Let \(\Omega\subseteq\CC\) be open and let \(f_k,f\colon\Omega\to\CC\) for \(k\ge1\). The sequence \((f_k)\) converges locally uniformly to \(f\) on \(\Omega\) if \(f_k\to f\) uniformly on every compact subset of \(\Omega\).
Let \(\Omega\subseteq\CC\) be open and let \(f_k,f\colon\Omega\to\CC\), with the convention \(\operatorname{dist}(A,\varnothing)=+\infty\) when \(\Omega=\CC\). Then \(f_k\to f\) locally uniformly on \(\Omega\) if and only if \(f_k\to f\) uniformly on every bounded set \(A\subseteq\Omega\) with \(\operatorname{dist}(A,\partial\Omega)>0\).
Let \(A\subseteq\Omega\) be bounded with \(\operatorname{dist}(A,\partial\Omega)=d>0\). Its closure \(\overline A\) is closed and bounded, hence compact, and every point of \(\overline A\) has distance at least \(d\) from \(\partial\Omega\). A point of \(\overline A\) lies in \(\overline\Omega\) but not in \(\partial\Omega\), so it lies in \(\Omega\); thus \(\overline A\) is a compact subset of \(\Omega\), and uniform convergence on \(\overline A\) gives uniform convergence on \(A\).
Conversely, let \(K\subseteq\Omega\) be compact. Then \(K\) is bounded, and \(z\mapsto\operatorname{dist}(z,\partial\Omega)\) is continuous and positive on \(K\) when \(\partial\Omega\neq\varnothing\), because \(\partial\Omega\) is closed and disjoint from \(K\), so it has a positive minimum on \(K\). Hence \(K\) is one of the sets \(A\).