Locally uniform convergence on an open subset of the complex plane is compact convergence

Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open, and let $f_n,f:\mathrm{\Omega }\to \mathbb{C}$ be continuous. The sequence $(f_n)$ converges locally uniformly to $f$ when each $a\in \mathrm{\Omega }$ has an open neighbourhood $V\subseteq \mathrm{\Omega }$ on which $f_n\to f$ uniformly. This is equivalent to uniform convergence on every compact subset of $\mathrm{\Omega }$, hence to convergence in the topology of compact convergence of The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$: uniform convergence on each compact subset of $X$.

Indeed, suppose first that convergence is uniform on compact subsets. Openness gives $r>0$ with the closed disc $\bar{D}(a,r)\subseteq \mathrm{\Omega }$ after shrinking an available ball; this closed disc is compact by Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, so convergence is uniform on the neighbourhood $D(a,r)$. Conversely, suppose convergence is uniform on a neighbourhood $V_x$ of every $x\in \mathrm{\Omega }$. For a compact $K\subseteq \mathrm{\Omega }$ and $\epsilon >0$, the sets $V_x$ cover $K$, and compactness in the ambient space (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it) gives a finite subcover. Taking the largest of the corresponding finitely many convergence thresholds makes $\mathrm{\mid }{f}_n-f\mathrm{\mid }<\epsilon$ throughout $K$. The empty compact set satisfies the uniform condition vacuously.