tiny extra edits in tw section

arxiv_cbd.tex

@@ -858,9 +858,9 @@ Before going further, let us recall some notions about treewidth.
   $(T,\beta)$, where $T$ is a rooted tree and $\beta:V(T)\to 2^{V(G)}$
   assigns a \emph{bag} to each of its nodes, such that
 \begin{itemize}
-\item for each $uv\in E(G)$, there exists $x\in V(T)$ such that
+\item for each edge $uv\in E(G)$, there exists $x\in V(T)$ such that
   $u,v\in\beta(x)$, and
-\item for each $v\in V(G)$, the set $\{x\in V(T):v\in\beta(x)\}$ is
+\item for each vertex $v\in V(G)$, the set $\{x\in V(T):v\in\beta(x)\}$ is
   non-empty and induces a connected subtree of $T$.
 \end{itemize}
 For nodes $x,y\in V(T)$, we write $x\preceq y$ if $x=y$ or $x$ is a descendant of $y$ in $T$.
@@ -877,13 +877,13 @@ how to prove that disk graphs of thickness $t$ are fractionally treewidth-fragil
 We first consider unit disk graphs. 
 By partitioning the plane with a random grid $\HH$ with square cells of side-length $2k$, any unit disk has probability $1/2k$ 
 of intersecting a vertical (resp. horizontal) line of the grid. Using a union bound, any disk has probability at most $1/k$ of intersecting 
-the grid. Using this probability distribution, we show that removing the disks intersected by the grid leads to a 
+the grid. Using this probability distribution, we show that removing the disks that intersect the grid leads to a 
 unit disk graph of bounded treewidth. Indeed, in such a graph any connected component corresponds to unit disks contained in the 
 same cell of the grid. Each cell has area bounded by $4k^2$, so there are at most $16tk^2/\pi$ disks contained in a cell. 
-This bounds the size of any connected component, and so the treewidth is also bounded. 
+This bounds the size of any connected component, and the treewidth is also bounded. 
 
-Note that the above distribution also works if we are given
-disks whose diameter lie in a certain range. That is, for any diameter $\delta$ with $1/c \le \delta \le 1$, applying the same process
+The preceding distribution also works if we are given
+disks whose diameter lie in a certain range. For any diameter $\delta$ with $1/c \le \delta \le 1$, applying the same process
 with a random grid of $2k\times 2k$ cells ensures that any disk is deleted with probability at most $1/k$, and the
 connected components have size at most $4tc^2k^2/\pi$. Dealing with arbitrary disk graphs (with any diameter $\delta$ in the range 
 $0< \delta \le 1$) necessitates deleting more disks. This can be handled by partitioning each $(2k\times 2k)$-cell in a quadtree-like manner. 
@@ -891,7 +891,7 @@ Now a disk with diameter between $\ell /2$ and $\ell$ (with $\ell =1/2^i$ for so
 in a $(2k\ell \times 2k\ell)$-cell of a quadtree. It is straightforward to see that each disk is deleted with probability at most $1/k$.
 To prove that the remaining graph has bounded treewidth, one should consider the following tree decomposition $(T,\beta)$. Here, the tree $T$ is obtained by linking the roots of the quadtrees used (as trees) to a new common root. 
 Then for a $(2k\ell \times 2k\ell)$-cell $C$, $\beta(C)$ contains all disks of diameter at least $\ell/2$ intersecting $C$.
-To see that such bag is bounded, consider the $((2k+1)\ell \times (2k+1)\ell)$ square $C'$ centered on $C$, and note that any
+To see that such bag is bounded, consider the $(2k+1)\ell \times (2k+1)\ell$ square $C'$ centered on $C$, and note that any
 disk in $\beta(C)$ intersects $C'$ on an area at least $\pi\ell^2/16$. This implies that $|\beta(C)| \le 16t(2k+1)^2 / \pi$.
 
 Let us now give a detailed proof in a more general setting.