I have a question on a proof in a paper on the orbital dimension of a permutation group.
Let $G \le S^\Omega$ be a permutation group. A base for $G$ is a subset $\Sigma \subseteq \Omega$ for which the pointwise stabilizer $G_{(\Sigma)}$ is the identity. The base size of $G$, denoted $b(G)$, is the smallest size of a base for $G$.
Suppose $G$ has orbitals $\{R_1,\ldots,R_t\}$ (these are the $G$-orbits on $\Omega \times \Omega$). Define the color $C(x,y)$ of an ordered pair $(x,y) \in \Omega \times \Omega$ to be the value of $i$ for which $(x,y) \in R_i$. A resolving set for $G$ is a subset $S = \{x_1,\ldots,x_k\} \subseteq \Omega$ such that for any $y,z \in \Omega, y \ne z$, we have $C(y,x_i) \ne C(z,x_i)$ for some $x_i \in S$. The orbital dimension of $G$, denoted $\mu(G)$, is defined to be the smallest size of a resolving set for $G$. It can be shown that every resolving set for $G$ is a base for $G$, whence $b(G) \le \mu(G)$.
Let $G$ be the action of $S_n$ on the 2-subsets of $\{1,\ldots,n\}$, and henceforth assume also that $n \equiv 1 \pmod 3$.
In Proposition 2.24 of [Bailey and Cameron, "Base size, metric dimension and other invariants of groups and graphs", preprint], it is shown that $b(G) = \frac{2}{3}(n-1)$. In Theorem 3.32, it is shown that a particular set $S$ of size $b(G)$ is not a resolving set for $G$, but augmenting this set $S$ by adding one more element to it gives a resolving set for $G$. From what I understand, this implies only that $b(G) \le \mu(G) \le b(G)+1$, whereas the authors conclude that $\mu(G)=b(G)+1$. How is this equality proved? I can see that $\mu(G)$ is one of $b(G)$ or $b(G)+1$. But to prove $\mu(G)=b(G)+1$, we need to show that any set $S$ of size $b(G)$ is not a resolving set for $G$.
I think the proof of Theorem 3.32 implicitly uses the fact that a base for $G$ that has cardinality $b(G)$ is unique (up to graph isomorphism). Using this fact and the fact that a resolving set for $G$ is a base for $G$, it follows that if $G$ had a resolving set $S$ of size $b(G)$, then this set $S$ would have to be the unique base for $G$, which can be shown to not be a resolving set, a contradiction. Uniqueness of minimum bases would thus establish that $\mu(G)=b(G)+1$. Is this what was assumed implicitly in the proof?