From d266c09389d8df8f8790703b70f0822a807eedc7 Mon Sep 17 00:00:00 2001
From: "jane.tan" <jane.tan@maths.ox.ac.uk>
Date: Mon, 14 Mar 2022 18:03:15 +0000
Subject: [PATCH] change to address referee comment

---
 arxiv_cbd.tex | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/arxiv_cbd.tex b/arxiv_cbd.tex
index 5f270e9..83e5749 100644
--- a/arxiv_cbd.tex
+++ b/arxiv_cbd.tex
@@ -778,7 +778,7 @@ Hence, the condition (c1) holds.
 Consider now a vertex $v_i$ and a clique $C$.  As we observed before, if $v_i\not\in V(C)$,
 then $p(C) \not\in h(v_i)$, and thus $h^\varepsilon(C)$ and $h(v_i)$ are disjoint (for sufficiently small $\varepsilon>0$).
 If $v_i\in C$, then the definitions ensure that $p(C)[i]$ is equal to the maximum of $h(v_i)[i]$,
-and that for $j\neq i$, $p(C)[j]$ is in the interior of $h(v_i)[j]$, implying
+and that for $j\neq i$, $p(C)[j]$ is in $h(v_i)[j]$, implying that
 $h(v_i)[j] \cap h^\varepsilon(C)[j] = [p(C)[j],p(C)[j]+\varepsilon]$ for sufficiently small $\varepsilon>0$.
 \end{proof}
 The \emph{treewidth} $\tw(G)$ of a graph $G$ is the minimum $k$ such that $G$ is a subgraph of a $k$-tree.
-- 
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