of a single subtree of $T-B$. Label that subtree as $T'$, and let $A$ denote the bag of $T'$ adjacent to $B$. Recall there is a partition $V(G)$ into $U_A, A \cap B, U_B$ where $|U_B \cap S| > \frac{1}{2}|S-(A \cap B)|$, since the edge $AB$ is oriented towards $B$. Hence $|U_A \cap S| < \frac{1}{2}|S-(A \cap B)|$. Also note the vertices of $G-B$ that only appear in the bags of $T'$ are exactly the vertices of $U_A$. Hence $C \subseteq U_A$, and $|U_A \cap S| > \frac{1}{2}|S-B|$.

So $\frac{1}{2}|S-B| < |U_A \cap S| < \frac{1}{2}|S-(A \cap B)|$. By our normalisation, $|A \cap B| = |B|-1$. So $|S-B| \geq |S-(A \cap B)|-1$. Thus $|S-(A \cap B)|-1 < 2|U_A \cap S| < |S-(A \cap B)|$, which is a contradiction as $|S-(A \cap B)|-1$, $2|U_A \cap S|$ and $|S-(A \cap B)|$ are all integers.

Now we provide a proof of the upper bound.

For any graph $G$, for all $c \in [\frac{1}{2},1)$, $${\textsf{\textup{bn}}}(G) \leq \frac{1}{1-c}{\textsf{\textup{sep}}}_c^*(G).$$

Say $\beta$ is an optimal bramble of $G$ with a minimum hitting set $H$. That is, $|H|={\textsf{\textup{bn}}}(G)$. For the sake of a contradiction, assume that $(1-c){\textsf{\textup{bn}}}(G) > {\textsf{\textup{sep}}}_c^*(G)$. So there is a $({\textsf{\textup{sep}}}_c^*(G),H,c)^*$-separator $X$. If $X$ is 