the vertices of $S-X$, and so merge two of them. This lowers the number of parts in the partition. As long as there are four or more parts, one of these operations can be performed, so repeat until at most three parts remain.

\[corollary:comptrick\] For every graph $G$ and for all sets $X, S \subseteq V(G)$ such that no component of $G-X$ contains more than two-thirds of the vertices of $S-X$, it is possible to partition the components of $G-X$ into at most two parts such that no part contains more than two-thirds the vertices of $S-X$.

This corollary follows by a very similar argument to Lemma \[lemma:comptrick\].

The following argument is similar to that provided in [@RS-GraphMinorsII-JAlg86].

\[lemma:sepleqtw\] For any graph $G$ and for all $c \in [\frac{1}{2},1)$, $${\textsf{\textup{sep}}}_c(G) \leq {\textsf{\textup{tw}}}(G)+1.$$

Fix $S \subseteq V(G)$ and let $k := {\textsf{\textup{tw}}}(G)+1$. It is sufficient to construct a $(k,S,\frac{1}{2})$-separator for $G$. $G$ has a normalised tree decomposition $T$ with maximum bag size $k$, by Lemma \[lemma:normal\]. Consider a pair of adjacent bags $X,Y$. Let $T_X$ and $T_Y$ be the subtrees of $T-XY$ containing bags $X$ and $Y$ respectively. Let $U_X \subseteq V(G)$ be the set of vertices only appearing in bags 