Nested-dissection grid fronts are bounded by a box separator and its exterior boundary

The root box has empty exterior boundary. If $B$ is a child of a box $C$, every grid neighbor just outside $B$ is either in $S_C$ or just outside $C$: distinct child boxes are separated by the removed coordinate hyperplanes. Thus ${\mathrm{\partial }}^{+}B\subseteq S_C\cup {\mathrm{\partial }}^{+}C$. Induction up the finite recursion tree, using that each ancestor separator is ordered after all vertices in its child boxes by [F1], shows that every vertex of ${\mathrm{\partial }}^{+}B$ is ordered after every vertex of $B$.

Fix $v\in S_B$ and a later filled neighbor $w$. If $w\in B$, then $w$ cannot lie in a child box because every child is ordered before $S_B$; hence $w\in S_B$. If $w\mathrm{\notin }B$, [F2] gives an original grid walk from $v$ to $w$ whose internal vertices were eliminated before $v$. The first vertex of this walk outside $B$ lies in ${\mathrm{\partial }}^{+}B$ and, by step 1.1, is ordered after $v$. It therefore cannot be an internal vertex of the walk, so it must be its endpoint $w$. Hence $w\in {\mathrm{\partial }}^{+}B$.

If the side lengths of $B$ are $m_1,\dots ,m_d\le s$, each of its at most $d$ coordinate hyperplanes contains at most $s^{d-1}$ vertices, so $\mathrm{\mid }{S}_B\mathrm{\mid }\le d{s}^{d-1}$. The exterior grid neighbors lie across the $2d$ faces, each with at most $s^{d-1}$ vertices, so $\mathrm{\mid }{\mathrm{\partial }}^{+}B\mathrm{\mid }\le 2d{s}^{d-1}$. The containment in step 2.1 therefore gives the stated column-size bound.

Steps 2.1 and 3.1 establish both the filled-neighbor containment and its numerical bound for every separator vertex, including the terminal one-vertex case.