plane_depth_sequencing: generalise Lemma 2.6 to arbitrary outer-face source; tighten the citation in coloring_nested_tire_graphs

In `plane_depth_sequencing/paper.tex`:
- Lemma 2.6 now allows any nonempty source S ⊆ V(G) whose vertices all
  lie on the boundary of the outer face of the chosen embedding,
  rather than only the outer-cycle case S = V(C).
- The proof is the same argument with S in place of C: at d=0 each
  S-vertex remains on the outer face after restriction; for d ≥ 1
  the BFS ball V_{<d}^S is connected and reaches the outer face
  via S.
- The original outer-cycle statement is preserved as a remark inside
  the lemma.
- Adds \label{lem:outerplanarity}.

In `coloring_nested_tire_graphs/paper.tex`:
- The proof of Lemma 1.7 drops the "extends verbatim" caveat and
  simply cites the generalised Lemma 2.6, noting that since the level
  source S is a single vertex (per the local Level-source definition)
  we may freely choose an embedding placing S on the outer face;
  outerplanarity is a graph property so the conclusion transfers.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>

papers/plane_depth_sequencing/paper.tex

@@ -107,7 +107,7 @@ A triangle $\{u, v, w\}$ in $G$ is an \emph{up triangle} if the multiset of dept
 \end{definition}
 
 \begin{remark}
-We now relate our terminology to existing terminology, namely $k$-outerplanar graphs \cite{baker1994}. The following definition and lemma show that the subgraph induced by any single depth level is outerplanar, i.e., $1$-outerplanar in the sense of Baker.
+We now relate our terminology to existing terminology, namely $k$-outerplanar graphs \cite{baker1994}. The following definition and lemma show that the subgraph induced by any single depth level relative to any source set on the outer face is outerplanar, i.e., $1$-outerplanar in the sense of Baker.
 \end{remark}
 
 \begin{definition}
@@ -115,19 +115,50 @@ A plane graph is \emph{outerplanar} if every vertex lies on the outer face. More
 \end{definition}
 
 \begin{lemma}
-Let $G$ be a graph with a plane embedding and outer cycle $C$. For each $d \geq 0$, the subgraph of $G$ induced by $V_d = \{v \in V(G) : \mathrm{depth}(v) = d\}$ is outerplanar.
+\label{lem:outerplanarity}
+Let $G$ be a planar graph with a plane embedding $\Pi$, and let
+$S \subseteq V(G)$ be a nonempty set of vertices, every one of which
+lies on the boundary of the outer face of $\Pi$.  For each $d \geq 0$,
+the subgraph of $G$ induced by
+\[
+    V_d^S := \{ v \in V(G) : \mathrm{dist}_G(v, S) = d \}
+\]
+is outerplanar.
+
+The special case $S = V(C)$, where $C$ is the outer cycle, recovers
+$V_d^S = V_d$ (depth-$d$ vertices as in Definition~2.1) and is the
+form most often used in the rest of this paper.
 \end{lemma}
 
 \begin{proof}
-Let $H = G[V_d]$ with the plane embedding inherited from $G$. It suffices to show every vertex of $H$ lies on the outer face of $H$.
+Let $H = G[V_d^S]$ with the plane embedding inherited from $\Pi$.  It
+suffices to show that every vertex of $H$ lies on the outer face of $H$.
 
-For $d = 0$, we have $V_0 = C$, so $H$ is outerplanar.
+For $d = 0$, $V_0^S = S$, and by hypothesis every vertex of $S$ lies
+on the boundary of the outer face of $\Pi$.  Removing the vertices and
+edges of $G \setminus H$ from the embedding only enlarges or merges
+face regions, so the outer face of $\Pi$ is contained in the outer face
+of $H$, and every vertex of $S$ remains on the outer face of $H$.
 
-For $d \geq 1$, let $U$ be the open subset of the plane obtained by removing all vertices and edges of $H$. We show every $v \in V_d$ lies on the boundary of the component $U_{\mathrm{out}}$ of $U$ containing the outer face of $G$.
+For $d \geq 1$, let $U$ be the open subset of the plane obtained by
+removing all vertices and edges of $H$.  We show every $v \in V_d^S$
+lies on the boundary of the component $U_{\mathrm{out}}$ of $U$
+containing the outer face of $\Pi$.
 
-Since every vertex in $V_{\leq d-1}$ has a shortest path to $C$ passing entirely through $V_{\leq d-1}$, the subgraph $G[V_{\leq d-1}]$ is connected and contains $C$. Its vertices and edges lie in $U$ (as they are not in $H$), and $C$ borders the outer face of $G$, so $G[V_{\leq d-1}]$ and the outer face of $G$ are connected within $U$, hence both lie in $U_{\mathrm{out}}$.
+Since every vertex in $V_{<d}^S := \bigcup_{e < d} V_e^S$ has a
+shortest path to $S$ passing entirely through $V_{<d}^S$, the subgraph
+$G[V_{<d}^S]$ is connected and contains $S$.  Its vertices and edges
+lie in $U$ (none belong to $H$), and $S$ borders the outer face of
+$\Pi$, so $G[V_{<d}^S]$ and the outer face of $\Pi$ are connected
+within $U$, hence both lie in $U_{\mathrm{out}}$.
 
-Now let $v \in V_d$. Since $\mathrm{depth}(v) = d \geq 1$, there exists $u \in V_{d-1}$ adjacent to $v$ in $G$. The edge $\{v, u\}$ is not an edge of $H$, so it lies in $U$. Since $u \in V_{d-1} \subset U_{\mathrm{out}}$ and $\{v,u\}$ is a connected subset of $U$ containing $u$, the entire edge lies in $U_{\mathrm{out}}$. The vertex $v$ is an endpoint of this edge but is not in $U$, so $v$ lies on the boundary of $U_{\mathrm{out}}$, i.e., on the outer face of $H$.
+Now let $v \in V_d^S$.  Since $d \geq 1$, there exists $u \in V_{d-1}^S$
+adjacent to $v$ in $G$.  The edge $\{v, u\}$ is not in $H$, so it lies
+in $U$.  Since $u \in V_{d-1}^S \subseteq U_{\mathrm{out}}$ and
+$\{v, u\}$ is a connected subset of $U$ containing $u$, the entire
+edge lies in $U_{\mathrm{out}}$.  The vertex $v$ is an endpoint of
+this edge but is not in $U$, so $v$ lies on the boundary of
+$U_{\mathrm{out}}$, i.e., on the outer face of $H$.
 \end{proof}
 
 \begin{definition}