I'm reading a paper which gives the following definition:
Let $K\subset\mathbb{R}^2$ be locally-connected and compact. Given $p,q\in K$, let $\mathrm{diam}(p, q)$ be the minimum diameter of a connected subset of $K$ containing both $p$ and $q$, and define the local-connectivity function $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ of $K$ by
\begin{equation} f(s) = \sup_{d(p,q)\leq s}\mathrm{diam}(p,q) \end{equation}
I'm struggling to understand how this makes sense - we do not assume $K$ is connected, so how can we assume that there is a connected subset of $K$ containing $p$ and $q$ in the first place? The example in my head is two disjoint closed discs in the plane, and taking $p$ in one disc and $q$ in the other. Am I missing something here? Thanks
You're not missing anything: $\operatorname{diam}(p,q)$ may not be defined for arbitrary $p$ and $q$ (or else you should define it to be infinite if there is no connected subset of $K$ containing them). As a result, $f(s)$ may not be defined for all $s$. However, in the paper you linked in the comments, the function $f$ is only actually being used for small values of $s$. In particular, the proof of Proposition 7.1(3) actually shows that $f(s)$ is well-defined for all sufficiently small $s$ (and in particular, if $d(p,q)$ is sufficiently small then there must exist a connected set continaing $p$ and $q$). This appears to be all that the paper ever really uses.
In fact, as far as I can tell the function $f$ is only ever used in the proof of Theorem 7.2, in which the set $K$ is assumed to be connected so $\operatorname{diam}(p,q)$ actually is defined for arbitrary $p$ and $q$.