Q$ and $M'\subseteq B^Q$ (Figure 15). Thus, $A^Q\subseteq K_0^Q$ and by condition (a3*) of Definition 2.4, we obtain that $K_0^Q$ is closed in $V'$ . As in the first case, we split into two cases depending on the type of the interval.

Case 2.1: $I \in{\mathcal{C}}_{V',M'}$ such that $I\cap M'=\emptyset$ .

Suppose that $I$ is a closed interval in $X$ . We claim that $V'(u_I)\cap M' = \emptyset$ , i.e., the descendants of $u_I$ are disjoint of $M'$ . Applying Proposition 2.3 to the closed interval $V'(u_I)$ and maximal comb $Q$ gives us that either $V'(u_I)\cap Q=\emptyset$ , $V'(u_I)\subseteq Q$ or $Q\subseteq V'(u_I)$ . If $V'(u_I)\cap Q=\emptyset$ , then we immediately obtain that $V'(u_I)\cap M'=\emptyset$ , since $M'\subseteq Q$ . If $Q\subseteq V'(u_I)$ , then $M'\subseteq V'(u_I)$ and consequently $P=M'\cap X\subseteq V'(u_I)\cap X=X(u_I)$ . This implies that $X(u_I)\neq I$ , which contradicts $I$ being closed in $X$ .

Thus, we may assume that $V'(u_I)\subseteq Q$ and $M' \not \subseteq V'(u_I)$ . Then, by Proposition 2.7, we have that $V'(u_I)=A^{V'(u_i)}\cup B^{V'(u_I)}$ where either $V'(u_I) \subseteq A^Q$ or $A^{V'(u_i)}=A^Q$ and $B^{V'(u_I)}\subseteq B^Q$ . For the first case, note that $A^Q\cap M'=\emptyset$ and therefore $V'(u_I)\cap M'=\emptyset$ . For the second case, note that since $M'\not \subseteq V'(u_I)$ , then $M'\not \subseteq B^{V'(u_I)}$ . This implies that $V'(u_I)\subseteq K_0\cup M'$ . Together with the fact that $I\cap M'=\emptyset$ and $I\subseteq V'(u_I)$ , we obtain that $I\subseteq K_0^Q$ . Since $K_0^Q$ is closed in $V'$ , we have that the common ancestor $u_{K_0^Q}=a(\!\min (K_0^Q),\max (K_0^Q))$ is an ancestor of the entire $I$ and therefore of $u_I$ . Hence, $V'(u_I)\subseteq K_0^Q$ , which implies that $V'(u_I)\cap M'=\emptyset$ . The fact that $I$ is closed in $V'$ now follows because $V'(u_I)=X(u_I)=I$ .

Figure 15. Maximal comb $Q$ and sets $K_0^Q$ and $K_1^Q$ .

Case 2.2: $I \in{\mathcal{C}}_{V',M'}$ such that $M'\subseteq I$ .

Let $I=K_0^I\cup M'\cup K_1^I$ be an interval in $V'$ containing $M'$ with $K_0^I\subseteq K_0$ and $K_1^I\subseteq K_1$ . Suppose that $\psi (I)=K_0^I\cup P\cup K_1^I$ is closed in $X$ . Since $K_0^Q\cup M'$ is closed, by the same argument of Case 1.2 (by considering $K_0^Q\cup M'$ instead of $M'$ ), we can show that if $x_I\lt \min (K_0^Q)$ or $y_I \gt \max (M')$ , then $u_I=a(x_I,y_I)=a(x_{\Psi (I)},y_{\Psi (I)})=u_{\Psi (I)}$ and consequently $I$ is closed in $V'$ .

Now suppose that $\min (K_0^Q)\leqslant x_I \leqslant \min (M')$ and $y_I=\max (M')$ . Since both $M'$ and $P$ are not closed intervals in their respective ground sets, we have that $x_I\neq \min (M')$ and consequently $x_{\Psi (I)}=x_I$ and $y_{\Psi (I)}=\max (P)$ . Hence, in this case, $K_0^I\subseteq K_0^Q$ and $K_I=\emptyset$ , which implies that $I=K_0^I\cup M'$ and $\Psi (I)=K_0^I\cup P$ . Because $Q$ is a maximal left comb with $M'\subseteq Q$ , then both sets $K_0^Q$ and $K_0^Q\cup M'$ are closed in $V'$ . Therefore, by Proposition 5.3 the intervals $K_0^Q$ and $K_0^Q\cup P$ are closed in $X$ . Fact 5.4 applied to $K_0^Q$ gives us that $a(z,y_{\Psi (I)})=a(z',y_{\Psi (I)})$ for every $z,z' \in K_0^Q$ . This implies that $u_{\Psi (I)}=a(x_{\Psi (I)},y_{\Psi (I)})=a(\!\min (K_0^Q),y_{\Psi (I)})$ , i.e., $K_0^Q\cup P$ and $\Psi (I)=K_0^I\cup P$ have $u_{\Psi (I)}$ as the same common ancestor. Since $K_0^Q\cup P$ and $K_0^I\cup P$ are both closed in $X$ , we obtain that $K_0^Q=K_0^I$ . Thus $I=K_0^Q\cup M'$ , which is closed in $V'$ .

Proof. The idea of the proof is to identify certain maximal combs in $V'$ with maximal combs in an edge $X$ . Because the structure of those maximal combs in $V'$ only depends on $H'$ , we will obtain the same for the corresponding combs in $X$ . Proposition 5.2 will be useful here, since by condition (a3*) and (b3*) of Definition 2.4 a comb can be defined by looking at certain closed subintervals. The proof is split into cases depending on the structure of the tree $T_{V'}$

Case 1: $M'$ is a closed interval in $V'$ .

We will construct a maximal comb in $X$ by looking at a maximal comb in $V'$ containing $M'$ . Write $K_0=\{x_1,\ldots,x_{k_0}\}$ , $M'=\{y_1,\ldots, y_{m^{\prime}}\}$ and $K_1=\{z_1,\ldots,z_{k_1}\}$ . There are two possibilities here:

Case 1.1: Either $M'\cup \{z_1\}$ is a closed interval in $V'$ or $M'\cup \{x_{k_0}\}$ is a closed interval in $V'$ .

Suppose without loss of generality that $M'\cup \{z_1\}$ is closed in $V'$ . In this case $M'\cup \{z_1\}$ is a left comb. Let $M'\cup \{z_1,\ldots,z_t\}$ be the maximal left comb obtained by extending $M'\cup \{z_1\}$ . We will assume during the entire proof that $t\lt k_1$ . For $t=k_1$ , the same proof work by removing any claims and sets involving $z_{t+1}$ . By condition (a3*) of Definition 2.4 and Definition 2.6, $M'\cup \{z_1,\ldots,z_t\}$ being a maximal left comb is the same as saying that the intervals $M'$ and $M'\cup \{z_1,\ldots,z_i\}$ are closed for every $1\leqslant i \leqslant t$ , but the interval $M'\cup \{z_1,\ldots,z_{t+1}\}$ is not closed.

Set $J=\{k_0+1,\ldots,k_0+r+t\}$ . Let $X$ be an edge of $H'$ with petal $P$ . We claim that $X_J$ is a maximal left comb in $X$ with $A^{X_J}\subseteq P$ . To see that consider the intervals

\begin{align*} J_i=\{k_0+1,\ldots,k_0+r+i\}, \quad 0\leqslant i \leqslant t+1. \end{align*}

In particular $J_t=J$ . Note that $X_{J_0}=M'\cap X=P$ and $X_{J_i}=(M'\cap \{z_1,\ldots,z_i\})\cap X$ for $1\leqslant i \leqslant t+1$ . Thus, by applying Proposition 5.2 with $I=M'$ and $I=M'\cup \{z_1,\ldots,z_i\}$ , we obtain that $X_{J_i}$ is closed in $X$ for $0\leqslant i \leqslant t$ and $X_{J_{t+1}}$ is not closed in $X$ . Hence, by condition (a3*) of Definition 2.4 and Definition 2.6, we have that $X_J=X_{J_t}$ is a maximal left comb. Since $P=X_{J_0}\subseteq X_{J_1}\subseteq \ldots \subseteq X_{J_t}=X_J$ are all closed intervals, we have that $A^{X_J}\subseteq P$ . Thus, $|A^{X_J}|\leqslant |P| =r$ and we have that either $X_J$ is a maximal comb of type 3 or type 4 depending on the size of $|X_J|=r+t$ . Because $t$ is a parameter that depends on the size of the maximal comb in $V'$ , i.e., on the structure of $H'$ , we conclude that the type of $X_J$ is independent of our choice of edge $X$ .

It remains to show that $a(P)\subseteq F(X_J)$ and $X_J$ is the smallest maximal comb containing $P$ with non-empty data colouring. For the first, note that $F(X_J)=a(X_J)$ because $X_J$ is of type $3$ or $4$ . Thus, $a(P)\subseteq a(X_J)=F(X_J)$ . For the latter, note that the only potential maximal comb smaller than $X_J$ containing $P$ is $P$ itself. However, if $P$ is a maximal comb, then it is a comb of type $2$ and therefore $F(P)=\emptyset$ . Hence, $X_J$ is the smallest maximal comb containing $P$ with non-empty colouring data.

Case 1.2: Both $M'\cup \{z_1\}$ and $M'\cup \{x_{k_0}\}$ are not closed in $V'$ .

By Definition 2.6, $M'$ is a maximal comb. Set $J=\{k_0+1,\ldots,k_0+r\}$ . Note that $X_J=P$ . By Proposition 5.2, the set $P=M'\cap X$ is closed in $X$ and $P\cup \{z_1\}=(M'\cup \{z_1\})\cap X$ and $P\cup \{x_{k_0}\}=(M'\cup \{x_{k_0}\})\cap X$ are not closed in $X$ . Thus, $P$ is a maximal comb in $X$ . It is clear that $A^{X_J}\subseteq X_J=P$ . Since $|P|=r$ and $P\cup \{z_1\}$ , $P\cup \{x_{k_0}\}$ are not closed, we have that $X_J=P$ is of type $1$ . Therefore, the type of $X_J$ does not depend on $X$ . Moreover, the fact that $X_J$ is of type 1 gives us that $a(P)=a(X_J)=F(X_J)$ . The minimality of $X_J$ is immediate from the fact that all combs with non-empty data has size at least $r$ .

Case 2: $M'\subseteq B$ for a maximal comb $I=A\cup B$ in $V'$ .

Suppose without loss of generality that $I=A\cup B$ is a maximal left comb. Let $A=\{x_{k_0-p-\ell +1},\ldots,x_{k_0-p}\}$ , $B\cap K_0=\{x_{k_0-p+1},\ldots,x_{k_0}\}$ and $B\cap K_1=\{z_1,\ldots,z_t\}$ . Set $J=\{k_0-p-\ell +1,\ldots,k_0+r+t\}$ . Let $X$ be an edge of $H'$ with petal $P$ (see Figure 16). Clearly, $X_J=I\cap X$ . We claim that $X_J$ is a maximal left comb with $P\subseteq B^{X_J}$ . By Definition 2.4 and 2.6 we have that $A$ is a closed interval in $V'$ , $A\setminus \{x_{k_0-p}\}$ and $I\cup \{z_{t+1}\}$ are not closed in $V'$ and

\begin{align*} \delta (x_{k_0-p},x_{k_0-p+1})&\gt \ldots \gt \delta (x_{k_0-1},x_{k_0})\gt \delta (x_{k_0},y_1)\gt \delta (y_1,y_2)\gt \ldots \gt \delta (y_{m^{\prime}-1},y_{m^{\prime}})\\[5pt] &\gt \delta (y_{m^{\prime}},z_1)\gt \delta (z_1,z_2)\gt \ldots \gt \delta (z_{t-1},z_t). \end{align*}

Figure 16. Case $2$ of Proposition 5.5.

Let $P=\{y_{i_1},\ldots,y_{i_r}\}$ . By Proposition 5.2, the set $A=A\cap X$ is closed in $X$ and the sets $A\setminus \{x_{k_0-p}\}=(A\setminus \{x_{k_0-p}\})\cap X$ and $X_J\cup \{z_{t_1}\}=(I\cup \{z_{t+1}\})\cap X$ are not closed in $X$ . Moreover, since $P\subseteq M'$ , we have that

\begin{align*} \delta (x_{k_0-p},x_{k_0-p+1})&\gt \ldots \gt \delta (x_{k_0-1},x_{k_0})\gt \delta (x_{k_0},y_{i_1})\gt \delta (y_{i_1},y_{i_2})\gt \ldots \gt \delta (y_{i_{r-1}},y_{i_r})\\[5pt] &\gt \delta (y_{i_r},z_1)\gt \delta (z_1,z_2)\gt \ldots \gt \delta (z_{t-1},z_t). \end{align*}

Thus, by Definition 2.4 and Definition 2.6, the interval $X_J$ is a maximal left comb in $X$ with $A^{X_J}=A$ and $|B^{X_J}|=r+t+p$ . Since $A\subseteq K_0$ , we obtain that $P\subseteq B^{X_J}$ . Note that to determine the type of $X_J$ we need to know the sizes of $X_J$ , $A^{X_J}$ and $B^{X_J}$ . None of this parameters depends on the choice of $X$ . Hence, the type of $X_J$ is independent of $X$ . Finally, because $|B^{X_J}|\geqslant |P|\geqslant r$ , we obtain that $|X_J|\geqslant r+1$ and consequently the comb $X_J$ is of type $3$ , $4$ or $5$ . If it is of type $3$ or $4$ , then $a(P)\subseteq a(X_J)=F(X_J)$ . If it is of type $5$ , then $a(P)\subseteq a(\max (A^{X_J})\cup B^{X_J})=F(X_J)$ .

Proof. Let $P, P'\subseteq M'$ be the petals of $X$ and $X'$ , respectively. By definition, $\chi (X)=\chi (X')$ implies that

\begin{align*} \sum _{I\in{\mathcal{I}}_X}\chi _0(I)=\sum _{I'\in{\mathcal{I}}_{X'}}\chi _0(I') \pmod{2}. \end{align*}

By the definition of colouring data, if $F(I)=\emptyset$ , then $\chi _0(I)=0$ . Thus, we may rewrite the equality above as

(4) \begin{align} \sum _{\substack{I\in{\mathcal{I}}_X\\[5pt] F(I)\neq \emptyset }}\chi _0(I)=\sum _{\substack{I'\in{\mathcal{I}}_{X'}\\[5pt] F(I')\neq \emptyset }} \chi _0(I) \pmod{2}. \end{align}

We claim that if $I=A\cup B$ is a maximal comb of $X$ with $F(I)\neq \emptyset$ , then either $I\cap P=\emptyset$ or $P\subseteq I$ . Note that, in the colouring defined in Subsection 3.3, whenever $F(I)=\emptyset$ , we have that $|I|\geqslant |P|=r$ . By Lemma 5.1, the daisy $H'$ satisfies one of the following conditions: Either $M'$ is a closed interval in $V'\;:\!=\;V(H')$ or there exists a maximal comb $Q=A^Q\cup B^Q$ such that $M'\subseteq B^Q$ . If $M'$ is closed in $V'$ , then by Proposition 5.2 the petal $P=M'\cap X$ is a closed interval in $X$ . Thus, Proposition 2.3 applied to the closed intervals $I$ and $P$ gives the desired result that either $I\cap P=\emptyset$ or $P\subseteq I$ . Now suppose that we are in Condition (2) of Lemma 5.1. By Proposition 5.5, we have that $P\subseteq B^{X_J}$ , where $B^{X_J}$ is the “teeth” part of the comb $X_J=A^{X_J}\cup B^{X_J}$ . Thus, Proposition 2.7 applied to the maximal combs $I$ and $X_J$ implies that $I\cap X_J=\emptyset$ , $I\subseteq A^{X_J}$ or $X_J\subseteq A \subseteq I$ . In the first two cases we obtain $I\cap P=\emptyset$ , while in the latter we have $P\subseteq I$ .

The idea of the proof of Proposition 5.6 is to show that there exists a bijection between $\{I\in{\mathcal{I}}_X\;:\: F(I)\neq \emptyset \}$ and $\{I'\in{\mathcal{I}}_{X'}\;:\: F(I')\neq \emptyset \}$ such that $X_J$ is sent to $X'_J$ and every $I\neq X_J$ is sent to an $I'\neq X'_J$ with $\chi _0(I)=\chi _0(I')$ . Hence, after some cancellation, we obtain from equation (4) that $\chi _0(X_J)=\chi _0(X'_J)$ . Based on the last paragraph, we construct such a bijection by splitting $\{I\in{\mathcal{I}}_X\;:\: F(I)\neq \emptyset \}$ into two parts:

Case 1: $I\in{\mathcal{I}}_X$ is a maximal comb of $X$ with $F(I)\neq \emptyset$ and $I\cap P=\emptyset$ .

We claim that $I\in{\mathcal{I}}_{X'}$ is a maximal comb in $X'$ of the same type and consequently $\chi _0(I)$ is the same in $X$ and $X'$ . Assume without loss of generality that $I\subseteq K_0$ . Let $x$ be the element preceding $\min (I)$ in $X$ (In the case that such $x$ does not exists, we simply take $x=\min (I)$ ). Let $y$ be the element after $\max (I)$ in $X$ . Similarly, define $x'$ as the element before $\min (I)$ in $X'$ and $y'$ as the element after $\max (I)$ in $X'$ . Since $I\subseteq K_0$ , clearly $x=x'$ . However, $y$ and $y'$ are not necessarily the same. By conditions (a3*) and (b3*) of Definition 2.4 and Definition 2.6, to prove that $I\in{\mathcal{I}}_{X'}$ is enough to check that $I\cup \{x\}$ , $L\subseteq I$ , $I\cup \{y\}$ are closed intervals in $X$ if and only if $I\cup \{x'\}$ , $L\subseteq I$ and $I\cup \{y'\}$ are closed intervals in $X'$ , respectively. Since $F(I)=\neq \emptyset$ , we have that $I$ is of type 1, 3, 4 or 5. Note that one can distinguish between this types by determining the size of the “handle” and ‘teeth” of $I$ . Thus, by checking the properties above, we also obtain that $I$ have the same type in $X$ and $X'$ .

For an interval $L\subseteq I$ , by Proposition 5.2 we have that $L=L\cap X$ is a closed interval in $X$ if and only if it is a closed interval in $V'$ . Another application of Proposition 5.2 gives us that $L=L\cap X'$ is a closed interval if it is closed in $V'$ . Hence, $L$ is closed in $X$ if and only if it is closed in $X'$ . Similarly, the same argument works for $I\cup \{x\}$ and $I\cup \{x'\}$ , because $x=x'\notin M'$ . Moreover, if $y\in K_0$ , then $y'=y\notin M'$ and we also obtain that $I\cup \{y\}$ is closed in $X$ if and only if $I\cup \{y'\}$ is closed in $X'$ . Hence, the only case remaining is when $y\notin K_0$ , i,e, $y=\min (P)$ and $y'=\min (P')$ .

We split the argument into two cases depending on the structure given by Lemma 5.1. Suppose that $M'$ is closed in $V'$ and let $u=a(\!\min (M'),\max (M'))$ be the common ancestor of $M'$ . By Fact 5.4, we have that $a(z,y)=a(z,y')=a(z,u)$ for every $z\in I$ . Therefore, the entire set $M'$ is descendant of the common ancestors of $I\cup \{y\}$ and $I\cup \{y'\}$ , which by condition (ii*) of Definition 2.2 implies that both sets are not closed. Now suppose that $M'\subseteq B^Q$ for some maximal comb $Q=A^Q\cup B^Q$ . Since $I$ is a closed interval in $X$ , then by Proposition 5.2 it is a closed interval in $V'$ . Thus, by Proposition 2.3, applied to $I$ and the maximal comb $Q$ , one of the following three possibilities holds: $I\cap Q=\emptyset$ , $I\subseteq Q$ or $Q\subseteq I$ . Clearly, the last possibility cannot hold, since $P\subseteq Q$ and $I\cap P=\emptyset$ . Suppose that $I\cap Q=\emptyset$ . By Proposition 5.2, the interval $Q\cap X$ is a closed interval in $X$ . Since $(I\cup \{y\})\cap (Q\cap X)=\{y\}\neq \emptyset$ and $\{y\}\neq Q\cap X$ , we obtain by Proposition 2.3 that $I\cup \{y\}$ is not closed in $X$ . Similarly, $I\cup \{y'\}$ is not closed in $X'$ .

Now we handle with the case that $I\subseteq Q$ . By Proposition 2.7, either $I\subseteq A^Q$ or $I=A\cup B$ is a comb with $A=A^Q$ and $B\subseteq B^Q$ . Since $I\subseteq K_0$ , we have that $Q$ is a maximal left comb. Let $K_0^Q=K_0\cap Q$ , $B^Q=\{z_1,\ldots,z_b\}$ and let $Q_{z_i}=A^Q\cup \{z_1,\ldots,z_i\}$ be the subcomb of $Q$ ending in $z_i$ . By condition (a3*) of Definition 2.4, we have that $A^Q$ and $Q_z$ are closed in $V'$ for every $z\in B^Q$ . Moreover, note that $\min (I)\in A^Q$ . Let $v=a(\!\min (A^Q),\max (A^Q)$ and $w=a(\!\min (I),y)$ be the common ancestor of $A^Q$ and $I\cup \{y\}$ , respectively. By Fact 5.4 applied to $A^Q$ , we have that $a(\!\min (I),x)=a(v,x)=a(\!\min (A^Q),x)$ for every $x\in M'$ . Thus, $X(w)=V'(w)\cap X=Q_y\cap X=K_0^Q\cup \{y\}$ , which implies that $I\cup \{y\}$ is a closed interval in $X$ if and only if $I=K_0^Q$ . Similarly, $I\cup \{y'\}$ is a closed interval in $X'$ if and only if $I=K_0^Q$ . Hence, $I\cup \{y\}$ is closed in $X$ if and only if $I\cup \{y'\}$ is closed in $X'$ .